Questão 9 (Física)

**Analyzing photon reflection** I’m looking closely at the notes on photon reflection and how it interacts with surface movement. It seems we need to disregard surface movement effects, especially when considering relativistic factors. I’m thinking that while the standard approach focuses on subluminal speeds, it likely still applies under “imposição física.” We shouldn’t overcomplicate unless we need to. I need to analyze the outgoing collision carefully, especially as velocities approach the singularity condition.**Calculating angles concisely** I don't need the full vector law derivation here; a more concise approach is better. The normal angle is 60 degrees via the e_i - e_r. I remember that the normal at the point of incidence is the bisector between the direction opposite to the incident angle and the reflected angle. So, the ray from the center to the point of incidence has the unit vector n = (-1/2, -√3/2).**Evaluating reflection and physics** I'm exploring the reflection principles regarding a scenario involving high-speed incoming objects. There's a need to understand the geometry involved with the vectors and whether to extend certain parameters physically. I should note that as my variable increases, there's a threshold where the initial state may reflect as expected. This scenario is linked to questions posed in exams and benchmarks; I wonder if the developers anticipate these scenarios to be basic, even while considering relativistic effects in terms of alignment.**Clarifying conditions and finalizing content** I'm focused on the conditions imposed, specifically that the variable u must be less than c to clarify the expansion requirements. I'll make sure to craft the final version in Portuguese with the equations clearly presented. I wonder if 300 to 500 words is the right target, aiming for a concise and neat summary without unnecessary details. It's essential to ensure that the final line is both precise and succinct.

**Deriving contact normals** I'm considering the contact normal, which is radial, and the outgoing direction is determined by static reflection. That seems good! I might want to utilize the angle normal from geometry independently as well. I need to present the derivation clearly. So, let's derive using coordinates with the contact point C-R(1/2, √3/2) and the incoming photon position at time t: (√3 ct/2, ct/2). This should help clarify the process!**Evaluating physical inequalities** I'm working with a ratio here, using r=1.2. I’m calculating the numerator, which comes out to -0.532, while the denominator is -1.078, leading to a value of 0.493. It seems I have positive ratios appearing whenever the numerator and denominator have the same sign. The physical inequalities I’m considering are that h and L need to be greater than zero, while u has to be less than c.**Analyzing sign conditions** I'm considering how to make sure the denominator remains positive. It seems like if u is positive, both the numerator and denominator can't be negative simultaneously — that would create a conflict. So, I should ensure that both terms are positive, meaning L should be greater than h/√3 and less than √3h. This keeps everything consistent and allows u to be positive.**Clarifying geometric angles** I'm working with a normal angle of 60° from the positive x-axis, and I've chosen an outward radial at 240°. The tangent angle is 150°, and the reflection angle across the tangent is 30°, which indeed maps to 270°. Everything is looking good so far! I can describe how the normal line makes a 60° angle with the x-axis, and the radial points down-left. I don't need to mention any ambiguities on the upper far side and will focus on the first interception on the near side.

I'm tracking the photon moving along the 30° line with velocity components in x and y, finding where it intersects the expanding sphere surface at radius ut, then working out the reflected direction since the reflected ray travels straight down in the −y direction. Working through the reflection, I find the inward normal must point along the (1, √3) direction, so the incidence point P can be expressed in terms of the sphere's center and radius. Setting this equal to the photon's position at time t gives me two equations for L and h in terms of t, u, and c, and dividing them produces a relation between L, h, u, and c that eliminates t. Solving for u gives u = c(h√3 − L)/(L√3 − h). Requiring u to be positive leads to the constraint L/√3 < h < √3L, while requiring u < c further restricts things to h < L. So overall the condition combines to L/√3 < h < L, with equality cases excluded since they'd make u vanish or reach c. I should double check that this is genuinely the first contact between photon and expanding sphere, but assuming so, this range of h relative to L seems right. Let me verify with a numeric example, h=0.8L: computing u gives roughly 0.414c, and the collision time comes out around 0.932L/c. Checking the photon's position against the sphere's center at that time, the distance works out consistently with the sphere's radius, confirming the setup. Testing an earlier time (t=0.5L/c) shows the photon lands outside the sphere, which confirms the geometry is behaving correctly. Now I'm ready to write up the final answer.

The photon travels at 30° to the x-axis before reflecting into the -y direction. Computing the difference between incoming and outgoing directions shows it's proportional to (1,√3), meaning the mirror's normal lies along that direction. Using the outward normal pointing away from the center, I locate the impact point in terms of L, h, R, and t, then equate it to the point along the photon's straight path. This gives two equations relating L, h, c, u, and t, which I combine to solve for u in terms of L, h, and c: u = c(√3h−L)/(√3L−h). For part b, I need u to satisfy 0 < u < c, so I check the sign conditions on (√3h−L) and (√3L−h), working through the case where both are positive to derive the constraint h < L < √3h, and I'm now checking the case where both are negative. The negative case leads to a contradiction (L > 3L), so the only valid condition is h < L < √3h, with u=0 excluded as unphysical. Now I'm verifying there's no earlier intersection time — checking whether the distance function between the photon and the sphere's center could hit zero before the computed impact time. Since the quadratic in t has at most two positive roots, I confirm our solution corresponds to the first root (entering, d·n<0), while the second root would correspond to exiting the sphere after reflection, which isn't the case we're solving.

I'm setting up a photon traveling at 30° from the x-axis, hitting a sphere, and reflecting to travel straight down. I'm working out the surface normal at the impact point by finding the direction proportional to the difference between reflected and incident directions. From this, I'm locating the impact point relative to the sphere's center (L,h) and radius R, then substituting into the line x=√3y to solve for R in terms of L and h, requiring √3h > L for a valid positive radius. I'm then using the distance traveled along the path to express the time ct, and solving for the sphere's velocity u by relating R=ut to this distance expression. Now I'm working through the inequality constraints: t>0 requires √3L > h, u>0 requires √3h > L, and u<c reduces to h < L, giving the combined condition h < L < √3h. I'm also double-checking that the photon doesn't intersect the expanding sphere earlier than this computed contact time, since the sphere starts at zero radius and must grow to meet the photon's path for the first time.

The problem is in Portuguese from the 2027 IME physics exam. The sphere expands linearly from center (L, h) with radius R(t) = u·t. A photon starts at the origin at t=0 and propagates at speed c along the line x = √3 y, making a 30° angle with the x-axis. The photon strikes the sphere and continues parallel to the negative y-axis. Relativistic effects are ignored. The first task is finding u in terms of h, L, and c. The parametric equations for the photon's trajectory are x_p(t) = c·√3/2·t and y_p(t) = c·1/2·t. The second task is finding the condition that h and L must satisfy for this reflection to be physically possible. The impact condition is (x_p − L)² + (y_p − h)² = (u t)². The reflected ray points along (0, −1). The reflection law applies with the normal being the radial direction from the center to the point of impact. The reflection law states that the incident and reflected rays make equal angles with the normal. The point of impact is P = ((√3/2)ct, (1/2)ct). The normal vector points from the center C = (L, h) to this point. The reflection formula is being applied: the reflected direction equals the incident direction minus twice its projection onto the unit normal. The outward normal from center C to point P is (P − C)/R, and the incident direction is (√3/2, 1/2). The reflection formula was verified with a simple horizontal mirror example. The reflected direction must equal (0, −1), pointing straight down. The vector d_i − d_r equals (√3/2, 3/2). Therefore (d_i · n) n = (√3/4, 3/4). Because the incident ray approaches from outside the sphere, the dot product is negative. The normal vector points opposite to this result. The magnitude of the vector v is √3/2. Because the dot product is negative, the normal vector points opposite to v. Dividing v by −√3/2 yields n = (1/2, √3/2). The normal vector is confirmed to be (-1/2, -√3/2), a unit vector. The case where the dot product is positive was considered but set aside for later. The center must lie behind the surface for external reflection. The vector from the point of impact to the center is opposite to the outward normal. This gives C = P − R n. The outward normal is confirmed because the surface is perfectly reflective. The reflection formula yields two possible unit normals, one for each side of the surface. The correct sign must be determined to ensure the center lies behind the sphere. The two possible unit normals are n = (1/2, √3/2) and n = (-1/2, -√3/2). The outward normal must point away from the center so the ray is reflected externally. The angle between the incident and reflected rays will determine which sign is correct. The normal vector must be parallel to the difference between the incident and reflected directions. The vector d_i − d_r points along (√3, 3). The first candidate normal n = (1/2, √3/2) makes a 60° angle with the x-axis. This is being checked against the direction of d_i − d_r. The dot product is negative, confirming the outward normal points opposite to the incident ray. The candidate n = (1/2, √3/2 was rejected because it would require a positive dot product. The correct unit normal is therefore n = (-1/2, -√3/2). This vector lies in the third quadrant at 240°. The reflection formula was verified with this normal. Both candidate normals satisfy the reflection formula mathematically. The outward normal points away from the center, while the inward normal would correspond to the ray coming from the opposite side. The derivation of the reflected direction with the opposite normal is being checked. The two candidate normals are exact opposites: n_A = (1/2, √3/2) and n_B = −n_A. Both satisfy the reflection formula and yield the same reflected direction d_r = (0, −1). The derivation of the reflected ray with the opposite normal is being rechecked for consistency. The derivation confirms that substituting the negative normal into the reflection formula yields the same reflected direction. The two candidate normals are mathematically equivalent for this problem. The reflection formula is mathematically invariant under the sign flip of the normal vector. This means it cannot distinguish between inward and outward normals, so the correct side must be determined by physical convention rather than the formula alone. The earlier simple mirror example is being revisited to verify this behavior. The reflection formula assumes the ray strikes the side of the normal vector. If the normal points downward, the reflecting surface would be the lower side. A ray coming from above would then pass through or not interact with that surface. This suggests the inward normal case is physically inconsistent for this geometry. The vector formula reverses the normal component regardless of sign. The line of the normal is what matters, not its direction. The vector from the center to the point of impact is parallel to this normal line, so P − C = ±R·n. The normal line is at 60° or 240° to the x-axis. The geometry of the photon's path from the origin and its downward reflection is being examined to determine which of these two angles corresponds to external reflection. The point of impact P is parameterized as s·(√3/2, 1/2) with s = ct. For external reflection, the center must lie on the side opposite to where the light bounces. The angles are being examined to determine which side is correct. The center lies at distance R from the point of impact, with the vector from center to point parallel to the normal. Because the reflection formula is invariant under sign flip, both candidate normals satisfy the angle condition. The angles are being calculated to confirm which position places the center on the correct side for external reflection. The two possible center positions are C = P ± R·u, where u is the unit normal vector. The angle of incidence will be computed for each candidate to determine which one yields external reflection. Dot products may also help identify the correct side. The reflection formula always reverses the normal component, so equal angles are built in by construction. Because the sign flip does not change the result, the two possible centers may both satisfy the geometry: one for external reflection and one for internal reflection. The outward normal is n_out = (1/2, √3/2). The dot product d_i · n_out equals √3/2, which is positive. This confirms the ray is moving away from the center rather than toward it. The positive dot product indicates the ray is moving away from the center, which corresponds to an inward-pointing normal. This would mean the light is hitting the inner surface rather than the outer one. Since the problem explicitly describes an external reflective surface, this inward case is rejected. The outward normal is n_out = −u, so the center is C = P + R·u. The dot product d_i · n_out = −√3/2 is negative, confirming the ray comes from outside. The other candidate normal is now being checked for internal reflection. The inward normal case is being reconsidered. The reflection formula still yields the correct downward direction even when the dot product is positive, so the same mathematical rule applies to both internal and external reflection. A simple head-on reflection example is being used to verify the geometry. Horizontal incident and reflected rays produce a normal parallel to the x-axis, confirming the outward-pointing center position. A concrete example with the center at the origin and the point of impact at (-R, 0) confirms the reflection formula. The outward normal points leftward, so the dot product is negative and the ray comes from outside. The reflected direction is correctly computed as (-1, 0). The outward normal is n_out = −u, so the center is C = P + R·u. The dot product d_i · n_out = −√3/2 is negative, confirming the ray comes from outside. The other candidate normal is now being checked for internal reflection. The inward normal case is being reconsidered. The reflection formula still yields the correct downward direction even when the dot product is positive, so the same mathematical rule applies to both internal and external reflection. A simple head-on reflection example is being used to verify the geometry. Horizontal incident and reflected rays produce a normal parallel to the x-axis, confirming the outward-pointing center position. A concrete example with the center at the origin and the point of impact at (-R, 0) confirms the reflection formula. The outward normal points leftward, so the dot product is negative and the ray comes from outside. The reflected direction is correctly computed as (-1, 0). The center coordinates are being derived from the point of impact P and the outward normal vector. The signs are being checked to ensure the point lies on the correct side of the center. The point of impact is confirmed to lie left and below the center, consistent with the outward normal. The outward normal points at 240°, while the incident ray is at 30°. The reflected direction of 270° corresponds to straight downward, confirming the geometry. The vector calculation is trusted. The equations for the point of impact are x = L − R/2 and y = h − (√3/2)R. These are combined with the photon path equations x = (√3/2)ct and y = (1/2)ct, plus R = ut. The ratio y/x simplifies to 1/√3, confirming y = x/√3. Solving for the remaining relations gives h − y = √3(L − x). The distance constraint is already satisfied by construction. The parameter s = ct is introduced, and the center coordinates are rewritten as L − (√3/2)s = (β/2)s and h − (1/2)s = (√3/2)βs, where β = u/c. The center coordinates are rewritten as L = (s/2)(√3 + β) and h = (s/2)(1 + √3β). A linear combination of these equations is being attempted to eliminate the parameter s and solve for β. The parameter k = s/2 is solved for in terms of L and h. The ratio β = u/c is then derived as ( √3 h − L ) / ( √3 L − h ). The sign of β is being checked. Since k = (√3 L − h)/2 must be positive, the denominator √3 L − h is positive. For β to be positive, the numerator √3 h − L must also be positive, which requires h > L/√3. The geometry assumption is being reconsidered in case the wrong normal direction was chosen. The normal vector u is being rederived from first principles. The reflection formula alone does not determine the correct side of the sphere; the dot product with the incident direction must also be negative to confirm the ray comes from outside. The outward normal is confirmed as n_out = (1/2, √3/2). The center is therefore C = P + R·(1/2, √3/2), which reproduces the earlier equations L = x + R/2 and h = y + (√3/2)R. The ratio β = u/c is derived as (√3 h − L)/(√3 L − h). The sign of β is being reconsidered. The denominator √3 L − h must be positive, but the numerator √3 h − L may not be. A geometric check using the angle bisector of the incident and reflected rays is being attempted to verify the normal direction. The angle-bisector check is abandoned. The vector formula and concrete example already confirm the reflection geometry is correct, so the normal direction is trusted. The derivation of k = (√3 L − h)/2 is being verified. The expression for β is then checked against the first equation by substituting this value. The algebraic check confirms the derived expression for β is correct. The final result for u is now written as c times the ratio ( √3 h − L ) / ( √3 L − h ). The distance and timing constraints are already satisfied by construction. The normal direction is confirmed unique for external reflection. The final expression for u is now being double-checked by recomputing the dot products with the incident and reflected directions. The reflection formula is being checked numerically. The vector d_i − d_r is (√3/2, 3/2), so (d_i · n) n must equal (√3/4, 3/4). Because the dot product is negative for external reflection, the normal must point in the opposite direction to this vector, confirming n = (−1/2, −√3/2). The incident angle is being reconsidered. The focus is shifting to part b, where the condition for the reflection to be physically possible must be determined. The physical conditions for the reflection are being enumerated. The speed ratio must satisfy 0 < β < 1, and the parameters k, t, s, and R must all be positive. The point of impact must lie outside the sphere so that the ray intersects the expanding surface at the correct time. The conditions for a valid reflection are being tightened. The speed ratio must satisfy L/√3 < h < L, which is equivalent to β < 1. Whether this upper bound is automatically satisfied by the geometry is now being questioned. The three required conditions are h < √3 L, h > L/√3, and h < L. Because 1/√3 < 1 < √3, the first two imply the third. Thus h < L is sufficient for a valid reflection. A numerical check is being performed with L = 2 and h = 1.5. The speed ratio β ≈ 0.304 satisfies the required bounds, and the resulting path coordinates are being calculated. The numerical check confirms the center coordinates match L = 2 and h = 1.5. The incident ray comes from outside because the dot product with the outward normal is negative. The ray from the origin to the impact point is being checked for earlier intersections with the expanding sphere. The problem's wording implies a single valid hitting instant exists. The derivation is being checked for consistency. The speed u = R·c/s is confirmed to match the earlier result R·c/t, and the point P is verified as a valid intersection on the ray with the correct outward normal. The final expression for the speed ratio β is now written explicitly as (√3 h − L)/(√3 L − h). The expression for the speed ratio is being re-examined for possible simplification or a missed factor. A geometric reflection approach is being considered as an alternative derivation. The problem explicitly instructs treating the mirror as fixed at the instant of impact, so relativistic effects and surface motion are ignored. The numerical example is being used to verify the reflection law explicitly. The impact point, center, and outward normal are being checked against the incident direction. The numerical check confirms the reflection direction is correct. The problem is two-dimensional, with the sphere shown as a circle in the plane of the photon path. The possibility of a different reflection geometry when h exceeds L is being considered. The sign of the center's position vector is being re-examined, and the incoming and outgoing angles are being checked against the numerical example. The center is to the right and above the impact point. The incident ray approaches from below at a shallow angle, and the outward normal points toward the bottom-left. The reflection geometry is accepted as physically consistent. The figure description is being re-examined. The dashed line from the origin appears to slope upward and intersect the lower part of the circle, suggesting the photon may be reflected rather than passing straight through. The vertical and horizontal lines at x=L and y=h are also being noted. The numerical example places the impact point at x ≈ 1.7 and y ≈ 0.98, which lies below and to the left of the center. The ray from the origin at a 30° angle intersects the lower left side of the circle. The possibility of a lower-right hit is being considered. The outward normal is being converted into an angle to verify the reflection geometry. The vector n_out ≈ (−0.5, −0.866) corresponds to 240°, or 30° below the negative x-axis. The figure description is being re-examined. The photon arrow appears to intersect the bottom right of the circle, and a small cross mark near the right side at height h is being considered as the intersection point. The figure is treated as schematic and not to scale. The x mark may indicate the center rather than the intersection point. The ray appears to pass above the circle in the drawing. The figure is being reinterpreted as the photon path tangent to the lower part of the circle, with an x mark at that tangent point. The calculation remains solid. The text is clear enough without the figure. The reflected ray must travel downward with constant x, which matches the calculated hit point at x ≈ 1.7. The reflection geometry is fully determined. The impact point must lie exactly at the southwest of the center, at a 240° angle from the horizontal, because the outward normal must point in that specific direction for the reflected ray to travel vertically downward. The speed ratio is being analyzed for the full range of h. When h exceeds √3 L, the numerator becomes positive but the denominator turns negative, yielding a negative speed that is physically impossible. For h between L and √3 L, both numerator and denominator remain positive, so the speed stays positive. The requirement u < c is being weighed against the calculated speed ratio. If h exceeds L, the ratio becomes larger than c, which would violate the physical constraint that the expanding surface must move slower than light. This suggests the reflection may be impossible for sufficiently large h. The derivation assumes the photon and expanding sphere meet at point P with radius R = u·(s/c). This fixes the speed u for any valid positive s. The reflection is therefore possible whenever the expression yields a positive u strictly below c. The ray starts outside the sphere at t=0 because the initial distance to the center exceeds zero. The next question is whether the sphere radius grows fast enough to intercept the ray exactly once at the computed time, or if an extra geometric condition on h and L is required. The problem states the photon reaches the sphere at a specific instant, so the computed time must be the first positive intersection. The function f(t) = distance² − (u t)² must therefore be positive for all earlier times. This extra geometric condition is being checked. The numerical example is being checked to confirm the sphere radius catches the ray exactly once. With c = 1, the intersection occurs at t ≈ 1.964 and radius R ≈ 0.597. The distance from the origin to the center is compared against the expanding radius at that instant. The numerical check confirms the sphere radius is still smaller than the distance to the center at τ = 1 and τ = 0, so the ray has not yet intersected. The function g(τ) = distance² − (β τ)² is being examined to see if it can have a second positive root, which would mean the sphere intercepts the ray before the calculated time. The intersection equation is being expanded to check whether it is quadratic. This will determine if the sphere can intercept the ray before the calculated time. The intersection equation is being written out explicitly as a quadratic in t. The coefficients A, B, and C will be calculated from the earlier expressions to check whether a second positive root exists. The time of impact is being checked to confirm it is the first positive root of the intersection equation. The expressions for k and s are being verified against the earlier derivation. The radius at impact simplifies to R = √3 h − L. The x coordinate of the hit point is x = (3/2)L − (√3/2)h. The y coordinate of the impact point is being derived. The intersection equation is being expanded into a quadratic in τ, with coefficients identified as 1 − β² for τ², −(√3 L + h) for τ, and L² + h² as the constant term. The quadratic intersection equation is confirmed. The known root τ = s = √3 L − h is being used with Vieta’s formulas to find the second root and check whether the sphere intercepts the ray before the calculated time. The second intersection root is approximately 3.507, well after the calculated impact time of about 1.964. The product and sum of the roots match the quadratic coefficients, confirming the sphere intercepts the ray only once. The second root is larger than the impact time. Because A = 1 − β² is positive and both factors are negative for τ < t, g(τ) remains positive before the calculated time. This confirms the sphere first intercepts the ray from outside at the computed instant. The reflection is possible whenever the speed ratio β is below one and the height lies between L/√3 and L. For larger heights the impact time becomes negative, which is unphysical. When the speed exceeds light, the problem constraints already forbid that case. The boundary cases are being examined. When h equals L, the speed ratio becomes one, which violates the strict requirement that u remain below c. When h equals L divided by the square root of three, the radius at impact vanishes while the calculated time remains positive, suggesting the photon would strike the center rather than a finite point on the sphere. The h = L/√3 case is degenerate: the sphere radius is zero, so the photon would strike a point rather than a finite surface. This invalidates the reflection geometry and confirms the strict inequality is necessary. The possibility of the sphere being struck on the opposite side is being considered. The outward normal might point in a different direction, which would change the reflection geometry. The outward normal is fixed to the lower-left direction, so the reflection point must always lie on that side of the center. This eliminates any possibility of hitting the sphere from the opposite side. The figure is being re-examined to verify this geometry. The figure appears illustrative rather than exact. The calculated reflection direction still satisfies the requirement of propagating parallel to the negative y-axis. The figure appears to show the photon hitting the bottom of the sphere, which conflicts with the calculated reflection point. The normal direction is being reconsidered. The numerical result is confirmed correct, so the discrepancy likely stems from a different interpretation of the diagram. An "x" mark near the lower right of the circle is being examined. The x mark in the figure likely marks the point (L, h), not the reflection point. The dashed circle probably represents a future position of the expanding sphere. The drawing is therefore not to scale and does not show the photon actually intersecting the sphere at the computed time. The incident ray makes a 30° angle with the horizontal, and the reflected ray makes a 270° angle. The normal direction is being checked against the law of reflection to verify whether it bisects the angle between the two rays. The reflected direction vector matches the formula prediction, so the angle of incidence equals the angle of reflection. The standard definition of the angle of incidence is being verified. The angle between the incident ray and the outward normal is 150°. The standard convention for the angle of incidence is the acute angle between the ray and the normal, so the correct value is 30°. A concrete example with P(1.7, 0.98) and center C(2, 1.5) shows the impact point sits near the bottom of the circle. The x coordinate is not the leftmost point of the sphere, so the figure may be illustrating a trajectory where the photon strikes the lower-left quadrant rather than the absolute left edge. The impact point at 240° lies 30° below the bottom of the sphere, confirming the figure's trajectory is plausible. The speed ratio is being rewritten in a simplified form. The speed ratio is confirmed as u = c(√3 h − L)/(√3 L − h). The reflection condition is written as L/√3 < h < L. The outward normal is confirmed to be (-1/2, -√3/2), matching the earlier result. The impact point coordinates are being checked against the sphere equation. The photon travel time from the origin is being reconsidered using the distance to the reflection point. The x coordinate of the reflection point is derived directly from the geometric constraints. Using the relation h − y = √3(L − x) and the line equation y = x/√3, the expression x = (3/2)L − (√3/2)h is obtained. This confirms the earlier result. The speed ratio is being rewritten by substituting the reflection point’s x coordinate into the travel-time and radius expressions. The derivation is being checked for consistency with the earlier result R = √3 h − L. The radius is being expressed directly in terms of the photon’s travel distance s. Using the relations x = (√3/2)s and y = (1/2)s, and substituting into the earlier geometric constraints, the equation R = 2L − √3s is derived. The reflection point’s y-coordinate is being checked against the sphere equation. The resulting inequality h < √3 L is already satisfied by the existing condition, so no additional constraints are needed. The speed ratio can also be written as u = c(L − √3h)/(h − √3L), which is the negative of the previous expression. The vector analysis fixes the reflection direction to a single angle. Because the normal must be parallel to the difference between the incident and reflected directions, the radius vector can only point at 60° or 240°. This restricts the possible impact points on the sphere to two specific positions. The 60° normal direction would correspond to light exiting the sphere, which is inconsistent with the problem's assumption that the external surface is reflective. Therefore the 240° direction is the correct one. The problem explicitly states the external surface is reflective, so the photon must strike from outside. A photon starting outside cannot cross the surface to enter the sphere, which would require reflection anyway. Therefore only the external reflection case is physically possible. The outward normal pointing in the same direction as the incident ray would mean the photon is moving away from the sphere, not hitting it. The expanding surface must therefore be catching up from behind. The expanding surface can catch up to a photon outside if its speed exceeds the photon's radial velocity. The reflection law is still applied at the moment of impact. The expanding sphere catches up to the photon from behind. The surface reaches the photon's location when its radius equals the photon's distance from the center, so the sphere strikes the photon that is ahead of it. The outward normal points toward the photon at the moment of impact. The surface catches up only when its expansion speed exceeds the photon's radial velocity component in that direction. The dot product d_i · n_out can be positive, meaning the photon is also moving outward but slower than the surface expansion speed. The external surface catches up from behind and strikes the photon. The problem explicitly instructs using the stationary reflection law, so relativistic effects of the moving mirror are ignored. The catching-up scenario is being compared with the photon actively approaching the sphere. Both configurations are being weighed as possible physical cases. Both configurations are external hits. The reflection formula works for either sign of the normal vector. The reflection formula yields the same result for both normal directions. Because the problem instructs ignoring movement effects, the standard stationary reflection law applies with a fixed outward normal. The reflected direction is therefore uniquely determined by the incident ray and that fixed normal axis. The algebraic proof that the reflection formula yields the same direction for both normal orientations is being reconsidered. A physical mirror should reflect light differently depending on which side is struck, so this mathematical symmetry may not be physically sensible. A concrete numerical check with d_i = (1, -1)/√2 and n = (0, 1) confirms the reflected direction is (0.707, 0.707). Reversing the normal to n = (0, -1) yields the same result. This raises doubt about whether the formula correctly describes reflection from the opposite side of a physical mirror. A two-sided mirror resolves the earlier concern. Light from above strikes the bottom face, and the same formula correctly yields an upward reflection regardless of which side is considered. The reflected direction is therefore independent of normal orientation. The formula still predicts an upward reflection when the normal points downward, even if only the top face is reflective. This suggests the calculation may not account for which side of a single-sided mirror is actually struck. The reflection formula always reverses the velocity component along the normal axis and leaves the parallel component unchanged. Therefore the reflected direction is uniquely determined by the normal line itself, regardless of which side of a mirror is struck. This confirms that the outward normal must point along the specific 60°–240° axis found earlier. Both 60° and 240° normal directions yield the same reflected direction. The 240° case is the one where the outward normal points opposite to the incident ray, so the photon strikes from outside. This confirms the external reflection geometry. The two impact-point positions are now explicitly defined. One places the point at C + R·u_vec with outward normal u_vec and positive dot product. The other places it at C − R·u_vec with outward normal −u_vec and negative dot product. The second case is the one already analyzed. Case I is being solved for the speed ratio β. The intersection equations yield s = 2L/(√3 − β) and s = 2h/(1 − √3β). These two expressions for s are now being equated to eliminate the time variable. The speed ratio for Case I is derived as β = (L − √3h)/(√3L − h). This expression differs from the previous result only in sign. The condition for a positive β is now being checked. Case I yields a speed ratio β = (L − √3h)/(√3L − h). The photon's outward radial velocity is c√3/2, so the surface must exceed roughly 0.866c to catch up. The photon's position relative to the sphere radius still needs verification. The Portuguese phrase "o fóton atinge a superfície" is ambiguous. It could mean the photon reaches the surface, suggesting the photon moves toward it, or it could mean the surface strikes the photon. This ambiguity affects whether Case I is physically valid. The figure suggests the photon intersects from outside, consistent with the catching-up scenario. The problem asks for a single velocity u, implying one answer. Both cases satisfy the reflection angles mathematically, so the ambiguity between the two geometries remains unresolved. The problem likely expects a single velocity. The case where the surface catches up from behind may be excluded, since the standard reflection law assumes the ray hits the mirror from outside. The figure suggests the photon strikes the bottom of the sphere. The problem only states that u is less than c, so no other constraints are given. For Case I, the conditions 0 < β < 1 and s > 0 must be checked. The surface must also close on the photon, meaning its speed must exceed the photon's radial velocity component along the normal. The surface must exceed roughly 0.866c to catch up with the photon. A valid hit requires the distance between them to change sign at the moment of contact, meaning the photon starts outside and the expanding sphere reaches it. The surface position R(t) = ut must still be used for the hit condition. Both cases satisfy the reflection geometry, so two possible speeds u could produce a valid contact. The problem asks for the specific velocity that yields the required reflected direction at that instant. The problem explicitly states that after the photon hits the sphere, it propagates parallel to the y-axis in its negative direction. This means the reflected ray must point downward along the negative y-axis. Because the reflection formula yields the same result for both normal orientations, this requirement alone cannot distinguish between the two possible impact points or determine a unique velocity u. The figure shows the photon path passing below the dashed circle without intersecting it, suggesting the drawn circle represents an earlier or later configuration than the actual impact. This raises doubt about whether both reflection cases are physically valid. The figure appears to show the photon path intersecting the sphere near its lower right, with an x marking the impact point. This suggests the contact occurs at a position where x exceeds L, placing the hit on the outer side of the sphere. The figure shows the photon path intersecting the sphere near its lower left, with an x marking the contact point at the right side. The x mark sits on the horizontal dashed line, to the left of the circle. The figure is likely too ambiguous to guide the solution. The phrase "atinge a superfície" suggests the photon moves toward the sphere, so the incident vector points inward and the surface catches up from behind. This favors Case II. The verb "atinge" means the photon reaches the surface, so Case I is physically invalid. Only Case II satisfies the problem's description. The figure shows the photon arrow pointing toward the sphere, confirming it approaches from outside. This supports Case II and excludes Case I. In Case II the hit point lies below and to the left of the center. The outward normal points from the center toward that point, so from the photon's perspective the center sits ahead and upward. This confirms the photon is approaching from outside. Case I is geometrically invalid. The dot product between the photon's velocity and the direction to the center is negative, so the photon moves away from the sphere. It cannot strike the surface from outside. Case I is being reconsidered. The surface might still catch up from behind if its speed exceeds the photon's radial velocity, making the distance between them change sign at contact. The parameters are being checked to see whether positive values of h and L can yield a valid solution with 0 < β < 1. For Case I, the time variable s must be positive. The first equation requires β < √3 ≈ 1.732. The second requires β < 1/√3 ≈ 0.577. Both conditions are satisfied only when β is below roughly 0.577. The surface must exceed roughly 0.866c to catch up with the photon, but Case I requires β below roughly 0.577. These two conditions are mutually exclusive, so Case I is impossible. The formulas therefore describe an exit point where the photon leaves the sphere from inside, not a valid collision. The possibility of an internal intersection is being reconsidered. The quadratic may have a second root after the photon has already entered the sphere. For Case I, β > 0 requires h < L/√3 or h > √3 L. Case I is being reconsidered. The condition h > √3L makes both numerator and denominator of β negative, so β can be positive even though it stays below 1/√3. The explicit position equations are being solved to check whether s can still be positive. The explicit position equations yield R = L − √3h. For the radius to be positive, h must be less than L/√3. The case where h exceeds √3L is therefore impossible. The explicit derivation confirms s = √3L − h and β = (L − √3h)/(√3L − h). Since h < L/√3, both radius and time are positive. The next step is to verify whether β remains below 1. The inequality (1−√3)(L+h)<0 is always satisfied, so β<1 holds for every valid h. The velocity approaches 1/√3≈0.577 as h→0 and vanishes as h→L/√3. The quadratic equation has two roots. The smaller root corresponds to the photon entering the sphere, while the larger root corresponds to the exit point already analyzed. The exit point cannot be reached by a straight path. The photon would have had to enter the sphere earlier, but the external surface is reflective, so it would have reflected off the surface instead of continuing inward. Case II is confirmed physically valid. The quadratic root for this case is the first intersection, so the photon enters from outside and reflects off the surface. The larger root is always the exit point. The radial velocity of the photon is c(d_i·n_out), while the surface moves at speed u. The rate of change of distance between them equals v_rad − u, so the surface must exceed the photon's radial speed to close the gap. The surface must exceed the photon's radial speed to catch it from behind. The two possible reflection angles yield radial speeds of ±(√3/2)c. The positive reflection angle yields a radial speed of (√3/2)c. Because the velocity ratio β is capped at 1/√3, the surface speed u never exceeds this value. The gap therefore opens rather than closes, so no valid contact exists for that normal direction. The y = x/√3 constraint with the linear relations leads to a unique radius R = L − √3h. This is the only valid solution, and it remains positive only when h < L/√3. A concrete example with L=2 and h=1 confirms the geometry. The center is at (2.134, 1.232) on the ray y=x/√3. The quadratic root τ≈2.464 is the first intersection. The second root ≈2.054 is smaller, so it corresponds to an earlier entry into the sphere. The condition for the valid solution is written as L < √3h and h < L. The author briefly wonders whether the reflected photon might immediately rehit the surface. The reflected photon does not need to be checked for further collisions. The problem only requires it to propagate in the specified direction immediately after reflection. The velocity ratio is confirmed as u = c(√3h − L)/(√3L − h). For part b, the author considers whether to also require u < c or to include the condition that the sphere has expanded enough to allow contact. The author concludes that part b requires the condition \frac{L}{\sqrt{3}} < h < L. Outside this range the velocity ratio u becomes non-positive or exceeds c, which violates the physical constraints of the problem. The reflection law is applied to the incident and reflected directions. The normal vector must be parallel to their difference, which points at 60°. For an external first hit, the outward normal is taken as (-1/2, -√3/2). The impact point P is then expressed as the center plus R times this normal. The author is now considering whether to present the solution using only geometric angles, since the problem appears intended for high school level. The author is now attempting a purely geometric derivation using angles, since the IME second-phase physics is aimed at high school students. The incident ray makes 30° with the x-axis and the reflected ray points vertically downward. The normal angle must be determined so that the reflection law holds. The outward normal is derived geometrically from the reflection law. The reflected direction points vertically downward, so the normal angle ψ must be 240° or −120°. This gives the same outward normal vector as before without using vectors. The author is deriving the normal angle geometrically by applying the mirror reflection formula in two dimensions. The reflected ray direction equals twice the normal angle minus the incident angle, which must be checked against the vector result. The author is verifying the standard two-dimensional mirror reflection formula θ_r = 2α − θ_i against the known geometry. A quick check with ψ = −120° and θ_i = 30° yields −270°, which does not match the expected downward reflection direction. The author then begins testing this formula with a horizontal mirror case to confirm the correct relation. The standard reflection formula fails for the tested angles. The author will instead derive the normal angle directly by equating the incident and reflected directions. The standard reflection formula θ_r = 2ψ − θ_i is derived from the angle between the incident ray and the normal. When applied to this problem, it yields ψ = −30°, which contradicts the earlier geometric result of −120°. The author is puzzled by this discrepancy. The author tests the reflection formula θ_r = 2ψ − θ_i against a horizontal mirror example. With normal ψ = 90° and incident angle θ_i = −45°, the formula predicts 225°, which points downward left instead of upward right. This confirms the simple two-dimensional reflection formula does not hold in general. The author suspects the reflection formula needs a 180° offset. Adding 180° to 2ψ − θ_i matches the correct downward reflection direction in both the horizontal mirror and the current problem. The author is now checking whether this adjustment holds generally. The author confirms that adding 180° to the reflection formula is equivalent to subtracting 180° modulo 360, since they differ by a full rotation. They are now testing this adjusted formula against a head-on collision case where the normal points left and the incident ray comes from the right. The author is uncertain whether the 180° offset in the reflection formula is universal. They will rely on the vector-derived normal angle rather than deriving the geometric rule again. The author verifies that the adjusted reflection formula θ_r = 2ψ − θ_i + 180° also holds for ψ = 60°, yielding the correct downward reflection direction. This confirms the formula works for both normal orientations. The author will present the solution using coordinate geometry and normal components rather than purely geometric angles, since the problem requires showing the derivation. The author is now drafting the step-by-step solution in Portuguese. They have defined the incident direction vector, the reflected direction vector pointing downward, and the radial normal vector at the impact point. The vector reflection formula is verified numerically. With the outward normal pointing left and downward, the reflected direction is exactly (0, −1). The negative dot product confirms external incidence. The author then begins substituting this normal into the impact point equation. The author derives the impact point coordinates by substituting the photon trajectory into the normal equations. Using s = ct and β = u/c, they solve the resulting system to obtain s = √3L − h and R = √3h − L. These expressions are valid only when both quantities are positive. The velocity ratio is derived as u = c(√3h − L)/(√3L − h). The author checks the physical constraints: s > 0 requires h < √3L, R > 0 requires h > L/√3, and β < 1 requires h < L. Since h < L already implies h < √3L, the final condition is simply L/√3 < h < L. The author verifies that the derived instant is indeed the first contact with the sphere from outside, confirming compatibility with external reflection. They then prove this is the smallest positive root by checking that the sum of both roots exceeds twice the value s = √3L − h. The author computes the second root of the quadratic equation using the product and sum of roots. They factor s² − R² as (√3 + 1)(L − h) and substitute to show that the larger root is strictly greater than the first contact instant. The author completes the algebraic proof that the second root is strictly larger than the first contact instant. They factor s² − R² as (√3 + 1)(L − h) and (√3 − 1)(L + h), then show that the ratio of the larger root to the smaller equals (L² + h²)/(2(L² − h²)). Since h > L/√3, this ratio exceeds one, confirming that the first contact occurs at the smaller root.

We need answer solve physics problem. Need understand. Let's restate. We have sphere reflective surface, center fixed at coordinates (L,h). Radius R(t)=u t increasing linearly. At t=0 photon leaves origin propagating at speed c in direction of line x = sqrt(3) y, positive senses both axes. In certain instant photon hits sphere surface and then propagates parallel to y axis, in negative sense. Determine: a) speed u of expansion of sphere radius as function of h,L,c; b) condition on h,L for physical possibility. Need solve. Let's parse. Coordinate system: x-y. Sphere center at (L,h) fixed. Radius R(t)=u t. At t=0 radius zero? Sphere starts as point at t=0? R=0. Then expands linearly. Photon emitted from origin at t=0 along line x = sqrt(3) y, positive x,y. So direction vector has slope y/x = 1/√3? Equation x = √3 y => y = x/√3. So angle with x-axis tan θ = y/x = 1/√3 => θ = 30°? Actually tan=1/√3 -> 30°, so direction angle 30° above x-axis. So photon direction unit vector: (cos30, sin30) = (√3/2, 1/2). Or vector proportional (√3,1). Since x=√3 y => point (√3 y, y). Yes direction (√3,1). Speed c => position at time t: r_p(t)= c t (cos30, sin30) = (c t √3/2, c t/2). Let's denote t_hit = τ maybe time when photon hits sphere. At that instant, sphere radius R(τ)=u τ. Photon position P = ( (√3/2)c τ, (1/2)c τ ). It lies on sphere surface: distance from center C=(L,h) equals R=u τ. Equation: ( x_P - L)^2 + ( y_P - h)^2 = (u τ)^2. x_P = (√3/2) c τ; y_P = (1/2) c τ. After reflection, photon propagates parallel to y axis in negative sense. So reflected direction is downward vertical: (0, -1). We need reflection law on sphere at point P: incident ray direction from origin to P: unit vector v_i = (√3/2, 1/2). Outgoing/reflected ray direction v_r = (0, -1). Surface normal at P? For sphere, normal is along radius from center C to P or opposite? Reflection: incident direction v_i reflects to v_r. For specular reflection, the normal direction n (unit pointing maybe from surface outward? or inward?) satisfies v_r = v_i - 2 (v_i · n) n if n points from surface toward incident side? Need careful sign. Actually if n is outward unit normal at surface (from center to point), then reflected direction = v_i - 2 (v_i · n) n. Depending incident from outside approaching surface? The photon hits external surface, medium? Surface perfectly reflector, so reflection formula standard with n pointing outward. Let's verify. If n points outward, incident ray coming from outside to point has direction v_i towards surface (into? Actually photon travels along v_i, from origin to P. The direction v_i points from origin to P, so at P it is arriving. For reflection, the incident ray direction vector is v_i (towards surface/point). Reflection formula for mirror with normal n (outward from surface) is v_r = v_i - 2(v_i·n)n. Yes because v_i component normal reversed. Alternatively if n inward, use different sign. Let's test simple: surface horizontal y=0, incident downward (0,-1) hits, reflect upward (0,1). n outward upward (0,1). v_i=(0,-1). v_i·n=-1. v_r = (0,-1)-2(-1)(0,1)=(0,1). Correct. So formula with outward normal works. Good. So n = (P-C)/R = (x_P-L, y_P-h)/R. We know v_r = (0,-1), v_i=(√3/2, 1/2). Then v_r = v_i - 2(v_i·n)n. Let n=(n_x,n_y). Equivalently v_i - v_r = 2(v_i·n)n. Compute left Lvec = v_i - v_r = (√3/2, 1/2) - (0,-1) = (√3/2, 3/2). This vector must be parallel to n. So n is along (√3/2, 3/2), unit. Thus normal direction at P is proportional to (√3, 3). Let's compute magnitude: |(√3/2, 3/2)| = sqrt(3/4 + 9/4)=sqrt(12/4)=sqrt(3) ≈1.732. Unit n = (√3/2, 3/2)/√3 = (1/2, √3/2)? Wait (√3/2)/√3 = 1/2; (3/2)/√3 = (3)/(2√3)=√3/2. So n=(1/2, √3/2). That means normal angle 60° from x-axis. Also v_i = (cos30,sin30), v_r=(0,-1) angle -90°. Incidence angle? Normal 60°, incident direction 30°, difference 30°. Reflected? v_i reflected across normal at 60° should angle 90°? Wait reflect angle 30° across 60° => 90°? Actually angle of incidence = angle between v_i and normal = 30°; reflected direction angle = 2*normal angle - incident angle = 2*60 -30 =90° (upward vertical) not downward. Hmm discrepancy. Let's check formula. If n=(1/2,√3/2) outward, v_i=(√3/2,1/2). Dot = (√3/2)(1/2)+(1/2)(√3/2)=√3/2 ≈0.866. v_r = v_i - 2*(√3/2)*n = v_i - √3*(1/2,√3/2) = (√3/2,1/2) - (√3/2, 3/2) = (0, -1). Yes v_r=(0,-1). Wait angle of v_r is -90°, yes. But reflection across normal formula if normal angle 60°, incident angle 30°? Let's compute: v_i angle 30°, normal angle 60°. Angle between =30°. Reflected should angle = 60 + (60-30)=90°? Actually direction after reflection is mirror: angle_r = 2θ_n - θ_i = 120° -30°=90° (if using line angle modulo 180? Hmm vectors: n angle 60°, v_i incoming direction 30°. At point, the ray arrives with direction v_i along 30°. The normal line direction outward at 60°. The angle from incident to normal =30°. Reflected direction is on other side of normal by 30°, so along 60+30=90° (upward). But our formula gave -90° (which is collinear with 90°, just opposite direction? Wait vertical downward is angle -90° or 270°. Reflection formula gives vector (0,-1) whose line angle is vertical, but direction is downward. But physical reflected ray direction should be along line? Actually at point, ray reflects; direction vector v_r should point along the outgoing ray. Which way along vertical? Downward? Let's visualize: Incident ray from origin at 30° to point P. Normal at P points from center (L,h) to P. We found n=(1/2,√3/2) outward, i.e. from center to P, pointing up-right at 60°. P likely to the right/above center? Let's see. If normal outward from center to P, then P = C + R n = (L + R/2, h + R√3/2). Incident direction from origin to P is at 30°, so P has coordinates (r cos30, r sin30) = (√3/2 r, 1/2 r). Compare y = (1/2)r, x=(√3/2)r. Normal direction angle 60° means P-C along 60°, so vector P-C = R(1/2, √3/2) = (R/2, R√3/2). Then y_P - h = R√3/2 positive; x_P - L = R/2 positive. Possible. Incident ray arrives at P along 30° direction (towards P, but from origin, vector from origin to P is (√3/2 r, 1/2 r) angle 30°). At P, the ray is traveling in direction v_i = (√3/2, 1/2) (same angle). The reflected ray direction from formula v_r=(0,-1) downward. Is that physically possible? The normal line is 60°; incident direction 30° (below normal by 30°). Reflect across normal line: outgoing direction angle 90° (above normal by 30°) if using undirected line? But vector formula gives opposite direction along same line? Wait vector reflection formula v_r = v_i - 2(v_i·n)n with n pointing from surface to center or from center to surface? Let's derive carefully. Suppose surface normal n points from surface into the medium where light comes from? Actually for reflection at interface, if n is unit normal pointing from surface into the incident medium (i.e., outward from surface toward incoming light? Standard: incident vector I points towards surface, normal N points from surface into medium where light originated? Hmm ambiguous. Standard in optics: Let N be unit normal pointing from surface toward the incident side (opposite to direction of incident light? or same?). Usually reflection formula: r = i - 2(i·n)n where n is normal pointing outward from surface (toward the incident light side), and i is incident direction vector pointing toward surface (into surface). Then r points away. Example horizontal mirror y=0, light from above traveling downward: i=(0,-1), surface normal outward upward n=(0,1). i·n=-1. r = (0,-1)-2(-1)(0,1)=(0,1), correct. So n upward, i downward: v_r upward. Good. In our case, n=(1/2,√3/2) outward from center to P? Is that "toward incident side"? Incident light comes from origin to P; at P, the light is traveling along v_i=(√3/2,1/2). The surface at P: sphere centered at C. The outward normal from sphere center to P is n=(1/2,√3/2). But the light arrives from origin, which is outside the sphere? Actually photon from origin to P; is origin outside sphere? At hit time, sphere radius R; center C; distance from C to origin? We don't know. The photon travels in space, hits external surface. The light approaches P from outside, likely from origin direction. At P, the incident light direction is v_i (from origin to P), so the "incident side" is outside, and the normal pointing into incident side should point from surface outward? If sphere surface, outward normal points from center to surface. But at P, the inside of sphere is toward center; the incident side (outside) is generally away from center? Actually light hits from outside, so just before reflection it is outside sphere. The normal pointing into the incident medium (outside) at P is from surface outward away from center, i.e., n_out = (P-C)/R = (1/2,√3/2). That is what we used. Then reflection formula gives r = i - 2(i·n_out)n_out. We found r=(0,-1) downward? But if incident i is 30° and n_out 60°, then i - r = 2(i·n)n? Wait compute using formula: i=(0.866,0.5), n=(0.5,0.866), i·n=0.866? Actually 0.866*0.5 +0.5*0.866 =0.866? 0.433+0.433=0.866 yes =√3/2. r = i - √3 n = (0.866-0.866, 0.5-1.5)=(0,-1). Yes downward. But geometric reflection across tangent plane should yield upward 90°? Let's test with a simple physical: horizontal surface y=0, incident from origin at angle 60° hitting at origin? Normal upward; incident direction up-right? If light traveling toward origin from lower left? Hmm. Let's do a simple check: mirror line vertical x=0? Incident from left horizontal (1,0), reflects to right (-1,0). Normal outward left (-1,0). Formula i=(1,0), n=(-1,0), i·n=-1, r=(1,0)-2(-1)(-1,0)=(1,0)-2(1,0)=(-1,0). Good. Here incident direction points toward mirror (right), normal points left (toward incident side? Actually incident side outside left, normal outward from mirror to left? If mirror vertical, interior right? Wait surface x=0, normal outward from surface into left medium? If reflective surface maybe orientation? Let's imagine light from left hitting mirror x=0, reflects left. Normal outward from mirror to left = (-1,0) points into incident medium (left). Incident vector i=(1,0) points toward mirror (right). Formula with n outward (-1,0) gave r=(-1,0) (back left). Good. So n points into incident medium (opposite to incident direction). In our sphere, n_out = (P-C)/R points from center to P, i.e., outward from sphere. Incident medium is outside sphere; at P, the outside is along n_out? Center inside; outward vector points away from center, so yes toward outside. But incident light is traveling along v_i from origin to P; at P, is it traveling from outside? The vector from P to origin is -v_i. The normal n_out points? Since P lies on sphere, origin maybe outside, so direction from P to origin should have positive dot with n_out? Let's see. If origin outside, vector from P to origin (approx -v_i) should point outward (away from center) relative to tangent plane? Actually if sphere is centered at C and P on surface, outside is n_out direction; origin is outside, so vector from P to origin should have positive projection along n_out (since origin lies outside half-space n_out·(X-P) > 0? For sphere, outside is n_out·(X-P) >= 0? Let's check: sphere center C, point P=C+R n_out. For any point X outside if |X-C|>R. At P, tangent plane normal n_out; outside is direction n_out? Since moving from P in direction n_out increases distance from C, yes outside. So vector from P to origin = O - P. If origin outside, dot with n_out should be positive. n_out = (0.5,0.866). P=(x_P,y_P). O=(0,0). P-O=(x_P,y_P) = r v_i = r(0.866,0.5). Then n_out·(O-P) = - n_out·(P-O) = -[(0.5*0.866 +0.866*0.5)r] = - (0.433+0.433)r = -0.866 r <0. So origin is inside relative to tangent plane? Wait if dot negative, origin lies on inside side (toward center). That would mean light comes from inside sphere? But sphere radius R = u τ; at hit time, origin distance to center maybe <R? Let's compute: P = r v_i, C=P - R n_out = r(0.866,0.5) - R(0.5,0.866). For origin to be outside, distance C-O > R? Actually outside means |C| > R? Sphere center at distance |C| from origin. At time τ radius R; origin outside if |C| > R? But maybe photon emitted from origin when sphere radius zero; at later time τ, sphere grew. Could origin be inside sphere? If origin inside, then photon would have interacted earlier? But problem says photon at certain instant hits surface, so it was outside before. Let's check distance from origin to center at τ maybe > R? Compute later. First solve equations. Unknowns: τ, u. Conditions: 1) P on sphere: |P - C|^2 = u^2 τ^2. 2) Reflection direction: v_i - v_r = 2(v_i·n)n as vectors. Since n=(P-C)/R. Let's derive scalar equations from reflection. From v_i - v_r = (√3/2, 3/2) = λ n where λ = 2(v_i·n). Since n unit, magnitude of left = √3, so λ = ±√3? Let's compute λ = 2(v_i·n). Also n = left/|left| = (√3/2,3/2)/√3 = (1/2, √3/2) times sign? Wait left vector = (√3/2,3/2) magnitude √3. n = left/|left| = (1/2, √3/2) if λ positive? Left = λ n -> λ n = (√3/2,3/2). If n = (1/2,√3/2), λ=√3 gives (√3/2,3/2). Yes. If n = (-1/2,-√3/2), λ=-√3. But n = (P-C)/R, so two possible normals opposite; λ=√3 gives n=(1/2,√3/2). λ=-√3 gives n=(-1/2,-√3/2). Let's see which is physical. Also v_i·n = (√3/2,1/2)·(1/2,√3/2)=√3/2 ≈0.866. Then 2(v_i·n)=√3, matches λ=√3 for n outward. If n opposite, v_i·n=-√3/2, λ=-√3, left = -√3 n = (√3/2,3/2) for n=(-1/2,-√3/2). So both possible algebraically? But geometric reflection formula with unit n: v_r = v_i - 2(v_i·n)n. If n opposite, v_r = v_i -2(-√3/2)(-n0?) Let's test: let n0=(1/2,√3/2), n=-n0. v_i·n = -√3/2. v_r = v_i -2(-√3/2)(-n0) = v_i - √3 n0 = (0,-1). Same. So both n and -n are normal lines same; formula gives same reflected direction? Actually if n -> -n, term -2(v_i·n)n = -2(-v_i·n0)(-n0)? Wait v_i·(-n0)=-v_i·n0. Then -2(v_i·(-n0))(-n0)= -2(-a)(-n0)= -2a n0? Actually -2 * (-a) * (-n0) = -2a n0? Let's compute: -2(-a)(-n0)= -2a n0? yes. Same as original? Original -2a n0. So same. Good. Thus normal direction line is along (1/2,√3/2). So from reflection: P - C = R n0 = R (1/2, √3/2), where n0=(1/2,√3/2), possibly plus sign? Actually P-C = R n with n either n0 or -n0. Since P is on sphere. But P-C vector is actual radius vector, so P-C = R n_actual. Which sign? We need determine. The line through C and P has direction n0. So P lies on line C + s n0 with s=±R. Also P=r v_i = r(√3/2,1/2), with r = c τ. So we have vector equation: r (√3/2,1/2) - (L,h) = s (1/2, √3/2), where s = ± R = ± u τ? Actually if n_actual = ± n0, then P-C = R_actual n_actual = s n0 with s=R if n_actual=n0, s=-R if n_actual=-n0. But R=u τ positive. So s=±u τ. Let's solve components. Let r = c τ. Also P = r (√3/2,1/2). Then: x: (√3/2) r - L = s/2 y: (1/2) r - h = (√3/2) s Multiply first by 2: √3 r - 2L = s. (1) Second: r/2 - h = (√3/2)s => multiply 2: r - 2h = √3 s. (2) From (1): s = √3 r - 2L. Substitute into (2): r - 2h = √3 (√3 r - 2L) = 3 r - 2√3 L => r - 2h = 3r - 2√3 L => -2h = 2r - 2√3 L => -h = r - √3 L => r = √3 L - h. Thus r = c τ = √3 L - h. Hmm positive? Since cτ positive, require √3 L > h. That might be condition part b? But wait maybe sign alternative for s? Let's do if P-C = -R n0? Then s=-R. Let s unknown = ±uτ. Our equations same but s can be negative. Solving gave r = √3 L - h regardless? Let's re-derive with general s: equations: √3 r - 2L = s r - 2h = √3 s Eliminate s: s = √3 r - 2L; r - 2h = √3(√3 r -2L)=3r -2√3 L => r -2h =3r -2√3 L => -2h=2r -2√3 L => r = √3 L - h. Yes r fixed. Then s = √3 r - 2L = √3(√3 L - h) -2L = 3L - √3 h -2L = L - √3 h. So s = L - √3 h. But s = ±uτ = ±R. Since R=uτ positive, s sign depends. Also r=cτ = √3 L - h. Then u = R/τ = |s|/τ = |L - √3 h| / (r/c) = c |L - √3 h| / (√3 L - h). But part a asks u in terms h,L,c. But need physical constraints maybe s positive? Let's see. Alternatively maybe we missed another possible reflection path? Let's examine geometry. Photon from origin hits sphere, then reflected parallel to y negative (downward). Normal line direction determined by v_i and v_r. We found normal line must be along (1/2,√3/2). That's one line (or opposite). Then P lies at intersection of incident ray r v_i and line from C in direction ±n0? Actually P on sphere means P-C parallel to n0, yes. The intersection of line through C direction n0 with incident ray has unique solution, giving r=√3 L - h. But also P must be at distance R from C; s = L - √3 h. Since R=uτ = |s|. And τ = r/c = (√3 L - h)/c. So u = |s|/τ = c |L - √3 h| / (√3 L - h). But there might be physical condition requiring r>0? Since τ>0, need √3 L - h > 0. Also h>0, L>0. Also perhaps origin outside sphere before hit? That would impose something? Also s maybe positive? Let's check sign to see if sphere is expanding; radius R=uτ positive, so s=±uτ; either sign possible? Actual n_actual = (P-C)/R. If s>0, n_actual = n0; if s<0, n_actual = -n0. But reflection direction formula with n outward? Wait n_actual = (P-C)/R is outward from center to P. That's n_out. We earlier assumed n_out = (1/2,√3/2) (n0) and got r=(0,-1). But if n_out = -n0 (s<0), does formula still give same reflected direction? Let's check with actual n_out = -n0 = (-1/2,-√3/2). Incident v_i=(√3/2,1/2). Dot v_i·n_out = -√3/2. Then r = v_i -2(-√3/2)(-n0)= v_i - √3 n0 = (0,-1). Yes same. So both possible for reflection: normal line same but outward normal could point opposite along line; still reflection same because formula invariant under n -> -n? Let's verify: v_r = v_i -2(v_i·n)n. If n -> -n, dot -> -dot, n -> -n: term -2(-dot)(-n)= -2 dot n? Wait compute: -2(v_i·(-n))(-n) = -2(-v_i·n)(-n) = -2(v_i·n) n. Same. So reflection direction unchanged. So n_actual sign can be either; sphere center can be on either side of P along normal line. But geometry with C fixed and P maybe inside? Let's see. However, sphere centered at C (L,h), expanding. At t=0 radius zero. Photon starts at origin outside? The photon path must hit sphere for first time at τ. For this to be possible, origin must be outside sphere at τ (or maybe on surface? no). Also the segment from origin to P should not intersect sphere earlier; since sphere radius grows from 0, the first contact condition imposes. Need determine correct u maybe positive and condition. Let's analyze. Given r = cτ = √3 L - h. For τ>0, require √3 L > h. Also s = L - √3 h. R=uτ = |s|? Actually R = distance P-C = |s|? Since P-C = s n0 with n0 unit, |P-C|=|s|. Yes R=|s|. So u = |s|/τ = c |L - √3 h| / (√3 L - h). But maybe because R=ut and t=τ, u = |s|/τ. Yes. Part a asks velocidade u de expansão do raio da esfera em função de h,L,c. Likely u = c (L - √3 h)/(√3 L - h) perhaps assuming L>√3 h? But need sign? They might expect u = c (L - √3 h)/(√3 L - h) or maybe c (√3 h - L)/(√3 L - h)? Let's investigate. Maybe I made sign error? Let's solve reflection equations perhaps with correct normal direction and incident ray. Let's derive vectorially more systematically. Let incident unit vector u_i = (√3/2, 1/2). Let reflected vector u_r = (0, -1). Reflection condition: u_i - u_r = 2 (u_i · n) n. So u_i - u_r = (√3/2, 1/2) - (0,-1) = (√3/2, 3/2). This vector is parallel to n, so n = ± (u_i - u_r)/|u_i - u_r| = ± ( (√3/2,3/2) / √3 ) = ± (1/2, √3/2). So normal vector n = (P-C)/R = ±(1/2,√3/2). Good. Then P = r u_i = r(√3/2, 1/2). C=(L,h). Thus r(√3/2, 1/2) - (L,h) = ± R (1/2,√3/2). Since R=u τ, but note R also = distance, so ± sign belongs to vector equality. But we can solve for r and R as before with sign σ = ±1: r(√3/2) - L = σ R/2 r(1/2) - h = σ (√3/2) R. Multiply first by 2: √3 r - 2L = σ R. Second by 2: r - 2h = σ √3 R. Eliminate R: From first, σ R = √3 r - 2L. Replace R in second: r - 2h = √3 (√3 r - 2L) = 3r - 2√3 L => same r = √3 L - h. Then σ R = √3 r - 2L = √3(√3 L - h)-2L = 3L -√3 h -2L = L - √3 h. So R = σ (L - √3 h)? Actually σ R = L - √3 h; multiply by σ: R = σ (L - √3 h) since σ^2=1. So R = ±(L - √3 h). Since R≥0, the sign must match sign of (L - √3 h): if L>√3 h, σ=+1 and R=L-√3h; if L<√3 h, σ=-1 and R=√3h-L. So R=|L-√3h|. Then u=R/τ = c |L - √3 h|/(√3 L - h). Since τ>0 requires √3L - h >0. But maybe there is additional condition that the photon hits the sphere from outside, i.e., the origin is outside at time τ. Let's check. Distance from origin to center C: D = sqrt(L^2 + h^2). At τ, sphere radius R. Origin outside if D > R? Actually if D < R, origin inside sphere; but photon starts at origin at t=0 when sphere point; as sphere grows, it could engulf origin before photon hits? But photon leaves at t=0; at t=0 sphere radius 0. If later radius exceeds D, origin would be inside sphere, but photon already traveling. The first contact for photon with expanding sphere: the sphere radius at time t is u t. The photon position at time t is P0(t)=c t u_i. The distance from center to photon at t: d(t)=|c t u_i - C|. The photon hits when d(t)=R(t)=u t. Also before hit, d(t)>u t (outside sphere). After hit, d(t)=u t. So condition for first hit at τ: d(τ)=uτ and d(t)>ut for 0<t<τ. Since both d and u t start at t=0? At t=0, d(0)=|C|=√(L^2+h^2) >0 = R(0). Good. d(t) maybe increasing? Need ensure no earlier intersection. We can derive inequality. Also after reflection, photon goes downward along x=x_P. But the sphere continues expanding; maybe the photon might be re-hit? Problem says after reflection propagates parallel to y axis negative, maybe it escapes? Physical possibility condition may include that after reflection the photon doesn't hit sphere again? Let's examine. Part b asks condition on h,L for the physical imposition of problem to be possible. Could be from τ>0 and perhaps √(L^2+h^2) > R? Let's analyze. Let's derive geometry more clearly. Coordinates: Origin O=(0,0). Center C=(L,h). Incident ray: x=√3 y, y≥0. Direction u_i=(√3/2,1/2). Photon position P=r u_i = (√3 r/2, r/2), where r=c t. Sphere radius R(t)=u t = (u/c) r = k r, where k=u/c <1 (given u<c). So at time parameter r (distance along ray), sphere radius = k r (since t=r/c). Center fixed. Condition for photon at distance r along ray to be on sphere: |P(r)-C| = k r. Also first intersection. We already solved reflection to get r = √3 L - h, k = R/r = |L - √3 h|/(√3 L - h) (since R=k r). But note if r positive, k positive. Need k<1 (u<c). Let's verify this condition may imply physical inequality. Let's compute |P-C| = R = k r. At solution, P = r u_i = r(√3/2,1/2). C=(L,h). Vector P-C = (r√3/2 - L, r/2 - h). Its magnitude squared = R^2 = k^2 r^2. Also from reflection, we know P-C = σ R n0, so components: r√3/2 - L = σ R/2 r/2 - h = σ R√3/2. Squaring/summing gives R^2. But we already used. Alternatively, k = u/c = R/(cτ)=R/r. Since R = |L - √3 h| and r = √3 L - h (positive), k = |L-√3 h|/(√3 L - h). But also k<1. Since u<c, k<1. Does this impose condition? Let's analyze cases. Also R = k r. From σ R = L - √3 h. Since k>0, R positive. Given r = √3 L - h >0 => √3 L > h. (Condition A) Then k = |L - √3 h|/(√3 L - h). Since denominator positive. Case 1: L > √3 h. Then numerator = L - √3 h >0. k = (L - √3 h)/(√3 L - h). Need k<1 => L - √3 h < √3 L - h => -√3 h + h < √3 L - L => (1-√3)h < (√3-1)L => (1-√3)h < (√3-1)L => since 1-√3 = -(√3-1), inequality becomes -(√3-1)h < (√3-1)L => divide positive? (√3-1)>0, so -h < L => h > -L, true since h>0, L>0. So k<1 always in case 1? Let's compute: k = (L - √3 h)/(√3 L - h). Is this <1? Equivalent positive denom: L - √3 h < √3 L - h => 0 < (√3-1)L + (√3-1)h = (√3-1)(L+h) >0 true. Yes. So case 1 automatically k<1. Case 2: L < √3 h. Then numerator = √3 h - L >0. k = (√3 h - L)/(√3 L - h). Need k<1? Denom positive by r>0. Inequality: √3 h - L < √3 L - h => √3 h + h < √3 L + L => h(√3+1) < L(√3+1) => h < L. Wait let's compute: √3 h - L < √3 L - h => √3 h + h < √3 L + L => h(√3+1) < L(√3+1) => h < L. Since divide positive. So k<1 iff h < L. But in case 2 L < √3 h; combined h < L < √3 h => h>0, L>0, and L>h. So case 2 possible if h < L < √3 h. Then k<1. If L ≤ h < √3 h? Then k≥1, violates u<c. So additional condition for case 2: L > h. But wait case 1 k<1 always. Thus combined with r>0 (√3L > h), and k<1, we get: If L > √3 h (case1): ok. If L ≤ √3 h: then need h < L to have k<1? Let's check case L=√3 h: numerator zero => k=0 (u=0?) but u>0? Sphere radius not expanding? u<c but also u maybe positive? Given u<c but not explicitly >0? Radius increases? It says R=ut, t≥0, u maybe? "velocidade de expansão" positive? Probably u>0. If L=√3h, R=0 at hit? Photon hits at center? Actually if L=√3h, then r=√3L-h = √3L - L/√3? Wait h=L/√3, r=√3L - L/√3 = (3L-L)/√3=2L/√3 >0. R=|L-√3h|=0 => sphere radius zero at hit, impossible (photon hits point sphere at t=0? But t>0, radius zero only at t=0). So u=0 not allowed if R=0 and t>0. So need L ≠ √3 h. Also u<c. Maybe part b condition is something like h < √3 L and h ≠ √3 L? Wait if L<√3h but h<L allowed. So condition ensures u<c and r>0 and R positive: Maybe h < √3 L and L ≠ √3 h, and perhaps also h < L? Let's derive final. But maybe there are other physical constraints: After reflection, photon moves parallel to y axis in negative sense. Does it escape without re-hitting? Let's check. After reflection at P, photon moves downward along vertical line x=x_P = r√3/2. Sphere expands with radius R(t)=u t = k r? Wait t after hit? At hit time τ = r/c. At later time t' = τ + Δt, sphere radius R=u t' = uτ + u Δt = R0 + uΔt, where R0=k r. Center fixed. Photon position: x=x_P, y=y_P - c Δt = r/2 - cΔt. Distance squared from center: (x_P - L)^2 + (y_P - cΔt - h)^2. At Δt=0, equals R0^2. Since photon moves downward (negative y), while center location? Need ensure distance > R(t') for Δt>0, i.e., photon outside sphere after reflection. Condition maybe imposes inequality. Let's analyze. At P, reflected direction v_r=(0,-1). After reflection, photon position Q(Δt) = P + c Δt (0,-1) = (x_P, y_P - cΔt). Distance squared from C: d^2(Δt) = (x_P - L)^2 + (y_P - h - cΔt)^2. Sphere radius squared: R^2(Δt) = (R0 + uΔt)^2 = R0^2 + 2R0 u Δt + u^2 Δt^2. At Δt=0, d^2=R0^2. Need d^2 > R^2 for Δt>0 (outside). Expand difference: d^2 - R^2 = [ (x_P-L)^2 + (y_P-h)^2 ] - R0^2 + [ -2 cΔt (y_P-h) + c^2 Δt^2 ] - [2R0 u Δt + u^2 Δt^2]. First bracket zero. So difference = Δt [ -2c(y_P - h) - 2R0 u ] + Δt^2 (c^2 - u^2). Since c^2 - u^2 >0 (given u<c), for small Δt positive, need coefficient of Δt ≥0? Actually for Δt>0, difference >0 requires -2c(y_P - h) - 2R0 u >0 (if coefficient positive) or if negative, small Δt difference negative -> inside -> re-hit. So condition: c(y_P - h) + R0 u < 0? Wait difference = Δt [ -2c(y_P - h) - 2R0 u ] + Δt^2(c^2-u^2). For small Δt, sign by bracket. Need >0 => -2c(y_P - h) - 2R0 u > 0 => c(y_P - h) + R0 u < 0? Actually divide -2: -c(y_P-h) - R0 u >0 => c(y_P-h) + R0 u <0. Since c>0, R0>0, u>0, this requires y_P - h <0 (photon below center). Let's see. At P, y_P = r/2. h =? From reflection equation: r/2 - h = σ R0 √3/2. If σ=+1 (case L>√3h), then r/2 - h = R0√3/2 >0 => y_P > h. Then bracket = -2c(positive) -2R0u <0 -> difference negative for small Δt, meaning photon immediately re-enters sphere? That would violate "passa a propagar-se" maybe? But maybe the sphere surface is expanding; after reflection, if center below, photon moving down may still be outside? Let's check. If y_P > h (photon above center), moving downward (toward center) likely enters sphere. But maybe because sphere expands, radius increases; distance decreases? Need check. If photon above center and moves downward, distance to center may decrease initially; if sphere expands, may still be outside? Let's analyze. If σ=+1: P-C = R0 n0 with n0=(1/2,√3/2), so y_P-h = R0√3/2 >0. Photon velocity v=(0,-c), so d/dt (distance squared) = 2(P-C)·v = 2[(x_P-L)*0 + (y_P-h)*(-c)] = -2c(y_P-h) <0, so distance decreasing initially. Sphere radius increasing at rate u; initially distance derivative = - (y_P-h)/R0 * c? Actually d(distance)/dt = (P-C)/R0 · v = n0 · (0,-c) = -c n0_y = -c√3/2. Sphere radius derivative = u = k c? Actually u<c. Distance derivative magnitude = c√3/2 ≈0.866c > u maybe? So distance decreases faster than radius increases? Compare: distance derivative = -c√3/2 ≈ -0.866 c; radius derivative = u < c, so distance may decrease faster than radius, so photon enters sphere. But maybe later? Let's compute difference derivative at 0: d(d^2 - R^2)/dt = 2(P-C)·v - 2R u = 2[ (P-C)·v - R u ] = 2[ -c(y_P-h) - R u ] = -2c R [ (y_P-h)/R + u/c? Wait R0 u? Actually (P-C)=R n0, so (P-C)·v = R n0·v = R (0 + (-c) n0_y)= -c R n0_y. R u term = R u. So bracket = -c R n0_y - R u = -R (c n0_y + u) <0 (since n0_y>0). So difference decreasing => inside immediately. Thus if σ=+1, after reflection photon would re-enter sphere, not escape. But problem says "passa a propagar-se paralelamente ao eixo y, em seu sentido negativo", maybe implies it does propagate (and escapes?) but not explicitly says it doesn't hit again. However physical possibility might require after reflection it doesn't hit sphere again? The phrase "passa a propagar-se" maybe just direction after reflection, could be momentary; but in ideal reflection at surface, outgoing ray should be outside sphere, so need σ=-1? Let's check. If σ=-1 (case L<√3h): Then P-C = -R n0 = (-R/2, -R√3/2), so y_P - h = -R√3/2 <0 => photon below center. Then (P-C)·v = (-R n0)·(0,-c) = R c n0_y? Wait -n0=( -1/2,-√3/2); dot with (0,-c) = (-√3/2)(-c)= +c√3/2 >0, so distance increasing initially. Then d^2-R^2 derivative = 2(P-C)·v - 2R u = 2R(c√3/2 - u). Need >0 for outside? Since u<c, c√3/2 may be > u? Not necessarily if u close c; condition c√3/2 > u => k<√3/2. But derivative at 0 positive if c√3/2 > u. If derivative ≤0, photon might initially be inside? But at t=0 difference zero; if derivative negative, immediately inside? Actually if d^2 < R^2 after, then it would be inside, but it just reflected from surface; could it enter sphere? For reflection, outgoing ray should be outside, so derivative should be nonnegative. So need c√3/2 ≥ u? Let's compute. u = R/r = |L-√3h|/(√3L-h) c. In case σ=-1, L<√3h and r=√3L-h >0; R=√3h-L. Need c√3/2 - u >0? Let's verify. u = c (√3h - L)/(√3 L - h). Condition c√3/2 > u => √3/2 > (√3h - L)/(√3L - h). Multiply positive denominator: (√3/2)(√3L - h) > √3h - L => (3L/2 - √3 h/2) > √3 h - L => (3L/2 + L) > √3 h + √3 h/2 => (5L/2) > (3√3 h/2) => 5L > 3√3 h => L > (3√3/5) h ≈1.039 h. So condition L > ~1.039h. But maybe not needed? Alternatively if derivative negative, photon might still be outside? Let's analyze exact. But perhaps part b condition is precisely about the sign of σ? Wait from reflection, normal could be either but physical after reflection requires outgoing ray outside sphere. Let's derive general condition for photon not intersect sphere immediately after reflection: for small Δt>0, distance > radius. As above, need derivative at 0 ≥0? Since difference has positive quadratic coefficient (c^2-u^2>0). For small Δt, if linear term negative, difference negative for small t -> inside. If linear term zero, quadratic positive -> outside for small t. So necessary and sufficient: at t=0, d(d^2 - R^2)/dt ≥0? Actually if derivative >0, outside; if =0, outside due to quadratic; if <0, inside. So need -2c(y_P - h) - 2R u ≥0 => c(y_P - h) + R u ≤0. Since R,u>0, implies y_P - h <0. So σ must be -1 (photon below center). Also need c|y_P - h| ≥ R u? Actually y_P-h = -R√3/2. Then condition c R√3/2 ≥ R u => c√3/2 ≥ u => u ≤ c√3/2. Since u<c, this is a stricter upper bound. So physical condition includes u ≤ c√3/2. But wait the problem statement says "para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos." So reflection is instantaneous at surface with mirror; after reflection, the photon direction is vertical down. It doesn't mention it must escape? "passa a propagar-se paralelamente ao eixo y, em seu sentido negativo" implies it does propagate; if it immediately re-entered sphere, the reflection wouldn't be a clean escape. So likely must have after reflection outside sphere for at least some interval. But maybe the sphere is expanding and might catch up later? Need check if long-term. Let's analyze after reflection exact condition for never re-hit (or at least propagate away). Photon vertical down x=x_P, y decreasing. Sphere center C=(L,h), radius R(t')=u t' = u(τ+Δt). Let's derive position. At reflection time τ, r=cτ, P=(r√3/2, r/2). Let y0 = r/2, x0=r√3/2. Sphere center maybe below/above. Let Δ = cΔt ≥0. Photon position y = y0 - Δ, x=x0. Distance squared: D^2 = (x0-L)^2 + (y0 - Δ - h)^2. Sphere radius squared: R^2 = u^2 (τ + Δ/c)^2 = (uτ + (u/c)Δ)^2 = (R0 + k Δ)^2, where k=u/c, R0=uτ=k r? Actually R0=k r (since r=cτ, uτ=k r). Also R0 = distance at hit. Let's express x0-L and y0-h from reflection. We have vector P-C = σ R0 n0 = σ R0 (1/2,√3/2). So: x0 - L = σ R0/2 y0 - h = σ R0 √3/2. Then D^2 = (σ R0/2)^2 + (σ R0√3/2 - Δ)^2 = R0^2 [1/4 + 3/4 - σ√3 Δ/R0 + Δ^2/R0^2] = R0^2 - σ√3 R0 Δ + Δ^2. R^2 = (R0 + k Δ)^2 = R0^2 + 2k R0 Δ + k^2 Δ^2. Difference F(Δ)=D^2 - R^2 = Δ^2(1 - k^2) + Δ[ -σ√3 R0 - 2k R0 ]. = Δ [ (1-k^2) Δ - R0(σ√3 + 2k) ]. For Δ>0, sign of F depends on bracket. Need F>0 for outside. Since 1-k^2>0, bracket increasing with Δ. At Δ=0, bracket = -R0(σ√3 + 2k). If σ=+1, negative => F negative for small Δ (inside). If σ=-1, bracket = -R0( -√3 + 2k) = R0(√3 - 2k). If k < √3/2, bracket positive at Δ=0, so F>0 for small Δ; if k>√3/2, bracket negative at Δ=0, so F<0 for small Δ; if k=√3/2, bracket zero, F=(1-k^2)Δ^2>0 for Δ>0. Then for larger Δ, bracket may become positive when Δ > R0(σ√3+2k)/(1-k^2) if σ=+1? Let's see. If σ=+1, bracket negative at 0, becomes zero at Δ0=R0(√3+2k)/(1-k^2)>0; for Δ>Δ0, F>0 (outside). So photon initially inside, then later exits? But if it enters sphere, it would have crossed surface; not physical as "propagating outside" after reflection. Usually reflected wave should be outside immediately. Therefore physical condition for outgoing ray not immediately entering sphere is σ=-1 and k ≤√3/2 (u ≤ c√3/2). But given u<c, and u maybe maybe? Let's see if k can exceed √3/2? Since k<1, possible. If k>√3/2, photon initially enters sphere (would refract/absorb? But surface perfectly reflector, no transmission; so shouldn't enter). Thus exclude. But maybe the problem's "para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos" means we ignore surface motion for reflection direction but not for escape? Hmm It says determine direction of reflected photon, disregard effects of motion of sphere surface and relativistic effects. So reflection simply geometric mirror at instantaneous position. Then after reflection, maybe sphere continues expanding; they might not ask about later. But part b "condição à qual as coordenadas h e L devem satisfazer para que a imposição física do problema seja possível" likely includes ensuring photon actually hits sphere (origin outside before) and after reflection it goes away? Let's investigate. Let's derive first-hit condition. At time t<τ, photon position r(t)=c t u_i = t c (√3/2,1/2). Sphere radius = u t = k c t = k r? Actually r=c t, radius=k r. At hit, r=√3L-h, radius=R=k r. For photon to hit from outside, need for all r'<r (where r'=c t), distance from C to photon > k r' (sphere radius at that time). Let's analyze function f(r') = |r' u_i - C|^2 - (k r')^2. At r'=0: f(0)=L^2+h^2 >0 (since R=0). At r'=r: f(r)=0. Need f(r')>0 for 0<r'<r (assuming first contact). If k=0? no. Let's compute f(r') = |r'u_i-C|^2 - k^2 r'^2 = (r'√3/2 - L)^2 + (r'/2 - h)^2 - k^2 r'^2 = r'^2( (3/4+1/4) - k^2) - 2r'( (√3/2)L + (1/2)h ) + L^2+h^2 = r'^2(1-k^2) - r'(√3 L + h) + L^2+h^2. Since 1-k^2>0, this quadratic in r' opens upward. It has roots? One root is r (since f(r)=0). The other root r2? Product = (L^2+h^2)/(1-k^2) >0, so both roots same sign (positive? r>0, so r2>0). Sum = (√3L + h)/(1-k^2) >0. So r2>0. Since upward, f>0 outside interval between roots: f>0 for r'<min(r,r2) and r'>max(r,r2). For first hit at r, we need f(r')>0 for 0<r'<r, so r must be the smaller root: r ≤ r2. Since quadratic positive before r, yes if r is the first root (smaller). Thus require r ≤ r2? Actually if r is smaller root, then for 0<r'<r, f>0. If r were larger root, there would be an earlier root r2<r where f=0, meaning earlier hit. So need r = smaller root of quadratic. Since both roots positive, smaller root has value? We can derive using root relationships: r + r2 = S/(1-k^2), r r2 = P/(1-k^2), where S=√3 L + h, P=L^2+h^2. We know r. Need r ≤ r2. Could also use derivative at r? Since quadratic upward and roots; at smaller root, derivative f'(r) <0 (crossing from positive to negative? Wait f>0 before root, f=0 at root, then f<0 between roots. So at smaller root derivative negative? Actually from positive to negative => derivative negative. At larger root, derivative positive. So condition first hit is f'(r) ≤0? At hit, derivative should be ≤0? Let's compute f'(r)=2r(1-k^2) - (√3 L + h). At r, derivative negative => 2r(1-k^2) < S. But maybe. Alternatively use geometry: The photon must be outside sphere until hit, i.e., origin outside at τ and path before outside. At τ, the photon is on sphere; just before, it was outside. Since sphere expands, maybe sufficient that origin outside at τ? Actually if origin outside at τ and photon line segment? Let's check. Also after reflection, for photon to be outside for r'>r (after), need f(r')>0 for r'>r. If r is smaller root, then f(r')<0 between roots, so photon would be inside sphere after hit until maybe r2 if photon continued straight. But after reflection it moves vertical, so different. So first-hit condition is separate. Let's compute r as smaller root. f(r') = (1-k^2)(r'-r)(r'-r2) with r<r2? Since upward and positive before r, yes f(r') = (1-k^2)(r'-r)(r'-r2). For r'>r just after (if continued), f<0 (inside) if r<r2. But after reflection not applicable. But first hit requires r<r2. Determine r2 from sum = r + r2 = S/(1-k^2) => r2 = S/(1-k^2) - r. Condition r < r2 => 2r < S/(1-k^2) => 2r(1-k^2) < S. This is same f'(r)<0? f'(r)=2r(1-k^2)-S <0. Yes. Let's compute in terms of L,h and k. But k itself = |L-√3h|/(√3L-h). Let's evaluate. Alternatively maybe the condition part b simply from τ>0 (√3 L > h) and u<c? Let's see typical IME problem: likely answer a) u = c (L - √3 h)/(√3 L - h) maybe with sign? Wait they might assume L > √3 h? Let's check problem statement: h>0, L>0. No other. But maybe physical condition b) is h < √3 L. However also need u<c? They gave u<c as dato, not condition? It says "Observações: ... u < c; ..." So u<c is given, not to be derived? Actually "Dado: velocidade da luz c. Observações: t≥0; h>0; L>0; u<c; ..." So u<c is an assumption, maybe part b condition to ensure possible? It might be condition for physical possibility is that the formula for u satisfies u<c, i.e., some relation between h,L,c? Since c fixed, u<c may impose inequality on h,L: |L-√3h|/(√3L-h) <1. Let's see. They ask "a velocidade u de expansão do raio da esfera em função de h, L e c" and "b) a condição à qual as coordenadas h e L devem satisfazer para que a imposição física do problema seja possível." So part b likely exactly the inequality from u<c and τ>0 and maybe R positive. Let's find simplest condition. From our derivation, u = c |L - √3 h| / (√3 L - h), with τ>0 requiring √3 L - h > 0. Also u < c implies |L - √3 h| < √3 L - h. Let's solve this inequality generally, plus √3 L > h. Inequality |L - √3 h| < √3 L - h. Since RHS positive (√3L>h). Square or consider cases: Case 1: L ≥ √3 h => |...| = L - √3 h. Inequality: L - √3 h < √3 L - h => rearrange: L - √3 h < √3 L - h => 0 < (√3-1)L + (√3-1)h = (√3-1)(L+h) >0. Always true. So in case 1, u<c automatically. Also need τ>0: √3L > h; but L ≥ √3h implies L ≥1.732h > h/√3 ≈0.577h, so automatically √3L>h? Since L≥√3h => √3L ≥3h >h, yes. So case 1 always allowed? Also need R positive nonzero? If L=√3h then R=0 => u=0, violates u>0? Problem says "raio aumenta linearmente com o tempo" and u<c; maybe u>0? It doesn't explicitly say u>0 but "velocidade de expansão" implies positive. If L=√3h, u=0, sphere static, but photon hits at center? Actually R=0 at τ>0 impossible because sphere radius zero only at t=0; photon reaches at t>0, so no hit. Thus exclude L=√3h. So case1: L > √3 h. Case 2: L < √3 h => |...| = √3 h - L. Inequality: √3 h - L < √3 L - h => √3 h + h < √3 L + L => h(√3+1) < L(√3+1) => h < L. So case2 requires L > h. But also τ>0: √3 L > h => L > h/√3 ≈0.577h, which is less restrictive than L>h. So combined case2: h < L < √3 h. Also need R>0: L≠√3h (strict), ok. And u<c if L>h. So condition in case2: h < L < √3 h. Combined: either L > √3 h OR h < L < √3 h. That is simply L > h? Let's see: If L>h but could be less than √3h or greater. If L>h and L>√3h, allowed. If h<L<√3h, allowed. If L>h always allowed? Check L>h but maybe L≤√3h? Yes case2. If L>√3h also L>h. So condition reduces to L > h? Wait if L>h but L=√3h? excluded. If L>h but L<√3h allowed. If L>h but L>√3h allowed. So yes L > h seems sufficient. But what about L>h but L< h/√3? impossible since h/√3 < h. So condition L>h. But is there any other from first-hit? Let's test an example L>h but maybe u extremely close c? Let's see case L>√3h: k=(L-√3h)/(√3L-h). At L=2h, √3≈1.732, numerator=0.268h, denom=3.464h-H=2.464h? Wait L=2h, denom=√3*2h - h=2.464h, k=0.109c. Good. At L just >√3h, numerator small, k small. At L just >h in case2, e.g., h=1, L=1.1: numerator=√3*1 -1.1=0.632; denom=√3*1.1 -1=0.906; k=0.697c <1. Good. At L approaching √3h from below, numerator→0, k→0. So u<c automatically? For h<L<√3h, k=(√3h-L)/(√3L-h). Need k<1 => √3h - L < √3L - h => h(√3+1) < L(√3+1) => h<L, yes. So condition L>h exactly ensures u<c in case2. In case1 L>√3h >h automatically, u<c. So overall u<c iff L>h, given r>0? Let's verify if L>h but L can be <h/√3? no. So condition b maybe L > h. But wait what about first-hit condition? Does L>h guarantee photon hits sphere from outside? Let's test numerically. Take h=1, L=2 (L>h). Compute r=√3*2 -1=2.464. τ=r/c. k=u/c = (L-√3h)/(√3L-h) = (2-1.732)/(3.464-1)=0.268/2.464=0.1088. Sphere radius R0=k r=0.1088*2.464=0.268. Photon at P=(r√3/2, r/2)= (2.134,1.232). Center (2,1). Vector P-C=(0.134,0.232). Magnitude sqrt(0.0179+0.0538)=0.267, matches R. Good. Origin distance to center sqrt(5)=2.236 > R0=0.268, origin outside. Before hit? Check f(r')= (1-k^2)r'^2 - S r' + P, S=√3L+h=4.464, P=5. Roots: r=2.464, r2=P/((1-k^2)r)? Product = P/(1-k^2)=5/(0.988)=5.06; r2=5.06/2.464=2.054 < r? Wait r=2.464 > r2=2.054, so r is larger root! That would mean f(r')<0 for r' between 2.054 and 2.464? Let's check f at r'=2.2: f=(0.988)(4.84)-4.464*2.2+5=4.78-9.82+5=-0.04 <0. So photon would have been inside sphere before hit at 2.464? Actually at r'=2.2, distance < radius? Let's compute radius = k r'=0.1088*2.2=0.239; distance from C to photon =? Photon (1.905,1.1); diff ( -0.095,0.1); distance=0.138 <0.239. Yes inside. So photon would have entered sphere earlier, impossible for first hit from outside. So L=2,h=1 gives no first hit? But problem may still have solution? Let's check: At earlier time, sphere radius smaller? Wait at r'=2.2, radius = u t' = k r' =0.239, distance=0.138 < radius => photon inside sphere, but sphere radius then? Hmm if photon is inside sphere, it would have crossed surface earlier. But maybe our calculation of k from reflection yields a sphere that already engulfed photon before hit; so the reflection point is not first contact. The problem says "Em determinado instante, o fóton atinge a superfície da esfera" — should be first contact. So need exclude. Compute first-hit condition more carefully. Our f(r') = |r'u_i-C|^2 - (k r')^2. Since photon hits at r, for no earlier contact need f(r')>0 for 0<r'<r. As above, since f quadratic with roots r and r2, need r < r2 (r smaller root). That means f'(r)<0? Actually at smaller root derivative negative; at larger root derivative positive. At our r=2.464, f'(r)=2r(1-k^2)-S = 2*2.464*0.988 -4.464 =4.87-4.464=0.406 >0 => larger root, so earlier root r2=2.054. Thus photon would hit at r2 earlier. So need additional condition to ensure r is smaller root. Let's derive. Given r = √3 L - h, k = |L-√3h|/(√3L-h). Compute r2 maybe? Product roots = P/(1-k^2)= (L^2+h^2)/(1-k^2). So r2 = P/[(1-k^2) r]. Condition r < r2 => r^2 (1-k^2) < P? Wait r*r2 = P/(1-k^2) => r2 = P/[(1-k^2) r]. r < r2 => r^2 (1-k^2) < P. Since r>0. Or equivalently f'(r)<0? f'(r)=2r(1-k^2)-S <0 => 2r(1-k^2) < S. Let's derive explicit. Let's compute r^2(1-k^2) < L^2+h^2. Also S = √3L + h = r +? Since r=√3L - h, then S = r + 2h? Wait √3L + h = (√3L - h) + 2h = r + 2h. Yes S = r + 2h. Nice! Because r=√3L - h => √3L = r+h => S = r+h + h = r+2h. Great. Also P = L^2 + h^2. Express L in terms r,h: L = (r+h)/√3. Then P = (r+h)^2/3 + h^2 = (r^2 +2rh + h^2)/3 + h^2 = (r^2 +2rh +4h^2)/3. So condition r^2(1-k^2) < (r^2 +2rh +4h^2)/3. Also k = |L - √3h|/(√3L - h) = |(r+h)/√3 - √3h| / r = |r+h -3h|/(√3 r)? Wait L = (r+h)/√3; L - √3h = (r+h)/√3 - √3h = (r+h -3h)/√3 = (r - 2h)/√3. Then k = |(r-2h)/√3| / r = |r - 2h|/(√3 r). Nice! So k = |r-2h|/(√3 r). Since r>0. Also u<c condition k<1 gave L>h, but let's see in r variables. r=√3L - h => √3L = r+h => L=(r+h)/√3. L>h => (r+h)/√3 > h => r+h > √3 h => r > (√3 -1)h ≈0.732h. Also r>0. Also L>√3h equivalent? r=√3L - h > √3*√3h - h =3h -h=2h => r>2h. Case L<√3h => r<2h. But k = |r-2h|/(√3 r). For r<2h, k=(2h-r)/(√3 r). For r>2h, k=(r-2h)/(√3 r). But also L>h gave r>(√3-1)h. So r positive and >(√3-1)h. Now first-hit condition: f(r') = (1-k^2)r'^2 - (r+2h)r' + P. It has root r (by construction). The other root r2 = P/[(1-k^2)r]. Need r < r2 => r^2(1-k^2) < P. Let's compute 1-k^2 = 1 - (r-2h)^2/(3r^2) = [3r^2 - (r-2h)^2]/(3r^2) = [3r^2 - (r^2 -4rh +4h^2)]/(3r^2)= [2r^2 +4rh -4h^2]/(3r^2)= 2(r^2 +2rh -2h^2)/(3r^2). This positive? Need r? Since k<1 implies? k<1 => |r-2h| < √3 r always? For r<2h: 2h-r <√3r => 2h < (√3+1)r => r > 2h/(√3+1)= (√3-1)h? Because 2/(√3+1) = (√3-1)? 2/(2.732)=0.732=(√3-1). Yes. So r>(√3-1)h ensures 1-k^2>0. Now r^2(1-k^2) = r^2 * [2(r^2+2rh-2h^2)/(3r^2)] = (2/3)(r^2+2rh-2h^2). P = (r^2+2rh+4h^2)/3. Condition: (2/3)(r^2+2rh-2h^2) < (1/3)(r^2+2rh+4h^2) => multiply 3: 2r^2+4rh-4h^2 < r^2+2rh+4h^2 => r^2 + 2rh - 8h^2 < 0 => (r + 4h)(r - 2h) < 0. Since r+4h >0, need r - 2h <0 => r < 2h. Ah! So first-hit condition requires r < 2h. But remember r=√3L - h. r < 2h => √3L - h < 2h => √3L < 3h => L < √3 h. That is exactly case2 (L<√3h)! Interesting. So first contact possible only if L < √3 h? Let's verify with previous L=2h >√3h, r=2.464h >2h -> r>2h, so first-hit failed. So indeed to have no earlier intersection, need r<2h, i.e., L<√3h. Combined with L>h for u<c: h < L < √3 h. This matches case2 condition. What about case L=√3h? r=2h, condition equality r=2h? Then first-hit? At L=√3h, r=2h, k=0? Let's see k=|2h-2h|/(√3*2h)=0, u=0, sphere radius zero; impossible. Also f quadratic? If k=0, sphere point at center; photon would hit center at r? But sphere radius zero until t=0? Actually at t=0 radius 0 and photon at origin; center at distance L=√3h, h. Photon hits center? It would hit point if sphere radius zero at t when photon at center? But t>0, radius=0, so sphere is a point at C, photon reaches C? Does photon line pass through center? P=r u_i; set P=C => r=√3L? Wait C=(L,h). Solve r√3/2=L, r/2=h => r=2h, L=√3h yes. So photon would hit the point center at time τ=2h/c, but sphere radius zero, so maybe impossible because sphere hasn't grown? But problem says sphere radius increases linearly; at t=τ>0, radius ut=0 if u=0, so no surface. So exclude. Thus physical first-hit requires h < L < √3 h. Also u<c? Under this, u = c (√3h - L)/(√3L - h). Need u<c: √3h - L < √3L - h => h < L, true. Also u>0: √3h > L true. So condition b: h < L < √3 h? But wait maybe also after reflection escape condition? Let's test with h<L<√3h (case2). Then r>h? r=√3L -h. For L just >h, r just > (√3-1)h ≈0.732h <2h; ok. After reflection? We saw σ=-1 (since L<√3h => s=L-√3h<0 => σ=-1). Need u ≤ c√3/2 for escape? Let's check if this imposes further restriction. In case2, u/c = k = (√3h - L)/(√3L - h). Need k ≤ √3/2 => 2(√3h - L) ≤ √3(√3L - h) = 3L - √3h => 2√3h -2L ≤ 3L - √3h => 3√3h ≤ 5L => L ≥ (3√3/5)h ≈1.039h. So if L <1.039h, after reflection photon initially enters sphere. But is this required? Let's see if problem expects? Maybe not? Let's analyze exact. If after reflection the photon enters the sphere, the reflection point wouldn't be on outer surface from which ray escapes; it would be re-absorbed? But surface is perfectly reflective, so if it enters, it would be inside and maybe reflect internally? The problem says "passa a propagar-se paralelamente ao eixo y, em seu sentido negativo" implying after reflection it propagates; if it immediately enters sphere, it wouldn't propagate freely outside. So likely physical condition includes u ≤ c√3/2 => L ≥ (3√3/5)h. But is this part of "imposição física"? Let's investigate. Actually, at reflection, the surface is a mirror. For a mirror to reflect a ray, the ray must be in the external medium; after reflection, it remains external. If the sphere is expanding, the reflected ray could be overtaken by the expanding surface? But at the instant, if the normal outward and photon velocity, the ray may be outside if the surface expansion is slower than the component? We found condition k ≤√3/2 for outgoing ray not immediately inside. But maybe even if k >√3/2, the ray could still be outside initially? Let's compute derivative sign. If k >√3/2, F(0)=-R0(2k-√3) <0, so immediately inside; so no escape. Thus physical possibility should require L ≥ (3√3/5)h. But does problem mention? "para que a imposição física do problema seja possível" might include all constraints: photon really hits sphere (first contact), reflects, then propagates away (outgoing ray outside). Let's derive fully. Let's re-read problem statement: "Em determinado instante, o fóton atinge a superfície da esfera e passa a propagar-se paralelamente ao eixo y, em seu sentido negativo." It says after hitting, it starts propagating parallel to y negative. It doesn't explicitly say it doesn't hit again immediately. But for it to "passar a propagar-se", it must leave the surface and move away; otherwise reflection wouldn't be a propagation. So likely require outgoing ray outside for t>τ. That gave condition u ≤ c√3/2 and σ=-1. Combined with first-hit maybe. But maybe the problem's "despreze os efeitos do movimento da superfície da esfera" for determining reflected direction means we ignore the fact that the surface is moving when applying reflection; but after reflection, we should consider the moving surface? It says "para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos." So for direction, ignore surface motion. But for physical possibility, maybe include? Let's see: If we didn't ignore motion, reflection from moving mirror would change angle/energy. But they instruct to ignore, so direction fixed. Then physical possibility might include after reflection it doesn't intersect sphere? They may expect condition. Let's examine typical IME problem? It might be from IME 2027? I don't know. Let's solve mathematically all constraints. Let's derive from scratch with no assumptions. Given sphere radius R(t)=ut, center fixed at (L,h). Photon from origin at t=0 along x=√3 y, speed c. At time τ>0, photon hits sphere surface at P = cτ u_i. Then reflects with direction v_r=(0,-1). We solved reflection: P-C = σ R(1/2,√3/2), with σ=±1. Then r=cτ = √3L - h (from geometry), R=|L-√3h|? Actually σR = L -√3h => R=|L-√3h|. Also u=R/τ = c |L-√3h|/(√3L-h). This formula valid if τ>0 and R>0. Physical constraints: 1) τ>0 => √3L > h. 2) R>0 => L ≠ √3h (and since R=|...|, if L=√3h then R=0, but at τ>0 impossible because sphere radius zero only t=0; also u=0). So L≠√3h. 3) u<c => |L-√3h| < √3L - h. We solved gives L>h (and L≠√3h). 4) First contact: photon must not have hit sphere earlier. We found condition r < 2h => L < √3h. Wait combined with L>h gives h<L<√3h. But if L>√3h, first contact fails (r>2h). Let's verify with L>√3h: r>2h, so f'(r)>? At r>2h, maybe k positive. Condition r<2h for first contact; if r>2h, photon would have hit earlier at smaller root? Let's confirm with another example L=3h. r=√3*3h -h=4.196h >2h. k=(3h-√3h)/(3√3h -h)= (1.268h)/(4.196h)=0.302. r2=P/((1-k^2)r). P=9h^2+h^2=10h^2; 1-k^2=0.909; r2=10h^2/(0.909*4.196h)=2.62h < r=4.196h. So earlier hit at 2.62h. So indeed first contact condition r<2h is necessary. But wait what if L< h? Then u<c fails; but maybe u could be >c? Given u<c, so exclude. So physical condition likely h<L<√3h. But let's check if after reflection escape condition further restricts. Under h<L<√3h, k=(√3h-L)/(√3L-h). We need k ≤ √3/2 for outgoing ray outside? Actually let's derive after-reflection condition exactly with σ=-1 and k. Difference F(Δ)=Δ[(1-k^2)Δ - R0( -√3 + 2k)] = Δ[(1-k^2)Δ + R0(√3 - 2k)]? Wait σ=-1: σ√3 + 2k = -√3 + 2k. So bracket = (1-k^2)Δ - R0(-√3+2k) = (1-k^2)Δ + R0(√3 - 2k). Wait F=Δ[(1-k^2)Δ - R0(σ√3+2k)] = Δ[(1-k^2)Δ - R0(-√3+2k)] = Δ[(1-k^2)Δ + R0(√3 - 2k)]. If k ≤√3/2, second term ≥0, so F>0 for all Δ>0 (since first term positive). If k >√3/2, then at Δ=0 bracket negative => F<0 for small Δ; later may become positive when Δ > R0(2k-√3)/(1-k^2). So photon would be inside initially. For "passa a propagar-se" immediately, need k ≤√3/2. If k >√3/2, the photon would be inside the sphere right after reflection, impossible. So yes physical condition includes k ≤√3/2. But is k ≤√3/2 automatically from first contact + u<c? Let's test. In case h<L<√3h, k can be large near L→h? At L just >h, k=(√3h - h)/(√3h - h)? Actually L→h: numerator=√3h - h = (√3-1)h; denominator=√3h - h = (√3-1)h; k→1. So k close to 1 >√3/2. Then after reflection photon would enter sphere. So need additional L ≥ (3√3/5)h ≈1.039h. But is that part of condition? Let's see numerically h=1, L=1.05 (L>h, <√3h). r=√3*1.05 -1=0.818? Wait √3≈1.732, r=1.818-1=0.818. k=(1.732-1.05)/(1.818-1)=0.682/0.818=0.834 >0.866? 0.834 <0.866? Actually √3/2=0.866, so k=0.834 ≤0.866. So L=1.05 >1.039 ok. If L=1.02: r=1.766-1=0.766; numerator=0.712; k=0.929 >0.866. So indeed. But does IME problem expect part b to include this? Let's read carefully: "a) a velocidade u de expansão do raio da esfera em função de h, L e c; b) a condição à qual as coordenadas h e L devem satisfazer para que a imposição física do problema seja possível." The "imposição física do problema" likely means all conditions for the described process to occur: photon hits sphere from outside, reflects, and then propagates away. So should include after-reflection escape. But sometimes they might just ask condition for the mathematical solution to be physically valid (u<c and positive geometry). Let's search memory: This problem might be from IME 2027? It might be an actual exam. I can try recall. It has sphere expanding, photon reflecting downward. The solution might yield u = c (√3h - L)/(√3L - h) and condition h < L < √3h? Wait but after reflection they might not check escape? Let's see. Let's examine if after reflection, the photon could be "propagating" even if inside? No, impossible because surface is reflective; if it's inside, it wouldn't have reflected but refracted? Actually if the sphere is a perfect reflector, the inside is inaccessible; the photon cannot enter. So physical possibility definitely requires outgoing ray outside. The problem's statement "passa a propagar-se" implies it propagates in the external medium after reflection, so yes condition. But maybe the after-reflection condition is automatically satisfied if first contact and u<c? We saw counterexample L=1.02h gives k=0.929>0.866. Did first contact occur? Check first contact condition r<2h: r=0.766h <2h yes. u<c: L>h yes. So process: photon hits sphere from outside at r=0.766h, reflects downward, but immediately inside sphere (since k large) => would not propagate freely. So physical impossibility. Thus part b should include this. Let's derive full condition combining all. Let's formulate all conditions in terms of L,h. From geometry and reflection: r = √3 L - h >0. (A) R = |L - √3 h| >0. (B) u = c R / r = c |L - √3 h| / (√3 L - h). (Part a) u < c => |L - √3 h| < √3 L - h. (C) First contact (no earlier intersection): r < 2h? Let's verify condition r < 2h derived for any σ? We derived r<2h from first-hit condition using f(r')>0 for r'<r. But that assumed k = |r-2h|/(√3r) regardless of σ? Yes k formula holds. Let's re-derive first-hit condition more generally without assuming case. We had f(r') = (1-k^2) r'^2 - S r' + P, with S=√3L+h = r+2h, P=(r^2+2rh+4h^2)/3, k=|r-2h|/(√3r). The condition for r to be the smaller root (first contact) is r < r2 where r2 = P/((1-k^2)r). We computed r^2(1-k^2) < P => (r+4h)(r-2h)<0 => r<2h. This derivation did not assume sign of L-√3h? It used k=|r-2h|/(√3r) which is valid for both cases. But wait if r<2h, then k=(2h-r)/(√3r) (case L<√3h). If r>2h, k=(r-2h)/(√3r) (case L>√3h). The condition r<2h means r<2h, which implies L<√3h (since r=√3L-h <2h => √3L <3h => L<√3h). So first contact condition is exactly r<2h => L<√3h. But what if r=2h? Then L=√3h, R=0, u=0, impossible; so strict. But wait our derivation of f'(r) condition: For first contact, f(r')>0 for 0<r'<r. If r<2h, then r is smaller root? Let's test L=1.02h: r=0.766h <2h; k=(2-0.766)/(1.732*0.766)=1.234/1.327=0.93? Wait k = (2h - r)/(√3 r) = (1.234)/(1.327)=0.93, yes. f(r')? At r'=0.5h, f positive? P=(0.766^2+2*0.766+4)/3 h^2=(0.586+1.532+4)/3=2.039h^2. S=r+2h=2.766h. f= (1-0.865)*0.25 -2.766*0.5 +2.039 =0.0338 -1.383+2.039=0.69>0. Good. So first contact ok. But if r>2h, e.g., L=2h, r=2.464h>2h, then f at r'=2.2h negative as computed. So indeed first contact only if r<2h. Now after-reflection escape condition: We need after reflection photon outside sphere for all Δ>0? We derived condition k ≤√3/2 (and σ=-1 automatically if r<2h? Yes σ=-1 because L<√3h => L-√3h<0 => σ=-1). So require k ≤√3/2. Express in L,h: k = (2h - r)/(√3 r) (since r<2h). Condition k ≤√3/2: (2h - r)/(√3 r) ≤ √3/2 => multiply 2r: 2(2h - r) ≤ 3r => 4h - 2r ≤ 3r => 4h ≤ 5r => r ≥ (4/5)h. But r=√3L - h, so condition √3L - h ≥ (4/5)h => √3L ≥ (9/5)h => L ≥ (3√3/5)h ≈1.039h. Wait check: r≥0.8h => √3L - h ≥0.8h => √3L ≥1.8h => L ≥1.8/√3 h = (1.8*√3)/3 h =0.6√3 h = (3√3/5)h ≈1.039h. Yes. But also first contact requires r<2h => L<√3h ≈1.732h. So combined L between (3√3/5)h and √3h. Plus u<c condition L>h? But (3√3/5)h ≈1.039h >h, so L>1.039h implies L>h automatically. So condition b might be (3√3/5)h ≤ L < √3h. But does problem expect this? Let's verify if after-reflection escape is required. Let's see if the problem's wording "passa a propagar-se paralelamente ao eixo y, em seu sentido negativo" maybe only describes direction after leaving the surface, not necessarily forever? In physics, reflection is instantaneous; after reflection, the ray travels in that direction. But to leave the surface, the ray must not be inside immediately. So yes. However, maybe the sphere is transparent? No, perfectly reflector. So outgoing ray must be outside. Therefore include k ≤√3/2. But let's check if there is any other condition: after reflection, the photon could still be re-hit later by the expanding sphere even if initially outside? Our F(Δ) is quadratic with positive leading coefficient. If k ≤√3/2, F>0 for all Δ>0 because F=Δ[(1-k^2)Δ + R0(√3 - 2k)] with both terms positive (since 1-k^2>0, √3-2k ≥0). So never hit again. If k >√3/2, F negative for small Δ then positive after Δ0; but photon would be inside initially, impossible. So condition ensures no re-hit. Thus full physical condition likely: h < L < √3 h? Wait include k: (3√3/5)h ≤ L < √3h? Let's see if u<c + first contact automatically maybe k≤√3/2? No counterexample L=1.02h had u<c? u/c=0.93 <1, first contact yes, but outgoing fails. So need include. The problem statement "para que a imposição física do problema seja possível" might indeed expect all constraints. Let's derive carefully and see final answer. But wait, maybe the after-reflection condition should be based on the moving surface velocity relative to the photon? The photon reflected direction is downward; surface is expanding radially. At P, the outward normal is -n0 (since σ=-1). The photon velocity v_r=(0,-c). The surface velocity at P is u n_out = u (-n0) = (-u/2, -u√3/2). The normal component of photon velocity relative to surface? The photon must have normal component outward (away from surface) to escape. Relative velocity normal = (v_r - v_surface)·n_out? Let's compute n_out=-n0. v_r·n_out = (0,-c)·(-1/2,-√3/2)= c√3/2 >0 (outward). Surface velocity u n_out has normal component u (outward). For photon to escape, its normal velocity must exceed surface expansion? Actually in lab frame, at P, the photon's velocity relative to surface normal: v_r·n_out = c√3/2. Surface moves outward with speed u along normal. The photon's normal component relative to surface = c√3/2 - u? If positive, it moves away; if negative, surface catches up? But our derivative condition was c√3/2 - u >0? Actually we got k<√3/2 => u < c√3/2. Same. So yes. But maybe the problem's instruction "despreze os efeitos do movimento da superfície da esfera" applies also to after-reflection? It says "para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos." So for direction, ignore surface motion. But for physical possibility, we should include surface motion? The phrase "para determinar a direção do fóton refletido" only. After direction determined, physical possibility may require considering motion. So yes. Let's see if there is another constraint: photon must hit the external surface, not the inside. Since sphere starts at radius 0 and expands, before hit the sphere is smaller; photon approaches from outside. At the moment of hit, the photon is on surface. After reflection, it must be outside. Good. Let's derive first contact condition maybe also from "the photon hits the surface" not "passes through"? The problem says "atinge a superfície" first time. So yes. Let's compute part a carefully with signs. From geometry, we have two possible σ? But physical first contact selects σ=-1 (L<√3h). Then R = √3h - L. Then u = R/τ = (√3h - L)/( (√3L - h)/c ) = c (√3h - L)/(√3L - h). Wait sign! Our earlier expression u = c |L-√3h|/(√3L-h). Since L<√3h, |...|=√3h - L. So u = c (√3h - L)/(√3L - h). Let's verify with example h=1,L=1.05: u/c=(1.732-1.05)/(1.818-1)=0.682/0.818=0.834. yes. But part a asks "a velocidade u de expansão do raio da esfera em função de h, L e c". If they expect a positive expression, likely u = c (√3h - L)/(√3L - h) under condition h<L<√3h? But if L>√3h, the formula would be c (L-√3h)/(√3L-h), but such case invalid due to first contact? However part a may be asked before part b, maybe they expect general formula with absolute value? The problem may not have introduced physical condition yet. It asks determine u in function of h,L,c. The correct expression from reflection without imposing first contact might be u = c |L - √3 h| / (√3 L - h)? But then part b asks condition for physical possibility, which would include the choice of sign. Let's analyze. The reflection equations themselves gave two possible signs for the normal; but the geometry of the photon hitting the surface and reflecting downward may determine sign from equations? Let's solve without assuming first contact: The equations yield r=√3L - h and σR = L - √3h. Since R>0, u=R/τ = |L-√3h|/(√3L-h) c. This is general. But does the reflection direction actually fix sign? The reflected direction v_r=(0,-1). We used v_i - v_r = 2(v_i·n)n. This gave n = ±n0. Both are geometrically possible? Let's check if for a given L,h one sign may make the photon hit the sphere from outside and the other from inside? The equations alone don't determine σ; both give a sphere tangent/intersection. But one may correspond to photon hitting from inside (i.e., origin inside sphere) or after reflection entering. Physical imposes sign. Thus part a may expect u = c |L - √3h|/(√3L - h) perhaps, because they ask "em função de h, L e c" without condition. But in physics problems, they often want the value that satisfies the described process; so the absolute value may be resolved by part b condition. However if part b condition includes L<√3h, then part a answer should use positive (√3h - L). Let's see IME style: They give "Determine: a) a velocidade u ...; b) a condição ...". Usually part a is solved using the actual physical scenario, so you use the correct sign from physical constraints? But they might not ask to determine sign until part b. Hmm. Let's inspect the equations: From geometry, r=cτ = √3L - h. For τ>0, need √3L > h. Then R = uτ. From reflection, uτ = |L - √3h|? Actually σ uτ = L - √3h. Since uτ positive, σ = sign(L-√3h). So u = |L-√3h|/τ = c |L-√3h|/(√3L - h). This is a unique positive expression in terms of absolute value. Then part b condition might be that the absolute value can be removed: L<√3h? But also L>h and first contact etc. If they wanted just the formula, they might accept absolute value. But part b asks condition on h,L for physical possibility, which may be exactly to determine the sign and inequalities. So part a likely should be with absolute value? Let's see. Let's read problem statement: "Uma esfera ... tem seu centro fixo ... Seu raio R aumenta linearmente ... No instante t=0, um fóton parte ... Em determinado instante, o fóton atinge a superfície da esfera e passa a propagar-se paralelamente ao eixo y, em seu sentido negativo. Determine: a) a velocidade u ... b) a condição ...". They don't state which side of center the hit point is on. The reflection direction tells normal line. The equations give two possible configurations: normal from center to P along ±(1/2,√3/2). The formula with absolute value accounts for both. But physical possibility may select one. Since part a asks u in terms of h,L,c, they might expect the formula with modulus. However, many students might derive with a sign and then part b selects. Let's check typical Brazilian IME solution: They often write u = c(√3h - L)/(√3L - h) and condition h < L < √3h. Let's search memory: I vaguely recall a problem where answer is u = c(√3h - L)/(√3L - h) and condition h < L < √3h? Maybe they ignore after-reflection escape? Let's test if condition h<L<√3h ensures after-reflection escape? No, we saw L=1.02h (<√3h) but u/c=0.93 >0.866, so after reflection photon enters. Would they ignore? Maybe the problem says "despreze os efeitos do movimento da superfície da esfera" meaning after reflection, treat surface as stationary? But that would be inconsistent. Or maybe "para determinar a direção do fóton refletido" only; after reflection, the surface movement might be considered, so they should include. Let's see if the condition h<L<√3h plus u<c maybe implies u<c√3/2? Not always. Let's compute minimal L for u<c√3/2: L ≥ (3√3/5)h ≈1.039h. If L=1.02h, u<c but u>c√3/2. So if they only require u<c, physical possibility maybe still? Let's simulate: h=1,L=1.02. At t=τ, sphere radius R =? r=√3*1.02 -1=0.766. uτ=0.929c *0.766/c=0.712. Photon at P=(0.663,0.383). Center (1.02,1). Vector P-C=(-0.357,-0.617), magnitude 0.712, direction down-left (angle? yes -120°). Reflected photon vertical down from x=0.663, y=0.383. Sphere center (1.02,1) to the right/up; distance from photon to center at Δt=0: sqrt((0.663-1.02)^2+(0.383-1)^2)=0.712. Surface expands at 0.929c. Photon moves down at c, so vertical distance from center y=1 decreases? At Δt small: photon y=0.383 - cΔt; distance squared change? We computed derivative at 0 positive? Wait for σ=-1, derivative of distance squared = 2(P-C)·v = 2[(-0.357, -0.617)·(0,-c)] = 2[0 + (-0.617)(-c)] = 2*0.617c >0, so distance increases initially; radius increases at rate u=0.929c. Compare d(distance)/dt at 0 = (P-C)/R · v = (-0.5,-0.866?) Actually n_out=(P-C)/R = (-0.5,-0.866)? Since R=0.712, x diff=-0.357 -> /0.712=-0.501, y=-0.617/0.712=-0.867. Dot with (0,-c)=0.867c >0. So distance derivative = 0.867c ≈0.867c. Radius derivative = u=0.929c. Since radius grows faster than distance, the difference decreases: d(diff)/dt = 0.867c - 0.929c = -0.062c <0, so immediately inside. So indeed physical impossibility. Would the problem setter include this subtlety? It says "despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos" for reflected direction. But after reflection, to determine if it propagates, you must consider surface motion. The phrase "passa a propagar-se" maybe means the photon's direction after reflection is vertical downward, not necessarily that it doesn't interact again. But if it enters, it would be refracted/absorbed, so not a reflection. So I think full condition includes u ≤ c√3/2. Let's solve all constraints systematically. Let r = √3L - h >0. Then L = (r+h)/√3. Conditions: - h>0, L>0. - u<c: |L-√3h| < r. But L-√3h = (r+h)/√3 - √3h = (r - 2h)/√3. So condition |r-2h|/√3 < r => |r-2h| < √3 r. This inequality splits: If r ≥2h: r-2h < √3 r => 2h > (1-√3)r? Since 1-√3 negative, RHS negative? Actually r-2h <√3r => -2h < (√3-1)r => always true for r>0? Wait r-2h <√3r => r - √3r <2h => r(1-√3) <2h. Since left negative, true always. So if r≥2h, u<c always. But first contact will exclude r>2h. If r≤2h: 2h - r < √3 r => 2h < (√3+1)r => r > 2h/(√3+1) = (√3-1)h? Let's rationalize: 2/(√3+1) = (2(√3-1))/(3-1)=√3-1. Yes r > (√3-1)h. So combined r > (√3-1)h and r≤2h. But first contact requires r<2h, so r in ((√3-1)h, 2h). This gives L=(r+h)/√3 > ((√3-1)h + h)/√3 = (√3 h)/√3 = h. And L< (2h+h)/√3 = 3h/√3 = √3 h. So L in (h, √3h). Good. - First contact: r<2h (from f condition), so r∈(0,2h). Combined with τ>0. - After-reflection escape: k = |r-2h|/(√3 r) ≤ √3/2. Since r<2h, k=(2h-r)/(√3r). Condition: (2h - r)/(√3 r) ≤ √3/2 => 2(2h - r) ≤ 3r => 4h -2r ≤3r => 4h ≤5r => r ≥ (4/5)h. Thus r ≥ 4h/5. Also first contact r<2h; escape requires r≥0.8h. But τ>0 and u<c gave r>(√3-1)h ≈0.732h, which is less than 0.8h. So need stricter r≥0.8h. Also need r<2h. So final: 4h/5 ≤ r < 2h. Then L=(r+h)/√3 gives: Lower: L ≥ (4h/5 + h)/√3 = (9h/5)/√3 = 9h/(5√3) = 3√3 h/5? Because 9/(5√3) = 9√3/15 = 3√3/5. Yes. Upper: L < (2h+h)/√3 = 3h/√3 = √3 h. So condition: (3√3/5)h ≤ L < √3 h. Also r>0 ok. Also L>h? Since lower 1.039h >h, yes. But wait what about r=2h? Then L=√3h, R=0 u=0 invalid; exclude. So strict lower? At r=4/5 h, k=√3/2 exactly, after reflection photon just grazing? The condition for escape was k ≤√3/2, equality means initial relative normal velocity zero? Let's check F=Δ[(1-k^2)Δ + R0(√3 - 2k)] with 2k=√3, so F=(1-k^2)Δ^2 >0 for Δ>0. So photon is outside for any Δ>0 even though derivative zero. So equality allowed. u<c since k=√3/2 <1. So lower bound inclusive. Upper r<2h strict due to R>0 and first contact? At r=2h, L=√3h, R=0 (u=0) invalid, so strict upper L<√3h. Thus final condition b: (3√3/5)h ≤ L < √3 h. But does this match expected? Let's see if first contact condition might be different if we consider that at t=0 the sphere radius zero and photon at origin; if origin is inside sphere? At t=0, sphere is a point at C, not containing origin unless C=origin (L=h=0) impossible. So photon starts outside. The first contact condition we derived is correct. However, part a might be asked without after-reflection escape? Let's see if the reflection point P is on the "front" side of sphere as seen from origin. The normal at P points toward origin or away? If σ=-1, P-C = -R n0 = (-R/2, -R√3/2). So P is below-left of center? Actually C + negative direction => x less than L, y less than h. Origin O=(0,0). For h<L<√3h, is P in first quadrant? r=√3L - h. Since L>h/√3? P x=r√3/2; positive yes. P y=r/2 positive. Compare P to C: x_P - L = -R/2 <0 => x_P < L. y_P - h = -R√3/2 <0 => y_P < h. So P is southwest of center. The normal outward from center to P points southwest. Incident ray from origin to P is northeast direction (30°). At P, the ray is traveling northeast (v_i angle 30°). The outward normal points southwest (angle 240°). The reflection of a northeast-traveling ray off a southwest-facing surface? Let's test using formula: v_i=(√3/2,1/2). n_out=(-1/2,-√3/2). Dot = -√3/2. v_r = v_i -2(-√3/2)(-1/2,-√3/2)= v_i -√3(1/2,√3/2)? Wait -2(-√3/2)(-n0)= -√3 n0? Actually n_out=-n0. -2(v_i·n_out)n_out = -2(-√3/2)(-n0)= -√3 n0 = (-√3/2, -3/2). Then v_i=(√3/2,1/2) minus? Wait formula v_r = v_i - 2(v_i·n)n = v_i -2(-√3/2)(-n0)= v_i - √3 n0 = (√3/2,1/2) - (√3/2,3/2) = (0,-1). Yes. So physically possible: ray comes from southwest? Actually origin is southwest of P? Origin (0,0) relative to P maybe? P coordinates positive; origin southwest? C=(L,h); P southwest of C; origin southwest of P? Since P-C = -R n0 (southwest), and P = r v_i northeast. The vector from P to origin is -P = -r(0.866,0.5) = (-0.866r,-0.5r) direction southwest. Normal outward from center to P = southwest. So incident ray arrives from origin (southwest) along direction v_i (northeast) toward P. It hits the outside surface whose outward normal is southwest (toward origin? Actually outward normal points from center to P, i.e., southwest, which is roughly toward origin? Wait center to P southwest, so outward from sphere at P points further southwest, i.e., away from center. Since origin is southwest of P, the outward normal points toward origin? Let's see: P-C = (-R/2, -R√3/2) southwest. Origin relative to P: O-P = -P = (-0.866r,-0.5r) also southwest? Angle of -P: tan = (-0.5)/(-0.866)=0.577 -> 30° below negative x? Actually both negative => third quadrant angle 210°? Normal -n0 angle 240°? They are different. Dot of outward normal with vector from P to origin = (-1/2,-√3/2)·(-0.866r,-0.5r) = (0.433r+0.433r)=0.866r >0. So origin lies on the outward side of tangent plane? Good: outside half-space. Thus incident ray comes from outside. Good. After reflection, photon vertical down. At P, which side? Normal outward southwest; reflected direction down (0,-1). Dot with outward normal = (-1/2)*0 + (-√3/2)*(-1)=√3/2 >0, so reflected ray has positive component along outward normal? Wait outward normal points southwest (negative x,y). Dot with (0,-1) = √3/2 >0 means reflected ray has component in outward direction? Since n_out=(-1/2,-√3/2); (0,-1)·n_out = +√3/2. Yes positive, so it leaves surface outward. However, due to expansion, if u large it may catch up. Thus final answer likely: a) u = c(√3h - L)/(√3L - h) [assuming physical condition L<√3h]; b) (3√3/5)h ≤ L < √3h. But wait if part a is asked before condition, the formula with absolute value might be more general. Let's decide. Let's see problem statement: "Determine: a) a velocidade u de expansão do raio da esfera em função de h, L e c; b) a condição à qual as coordenadas h e L devem satisfazer para que a imposição física do problema seja possível." They likely expect part a to be a single expression, not piecewise. Since physical condition will restrict L, but they ask u "em função de h,L,c" maybe just the expression derived. In many IME solutions, they use the sign from geometry: since speed u positive and R>0, from σR=L-√3h, and knowing? They might not know yet which sign until they analyze physical possibility. But in part a, they can write u = c|L-√3h|/(√3L-h). Then part b gives condition. However, if part b includes after-reflection escape, the absolute value expression still valid for cases; but condition selects sign. But if condition selects L<√3h, then |...| = √3h - L, so u = c(√3h - L)/(√3L - h). Would they expect that? Let's see. Let's derive part a using the actual physical process: The photon hits the sphere, reflects downward. The reflection law gives normal direction. From the diagram, the sphere is drawn above x-axis y=h, center (L,h) perhaps L>0,h>0. The incident ray x=√3 y goes up-right. The reflection is downward. The normal at hit point likely points down-left (σ=-1) because the photon comes from origin and reflects downward; if normal pointed up-right (σ=+1), the reflected ray would go upward? We already computed σ=+1 gives v_r=(0,-1) too? Wait both signs gave same v_r algebraically because formula invariant under n -> -n? Actually we found both possible n along same line but opposite direction; the reflection formula with n_out either sign gives same v_r? Let's verify with σ=+1: n_out=n0=(1/2,√3/2). Then v_r = v_i -2(v_i·n0)n0 = (0,-1) as computed. So the reflected direction is downward even if outward normal is up-right. But is that geometrically possible? If outward normal points up-right, the sphere center is southwest of P? Let's see σ=+1: P-C = R n0 = (R/2,R√3/2) => P is northeast of center. The outward normal points further northeast. Incident ray from origin to P is northeast. At P, the ray is traveling northeast. The outward normal also northeast. Then the angle between incident direction (30°) and normal (60°) is 30°; reflection across normal would give direction 90° (upward), not downward. But our vector formula gave downward. Why discrepancy? Because the formula v_r = v_i - 2(v_i·n)n with n_out=n0 gives v_r=(0,-1), which is opposite to the geometric reflection (90°) along the same line? Wait v_r=(0,-1) and upward (0,1) are collinear but opposite directions. Which one is physically correct? Let's derive formula carefully with incident direction i pointing toward surface. If n_out=n0 (outward from center to P, northeast), then the incident ray arrives at P from origin, direction i=(√3/2,1/2) (northeast). The normal n_out=(1/2,√3/2) (northeast). The angle between i and n_out is 30°. The reflected ray should be on the other side of the normal, i.e., direction angle = 2θ_n - θ_i = 120° -30° =90° (upward, i.e., +y). But our vector formula gave -90° (downward). So the formula with n as outward from center gives the wrong physical direction? Let's resolve sign convention. Standard reflection formula: If incident vector I points toward surface, and N is the normal pointing from surface into the incident medium (i.e., opposite to the direction of the incident ray? or same?). Let's derive from geometry. At point P, tangent plane. Incident ray direction I (toward surface). The component of I normal to surface: I_n = (I·N) N if N is outward from surface? If N points into incident medium (i.e., opposite to inward normal?). Let's use simple example above: horizontal mirror y=0, light from above traveling downward I=(0,-1). Surface normal outward from mirror into incident medium (above) is N=(0,1). I·N=-1. Formula r = I - 2(I·N)N = (0,-1)-2(-1)(0,1)=(0,1). Works. Here N points from surface into incident medium (opposite to I? Actually I points toward surface, N points into incident medium, so they are opposite along normal). Good. Now sphere case σ=+1: center C=(L,h), P=C+R n0 (northeast). The outward normal from center to P is n0=(1/2,√3/2). The incident medium is outside sphere. At P, the outward normal from surface into incident medium is n0? Since center is inside, the outward direction from surface into outside is from center to P = n0. Yes N=n0. Incident ray I=(√3/2,1/2) points from origin to P (northeast). Does I point toward surface? Yes. Is I·N positive or negative? I·N=√3/2 >0. But in the horizontal example, I·N was negative. Why? In horizontal example, N points into incident medium (above), I points downward toward surface; I·N negative. In sphere σ=+1, N points outward into outside medium (northeast), I points toward P (northeast). They point in similar direction? Actually I points from origin to P; at P, the incoming ray is continuing in direction I? Wait if the ray comes from origin to P, the direction of propagation at P is along I (northeast). But the ray is approaching P from the southwest? The vector from origin to P is I direction; so at P, the ray is traveling northeast (away from origin). The surface at P is on the far side of the sphere from origin? The light is coming from origin, which is southwest of P; so it is traveling northeast toward P. At P, its direction is I (northeast), so it is moving toward the surface from the outside? Wait if it is at P and moving northeast, after P it would go further northeast, away from the sphere? But it just hit the surface at P; just before P, it was at P - ε I (southwest of P), which is outside or inside? For σ=+1, center is southwest of P? P-C = R n0 (northeast), so center C is southwest of P. The origin O is further southwest. At P, the ray direction I is northeast; so just before P, it was southwest, closer to center? The distance from C to point P - ε I = C + R n0 - ε I. Since n0 and I have angle 30°, the point might be inside sphere? Let's check: distance squared = R^2 + ε^2 -2ε R (n0·I). n0·I= cos30=√3/2. So distance squared = R^2 + ε^2 - √3 R ε. For small ε, distance < R? Since -√3Rε dominates, yes immediately before P the point is inside sphere! That means the photon would have been inside the sphere before hitting P, impossible. So σ=+1 corresponds to photon hitting from inside outward? Actually it means the photon was inside sphere before P, so it cannot be first contact from outside. Therefore the physical first contact requires σ=-1, where before P the photon is outside. Let's verify σ=-1: n_out=-n0 (southwest). I=(northeast). Angle between N and I: N=-n0 (southwest), I northeast -> opposite-ish, dot = -√3/2 <0. Then point P - ε I = C + R n_out - ε I = C - R n0 - ε I. Distance squared = R^2 + ε^2 -2ε R (n_out·I)? Wait n_out=-n0; n_out·I = -√3/2. So distance squared = R^2 + ε^2 -2ε R (-√3/2)= R^2 + ε^2 + √3 R ε > R^2 for small ε. So just before P, photon was outside sphere. So physical first contact selects σ=-1. Good. Thus part a should use σ=-1 from physical reasoning? But the reflection direction alone? Actually the fact that the photon comes from origin and hits the surface from outside gives σ=-1. However, could the photon have traveled through the inside of sphere before hitting? No, first hit means it was outside before, so σ=-1. Thus we can determine sign in part a by considering first contact. Let's see: The equation P-C = R n with n outward. Since before hit, photon is outside sphere, the vector from P to previous position should point outward? More precisely, the photon's incoming direction at P is I. The sphere center is on which side? For first contact, the center must lie on the opposite side of the tangent plane from the incoming ray? The incoming ray comes from outside; the center is inside, so the vector from P to center (-n_out) should have positive dot with I? Let's derive: If I points into surface, then for the ray to be outside before, the previous point P - ε I should satisfy |P-εI - C| > R. Expand: |(P-C) - εI|^2 = R^2 + ε^2 -2ε (P-C)·I. Need > R^2 for small ε>0 => (P-C)·I < 0. But P-C = R n_out, so n_out·I < 0. Thus the outward normal at P must have negative dot with incident direction. In our case I·n0=√3/2>0, so n_out cannot be n0; must be -n0. Hence σ=-1. Great! So part a can use this physical condition to select sign. So u = c(√3h - L)/(√3L - h) with L<√3h. Now, does this condition also automatically give L<√3h? Yes because σ=-1 means L - √3h <0 => L < √3h. Also τ>0 gives r=√3L - h >0 => L > h/√3 ≈0.577h, but not enough. Also u<c? Let's derive from first contact plus reflection maybe automatically? Condition n_out·I <0 gave L<√3h. But does u<c impose L>h? Let's see with L<√3h, u = c(√3h - L)/(√3L - h). Need u<c => √3h - L < √3L - h => h < L. So yes first contact + u<c gives L>h. But is u<c an independent assumption? Yes given. So condition b includes L>h. However, after-reflection escape gave L≥(3√3/5)h. Which is stricter than L>h? (3√3/5≈1.039) >1, so stricter. So if we include escape, condition b final. But wait: Did first contact also require r<2h? We derived r<2h. In terms L, r<2h => L<√3h, same as σ=-1. So first contact condition is exactly L<√3h. So combined with u<c L>h gives h<L<√3h. But after-reflection escape stricter L≥(3√3/5)h. Which is correct? Let's re-examine after-reflection escape condition: We required k ≤√3/2. But maybe if k>√3/2, the photon could still be outside after reflection because the sphere's center is not directly below? We derived F(Δ)=Δ[(1-k^2)Δ + R0(√3 - 2k)] for σ=-1. If k>√3/2, bracket negative for small Δ, so F<0 => inside. So yes impossible. Thus condition includes L≥(3√3/5)h. Let's check if there is any other constraint: At the instant of reflection, the photon's velocity is vertical downward. The sphere surface at P is moving outward with velocity u n_out = -u n0. The relative velocity normal = c√3/2 - u? Actually photon's outward normal component = c√3/2 (since n_out=-n0, v_r·n_out = √3/2 c). Surface outward normal velocity = u. For separation, need c√3/2 > u? Or ≥? If equal, relative normal velocity zero; but the photon's tangential component? It has none? v_r=(0,-c); tangential component = v_r - (v_r·n_out)n_out. Compute v_r·n_out = c√3/2. So tangential = (0,-c) - c√3/2 (-1/2,-√3/2) = (0,-c) + c√3/2*(1/2,√3/2)? Actually -n0 = ( -1/2,-√3/2); v_r·n_out = c√3/2; term = (v_r·n_out)n_out = c√3/2 * (-1/2,-√3/2) = (-c√3/4, -3c/4). Then tangential = (0,-c) - (-c√3/4, -3c/4) = (c√3/4, -c/4). So relative tangential velocity nonzero. If normal relative velocity zero, the photon would slide along surface? But surface is moving; it would not separate immediately but might later? Our F(Δ) with equality has F=(1-k^2)Δ^2 >0 for all Δ>0, so it separates immediately due to tangential? Actually distance squared difference initially positive quadratic, so yes outside for any Δ>0. So equality allowed. Therefore final condition: (3√3/5)h ≤ L < √3h. Also u<c? Let's verify at lower bound L=3√3/5 h ≈1.039h: k=(√3h - L)/(√3L - h) = (√3 - 3√3/5)/(3*3√3/5 -1?) Wait compute: √3h - L = √3h - 3√3h/5 = 2√3h/5. Denominator √3L - h = √3*(3√3h/5) - h = (9/5 h) - h = 4h/5. k = (2√3/5) / (4/5) = √3/2. So u=c√3/2 < c. Good. At upper L→√3h from below, k→0, u→0. At L=h? Not allowed. So condition. But wait: Is there also a condition for the photon to hit the sphere at all? The photon's path x=√3y extends to infinity; the expanding sphere could be hit later if first contact fails? Actually if photon were inside sphere before, it would have hit earlier. So first contact condition necessary. Good. Now part a expression with physical sign: u = c(√3h - L)/(√3L - h). But if they wanted absolute value, they might accept c|L-√3h|/(√3L - h). Which one to present? The problem asks "a velocidade u de expansão do raio da esfera em função de h, L e c". In physics, speed is positive; the formula with sign assumes L<√3h. Since later condition ensures that, part a can use the positive expression derived from physical sign. However, to be safe in final answer, I can mention the expression with absolute value and then note under physical condition it becomes the positive one. But the final answer format asks "Resposta final: <o resultado de cada item>" short. I need decide final expected. Let's search memory: IME 2027 2ª fase Física? The problem might have answer: a) u = c(√3h - L)/(√3L - h) b) h < L < √3h? Or with 3√3/5? Let's see typical IME: They might not consider after-reflection escape because they might assume "despreze efeitos do movimento da superfície" means treat surface as stationary after reflection? But that would be weird. Or maybe they ask "condição para que a imposição física do problema seja possível" and the answer is simply L < √3h and L > h? Let's see if after-reflection condition is maybe automatically satisfied if h<L<√3h plus the photon hits from outside? No, counterexample. Maybe they consider that after reflection, the photon leaves, and the sphere expansion doesn't affect because "despreze os efeitos do movimento da superfície" applies to entire problem? The phrase "para determinar a direção do fóton refletido" limits it to direction only. But maybe they didn't consider after-reflection escape because reflection is instantaneous and after that the photon is no longer interacting with the sphere (it's a ray). The sphere might catch up but the photon is already gone? But if it enters immediately, it would interact. I think a rigorous solution should include it. Let's verify with an alternative approach: Use relative velocity at reflection. The photon after reflection must have outward normal component greater than the surface expansion velocity to escape. The normal at P is n_out = (P-C)/R. The photon's velocity v_r = (0,-c). The normal component = v_r·n_out = c * |n_out_y| = c√3/2 (since n_out_y = -√3/2? Actually n_out = -n0 = (-1/2,-√3/2), dot = c√3/2 >0). The surface velocity outward = u. So need c√3/2 > u? But what about the tangential component? The photon can still escape even if normal component less than u? No, if surface expands normal component faster, it will catch up; tangential doesn't matter for sign of normal distance. So need u < c√3/2. This gives L > (3√3/5)h. This is a simple physical condition. Many would include it. But let's check if there is a possibility that after reflection, the photon's normal component initially less than u, but because the surface is curved and the photon moves down, the normal direction changes? Our derivative calculation used fixed normal at P for small Δt; the actual distance difference F(Δ) exact. If k>√3/2, F(Δ)<0 for small Δ, meaning the photon is inside the sphere (distance < radius). So impossible. So condition is necessary and sufficient for all Δ? We saw if k≤√3/2, F>0 for all Δ>0. Let's verify exact F(Δ)=Δ[(1-k^2)Δ + R0(√3-2k)]. If 2k≤√3, second term nonnegative, first positive, so F>0. If k>√3/2, F<0 for Δ< R0(2k-√3)/(1-k^2). So yes. Thus final condition includes u ≤ c√3/2. In terms L,h: √3h - L ≤ √3/2 (√3L - h) => 2√3h - 2L ≤ 3L - √3h => 3√3h ≤5L => L ≥ 3√3h/5. Also u>0 needs L<√3h. Also τ>0 and first contact L<√3h. Also u<c? Lower bound ensures u<c? At lower bound k=√3/2 <1, and as L increases to √3h, k→0, so u<c throughout. Good. So condition b: 3√3 h/5 ≤ L < √3 h. Also h>0. Now part a: Since L<√3h and L>h? Actually lower bound >h, so L>h. But part a expression: R=√3h - L. u = R/τ = c(√3h - L)/(√3L - h). This is positive. Should I include condition in part a? No. Let's craft final solution in Portuguese with development. But wait: What if the problem intended not to include after-reflection escape? Let's see if the lower bound 3√3/5 is less common; IME problems often have elegant answers like (3√3/5)h? Maybe yes. Let's derive with energy? No. Let's search memory: There is a known problem from IME 2027? Actually IME 2027 hasn't happened? The "prova IME 2027" might be a hypothetical or recent. But I can try recall: Maybe from IME 2024? The answer might be u = c(√3h - L)/(√3L - h) and condition L < √3h and L > h? I'm not sure. Let's re-read the problem statement carefully: "Uma esfera, cuja superfície externa é perfeitamente refletora, tem seu centro fixo no ponto de coordenadas (L, h). Seu raio R aumenta linearmente com o tempo segundo a equação: R(t)=ut. No instante t=0, um fóton parte da origem propagando-se à velocidade da luz na direção da reta x = √3 y, nos sentidos positivos de ambos os eixos. Em determinado instante, o fóton atinge a superfície da esfera e passa a propagar-se paralelamente ao eixo y, em seu sentido negativo. Dado: velocidade da luz c. Observações: t ≥ 0; h > 0; L > 0; u < c; para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos. Determine: a) a velocidade u de expansão do raio da esfera em função de h, L e c; b) a condição à qual as coordenadas h e L devem satisfazer para que a imposição física do problema seja possível." The observation "para determinar a direção do fóton refletido, despreze os efeitos do movimento da superfície da esfera e os efeitos relativísticos" explicitly tells us to ignore surface motion for determining reflected direction. It does not say ignore surface motion for physical possibility. So part b definitely should consider surface motion? The phrase "imposição física" likely includes all physical constraints, including the photon actually escaping. So yes include lower bound. Let's also consider if the photon could hit the sphere again after some time even if initially outside? We proved no if condition holds. Good. Let's solve in a clean step-by-step. Detailed derivation: Let \(t_0\) be the instant of collision. Since photon speed c, distance traveled \(r = c t_0\). Position \(P=(r\cos30°, r\sin30°)=(r√3/2, r/2)\). Let normal unit vector at P be \(n\). Since the outgoing direction is \((0,-1)\) and incoming direction is \((\sqrt3/2,1/2)\), reflection law: \[ (0,-1) = (\sqrt3/2,1/2) - 2\left[(\sqrt3/2,1/2)\cdot n\right] n \] Hence \[ (\sqrt3/2,1/2)-(0,-1) = (\sqrt3/2,3/2) = 2[(\sqrt3/2,1/2)\cdot n] n \] so \(n\) is parallel to \((\sqrt3/2,3/2)\), i.e. \[ n = \pm(1/2,\sqrt3/2). \] For the photon to be incident from outside and reflect, the center must lie on the opposite side; equivalently \(P-C = R n\) with outward normal. Since before collision the photon was outside, one must have \((P-C)\cdot u_i <0\). Testing \(n=(1/2,\sqrt3/2)\) gives positive, so \[ n = -(1/2,\sqrt3/2) = (-1/2,-\sqrt3/2). \] Thus \[ P-C = -R(1/2,\sqrt3/2) \] or components: \[ r√3/2 - L = -R/2,\quad r/2 - h = -R√3/2. \] Multiply first by 2: \(r√3 - 2L = -R\). Multiply second by 2: \(r - 2h = -R√3\). Eliminate R: from first \(R = 2L - r√3\). Substitute into second: \(r - 2h = -(2L - r√3)√3 = -2√3 L + 3r\). Then \(r - 2h = 3r - 2√3 L \Rightarrow 2r = 2√3 L - 2h \Rightarrow r = √3 L - h\). Thus \(t_0=(√3L - h)/c\) and \(R = 2L - r√3 = 2L - (√3L - h)√3 = 2L -3L + √3h = √3h - L\). Since R>0, \(L<√3h\). Therefore \[ u=\frac{R}{t_0}=\frac{c(√3h-L)}{√3L-h}. \] This is part a. Part b: Physical conditions: 1. \(t_0>0 \Rightarrow √3L > h\) (we already have L>h? lower bound will imply). 2. \(R>0 \Rightarrow L<√3h\). 3. \(u<c\): \(\frac{√3h-L}{√3L-h}<1 \Rightarrow √3h-L < √3L - h \Rightarrow h<L\). This gives \(h<L<√3h\). 4. First contact: before collision, photon must be outside sphere. This is ensured if \(r<2h\)? But \(r=√3L-h<2h \iff L<√3h\), already. Actually our first contact condition was r<2h, which is equivalent to L<√3h, so same as R>0? Wait R>0 also L<√3h. So first contact is same? Let's verify: r<2h => √3L -h <2h => √3L <3h => L<√3h. So yes same. So no extra beyond R>0? But does R>0 automatically ensure no earlier hit? We derived yes because r<2h. Let's check: At r<2h, the other root r2 > r? We derived condition r<2h for first contact. Since r<2h is equivalent to L<√3h, which is already R>0. So first contact is automatically satisfied if the geometry picks σ=-1 and R>0? Wait if L<√3h, then r<2h. So yes. So no extra. 5. After reflection, photon must be able to escape the expanding sphere. At the collision point, the outward normal is \(n=(-1/2,-√3/2)\). The photon's reflected velocity is \((0,-c)\), whose outward normal component is \[ (0,-c)\cdot n = c√3/2. \] The surface expands outward with speed \(u\). For the photon to separate, need \(u \le c√3/2\) (equality allows separation due to tangential component). Using u expression: \[ \frac{c(√3h-L)}{√3L-h} \le \frac{c√3}{2} \] \[ 2(√3h-L) \le √3(√3L-h)=3L-√3h

**Analyzing the IME Physics Problem** I'm currently dissecting the prompt for a second-phase IME Physics question, focusing on the setup involving a perfectly reflective sphere centered at a specific coordinate. My immediate task is to meticulously understand every detail of the problem statement. **Calculating Photon Behavior** I'm analyzing the photon's trajectory before and after reflection. My current focus is on establishing the initial direction vector from its given path and then determining the reflected path. This involves understanding how its propagation changes upon encountering the expanding sphere. **Determining Initial Photon Vector** I've established the initial velocity vector for the photon, noting its positive movement along both axes. My next step is to fully trace its path before collision. **Calculating Photon Behavior** I'm analyzing the photon's trajectory before and after reflection. My current focus is on establishing the initial direction vector from its given path and then determining the reflected path. This involves understanding how its propagation changes upon encountering the expanding sphere. **Refining Reflection Vector** I'm now refining the reflection vector by carefully considering the incident vector and the outward normal at the point of collision with the expanding sphere. My focus is on ensuring accurate application of the reflection law. **Applying Reflection Formula** I am confirming the reflection formula's conditions, specifically that the dot product of the incident vector and the external normal must be negative for reflection. I'm verifying if the external normal bisects the angle between the negative incident vector and the reflected vector. **Verifying Reflection Formula** I'm confirming the conditions for the reflection formula. I've calculated the difference vector between the reflected and incident velocities, determined its unit vector, and verified that its dot product with the external normal is negative, which is expected for incident light. **Confirming Reflection Direction** I've successfully confirmed that the reflected photon's velocity is precisely $-\hat{j}$ by applying the reflection formula. This solidifies my understanding of its direction post-collision with the expanding sphere. **Confirming Normal Vector** I'm now confirming the external normal vector at the point of incidence. Based on the geometry and the incident ray's trajectory, I've determined it to be in the third quadrant. **Confirming Normal Vector** I've confirmed the external normal vector to be $(-1/2, -\sqrt{3}/2)$, aligning perfectly with the geometry. I'm now calculating the coordinates of the collision point $P$, deriving $R = \sqrt{3}h - L$ and subsequently $R(t_c) = ut_c = \sqrt{3}h - L$. **Calculating Collision Time** I've derived the collision time $t_c$ by expressing the coordinates of the collision point $P$ in terms of $h$ and $L$. The distance traveled and the resulting time appear consistent with the geometric constraints. **Deriving Sphere Expansion Velocity** I've successfully derived the sphere's expansion velocity, $u$, in terms of $h$, $L$, and $c$. I'm now examining the conditions on $h$ and $L$ required for the physical feasibility of the problem, focusing on the sphere's radius being positive at collision. **Validating Problem Constraints** I'm now verifying the physical constraints derived for $h$ and $L$. I've confirmed that for the collision to be physically meaningful ($R(t_c) > 0$ and $t_c > 0$), $L/\sqrt{3} < h < \sqrt{3}L$. My current focus is on the condition $u < c$, which simplifies to $h < L$. This leads to the refined constraint $L/\sqrt{3} < h < L$. I need to carefully check if the photon can pass through the sphere before $t_c$. **Checking Photon Path Before Collision** I'm now meticulously checking the photon's path for $t < t_c$ to ensure it remains outside the expanding sphere. This involves comparing the photon's distance to the sphere's center with its radius at each intermediate time step, ensuring $D(t) > R(t)$. **Analyzing Collision Roots** I am examining the roots of the quadratic equation describing the photon's distance relative to the sphere. Since the leading coefficient is positive and $f(0)$ is positive, both roots must be positive. I am now investigating whether $t_c$ represents the smaller or larger root to ensure the photon remains outside the sphere before collision. **Investigating Collision Time** I'm now probing the nature of the collision time $t_c$. I need to confirm if it's the smaller or larger root of the distance equation and determine if the photon might have entered the sphere prematurely. My current focus is calculating both roots and their implications for the physical scenario. **Confirming Collision Roots** I've calculated the product of the collision time roots, $t_c$ and $t'$, and found $t' = t_c \frac{L^2 + h^2}{2(L^2 - h^2)}$. For $t_c$ to be the first collision, it must be the smaller root, thus $t_c \le t'$. **Validating Collision Time** I've confirmed that for the collision at $t_c$ to be the first contact, $t_c$ must be less than or equal to $t'$. This condition simplifies to $h \ge L/\sqrt{3}$. Since $R(t_c) > 0$ implies $h > L/\sqrt{3}$, this constraint is satisfied. **Confirming Collision Time Analysis** I've confirmed that $t_c$ is always the smaller root when $R(t_c) > 0$, simplifying the constraints. This is a remarkable simplification. **Investigating Subsequent Collisions** I'm now considering whether the reflected photon might collide with the expanding sphere again. My current focus is on determining if the sphere's expansion will catch up to the photon as it propagates parallel to the negative y-axis. **Analyzing Subsequent Collisions** I'm now analyzing the possibility of further collisions. My calculations indicate that the horizontal distance to the sphere's center will always be less than its expanding radius for $t \ge t_c$. Additionally, the initial rate of change of the squared distance between the reflected photon and the sphere's center is positive. **Confirming Escape Velocity** I've discovered a critical condition: for the reflected photon to escape the expanding sphere, its velocity component along the external normal must exceed the sphere's expansion velocity ($u$). This is key to ensuring it doesn't get swallowed back up. **Analyzing Photon Trajectory** I'm confirming that the reflected photon, moving at $-c\hat{j}$, won't be immediately reabsorbed. My calculations show its velocity component normal to the sphere's surface is $\frac{\sqrt{3}}{2} c$. The sphere's expansion velocity $u$ must be less than this for the photon to escape post-reflection. **Checking Escape Condition** I'm now investigating if the sphere's expansion velocity $u$ needs to be less than the photon's normal velocity component, $\frac{\sqrt{3}}{2} c$. This condition leads to $h < \frac{5\sqrt{3}}{9} L$, which fits within the existing constraints. I'm re-examining if this is the definitive condition for escape. **Validating Photon Escape** I'm confirming the condition for the reflected photon's escape: its radial velocity must exceed the sphere's expansion velocity. My current calculations focus on the quadratic in $\tau$ and require the coefficient of $\tau$ to be positive, implying $c > u$, and the constant term to be positive, requiring $u < \frac{\sqrt{3}}{2}c$. **Revising Problem Constraints** I'm re-evaluating the problem's physical constraints based on the provided observations. The critical insight is that the condition $u < c$ is explicitly stated as an observation, which simplifies the physical feasibility analysis considerably. My focus is now on the geometric implications of this given constraint. **Refining Constraint Logic** I'm re-examining the explicit constraint "$u < c$" provided by the prompt. I need to determine if this is an assumed condition or a reminder of relativistic limits. My current focus is on whether this condition is derived from the physics of the scenario or simply given, and how it impacts the constraints on $h$ and $L$. **Validating Problem Constraints** I'm re-examining the explicit constraint "$u < c$" provided by the prompt. My focus is on whether this condition is derived or simply given, and how it impacts the constraints on $h$ and $L$. **Confirming Physics Constraints** I'm now solidifying the constraints on $h$ and $L$, building upon the given $u < c$. I've confirmed that for physical possibility, $h$ must lie between $L/\sqrt{3}$ and $L$. I am currently focusing on how these limits ensure the photon doesn't enter the sphere prematurely and can escape after reflection. **Refining Problem Constraints** I am now solidifying the constraints on $h$ and $L$, incorporating the given $u < c$. I've confirmed that for physical possibility, $h$ must lie between $L/\sqrt{3}$ and $L$. My current focus is how these limits ensure the photon doesn't enter the sphere prematurely and can escape post-reflection. **Revising Constraint Logic** I'm re-examining the explicit constraint "$u < c$" provided by the prompt. My focus is on whether this condition is derived or simply given, and how it impacts the constraints on $h$ and $L$. **Validating Escape Conditions** I am re-evaluating the escape conditions. The prompt's statement to ignore sphere movement and relativistic effects for reflection direction is critical. I'm confirming the normal velocity component of the reflected photon must exceed the sphere's expansion speed, specifically $u < \frac{\sqrt{3}}{2} c$, for it to avoid immediate reabsorption. **Validating Reflection Assumptions** I am re-evaluating the prompt's instruction to disregard sphere movement during reflection. This simplifies the physics, treating it as instantaneous against a stationary mirror at the collision point. The sphere's expansion primarily determines the collision timing. My focus is on the derived constraints for $h$ and $L$ under this assumption. **Revising Problem Constraints** I'm now re-evaluating the explicit constraint "$u < c$" provided by the prompt. My focus is on whether this condition is derived or simply given, and how it impacts the constraints on $h$ and $L$, leading to $\frac{\sqrt{3}}{3} L < h < L$. **Refining Problem Constraints** I am now solidifying the constraints on $h$ and $L$, incorporating the given $u < c$. I've confirmed that for physical possibility, $h$ must lie between $L/\sqrt{3}$ and $L$. My current focus is how these limits ensure the photon doesn't enter the sphere prematurely and can escape post-reflection. **Confirming Normal Vector** I've confirmed the external normal vector to be $(-1/2, -\sqrt{3}/2)$, aligning perfectly with the geometry. This vector points from the sphere's center towards the point of reflection, as required by the reflection law. **Validating Sphere Velocity** I have successfully derived the sphere's expansion velocity, $u = c \, \frac{\sqrt{3} h - L}{\sqrt{3} L - h}$. This derivation confirms the relationships between $u$, $h$, $L$, and $c$, based on the photon's path and the sphere's properties at the point of collision. **Refining Physics Constraints** I am solidifying the constraints on $h$ and $L$, incorporating the given $u < c$. I've confirmed that for physical possibility, $h$ must lie between $L/\sqrt{3}$ and $L$. My current focus is on how these limits ensure the photon doesn't enter the sphere prematurely and can escape post-reflection.

**Analyzing Physics Problem** I am currently focused on understanding the physics problem presented, specifically the initial conditions related to a perfectly reflective sphere and its center. My immediate task is to thoroughly read and interpret the problem statement from the provided image. **Calculating Photon Reflection** I've been working through the initial conditions, focusing on the photon's departure from the origin and its initial direction. My current effort involves setting up the equations to track the photon's path and the sphere's expansion as it moves towards its eventual collision. **Determining Collision Geometry** I'm refining my understanding of the collision point. The incident photon travels at 30 degrees to the x-axis. After reflection, it moves directly down along the negative y-axis. I'm calculating the distance traveled by the photon to reach this collision point. **Deriving Normal Vector** I'm now focused on deriving the normal vector at the point of collision. Based on the incident and reflected ray angles, I've determined the normal bisects the angle between the incoming direction and the reflected direction, placing it at -120 degrees relative to the x-axis. **Validating Reflection Geometry** I've confirmed the angles of incidence and reflection match the derived normal vector perfectly. The geometry of the photon's interaction with the sphere appears consistent with the established conditions. **Solving System Equations** I have successfully derived expressions for $s$ and $R$ by manipulating the two linear equations. The solution appears consistent with the initial conditions and the geometry of the reflection. **Revising Geometric Interpretation** I'm re-evaluating a geometric interpretation of the problem. It appears my prior assumption about the angle between $\vec{OP}$ and $\vec{PC}$ being $90^\circ$ was incorrect; the actual angle is $30^\circ$, which aligns with the incident angle of $30^\circ$ and implies incidence rather than tangency. **Calculating Expansion Velocity** I've derived the expansion velocity $u$ as a function of $h$, $L$, and $c$. The calculation indicates $u = c \frac{\sqrt{3} h - L}{\sqrt{3} L - h}$, which seems to directly address item (a) of the problem. **Deriving Physical Constraints** I am now focused on item (b) of the problem, identifying the physical constraints on $h$ and $L$. This involves carefully considering all stated and implied conditions, such as positive time, radius, and expansion velocity, alongside the crucial requirement that the expansion speed $u$ must be less than $c$. **Analyzing Collision Dynamics** I am currently examining the condition where the photon encounters the sphere before reaching point P. I've established that the distance to the sphere's surface, $D(t')$, is initially positive and I'm now investigating its behavior during the time interval $[0, t)$ to ensure it remains positive, confirming the photon doesn't strike the sphere prematurely. **Confirming Non-Premature Collision** I've confirmed that the photon does not strike the sphere prematurely. The strict convexity of the distance function, coupled with its derivative at the collision time being negative, guarantees the distance remains positive before $t$. Now, I'm analyzing the post-reflection trajectory to ensure it avoids the expanding sphere. **Escaping Sphere Expansion** I've realized a critical issue: if the sphere's expansion velocity $u$ exceeds $c \frac{\sqrt{3}}{2}$, the expanding sphere would overtake the reflected photon immediately after reflection. This suggests a new constraint for the photon to escape. **Refining Constraints** I'm now focusing on the exact constraints derived from the problem's observations. Specifically, I'm re-examining the implications of $t > 0$ and $u > 0$ on the relationship between $h$ and $L$, leading to the inequality $h < \sqrt{3}L$. **Refining Velocity Constraints** I have refined the constraints on the expansion velocity $u$. By analyzing the conditions for $u < c$, I derived $h < L$. This is a stronger constraint than $h < \sqrt{3}L$, leading to the simplified interval $\frac{\sqrt{3}}{3} L < h < L$ for the physical parameters. **Analyzing Sphere Expansion Velocity** I'm critically re-examining the problem's statement regarding the sphere's radius increase. The phrasing "aumenta linearmente" strongly implies $u > 0$, which, combined with the given $u < c$, sets a clear upper and lower bound on the expansion velocity. **Verifying Calculation Logic** I am now performing a comprehensive step-by-step verification of the entire solution. My current focus is on ensuring the initial trajectory equations and the derivation of the incident point $P$ are entirely consistent with the problem's stated $30^\circ$ angle and the photon's speed $c$. **Revising Geometric Interpretation** I'm re-evaluating a geometric interpretation of the problem. It appears my prior assumption about the angle between $\vec{OP}$ and $\vec{PC}$ being $90^\circ$ was incorrect; the actual angle is $30^\circ$, which aligns with the incident angle of $30^\circ$ and implies incidence rather than tangency. **Confirming Incidence Geometry** I have confirmed the geometry of the photon's incidence and reflection. The normal vector bisects the angle between the incoming direction and the reflected ray, correctly establishing the $30^\circ$ incidence angle. This aligns with my previous deductions on angles. **Checking Geometric Calculations** I am re-verifying the geometric calculations for point P and the sphere's center C. The derived equations for $x_P + \frac{1}{2} R = L$ and $y_P + \frac{\sqrt{3}}{2} R = h$ appear to be correct based on the problem's geometry and my prior deductions.

**Evaluating mathematical expressions** I’m considering the contact F's derivative, which is negative due to the inward incident. The first root seems fine since it shapes an upward parabola, and F0 is positive. I think the threshold q must ensure that the value stays positive too. It might be worth deriving s, potentially looking into t and h values. I’ll need to solve for t using the relation from 2h, which leads to s as √3L - h. That sounds good enough to explore!