Questão 8 (Matemática)

**Parsing rectangle geometry** I need to solve a problem involving rectangle ABCD, where AB is 10 and BC is 6. There's a point E on AB and point F on BC, with rectangle EFGH inscribed inside ABCD. I must determine the coordinates for E, F, G, and H. It seems like E and F are adjacent, and I’ll start by assigning coordinates and setting up equations based on the relationships between these points, ensuring they fit within the defined rectangle.**Analyzing rectangle dimensions** I need to determine the coordinates for the interior points E and F, focusing on their conditions. The values for x range from 0 to 10, and for y, it ranges from 0 to 6. To find valid positions, I need the equation \(y(6-y) \leq 9\) to hold, leading us to certain restrictions on x values. Looking at the resulting lengths and areas, I wonder if the inscribed rectangle EFGH can be validly defined within the bounds, ensuring it's non-degenerate with necessary side lengths calculated.**Exploring rectangle dimensions** I'm analyzing potential dimensions for a rectangle inscribed in a 10x6 area. To derive the area using x values, I consider the equation formed by the cross product of two sides. Expanding the equation gives me relationships between side lengths. I think I can express the area in terms of x and y, possibly reaching a fixed area of 30. Each dimension's orientation affects the rectangle's configuration, and it might be unique depending on how I approach it. I need to solve this!**Analyzing rectangle dimensions** I'm examining a rectangle whose vertices touch all sides, with its dimensions defined by the equations involving angles. I have constraints on the area and need to explore solutions for the lengths that fit the requirement of an area of 30. The conditions involve finding points for valid coordinates and using orthogonality. I want to check the potential lengths as parameters and relate them back to the rectangle’s fixed dimensions. This requires some careful solving to explore side lengths and relationships!**Solving for dimensions and angles** I'm working through a mathematical problem involving the sum of squares and some trigonometry. I start with equations for sides p and q, and it involves careful handling of terms. I substitute and end up with a quadratic equation to solve for S. After deriving S values, I establish that S must be greater than 60. I confirm dimensions match, then calculate angles with trigonometric identities. Finally, I check the area to ensure everything aligns with the problem's requirements. I want to be thorough!**Finding area via vector algebra** I’m thinking about how to find the area of a geometric shape using algebra and vector analysis. I want to present a neat proof involving sides p and q, and angle θ, but I need to ensure my angle conventions and side directions are clear. Given the coordinates, I establish that EF points down-right and FG points down-left, with both components being positive within their defined range. I also express the outer width and height in relation to p and q. It looks like I'm on the right track!

I'm setting up rectangle ABCD with dimensions 10 by 6, placing the inscribed rectangle EFGH with vertices on each side, and using symmetry to parametrize E's position on AB as distance x from A, with corresponding distances on the other sides. Checking the perpendicularity of adjacent sides using coordinates, I derive that x(10−x) = y(6−y) must hold. I also verify that the diagonals of EFGH bisect each other at the center of ABCD, confirming the configuration must be centrally symmetric with x+g=10 and y+h=6. For part (a), since y(6−y) ranges over (0,9] for y∈(0,6), I need x(10−x) ≤ 9, which gives (x−1)(x−9) ≤ 0, so x must lie in [1,9]. I check that for each such x, a valid y exists satisfying the equation. Now for part (b), I want to express the area of EFGH as the area of ABCD minus the four corner triangles, setting up expressions for each triangle's area in terms of x and y. Setting the area equal to 30 gives me 3x + 5y − xy = 15, which combined with the equal-area condition from part (a) factors nicely into (x−5)(y−3) = 0. Testing x=5 leads to a negative discriminant, so that case gives no real solution, meaning I need to check y=3 instead. With y=3, I get x² − 10x + 9 = 0, so x = 1 or 9, which both give the same rectangle sides. Checking x=1, y=3: the diagonal lengths come out to 3√10 and √10, and the area confirms correctly at 30. So the sides of the rectangle are √10 and 3√10.

I'm setting up coordinates for rectangle ABCD with points E, F, G, H on its sides, trying to work out the inscribed rectangle EFGH's vertex positions. Using the parallelogram relation G = F + H - E, I find the shape is centrally symmetric about the rectangle's center (5,3). Then I impose perpendicularity between EF and FG to get the rectangle condition x(10-x)=y(6-y), and I'm checking this against the equal-diagonals condition as an alternative approach. Both methods agree on the same constraint, so I'm solving x²-10x+9≥0 given y(6-y) has max value 9, which restricts x to (0,1]∪[9,10). I'm also considering whether other vertex orderings around the rectangle could give additional valid configurations. For part b, I'm computing the area by subtracting the four corner triangles from the rectangle's total area of 60, using triangle legs x, 6-y, 10-x, and y, which simplifies to 60-6x-10y+2xy.

I'm setting up coordinates for the rectangle with E on AB and F on BC, then working out where G lands on CD by moving perpendicular to EF from F. Checking where H falls on DA, I find H=(0,6-y), and requiring a real solution leads to a constraint on x: either x≤1 or x≥9. For the area condition, I'm expressing the rectangle's area as EF times FG, substituting y² in terms of x to get an equation to solve for area 30. Alternatively, I compute the area by subtracting the four corner triangles from the full rectangle ABCD, which simplifies to 3x+5y-xy=15, factoring into (5-x)y=3(5-x), giving x=5 or y=3 — checking whether x=5 is valid. So y=3 forces x=1 or x=9 after solving the discriminant condition, and verifying both give consistent rectangle dimensions of √10 by 3√10, confirming the area is 30 in both cases.

The problem involves a rectangle ABCD with sides AB = 10 and BC = 6. An inscribed rectangle EFGH has vertices on the boundary of ABCD. E is on side AB and F is on side BC. The positions of G and H are not specified. The sides of EFGH may not be parallel to ABCD. A coordinate system is being set up with A at (0,0), B at (10,0), C at (10,6), and D at (0,6). E is placed on AB at (x, 0) where 0 < x < 10. F is placed on BC at (10, y) with 0 < y < 6. The positions of G and H remain unclear because the problem only specifies E and F. It is assumed that the other vertices of EFGH also lie on the sides of ABCD. The vertices of EFGH are assumed to lie on opposite sides of ABCD: E on AB, F on BC, G on CD, and H on DA. The cyclic order is likely E-F-G-H. The rectangle condition requires perpendicular adjacent sides and equal opposite sides. The coordinates are now fixed: E(x, 0), F(10, q), G(r, 6), and H(0, s). The rectangle condition will be enforced by requiring perpendicular adjacent sides or equal opposite sides. The midpoint of diagonal EG is ((p+r)/2, 3). The midpoint of diagonal FH is (5, (q+s)/2). For EFGH to be a parallelogram these midpoints must coincide. This yields p + r = 10 so r = 10 − p. The midpoint condition also yields s = 6 − q, so H(0, 6 − q). The vertices are now fully determined. The next step is to enforce the rectangle condition by requiring adjacent sides to be perpendicular. The vectors EF and FG are computed. Their dot product must be zero for perpendicularity. The remaining sides GH and HE are verified to be negatives of EF and FG, confirming the quadrilateral is a parallelogram. The perpendicularity condition yields q(6 − q) = p(10 − p). This leaves two variables, so the rectangle is not uniquely determined by x alone. The problem likely asks for all possible values of x. The problem asks for possible values of x, so not all positions of E are valid. The missing constraint is being reconsidered. Perhaps the inscribed rectangle does not have one vertex on each side of ABCD, or perhaps its sides follow a different rule. The maximum value of p(10−p) is 25, while the maximum of q(6−q) is only 9. Therefore p(10−p) must be at most 9 for a valid rectangle to exist. This yields x(10−x) ≤ 9, or equivalently x² − 10x + 9 ≤ 0. Solving this inequality gives 1 ≤ x ≤ 9. The valid range for x is (0, 1] ∪ [9, 10). The endpoints are included because the rectangle remains non-degenerate. For example, when x = 1, q = 3 yields G(9, 6) and H(0, 3), and the dot product of adjacent vectors EF and FG is zero, confirming perpendicularity. All vertices are distinct from the corners of ABCD. The problem does not explicitly forbid G or H from coinciding with the corners of ABCD. However, as x approaches 0, G approaches C and q approaches 0 or 6, which may violate the strict interior placement requirement for a proper inscribed rectangle. The next step is part b: with area 30, determine the side lengths of EFGH. The area will be expressed in terms of x. The side lengths of EFGH are being derived from the coordinates. EF has length sqrt((10−p)² + q²) and FG has length sqrt(p² + (6−q)²). The area is |EF|·|FG| because the sides are perpendicular. A second equation is needed to solve for the side lengths when the area equals 30. The area of EFGH was computed as |(10−p)(6−q) + pq|, which expands to 60 − 10q − 6p + 2pq. This expression is too unwieldy for a contest problem, so the initial assumption about the positions of G and H may be incorrect. The problem statement is being re-read carefully. The problem's wording confirms that x must be chosen so that a valid inscribed rectangle exists. For part b, the side lengths of EFGH are being sought when its area equals 30. The approach is shifting to examine the corner triangles cut off by the inscribed rectangle, rather than relying solely on the earlier coordinate-based area formula. The segments from each corner of ABCD to the vertices of EFGH are being examined. For example, BE equals 10 minus p. The lengths of the four right triangles cut off by the inscribed rectangle are being compared. The right triangles cut off by the inscribed rectangle are now identified. At corner B, triangle BEF has legs BE = 10 − p and BF = q. At corner C, triangle FCG has legs CF = 6 − q and CG = p. The slopes of EF and FG are being examined. EF has slope q/(10−p) and FG has slope −(6−q)/p. The perpendicularity condition is being checked by requiring their product to equal −1. The perpendicularity check confirms the earlier condition q(6−q) = p(10−p). The corner triangles are not similar, since their legs do not satisfy a proportional relationship. The right triangles cut off at corners A and D are now identified. At A, triangle HAE has legs AH = 6 − q and AE = p. At D, triangle HDG has legs DH = q and DG = 10 − p. The right triangles cut off at corners A and C are congruent. Both have legs of length x and 6 − q. The right triangles cut off at corners B and D are also congruent. Both have legs of length 10 − x and q. The next step is to examine whether the four corner triangles are similar in pairs, using the rectangle's angles or slopes. The rectangle angle confirms the setup is consistent. To find side lengths for area 30, the substitution u = x and v = 6 − q is being introduced. The area of the inscribed rectangle is being expressed in terms of u and v. The four corner triangles contribute a total area of uv + (10−u)(6−v). Subtracting this from the outer rectangle’s area of 60 gives the inner area as 60 − uv − (10−u)(6−v). The inner rectangle area simplifies to 6u + 10v − 2uv, which equals 30. Dividing by 2 gives the equation 3u + 5v − uv = 15. The next step is to solve this for one variable or to find the side lengths directly. The rectangle sides are being identified as the hypotenuses of the corner triangles. The side lengths are being denoted as l and w, with the area constraint l·w = 30. The inner rectangle’s sides are the hypotenuses of the corner triangles. Side 1 is sqrt((10−x)² + q²), and side 2 is sqrt(x² + (6−q)²). Their product equals 30. The angle that EF makes with the horizontal is being examined to find a relationship between x and q. The angle θ that EF makes with the horizontal is being calculated as arctan(q/(10−x)). Because adjacent sides of the inscribed rectangle are perpendicular, FG must make a complementary angle with the vertical. The next step is to examine the angles of triangle A or the direction of HE to find a relationship between x and q. The direction vector of FG is being compared to the 90-degree rotations of EF. The vector (-x, 6-q) does not match a simple scaled rotation of (10−x, q), so the perpendicularity condition must be satisfied by another relationship. The direction vector of FG is now being compared to the 90-degree rotations of EF. Rotating EF counterclockwise gives (-q, 10−x), and since FG must be parallel to this vector, the ratio x/q equals k. Substituting into the second component yields 6−q = k(10−x), which simplifies to q(6−q) = x(10−x). The rotated direction vector of EF has length equal to the side length |EF|. Because FG is parallel to this vector and scaled by k = x/q, the length of FG equals k times that same side length. This confirms the proportion between the two sides of the inscribed rectangle. The area formula side1²·k = 30 was derived but does not seem immediately useful. The approach is shifting to express the side lengths using the angle θ and its sine and cosine. The possibility that the corner triangles are similar is being reconsidered. The algebraic approach is being pursued. The area equation 6u + 10v − 2uv = 30 is being rearranged to isolate u: 2u(3 − v) + 10v = 30. The algebraic manipulation of the area equation 2u(3−v) = 10(3−v) is being completed. Dividing by (3−v) would yield u = 5, but that contradicts the constraint u(10−u) = v(6−v), since 25 exceeds the maximum value of v(6−v). Therefore 3−v must equal zero, giving v = 3. The algebraic derivation is confirmed correct. When v = 3, the area equation reduces to 0 = 0, so it is automatically satisfied for any valid value of u. Because u = 5 would violate the constraint u(10−u) = v(6−v), the only viable case for area 30 is v = 3. The constraint u(10−u) = 9 yields u = 1 or 9. These correspond to x = 1 or 9 with q = 3. The area equation collapses to an identity when v = 3, so the area is not necessarily 30 in that case. The area formula is being rechecked. The four corner triangles contribute a total area of uv + (10−u)(6−v). Expanding that product gives 60 − 10v − 6u + uv, so the inner rectangle area is 6u + 10v − 2uv. The inner rectangle area is exactly 30 whenever v = 3, because the u terms cancel. Only u = 1 or 9 satisfy the existence condition for that value of v, so the area equals 30 precisely at those two points. The side lengths are being calculated numerically for the case u = 1 and v = 3. The length of EF is sqrt(90), or 3√10. The vector for FG is being determined. The side lengths for u = 1 and v = 3 are confirmed: |EF| = 3√10 and |FG| = √10, giving area 30. The same holds for u = 9 and v = 3. The conclusion that no other solutions exist is being questioned. The algebraic derivation is confirmed: the only rectangles with area 30 have sides √10 and 3√10. The earlier question about whether other solutions exist is resolved. Rectangles with area not equal to 30 can still exist for values of x where x(10−x) < 9, but those are not the target case. A numerical check with u = 0.5 confirms that the area is not 30. The two possible values of v yield areas of approximately 48.6 and 11.4, so only the boundary cases u = 1 or 9 produce the target area. The assumption that each vertex of the inscribed rectangle lies on a different side of the outer rectangle is being questioned. The possibility of alternative configurations, such as two vertices on one side or opposite sides, is being considered. The vertices E and F are adjacent sides of the inscribed rectangle, so G and H must lie on the remaining two sides. This confirms the earlier assumption that each vertex sits on a different side of the outer rectangle. The possibility of placing two vertices on the same side is being examined. If E and H both lie on AB, then HE is a side of the rectangle. The next step is to determine whether such a configuration can still be inscribed with the other vertices on the boundary. Because EF must be perpendicular to FG, and F lies on BC, the adjacent vertex H cannot also lie on AB. The rectangle’s vertices must therefore occupy distinct sides of the outer rectangle. The possibility of placing G on DA or on the same side as E is being considered. A more general vector or coordinate approach without fixed vertex assignments is being explored. The standard configuration with one vertex on each side of the outer rectangle is being accepted as the intended setup. The possible values of x are being derived from the constraint x(10−x) ≤ 9, which yields 0 < x ≤ 1 or 9 ≤ x < 10. The maximum value of q(6−q) is noted to be 9. The possibility of placing vertices on opposite sides or in non-cyclic order is briefly considered but set aside. The boundary values x = 0 and x = 10 are excluded because they would place vertices on corners rather than open sides. For x in (0, 1) or (9, 10), two distinct rectangles exist. At x = 1 or x = 9, only one rectangle is possible. The possibility of alternative configurations for part b) is being explored. Placing G on DA and H on CD is considered, but it may not yield a valid inscribed rectangle. A coordinate approach is being set up with E(x, 0) on AB and F(10, y) on BC. The remaining vertices G and H must lie on the boundary of the outer rectangle. The coordinate approach is being formalized with vectors. The condition that angle EFG be 90 degrees is being translated into a perpendicularity requirement between vectors FE and FG. The vector from E to F is (10−x, y). Rotating it 90 degrees gives two possible directions for the adjacent side. The next vertex G is placed at F plus k times that rotated vector, where k is positive. The vector approach confirms that any rectangle with one vertex on AB and one on BC can be generated by rotating the connecting vector 90 degrees and scaling it. The next step is to check which of the two possible rotations places the remaining vertices G and H on the boundary of the outer rectangle. The first rotation case is being worked out. With the vector rotated 90 degrees counter-clockwise, G and H are placed at (10−ky, y+k(10−x)) and (x−ky, k(10−x)). The next step is to check which values of k keep both points on the boundary of the outer rectangle. The first rotation case is being solved. Setting H on the left side gives k = x/y, which fixes G at (10−x, y + (x/y)(10−x)). The remaining step is to check that G lies on the top side. The first rotation case is fully solved. Requiring G to lie on the top side yields k = (6−y)/(10−x), which equals x/y from the left-side condition. This confirms the constraint x(10−x) = y(6−y). The remaining vertices therefore land exactly on the boundary, and the derivation is complete. The first rotation case is fully solved. Requiring G to lie on the top side yields k = (6−y)/(10−x), which equals x/y from the left-side condition. This confirms the constraint x(10−x) = y(6−y). The remaining vertices therefore land exactly on the boundary, and the derivation is complete. The counter-clockwise rotation sends G toward the top side and H toward the left or top. This suggests the first rotation case may not be the only valid configuration. A systematic check of possible side assignments for the remaining vertices is underway. Since E and F already occupy the bottom and right sides, H and G cannot lie on those sides in a non-degenerate configuration. The first rotation case places H on the left boundary and G on the top boundary. The first rotation case is being ruled out. H cannot lie on the bottom or right side, and G cannot lie on the right side. This leaves only two possible boundary positions for each vertex: H on the left side and G on the top side. The systematic side-assignment check is underway. Case 1 places H on the left and G on the top, which matches the previously solved configuration. Case 2 places both H and G on the left side. Case 2, where both H and G lie on the left side, is impossible because it would require x = 10. Case 3, with both vertices on the top side, is also impossible since it forces y to zero. Case 4 places H on the top boundary and G on the left boundary; solving for k yields k = 6/(10−x), and substituting into G_x = 0 gives y = (5/3)(10−x). The next step is to verify that this value of y lies in the open interval (0, 6). Case 4 is impossible. The derived value of y = (5/3)(10−x) exceeds 6 for x < 6.4, so it violates the boundary constraint. Also, H_x = x − ky = x − 10 < 0, placing H outside the rectangle. The counter-clockwise rotation is now fully ruled out. The only valid configuration remains the first case already solved. The clockwise rotation direction is being examined next. The clockwise rotation is ruled out. For any positive k, the second vertex H lands below the bottom side of the rectangle, so it cannot lie on the boundary. Negative values of k are equivalent to the counter-clockwise rotation with positive scaling, so they do not yield any new configurations. The clockwise rotation is ruled out. For any positive k, the second vertex H lands below the bottom side of the rectangle, so it cannot lie on the boundary. Negative values of k are equivalent to the counter-clockwise rotation with positive scaling, so they do not yield any new configurations. The clockwise-orientation possibility is being reconsidered. A clockwise rectangle might exist if the adjacent side is added in the opposite direction, or if H is treated as the second vertex instead of G. The parametrization is confirmed to cover all possible rectangles with EF as one side. Each rectangle has exactly two valid orientations, corresponding to the two half-planes on either side of the segment EF. <summary goes here> The counter-clockwise rotation places the rectangle inside the outer boundary, while the clockwise rotation places it outside. The possibility that E and F are not adjacent vertices is now being considered. The naming convention EFGH likely implies consecutive vertices around the rectangle, similar to ABCD. This suggests EF is a side rather than a diagonal. The consecutive-vertex assumption is being reconsidered. If EF is a diagonal, the rectangle would have area 30 with sides √10 and 3√10. The consecutive-vertex assumption is now accepted. The question asks for possible values of x, which confirms that not all points in the interval (0, 10) are valid. The idea that the inscribed rectangle might have sides parallel to the outer rectangle is dismissed. If E lies on AB and F lies on BC, the sides cannot be parallel. The possible values of x are 0 < x ≤ 1 or 9 ≤ x < 10. The sides of the inscribed rectangle measure √10 and 3√10. The endpoints x = 1 and x = 9 are valid because they yield area 30. The solution will be written in Portuguese with step-by-step development. The outer rectangle is placed in a Cartesian coordinate system with A(0,0), B(10,0), C(10,6), and D(0,6). The geometric derivation is being pursued without heavy coordinate algebra. The condition x(10−x) = y(6−y) is being justified through similar triangles and the segment lengths AE = x, EB = 10−x, BF = y, and FC = 6−y. The geometric derivation is being pursued through the four corner right triangles. The angle at F in triangle EBF is being defined so that the rectangle's sides remain perpendicular to EF. The solution will use Cartesian coordinates. The derivation of the constraint x(10−x) ≤ 9 is being revisited to ensure it does not rely on hidden assumptions. The earlier derivation assumed G and H lie on specific sides, then used the parallelogram property to fix their positions. The perpendicularity condition then produced the product equation. This confirms the constraint is valid under those side-assignment assumptions. The vector method confirmed the only valid configurations, but the solution will be presented using similar triangles. The corner triangles at B and C are being checked for similarity with the example values x = 1, y = 3, v = 3, u = 1. The corner triangles at B and C are similar for the example values x = 1 and y = 3. A numerical check with x = 0.5 and y ≈ 0.938 confirms this ratio holds approximately. The general condition is being derived from these proportional relationships. The derivation is complete. The proportion (10−x)/y = (6−y)/x is equivalent to x(10−x) = y(6−y), confirming the constraint arises from the similarity of the corner triangles. The rotation method was considered but set aside as too lengthy. The answer is already clear, so the focus returns to part a). That part asks for possible values of x, not the area. A trigonometric proof is being explored. EF makes angle θ with the base, so tanθ = y/(10−x). The adjacent side FG must make the same angle with the vertical. This geometric relationship will yield the constraint x(10−x) = y(6−y). The solution will use coordinate geometry and vectors. The vector from E to F is (10−x, y). A perpendicular vector is obtained by rotating 90° anti-clockwise: (−y, 10−x). The vector from F to G is then k(−y, 10−x) for some positive k. The coordinates of G and H are derived from the perpendicular vector. Setting H on AD and G on CD yields k = x/y and k = (6−y)/(10−x). Equating these expressions gives x(10−x) = y(6−y). The maximum value of y(6−y) is 9, so x(10−x) must be at most 9. Solving the resulting inequality yields the valid intervals 0 < x ≤ 1 or 9 ≤ x < 10. The coordinates of G and H remain inside their respective sides for these values, and the scaling factor k stays positive. The area of the inscribed rectangle is being derived from the vector method. Since the side FG is k times the length of EF, the area equals k times the square of |EF|. The item b) will use this expression to find the side lengths when the area is 30. The side lengths of the inscribed rectangle are being derived from the vector method. The length of EF is \sqrt{(10-x)^2 + y^2}, and the length of FG is k times that same value. Since k equals both x/y and (6-y)/(10-x), the area can be expressed as k times the square of |EF|. This expression will be used to solve for the side lengths when the area is 30. The area formula is being rewritten as (x/y) times the square of |EF|^2. The coordinates of G and H are fixed as (10−x, 6) and (0, 6−y). The shoelace formula will be applied to these four points to solve for x and y when the area equals 30. The shoelace formula is being applied to the four vertices E(x,0), F(10,y), G(10−x,6), and H(0,6−y). The cross-products are being expanded to derive the area expression. The shoelace expansion is being simplified. The area expression is being reduced to a sum of constants and terms involving x and y. The shoelace expansion yields the area expression 60 − 6x − 10y + 2xy. A quick check with x = 1 and y = 3 confirms this matches the required area of 30. The author then notices this differs from an earlier expression they had derived, and begins to trace the source of the discrepancy. The author traced the discrepancy between the two area formulas to a variable substitution. The earlier expression 6u + 10v − 2uv used v = 6 − y, while the shoelace formula uses the raw y coordinate. Both expressions evaluate to 30 for x = 1 and y = 3. The author realized the earlier area formula 6u + 10v − 2uv was incorrect. Substituting v = 6 − y into the correct expression 60 − 6x − 10y + 2xy yields exactly that same result, confirming the discrepancy was a sign error in the first calculation. The author realized the earlier area formula 6u + 10v − 2uv was actually correct. Substituting v = 6 − y into the shoelace expression 60 − 6x − 10y + 2xy yields exactly that same result, so the discrepancy was a sign error in the first calculation. The author discovered an inconsistency in the variable definitions. The shoelace formula uses y directly, while the earlier triangle subtraction used v = 6 − y. Setting the area equal to 30 yields two different equations, so the definitions need clarification. A quick re-derivation of the triangle subtraction method is being attempted to confirm the correct expression. The author realized the shoelace area formula 60 − 6x − 10y + 2xy is valid for any polygon with those four vertices, regardless of whether the shape is a rectangle. The perpendicularity condition x(10−x) = y(6−y) is what restricts the configuration to a rectangle. The four corner triangles are being identified. The triangle at A has legs x and 6−y. The triangle at B has legs 10−x and y. The triangle at C has legs x and 6−y. The corner triangle at D has legs 10−x and y, so its area is (1/2)(10−x)y. Adding the four corner triangles gives x(6−y) + (10−x)y = 6x + 10y − 2xy. Subtracting this from the total rectangle area of 60 yields exactly the same expression as the shoelace formula: 60 − 6x − 10y + 2xy. This confirms the earlier variable substitution was correct. The author noticed that setting the area equal to 30 yields u = 5 or v = 3. When v = 3, the area formula always equals 30 regardless of x, so this condition does not uniquely determine a rectangle. The author is questioning whether this result is meaningful. The author confirmed that when y = 3, the area expression simplifies to exactly 30 for any value of x. The quadrilateral EFGH therefore has constant area 30 regardless of x, provided the points are fixed at E(x,0), F(10,3), G(10−x,6), and H(0,3). The author realized that when y = 3, the diagonal HF is horizontal and both diagonals share the same midpoint (5, 3). This confirms the quadrilateral is always a parallelogram with area 30. It becomes a rectangle only when x(10−x) = 9, which occurs at x = 1 or x = 9. The author checked x = 5 as a concrete example. The vectors EF and FG have a dot product of −16, confirming the sides are not perpendicular. This confirms the quadrilateral is a parallelogram with area 30 but not a rectangle. The author calculated the exact area for x = 5. The side lengths are both √34, and the angle between them is arccos(−8/17), giving sin = 15/17. The area equals 34 × (15/17) = 30, confirming the parallelogram property. The author noted that the problem requires a rectangle, so only x = 1 or x = 9 satisfy the perpendicularity condition. In those cases the area is exactly 30. The author is checking whether other rectangles with area 30 exist for values of y other than 3. They note that the four vertices always form a parallelogram, since both diagonals share the same midpoint (5, 3). The author derived the length of diagonal FH as sqrt(100 + 4(y−3)²). For a rectangle, the two diagonals must be equal, which yields the equation (10−2x)² − 4(y−3)² = 64. They also noted that the area can be computed via vector cross products. The author verified the area formula 60 − 6x − 10y + 2xy gives positive values for x = 5, y = 3 and x = 0.5, y ≈ 0.938. They noted that an earlier incorrect variable substitution had produced a different result. The area for the second y value is approximately 11.442. The author then set the area expression equal to 30 and multiplied both sides by −1, arriving at the equation 6x + 10y − 2xy = 30. The author factored the area equation into (x − 5)(y − 3) = 0. This means the area equals 30 if and only if x = 5 or y = 3. They then checked whether x = 5 could still yield a rectangle, since the perpendicularity condition requires x(10 − x) = y(6 − y). The author confirmed that x = 5 cannot yield a rectangle, since the perpendicularity condition would require y(6 − y) = 25, which is impossible. The only valid rectangles with area 30 are those where x = 1 or x = 9 and y = 3. For x = 1, the side lengths are EF = 3√10 and FG = √10. The author verified that the second rectangle case (x = 9, y = 3) also has side lengths √10 and 3√10. The factoring result confirms these are the only possible rectangles with area 30. They also noted that both configurations lie inside the original rectangle. The author determined the set of valid x values for part a) is (0, 1] ∪ [9, 10). This follows from the constraint that y(6−y) = x(10−x) must have a solution with 0 < y < 6. They also confirmed the side lengths are √10 and 3√10. The author is now drafting the final solution in Portuguese. They have begun by setting up the coordinate system for ABCD and defining E(x,0) and F(10,y), then noting that G and H must lie on AD and CD respectively to keep the inscribed quadrilateral inside the rectangle. The author derived the perpendicularity condition by setting the dot product of vectors EF and FG to zero. This yields y(6−y) = x(10−x). They then noted that for a valid solution with 0 < y < 6, it is necessary and sufficient that x(10−x) ≤ 9, which is the maximum value of y(6−y). The author is exploring an alternative approach to part b) by using the fact that the area equals 30. They note that |EF|·|FG|=30, so |EF|²·|FG|²=900. They then try to expand this product of squared side lengths and use the perpendicularity condition y(6−y)=x(10−x) to find a relationship between the two lengths. The author abandoned the alternative approach of adding squared side lengths, finding it too unwieldy. They confirmed that factoring the area equation into (x − 5)(y − 3) = 0 is the cleaner and more direct way to show that only x = 1 or x = 9 yield rectangles with area 30. They also briefly considered rewriting the area expression as 2(x − 5)(y − 3) plus a constant. The author derived the area formula as 30 plus twice the product (x−5)(y−3). This shows that the area equals 30 if and only if that product is zero, confirming the earlier factoring result. The author is finalizing the Portuguese solution. They will briefly justify why G and H must lie on AD and CD, rather than simply stating it as the only configuration.

We need answer in Portuguese, solve. Need parse problem. Need be careful. We have rectangle ABCD with sides m(AB)=10 and m(BC)=6. Rectangle EFGH is inscribed in ABCD, with E belonging to side AB and distinct from A and B, and F belonging to side BC and distinct from B and C. a) If x represents distance from A to E, determine possible values of x. b) If area of EFGH is 30, determine measures of its sides. We need infer configuration. Rectangle ABCD typical labeling maybe A-B-C-D in order, AB horizontal length 10, BC vertical length 6? Usually rectangle ABCD vertices consecutive, AB and CD lengths 10, BC and DA lengths 6. E on AB (bottom? top?) distinct endpoints, F on BC adjacent side from B to C. Rectangle EFGH inscribed in ABCD, with E on AB, F on BC. Since E and F are on adjacent sides meeting at B? Wait AB and BC share B; E on AB, F on BC. For a rectangle inscribed with E on AB, F on BC, the other vertices G and H must lie on CD and DA respectively? Need verify. A rectangle EFGH with consecutive vertices E, F, G, H, with E on AB, F on BC. If EF is side, then EF not necessarily along sides; F is on BC adjacent at B. Then G must be on CD? H on DA? Since rectangle inside big rectangle, likely vertex E on AB, F on BC, G on CD, H on DA, one on each side, in order. But image may show? Let's analyze. Given rectangle ABCD dimensions 10x6. Rectangle EFGH inscribed in ABCD, with E on AB distinct from A and B, F on BC distinct from B and C. So E not endpoints on AB; F not endpoints on BC. This leaves G,H must lie on CD and DA? Since a rectangle inscribed in a rectangle with one vertex on each side if nondegenerate? Need consider possible orientations: sides of inscribed rectangle could connect E-F, F-G, G-H, H-E, with maybe E and F adjacent, so EF is a side. If E on AB, F on BC, then EF segment from AB to BC; possible for rectangle tilted. The other vertices might lie on CD and DA. Yes "inscrito" means each vertex lies on a side? Probably vertices on sides of ABCD (not necessarily distinct? one per side maybe). E on AB, F on BC, G on CD, H on DA. Need find possible values of x = m(AE). Then area 30 determine side measures. This is classic inscribed rectangle problem. Let's set coordinates. Let A=(0,0), B=(10,0), C=(10,6), D=(0,6). E on AB: E=(x,0), x in (0,10) because distinct from A (0) and B (10). F on BC: F=(10,y), y in (0,6) because distinct from B/C. Rectangle EFGH has E,F,G,H consecutive. Since EF is side? Given rectangle EFGH, vertices in order E-F-G-H. E on AB, F on BC, so EF is a side connecting points on adjacent sides at B. The opposite side GH is parallel to EF. The other vertices G,H must lie on CD and DA maybe. Need determine. Let's denote vector v = F - E = (10-x, y). Side EF length L = sqrt((10-x)^2 + y^2). Since rectangle, adjacent side vector w must be perpendicular to v. Let w = (-y, 10-x) or (y, -(10-x)) maybe, scaled by t? Actually length of other side unknown. Let H = E + w? Since rectangle EFGH with consecutive E,F,G,H, adjacent sides: EF and FG, so vector FG = w. Then G = F + w; H = E + w. Also H must lie on DA (x=0 line? Actually DA vertical x=0, y between 0 and6) and G on CD (y=6 horizontal, x between 0 and10). w must be chosen such that F+w lies on y=6 (CD), E+w lies on x=0 (DA). Let w = (a,b). Conditions: H = E + w = (x+a, b) lies on DA: x+a = 0 => a = -x. G = F + w = (10+a, y+b) = (10-x, y+b) lies on CD: y+b = 6 => b = 6 - y. Thus w = (-x, 6-y). Also w perpendicular to v = (10-x, y): v · w = (10-x)(-x) + y(6-y) = -x(10-x) + y(6-y) = 0? Wait for rectangle, adjacent sides perpendicular: EF · FG = 0 => v · w =0. Thus -x(10-x) + y(6-y)=0 => x(10-x) = y(6-y). Also w length is other side. This equation. Need find possible x? But x and y variables with y in (0,6), x in (0,10). Equation determines possible pairs. But maybe for any x? Let's solve: x(10-x) = y(6-y). Left ranges 0 to 25 (at x=5), right y(6-y) ranges 0 to 9 (at y=3) for y in (0,6). Thus for every x in (0,10), left maybe 0 to25; right max 9. Wait x can be 1? left=9? x(10-x)=9 at x=1 or9; right=9 at y=3. For x=5, left=25 >9 impossible. Thus x cannot be too large. There is constraint. But wait perhaps orientation of w could be opposite? Let's check. If rectangle EFGH inscribed not necessarily with E,F,G,H in order? It likely says rectangle EFGH inscribed in ABCD, with E on AB, F on BC. The order might be E,F,G,H. Yes. But there may be another possible choice: vector w could be (x, -(6-y))? Let's derive generally. Conditions: H lies on DA: E + w has x=0 -> since E=(x,0), a=-x. G lies on CD: F + w has y=6 -> since F=(10,y), b=6-y. If w=(a,b). So w=(-x,6-y) uniquely if H on DA and G on CD. No other. Perpendicular condition: (10-x, y) · (-x, 6-y)=0 => -x(10-x) + y(6-y)=0 => x(10-x)=y(6-y). So yes. But wait if H on DA and G on CD, then the side GH = w? Actually H=E+w, G=F+w, so GH = F-E = v, parallel and equal to EF. Good. Thus for any E (x), F (y) satisfying equation, rectangle exists. Part a asks possible values of x. Need determine range? x distinct from 0,10 and y in (0,6) satisfying. Equation: x(10-x) = y(6-y). Since y ∈ (0,6), RHS ∈ (0,9] (at y=3 max 9; as y→0 or6, 0). Thus LHS ∈ (0,9]? x(10-x) ∈ (0,9] for x in (0,10)? Wait x(10-x) at x=0 is0, x=5 is25, x=10 is0. It takes all values from 0 to25; but to equal RHS ≤9, require x(10-x) ≤9. Solve 10x - x^2 ≤9 => x^2 -10x +9 ≥0 => (x-1)(x-9) ≥0 => x ≤1 or x ≥9. Since x∈(0,10), possible x ∈ (0,1] ∪ [9,10)? But at x=1 or9 equality 9; y=3. At x=0 or10 left=0, y→0 or6 (not allowed y distinct endpoints), but x=0/10 not allowed E distinct from A/B. So possible values: 0 < x ≤ 1 or 9 ≤ x < 10. But wait at x=1 left=9, y=3; y in (0,6), okay. At x=9 left=9, y=3. At x=0+ small left small positive, y small positive? yes. At x=10- small left small positive, y near? if x=9.9, left=0.99, solve y(6-y)=0.99 has two roots? y = 3 ± sqrt(9-0.99)=3±sqrt(8.01) ≈3±2.830 =0.170 or5.830, both in (0,6). So yes. But is there any other possible orientation where G and H not on CD/DA? The statement "retângulo EFGH é inscrito em ABCD, com E pertencente ao lado AB e distinto de A e B, e F pertencente ao lado BC e distinto de B e C." It doesn't explicitly say G on CD and H on DA, but "inscrito" implies vertices on sides? In a rectangle inscribed in a rectangle, each vertex on a side; but could also be that one side of inscribed rectangle lies along AB? No, E distinct endpoints, F distinct endpoints. Since rectangle has 4 vertices, and E,F are on AB, BC not adjacent? They are adjacent sides at B. The remaining two vertices must be on remaining sides CD, DA due to convexity and distinctness? Likely. But maybe there is alternative: Rectangle EFGH could have E on AB, F on BC, and G,H maybe on CD and DA but not necessarily one-to-one? Actually if E,F on AB,BC, then the sides of ABCD: AB contains E; BC contains F; CD and DA each one vertex maybe. Could G lie on BC too? No, BC only F distinct from B,C, so G not on BC. Could H lie on AB? E distinct, but maybe H also on AB? But rectangle degenerate? Not likely; "inscrito" usually one vertex per side. We'll assume standard. Part b: area of EFGH is 30, determine measures of its sides. Area = |EF| * |FG| = |v| * |w| = 30, with v=(10-x, y), w=(-x,6-y). Since perpendicular, area = |v| |w|. Need find side lengths? There may be multiple rectangles? Need solve. We know v length L1 = sqrt((10-x)^2 + y^2). w length L2 = sqrt(x^2 + (6-y)^2). Conditions: x(10-x)=y(6-y) and L1 L2 =30, with x ∈ (0,1]∪[9,10), y∈(0,6). But maybe due symmetry multiple? Need determine sides. Let's set up. Let v = (a, y) where a=10-x >? If x∈(0,1], a∈[9,10); if x∈[9,10), a∈(0,1]. Let y? Equation x a = y(6-y). Let a=10-x. Then x a = y(6-y). Also w=(-x,6-y) = (-x, a?) No, 6-y =? w = (-x, 6-y). Let b=6-y. Then y(6-y)=yb = xa. Lengths: L1 = sqrt(a^2 + y^2), L2 = sqrt(x^2 + b^2), where a=10-x, b=6-y, and x a = y b. Also since w = (-x, b) length L2. We have L1 L2 =30. Could maybe express using equality xa = yb. Let's solve. Let xa = yb = k. Since x,a positive, y,b positive, k>0. Also x+a =10, y+b=6 (since a=10-x, b=6-y). Product k = xa; for given k, x and a are roots of t^2 -10t + k=0; y,b roots of t^2 -6t + k=0. Since discriminants nonnegative: 100 -4k ≥0 => k≤25; 36 -4k ≥0 => k≤9. Also x,a positive => 0<k≤25? Since x in (0,10), a>0, product k ∈ (0,25), but with y∈(0,6), k∈(0,9]. So 0<k≤9. Also x=5±sqrt(25-k); a=5∓sqrt(25-k). y=3±sqrt(9-k); b=3∓sqrt(9-k). Need signs. Then L1 = sqrt(a^2 + y^2). But a and y not independent: a = 10-x. There may be relation using xa = yb, but not directly. We can maybe use vector lengths and dot? Let's find. Let v=(a,y), w=(-x,b). Area = |v||w| =? Since perpendicular, |v||w| = |det(v,w)|? Actually for perpendicular, |v||w| = |det(v,w)| = |a b - (-x) y|? Wait det = a*b - (-x)*y = ab + xy = (10-x)(6-y)+xy = 60 -10y -6x + xy + xy? Compute: (10-x)(6-y) =60 -10y -6x + xy. + xy =60 -10y -6x + 2xy. Not simple. Alternatively |v||w| = |v| * |w| = area. But since v and w perpendicular, area = |v×w| = |det(v,w)| = |a*b - (-x)*y| = |ab + xy| = ab+xy = (10-x)(6-y)+xy =60 -10y -6x +2xy. Is that equal to? Also from xa = yb, maybe express xy? Let's solve. Actually area of inscribed rectangle EFGH in big rectangle has formula? For a rectangle inscribed with sides tilted, area =? It might be (10-x)(? )? Let's derive using coordinates: Area = |EF| * |FG|. Since EF perpendicular FG, could compute determinant: Area = |det(v, w)| = |v_x w_y - v_y w_x| = |(10-x)(6-y) - y(-x)| = (10-x)(6-y)+ xy = (10-x)(6-y)+xy. Let's denote S = 60 -10y -6x +2xy. Need S=30. Also condition x(10-x)=y(6-y). Let a=10-x, b=6-y. Then xa=yb, and S = a b + x y? Since (10-x)(6-y)+xy = a b + x y. Need S=30. Also x+a=10, y+b=6. There may be relation S =? Since xa=yb=k, maybe S = ab + xy. But ab? From xa=k and yb=k, maybe express ab =? ab = (a)(b). We know yb=k => b=k/y. xa=k => a=k/x. Then ab = k^2/(xy). Also xy? From xa=yb => x/y = b/a? Hmm. Maybe solve more directly using geometry. Inscribed rectangle sides are parallel to? Since EF connects AB to BC; maybe the rectangle sides have slopes. Also big rectangle dimensions 10 and 6. Area 30. There may be a simple answer: sides 5 and 6? But if one side vertical/horizontal? Let's test. If x=1, y=3? Then v=(9,3) length sqrt(90)=3√10, w=(-1,3) length √10, area=30. Ah! Area = 3√10 * √10 = 30. Sides lengths: 3√10 and √10? But question asks determine measures of its sides (possibly multiple). Let's test with x=9? If x=9, then a=1, equation xa=9 = yb => y=3,b=3? y=3 => RHS=9, yes. Then v=(1,3), w=(-9,3)? Wait w=(-x,b)=(-9,3), length sqrt(90)=3√10, L1=√10? Area = √10 * 3√10=30. Same side lengths swapped? Actually side lengths: EF = sqrt(1+9)=√10, FG = sqrt(81+9)=√90=3√10. Same. But there may be other solutions? Let's solve equation S=30 and xa=yb. Let x,a positive with sum 10, product k = xa. Let y,b positive with sum 6, product k = yb. So for a given k, x,a determined up to swap? Since sum fixed, roots: x = (10 ± sqrt(100-4k))/2 = 5 ± p, where p = sqrt(25-k). a=5∓p. y=3 ± q, b=3∓q, q=sqrt(9-k). Sign choices correspond to whether x is larger or smaller; but E distinct anyway. There are two possible x: 5+p and 5-p; but k≤9 so p≥sqrt(16)=4? Actually p = sqrt(25-k) ∈ [sqrt(16),5] = [4,5] for k∈[0,9]. So x can be 5±p ∈ [0,10]; specifically x=5+p ∈ [9,10), x=5-p ∈ (0,1]. Corresponds to our intervals. Similarly y=3±q with q=sqrt(9-k) ∈ [0,3]? k∈(0,9], q∈[0,3). y ∈ (0,6), two choices y=3±q, both in (0,6) for k<9; if k=9 q=0 y=3. Now area S = ab + xy = (5∓p)(3∓q) + (5±p)(3±q) depending sign pairing. We need S=30. Let's compute depending on which signs. Because x and a are complementary; y and b complementary. Let x=5+s p, a=5-s p; y=3+t q, b=3-t q, where s,t ∈ {+1,-1}. But product k = xa = yb = (5+s p)(5-s p)=25-p^2 = k? Wait p^2=25-k, yes product k. Similarly q^2=9-k. Note p^2 - q^2 = (25-k)-(9-k)=16, so p^2 - q^2=16, i.e. p^2 = q^2 +16. Also p∈[4,5], q∈[0,3]. Area S = ab + xy = (5-s p)(3-t q) + (5+s p)(3+t q). Compute: First: 15 -5t q -3s p + s t p q Second: 15 +5t q +3s p + s t p q Sum =30 + 2 s t p q. Wow! So S = 30 + 2 s t p q. Since p,q≥0. For S=30, need 2 s t p q =0 => s t p q =0. Since q=0 (k=9) implies p=4? p^2=16+0=16 => p=4; yes k=9. Or p=0 impossible? p≥4. Thus only q=0 => y=3, b=3, k=9, p=4, x=5±4 =1 or9. Then area =30. So unique k=9, y=3, x=1 or9. Thus side lengths: EF = sqrt(a^2+y^2) with a=10-x: if x=1,a=9,y=3 => L1=√(81+9)=√90=3√10; FG = sqrt(x^2+b^2)=√(1+9)=√10. If x=9,a=1,y=3 => L1=√10, FG=√90=3√10. Same multiset {√10, 3√10}. So answer b: √10 and 3√10 (measures of sides). Part a possible values: 0 < x ≤ 1 or 9 ≤ x < 10? Need check endpoints x=1 and 9 included? E distinct from A and B: A distance 0, B distance 10. At x=1 or9, E is not A/B yes. At x=1, y=3, F=(10,3) not B/C (B y=0, C y=6), yes. At x=9, y=3. At endpoints x=0 or10 not allowed; x=0 (A) and x=10 (B) excluded. But x=0 also equation gives y=0/6 (F B/C) excluded. So possible values: 0 < x ≤ 1 or 9 ≤ x < 10. However check if x exactly 1 and x exactly 9 included? Yes, because y=3. x=1 and x=9 valid: E distinct from A (0) and B (10), F=(10,3) distinct from B (10,0) and C (10,6). Good. But wait part a asks "Se x representa a distância do ponto A ao ponto E, determine os possíveis valores de x." The wording: E pertencente ao lado AB e distinto de A e B. x distance from A to E. Since AB length 10, x ∈ (0,10). Condition for existence? Need determine possible values de x. We found due y ∈ (0,6) and rectangle possibility, x cannot be middle? Actually x ∈ (0,1] ∪ [9,10). Need verify if for x in (0,1] or [9,10) there exists y. Equation x(10-x)=y(6-y). Let LHS = k ∈? For x ∈ (0,1], LHS from 0+ to 9 (at x=1) (increasing? x(10-x) derivative 10-2x >0 for x<5, so yes 0 to9). For x ∈ [9,10), LHS from 9 down to 0 (derivative negative for x>5). Need k ∈ (0,9]? At x=1 k=9; at x=9 k=9; at x=0+ k→0; at x=10- k→0. For each k in (0,9], equation y(6-y)=k has two solutions y=3±√(9-k), both in (0,6) for k<9; for k=9 y=3. So yes. Thus answer a: 0 < x ≤ 1 ou 9 ≤ x < 10. Maybe can express as x ∈ (0,1] ∪ [9,10). But maybe they expect "0 < x ≤ 1 ou 9 ≤ x < 10" in terms of distances. Need check if x=0? A, not allowed; x=10 B not allowed. But if x=1, maybe E? yes. Let's ensure no alternative interpretation where x could be any (0,10)? Suppose rectangle EFGH inscribed but maybe F on BC, E on AB, and other vertices not necessarily on CD/DA? But inscribed in rectangle means all vertices on boundary. Since E,F on AB,BC, G,H must be on boundary but could be on CD and DA exactly due sides length. I think standard. However let's be cautious. There may be another possibility: Rectangle EFGH could have E on AB, F on BC, and then side FG goes from F to G, side GH, side HE. If the rectangle is oriented such that G and H lie on AB and BC? No, vertices distinct? Let's see if G could be on CD and H on AD as we assumed. But also could be that E on AB, F on BC, and the side EF is a diagonal? No, EF connects adjacent sides at B; if E,F are adjacent vertices, then EF is a side. The other vertices must be on opposite sides CD,DA due convexity; yes. But wait if E and F are not adjacent? In rectangle EFGH, vertices named consecutively? Standard: "retângulo EFGH" implies vertices in order E-F-G-H, so E and F adjacent. So okay. Let's test with area 30: If our area formula S = ab + xy = (10-x)(6-y)+xy. For x=1,y=3: S=9*3 +3 =30. For x=9,y=3: S=1*3+27=30. Good. Now, part b "determine as medidas de seus lados" maybe could be asked "as medidas" plural; answer √10 and 3√10. Could include no matter orientation. But let's check if there might be two non-congruent inscribed rectangles with area 30? Our derivation suggests unique multiset of side lengths, but maybe two possible configurations (x=1 or x=9) correspond to mirror images; side lengths same. But maybe if area 30, sides are 5 and 6? Wait area 30 with sides 5 and6 not possible? Let's test if a rectangle inscribed with sides 5 and6? The big rectangle has dimensions 10 and6; a 5x6 rectangle could be placed vertically? But E on AB, F onBC maybe? Let's explore: If one side of inscribed rectangle is horizontal? Not possible because E and F on adjacent sides; a horizontal side could connect AB to BC? It would be top/bottom edge? Actually if rectangle has side EF vertical? E=(x,0), F=(10,y), slope nonzero; for EF vertical would require y=0? no. Horizontal would require? E,F have same y: y=0? F on BC y=0 at B, not allowed. So no. But maybe inscribed rectangle could be oriented differently: E on AB, F on BC, G on CD, H on DA, but sides not necessarily EF and FG? Maybe rectangle named EFGH with E and F opposite? No, convention vertices in order. Let's verify with another method: In general, an inscribed rectangle in an outer rectangle with sides a,b has area formula? If vertices (x,0),(a,y),(a-x,b),(0,b-y)? Wait our coordinates: E=(x,0), F=(a,y), G=(a-x,b), H=(0,b-y) with b=6? We found G=(10-x,6), H=(0,6-y). Yes. Perpendicular condition x(a-x)=y(b-y). Area =? We computed S = (a-x)(b-y)+xy. Let's derive in general: rectangle vertices E=(x,0), F=(a,y), G=(a-x,b), H=(0,b-y). This is a rectangle if (a-x)(b-y)=xy? Check perpendicular: v=(a-x,y), w=(-x,b-y); dot = -(a-x)x + y(b-y) = -x(a-x)+ y(b-y)=0 => x(a-x)=y(b-y). So area = |v||w| = sqrt((a-x)^2+y^2)*sqrt(x^2+(b-y)^2). But also since perpendicular, area = |det(v,w)| = (a-x)(b-y)+xy. For a=10,b=6: S = (10-x)(6-y)+xy = 60 -10y -6x +2xy. Condition x(10-x)=y(6-y) => 10x -x^2 =6y -y^2 => x^2 -10x = y^2 -6y? Actually 10x -x^2 -6y +y^2=0. Can express x^2 -10x = y^2 -6y? No. Maybe S simplification? Let's try using condition: x(10-x)=y(6-y) => 10x -x^2 =6y -y^2 => x^2 -10x +6y -y^2=0 => (x-5)^2 -25 + (y-3)^2 -9? Hmm. Area S = (10-x)(6-y)+xy =60 -10y -6x +2xy. Is there a relation S = ? From condition: 6y - y^2 =10x -x^2 => y^2 =6y -10x +x^2. Then xy? Could maximize? Let's solve S=30 maybe via inequalities? We can derive S = ? Maybe S = 30 + something? Let's express in terms of deviations u=x-5, v=y-3. Condition x(10-x)=y(6-y): (5+u)(5-u) = (3+v)(3-v) =>25-u^2 =9-v^2 => u^2 - v^2 =16. Area S = (10-x)(6-y)+xy = (5-u)(3-v)+(5+u)(3+v)? Wait 10-x=5-u, 6-y=3-v, xy=(5+u)(3+v). Sum = (5-u)(3-v)+(5+u)(3+v) = 30 + 2uv? Let's compute: (15 -5v -3u + uv) + (15 +5v +3u + uv)=30+2uv. Yes. Condition u^2 - v^2 =16. Area=30+2uv. Set 30 => uv=0. Since u^2 - v^2 =16 => if u=0 then -v^2=16 impossible; so v=0 => u^2=16 => u=±4. x=5±4 =1 or9; y=3. Nice. This is elegant. Part b answer sides? Compute sides: EF length = sqrt((10-x)^2+y^2) = sqrt((5∓? )?). If x=1, u=-4, y=3,v=0: EF vector (9,3), length √90=3√10; FG=(-1,3), length √10. If x=9,u=4: vectors (1,3) and (-9,3), sides √10 and 3√10. So side measures: √10 e 3√10. Now, we need produce final answer in Portuguese, show development. The user says "Resolva a questão como um candidato faria na prova: mostre o desenvolvimento em português, item por item quando a questão tiver itens (a, b, c…). Termine com uma linha neste formato: Resposta final: <o resultado de cada item, de forma curta> Por exemplo, "Resposta final: a) 3/2; b) 5√2"." So final should include solution and final line. Need maybe include conventions text? Not needed. Need solve based on image. We should write a well-organized solution. Let's formulate: - Set coordinates: A=(0,0), B=(10,0), C=(10,6), D=(0,6). Let E=(x,0), 0<x<10. Let F=(10,y), 0<y<6. Since ABCD rectangle, G on CD and H on DA. For inscribed rectangle EFGH, if EFGH is a rectangle, adjacent sides EF and FG perpendicular, and G,H positions. Need be precise: Does the problem guarantee G on CD and H on DA? We can say como EFGH está inscrito em ABCD e E,F estão em lados adjacentes, os outros vértices G e H devem estar sobre CD e DA, respectivamente. Then derive. Let's write: Seja E=(x,0), F=(10,y), G=(10-x,6)? Wait we can derive: Since G on CD (y=6), and H on DA (x=0). The vector FG must be perpendicular to EF and parallel to EH? Actually we can determine: Since EFGH is rectangle, \vec{FG} = \vec{EH} and \vec{GH}=\vec{EF}. Let H=(0,h), G=(g,6). From \vec{EH}=( -x, h), \vec{FG}= (g-10, 6-y). Equality gives -x = g-10 => g=10-x; h=6-y. Thus H=(0,6-y), G=(10-x,6). Or via perpendicular condition. Good. Then condition EF ⟂ FG: (10-x, y) · (-x, 6-y)=0 => -x(10-x)+y(6-y)=0 => x(10-x)=y(6-y). This equation. Part a: Since y(6-y) ∈ (0,9] for y∈(0,6) (max 9 at y=3, limits 0 at 0,6), need 0 < x(10-x) ≤9. Solve 10x - x^2 ≤9 => x^2 -10x +9 ≥0 => (x-1)(x-9) ≥0 => x ≤1 or x ≥9. With 0<x<10, x∈(0,1]∪[9,10). Check endpoints: x=1? valid; x=9? valid; x=0,10 excluded by E distinct. So possible values: 0 < x ≤ 1 ou 9 ≤ x < 10. Alternative: Because y∈(0,6), y(6-y)>0 and ≤9. Also x(10-x)>0 because x∈(0,10). So x(10-x)≤9. Good. Part b: Use area. Let set u=x-5, v=y-3. Equation x(10-x)=y(6-y) becomes (5+u)(5-u)=(3+v)(3-v) =>25-u^2=9-v^2 => u^2 - v^2=16. Area S = m(EF)*m(FG) = |det((10-x,y), (-x,6-y))| = (10-x)(6-y)+xy. Now (10-x)(6-y)+xy = (5-u? Wait 10-x=5-u, 6-y=3-v, xy=(5+u)(3+v). Actually yes; but since x=5+u, 10-x=5-u; y=3+v, 6-y=3-v. Then S=(5-u)(3-v)+(5+u)(3+v)=30+2uv. Need S=30 => uv=0. With u^2 - v^2=16, uv=0 gives v=0 (since u=0 impossible). Thus v=0 => y=3, u^2=16 => u=±4. Therefore x=5±4 =1 or9; y=3. Then side lengths: For x=1, EF = sqrt((10-1)^2+3^2)=sqrt(81+9)=sqrt90=3√10; FG = sqrt(1^2+3^2)=√10. For x=9, same sides swapped. Hence medidas dos lados: √10 e 3√10. Need ensure area formula: We used determinant, but if v and w perpendicular, area = |v||w| = |det(v,w)| because perpendicular. Or simply compute area = |EF|*|FG|. We can state "Como os lados são perpendiculares, a área é o módulo do determinante..." Let's detail. Maybe there is another way: The rectangle side lengths are sqrt((10-x)^2+y^2) and sqrt(x^2+(6-y)^2). Use area condition. But our determinant simplification is elegant. Let's check if area formula sign: det(v,w) = (10-x)(6-y) - y(-x) = (10-x)(6-y)+xy positive? (10-x)>0, (6-y)>0, positive; plus xy positive, so positive. So S = determinant = (10-x)(6-y)+xy. Good. Now, ensure part a answer maybe "0 < x ≤ 1 ou 9 ≤ x < 10" but in final line perhaps "a) 0 < x ≤ 1 ou 9 ≤ x < 10; b) √10 e 3√10". Need maybe use "m(EF)=..." etc. Let's consider if there is any possibility that x=0 or10? Problem says E distinto de A e B, so x cannot be 0 or10. But in possible values, x=0 would make E=A not allowed; x=10 E=B not allowed. So exclude. Let's verify if x can be exactly 1? At x=1, equation: x(10-x)=9; y(6-y)=9 => y=3 (unique because discriminant zero). So F=(10,3); distinct from B (10,0) and C (10,6), yes. Good. At x=9, same. Part b side lengths: If x=1, E=(1,0), F=(10,3), G=(9,6), H=(0,3). Sides: EF vector (9,3) length 3√10; FG vector (-1,3) length √10. If x=9, E=(9,0), F=(10,3), G=(1,6), H=(0,3): EF vector (1,3) length √10; FG vector (-9,3) length 3√10. So measures of sides √10 and 3√10. The problem asks "determine as medidas de seus lados" maybe if orientation matters? Usually rectangle sides unordered; so both. Could write "10√? no." Let's ensure no other rectangle with area 30 maybe with x= something else? Our derivation equating uv=0 used area formula and condition, but let's verify no missing case where uv not zero yet S=30 due determinant absolute? Since S=30+2uv. uv could be negative? But u^2-v^2=16, so u and v can have same sign? If uv negative possible? Let's solve uv=0 only; because area=30 => 30+2uv=30 => uv=0. Yes no other. But wait condition u^2 - v^2 =16. If uv negative, e.g. u=5, v= -? u^2 - v^2=25-v^2=16 => v^2=9 => v=±3, uv=15 or -15. Area=30+2uv =60 or0? If uv=-15 area=0 impossible for nondegenerate? Actually area positive? It can't be 0 because rectangle nondegenerate. But equation area=30+2uv =0? No, area formula positive? Wait if uv=-15, area=30-30=0, not valid. But uv could be? Solve area =30+2uv. Since area positive, uv > -15. But could uv=-5 => area20 etc. But area=30 requires uv=0. So yes. However is it guaranteed S=30+2uv with u,v as defined? Let's double-check: x=5+u, y=3+v. 10-x=5-u. 6-y=3-v. xy=(5+u)(3+v). S = (5-u)(3-v) + (5+u)(3+v) = 15 -5v -3u + uv +15 +5v +3u + uv =30 +2uv. yes. Equation x(10-x)=y(6-y): x(10-x) = (5+u)(5-u)=25-u^2. y(6-y) = (3+v)(3-v)=9-v^2. =>25-u^2 =9-v^2 => u^2 - v^2 =16. yes. Then area=30 => 30+2uv=30 => uv=0. Since u^2 - v^2 =16. If u=0 impossible because -v^2=16. If v=0 => u^2=16 => u=±4. Good. Now, part a solve elegantly: From x(10-x)=y(6-y). Since y∈(0,6), y(6-y) ∈ (0,9]. Hence 0 < x(10-x) ≤9. Solve: x(10-x) ≤9 => x^2 -10x +9 ≥0 => (x-1)(x-9)≥0. As x∈(0,10), x≤1 or x≥9. Combined with E distinct => 0<x≤1 or 9≤x<10. At x=1 or9, y=3 valid; at x near 0 or10 also valid? Need ensure y∈(0,6) for these x. For x in (0,1], LHS positive up to9; equation y(6-y)=k has two solutions in (0,6) for 0<k<9, one solution at k=9. Similarly x∈[9,10). So yes. Potential issue: At x=1 or9, F has y=3 not endpoints; okay. At x=0.5, k=4.75; y=3±sqrt(9-4.75)=3±sqrt4.25 =3±2.062, both in (0,6): 0.938,5.062. So F distinct. Good. Let's maybe mention "com y∈(0,6), a função y(6-y) tem máximo 9 e tende a 0 nos extremos; portanto 0<y(6-y)≤9." Then solve. Now write final answer in Portuguese with appropriate detail. Need obey user: "Resolva a questão como um candidato faria na prova: mostre o desenvolvimento em português, item por item quando a questão tiver itens (a, b, c…). Termine com uma linha neste formato: Resposta final: <o resultado de cada item, de forma curta> Por exemplo, "Resposta final: a) 3/2; b) 5√2". Se a questão tiver uma resposta só, escreva só o resultado." So final should include solution and final line. We can include equations. Use LaTeX maybe. Let's craft final with sections. Potential need to mention "coordenadas cartesianas" because convenções. Use A(0,0), etc. Let's write: "Considere A=(0,0), B=(10,0), C=(10,6) e D=(0,6). Como E∈AB, escrevemos E=(x,0) com 0<x<10. Como F∈BC, F=(10,y) com 0<y<6. Para o retângulo EFGH inscrito, os outros vértices estão em CD e DA; seja H=(0,h), G=(g,6). Do paralelismo/igualdade dos lados opostos, \vec{FG}=\vec{EH}: (g-10,6-y)=(-x,h) => g=10-x, h=6-y. Assim H=(0,6-y), G=(10-x,6)." Then "A condição de perpendicularidade EF⊥FG é (10-x,y)·(-x,6-y)=0 => ..." Maybe derive area formula: "Além disso, a área é S = |det((10-x,y),(-x,6-y))| = (10-x)(6-y)+xy." Then part a and b. Let's ensure final line exactly "Resposta final: a) 0 < x ≤ 1 ou 9 ≤ x < 10; b) √10 e 3√10" maybe include "m(EF)=√10 e m(FG)=3√10 (ou vice-versa)". Since asks medidas dos lados, can write "as medidas são √10 e 3√10". In final line, perhaps "a) \(0<x\le1\) ou \(9\le x<10\); b) \(\sqrt{10}\) e \(3\sqrt{10}\)". Good. Let's think if "possíveis valores de x" could be expressed as "0 < x ≤ 1 ou 9 ≤ x < 10" but maybe include equality at 1 and 9? At x=1, E is at distance 1 from A; yes distinct. At x=9, distance 9; yes. At x=0,1,9,10? 0 and10 excluded. So correct. Let's maybe check if x=1 or9 leads to F at y=3, which is midpoint of BC; okay. Now, possible issue: If x=1, E is on AB, F on BC; but G=(9,6), H=(0,3). H on DA (x=0, y=3), yes between D(0,6) and A(0,0); G on CD (y=6, x=9) between C(10,6) and D(0,6). Good. Let's verify rectangle side measures from EFGH: For x=1, EF length sqrt(9^2+3^2)=sqrt90=3√10; FG length sqrt(1^2+3^2)=√10. For x=9, EF=√10, FG=3√10. Same. Could there be rectangles with area 30 where the sides are not "√10 and 3√10" but maybe length and width swapped; same. Now, maybe the problem wants "determine as medidas de seus lados" maybe if area 30 and x possible? It asks conditional: If area of EFGH is 30, determine measures of its sides. This is part b independent of a? It might also use x? They might expect solving both items; our b uses a's formula? Actually part b asks area condition; we solve. Good. Let's formulate final response elegantly: ``` Sejam A=(0,0), B=(10,0), C=(10,6), D=(0,6). E=(x,0), 0<x<10; F=(10,y), 0<y<6. ... a) Da perpendicularidade, x(10-x)=y(6-y). Como 0<y<6, 0<y(6-y)≤9. Logo 0<x(10-x)≤9 => x^2-10x+9≥0 => (x-1)(x-9)≥0 => x≤1 ou x≥9. Com 0<x<10: 0<x≤1 ou 9≤x<10. b) Seja u=x-5, v=y-3. ... ``` Need maybe prove "y(6-y)>0" and ≤9: Because y ∈ (0,6), product positive; max at y=3 is9. Good. Let's produce final answer with desired oververbosity 5. Keep concise but show development. Potential final line: "Resposta final: a) \(0<x\le1\) ou \(9\le x<10\); b) \(m(EF)=\sqrt{10}\) e \(m(FG)=3\sqrt{10}\) (ou vice-versa)." Let's ensure final answer line is last. We'll include maybe not too many extra lines. Let's write now.

**Exploring Rectangle Inscriptions** I'm currently analyzing the geometric constraints for inscribing a smaller rectangle within a larger one. The key challenge lies in defining the conditions for the vertices of the inscribed rectangle to lie on the sides of the outer rectangle. **Defining Rectangle Vertices** I'm currently establishing a coordinate system to represent the vertices of the outer rectangle ABCD and an inscribed rectangle EFGH. My focus is on defining the coordinates of E on AB, and F on BC, to then determine the range of possible distances from A to E. **Analyzing Vertex Placement** I've confirmed that an inscribed rectangle, by definition, requires its vertices to lie on all four sides of the outer rectangle. I'm now exploring the implications of this for the corner triangles, specifically that they must be right-angled and similar. This suggests a strong relationship between their angles. **Proving Triangle Congruence** I've established that the corner triangles formed by the inscribed and outer rectangles are indeed similar. Further analysis shows that adjacent pairs of these triangles are also congruent due to equal hypotenuses, which simplifies the relationships between their side lengths. **Defining Side Ratios** I've begun parameterizing the side lengths of the corner triangles. Using an angle $\theta$, I'm expressing the segments of the outer rectangle's sides in terms of the inscribed rectangle's side lengths and this angle, which should allow for direct calculation. **Establishing Side Relationships** I've derived a crucial relationship: $y(6 - y) = x(10 - x)$. This equation directly links the segments of the outer rectangle's sides created by the vertices of the inscribed rectangle, confirming its geometric validity. **Solving for Possible x** I've successfully derived the quadratic equation relating the segments of the outer rectangle's sides and used the discriminant to find the possible range for $x$. The conditions are now established for a real solution for $y$ to exist, ensuring valid inscribed rectangle configurations. **Validating Segment Solutions** I've confirmed that for $x$ in $(0,1) \cup (9,10)$, both resulting $y$ values fall within the $(0,6)$ range. This implies valid positions for vertex $F$ on side $BC$ given the constraints on $x$, excluding endpoints as required. **Confirming x Bounds** I've confirmed that values of $x$ in $(0, 1] \cup [9, 10)$ guarantee at least one valid point $F$ on side $BC$. If $1 < x < 9$, the discriminant is negative, meaning no real $y$ exists, so inscription is impossible. Thus, the valid range for $x$ is indeed $(0, 1] \cup [9, 10)$. **Verifying Vertex Constraints** I've confirmed that an inscribed rectangle's vertices must indeed lie on all four sides. My current focus is on ensuring the vertex $G$ specifically lands on side $CD$, and investigating the implications if the perpendicular line to $EF$ strikes before reaching the intended side. **Confirming Vertex Location** I've definitively proven that vertex $G$ must lie on side $CD$ and vertex $H$ on side $DA$. My prior concern about $G$ potentially landing on $AD$ was unfounded, as it would force $H$ outside the bounds of the outer rectangle. **Confirming Rectangle Setup** I've confirmed that an inscribed rectangle must have one vertex on each side of the outer rectangle. This confirms our current general configuration is indeed unique and covers all valid cases. **Deriving Area Equation** I've derived a linear equation relating $x$ and $y$ from the sum of the corner triangle areas: $6x + 10y - 2xy = 30$. This, combined with the similarity relationship $y(6-y) = x(10-x)$, forms a system of two equations for $x$ and $y$. I am now exploring how to solve this system. **Finding Rectangle Solutions** I've discovered a key factorization of the area equation, $2(x-5)(y-3) = 0$, simplifying the system. This leads to two possibilities: $x=5$ or $y=3$. Testing $x=5$ yields no real solutions for $y$, indicating that case is impossible. **Solving for Valid Rectangles** I've resolved the system of equations derived from the area and similarity relationships. The only valid solutions occur when $y=3$, leading to $x=1$ or $x=9$. These solutions yield inscribed rectangles with sides $\sqrt{10}$ and $3\sqrt{10}$, confirming their area of 30. **Revisando Segmentos de Lados** Estou revisando minhas deduções para o item "a)" em particular. A minha análise das restrições de $x$ como a distância de $A$ a $E$ em $\overline{AB}$ e $y$ como a distância de $B$ a $F$ em $\overline{BC}$ parece correta, garantindo $0 < x < 10$ e $0 < y < 6$. **Validating Similarity Conditions** I've confirmed that the corner triangles are similar and derived the crucial relationship $y(6-y) = x(10-x)$. I've also analyzed the discriminant of the resulting quadratic in $y$, establishing that $x$ must be less than or equal to 1 or greater than or equal to 9 for real solutions to exist. **Validating Solution Bounds** I've confirmed that the bounds for $x$ are indeed $0 < x \le 1$ or $9 \le x < 10$. My analysis indicates that $x=1$ and $x=9$ are valid endpoints, leading to a unique $y=3$ solution. These values ensure the vertices lie strictly between the outer rectangle's vertices. **Calculating Rectangle Areas** I've confirmed the area of the outer rectangle $ABCD$ is 60. I'm now calculating the area of the inscribed rectangle $EFGH$ using the derived relationships between $x$ and $y$, specifically when the area is given as 30. **Solving for Inscribed Rectangles** I've successfully factored the area equation, revealing that $y$ must be 3. This, when combined with the similarity condition, yields the two possible solutions for $x$: 1 and 9. Both scenarios lead to inscribed rectangles with sides $\sqrt{10}$ and $3\sqrt{10}$, confirming the area requirement.

**Analyzing Geometry Problem** I am currently focusing on understanding the geometric constraints presented by the rectangle $ABCD$ with given side lengths of 10 and 6. My next step is to identify how the additional information about a second rectangle will interact with the initial figure. **Exploring Inscribed Rectangle Constraints** I'm now investigating how the vertices of the inscribed rectangle $EFGH$ are positioned on the sides of $ABCD$. Specifically, with $E$ on $AB$ and $F$ on $BC$, I need to determine the placement of $G$ and $H$ to ensure $EFGH$ remains a rectangle. This is key to defining the possible values for $x$ and calculating side lengths. **Proving Center Alignment** I am now rigorously proving that the center of an inscribed rectangle $EFGH$ must coincide with the center of the outer rectangle $ABCD$. My coordinate-based derivation confirms that for $E$ on $AB$ and $F$ on $BC$, the positions of $G$ and $H$ are constrained such that the centers align. This validates my initial assumption. **Confirming Orthogonality Condition** I've derived the condition $x(10 - x) = y(6 - y)$ by ensuring $\angle HEF$ is 90 degrees. I'm now verifying if this same orthogonality condition arises from examining other angles within the inscribed rectangle, which would confirm consistency. **Verifying Triangle Congruence** I've re-examined the corner triangles, realizing that $\triangle AEH$ and $\triangle CGF$ are congruent, as are $\triangle BFE$ and $\triangle DHG$. This realization confirms the condition $x(10-x) = y(6-y)$ arises directly from the requirement that opposite sides are equal, not just from the 90-degree angle constraint. **Confirming Parallelogram Property** I've just realized the key: if $EFGH$ is a parallelogram and has one right angle, it must be a rectangle. This simplifies my approach as the condition $x(10-x) = y(6-y)$ is indeed sufficient because it forces the shape to be a rectangle, not just a parallelogram. **Deriving Solution Constraints** I've discovered a critical inequality: for a valid inscribed rectangle, $x(10-x)$ must be less than or equal to 9. Testing values, I found that this constraint on $x$ restricts it to $(0, 1] \cup [9, 10)$, significantly narrowing the possibilities. I'm now verifying the implications of this range. **Exploring Alternative Inscriptions** I am now questioning my assumption that vertices $G$ and $H$ must lie on specific sides. I need to confirm if there are other valid ways to inscribe a rectangle $EFGH$ within $ABCD$, especially considering whether multiple vertices could fall on the same side of the outer rectangle. **Revisiting Vertex Placement** I am now re-evaluating all possible positions for the vertices of the inscribed rectangle. My previous derivations assumed a specific vertex-to-side mapping, and I'm exploring if alternate configurations, like multiple vertices on a single side of the outer rectangle, can still form a valid inscribed rectangle. **Reconsidering Vertex Assignments** I am now meticulously re-examining all potential vertex placements. The assumption that vertices must map to specific sides needs rigorous validation, as alternate configurations might exist where multiple vertices land on a single outer rectangle side. **Confirming Rectangle Case** I've confirmed that when $x=1$, $y=3$, the inscribed shape $EFGH$ forms a perfect rectangle with side lengths $3\sqrt{10}$ and $\sqrt{10}$. This specific instance yields an area of 30, aligning with a potential solution to item (b). **Validating Triangle Relationships** I've confirmed that the similarity of the corner triangles $\triangle BFE$ and $\triangle AEH$ is the key. The consistent ratio between their corresponding legs dictates the relationship between $x$ and $y$, leading to the equation $x(10-x) = y(6-y)$, which I've already established. This reinforces the geometric constraints. **Solving Parameterized System** I've successfully solved the linear system for $x$ and $y$ in terms of $r$, yielding expressions like $x = \frac{6r - 10}{r^2 - 1}$. I've also realized that positivity constraints on $x, y, r$ automatically ensure $x < 10$ and $y < 6$, simplifying condition verification significantly. **Analyzing Parameter Ranges** I am now analyzing the parameter $r$ across different cases. I've determined that for $r > 5/3$, the valid range for $x$ is $(0, 1]$. I'm now focusing on the case where $0 < r < 1$, where the conditions on $x$ and $y$ require $r < 3/5$. **Solving Parameterized System** I've successfully solved for $x$ and $y$ in terms of parameter $r$, yielding expressions like $x = \frac{6r - 10}{r^2 - 1}$. Positivity constraints on $x$, $y$, and $r$ automatically ensure $x < 10$ and $y < 6$. **Analyzing Parameter Ranges** I'm now examining the parameter $r$. For $r > 5/3$, $x$ is in $(0, 1]$. In the case $0 < r < 1$, $x$ and $y$ conditions lead to $r < 3/5$. **Confirming Solution Consistency** My analysis using both the quadratic constraint and the similarity ratio confirms the same valid range for $x$: $(0, 1] \cup [9, 10)$. I am now investigating if an area of 30 is exclusively achieved at $x=1$ or $x=9$. **Expressing Area in Terms of Parameter** I've derived the area $S$ in terms of $r$: $S = \frac{(6r - 10)(10r - 6)(r^2 + 1)}{(r^2 - 1)^2}$. Setting $S=30$ simplifies to $\frac{(3r - 5)(10r - 6)(r^2 + 1)}{(r^2 - 1)^2} = 15$. **Solving Quadratic Equation** I've successfully derived a quadratic equation for the parameter $u$, which represents $r + 1/r$. This equation, $15u^2 - 68u + 60 = 0$, has a discriminant of $32^2$, leading to clean solutions for $u$. **Solving Quadratic Equation** I have found the roots for $u = r + 1/r$ to be $10/3$ and $6/5$. Since $6/5$ is extraneous, only $10/3$ remains, leading to valid $r$ values of 3 and $1/3$. I am now exploring if a simpler, trigonometric approach can bypass the parameter $r$ entirely. **Exploring Trigonometric Solution** I am now exploring a trigonometric approach to solve for the side lengths $a$ and $b$ of the inscribed rectangle. I've set up a system of equations based on the angles and side lengths, and I'm investigating how squaring and manipulating these equations reveals relationships between $a$, $b$, and $\theta$. **Solving System Directly** I am now directly solving the system of equations derived from the geometric constraints and the target area. This approach seems more streamlined than parameterization and avoids trigonometric complexities. I am confident this path will lead to a definitive solution for $x$ and $y$. **Discovering Algebraic Shortcut** I've made a phenomenal breakthrough! A simple manipulation of the area calculation for the inscribed rectangle reveals that its area is precisely $30 + 2(x - 5)(y - 3)$. This unexpected simplification dramatically alters the problem's landscape, and I'm now exploring its implications for solving for $x$ and $y$. **Finding Algebraic Shortcut** I've discovered a crucial simplification: the area of the inscribed parallelogram is exactly half the area of the outer rectangle when either $x=5$ or $y=3$. Since $x$ cannot be 5, this forces $y=3$, leading to solutions $x=1$ or $x=9$. I'm now meticulously re-verifying all details of this elegant solution. **Confirming Solution Logic** I've confirmed that the area of the inscribed parallelogram is half the outer rectangle's area when $x=5$ or $y=3$. Since $x \neq 5$, $y=3$ must hold, leading to solutions $x=1$ or $x=9$. I am now meticulously re-verifying all aspects of this elegant solution. **Verifying Triangle Congruence** I've re-examined the corner triangles, confirming that $\triangle AEH \cong \triangle CGF$ and $\triangle BFE \cong \triangle DHG$. This establishes the congruence of opposite corner triangles and directly supports the derived relationship between the segments of the outer rectangle's sides. **Analyzing Similarity Relationship** I have confirmed that the similarity of the corner triangles, specifically $\triangle AEH \sim \triangle BFE$, directly yields the crucial relationship $x(10 - x) = y(6 - y)$. This equation is the bedrock for determining valid configurations of the inscribed rectangle. **Confirming Parameter Constraints** I've verified that for a real solution for $y$, $x$ must fall within $(0, 1]$ or $[9, 10)$. Furthermore, I've confirmed that for any $x$ in these ranges, the resulting $y$ values are always within $(0, 6)$, ensuring valid inscribed rectangles. **Revisiting Solution Logic** I've confirmed that the area constraint simplifies to $(x-5)(y-3)=0$. Since $x \ne 5$, this forces $y=3$, which in turn leads to $x=1$ or $x=9$. I'm now re-verifying all conditions for these specific solutions.