Questão 10 (Matemática)

**Exploring geometric concepts** I'm pondering the "prisma quadrangular reto" and whether it's based on a nonconvex shape. The area formula is still applicable, so that’s interesting. And when I think about "corpo esfera," I realize it refers to a contained ball. It's fun to connect those ideas! I wonder how these shapes interact in geometry and what properties they share. Let’s keep exploring this topic together!

**Solving for prism volume** I need to focus on maximizing the volume of a right quadrilateral prism contained within a sphere of radius 6. One face's vertices are on the sphere, which tells me this face could be the base. The prism's height will be perpendicular to this base. The corners of the face being cocircular indicates that their arrangement fits within a circle with a radius that doesn't exceed 6. Additionally, I must ensure that all vertices remain inside the sphere as they are translated along the normal by height.**Maximizing prism volume** I'm analyzing a right quadrilateral prism with constraints related to its vertices. Given that the vertices share the same xy-coordinates, I realize there's a boundary condition for corners to remain within the sphere. I need to maximize the area of a quadrilateral confined to a circle's radius while ensuring the height stays valid within the sphere's limits. After deriving formulas and optimizing, I find the volume, ensuring all vertices are confirmed on the sphere, and the face aligns appropriately. It's important for the prism structure to extend inward correctly!**Clarifying the prism structure** I'm realizing that the opposite base of the prism can also be situated on the sphere, specifically at the point where \(d - h = -d\) for optimization. This suggests that all vertices can indeed remain on the sphere. It seems I'm looking at a square prism structure, where both bases are aligned on circle latitudes, with vertices completely inscribed within the sphere. This understanding feels like a good breakthrough in visualizing the prism's configuration!**Evaluating geometric properties** I'm working through the geometry of a cube and a sphere, wondering about the radius and volume. I calculated that if a cube has a side of 4√3, its space diagonal equals 12, leading to a sphere radius of 6. The volume of that cube is 192√3. However, I’m thinking about how to classify the faces of a prism, especially when it comes to vertices on a sphere. The phrasing could imply a lateral face, so I need to consider all face types for optimization.**Exploring prism geometry** I'm working on a quadrilateral prism with vertices on a sphere. If a face is a rectangle, and considering it's a right prism, I realize all lateral faces are rectangles too. The dimensions involve the prism's height and base side length, with the diagonal relating to the circle's radius. I need to maximize the area of the base multiplied by the height, ensuring the entire prism fits within the sphere. I’ll analyze how the coordinates of the rectangle’s vertices relate to the sphere.**Analyzing sphere and prism geometry** I'm examining how the sphere's center projects onto the face plane of a prism. For the rectangle to be non-degenerate, its center must align with the projected center of the sphere, ensuring all four vertices are equidistant. The sphere's center must satisfy certain coordinates, leading to an equation involving the base dimensions and height. I'm setting the extrusion of the prism in the z direction, with the base resting in the xy plane, so the base remains in a horizontal orientation.**Interpreting prism faces** I'm interpreting the phrase "uma das faces do prisma" in relation to a quadrangular prism. It includes both bases and lateral faces. I need to mention cases to show that both yield the same upper bound. For the lateral faces, I should consider that their shape is necessarily rectangular. It's an interesting way to explore the geometry of prisms!**Optimizing quadrilateral areas** I'm considering how to optimize the area of a quadrilateral inscribed in a disk. If I take a designated face as the base and measure distance from the sphere center, I notice that this distance must be less than 6. The circumcircle radius helps me find the area constraints. Essentially, the area is limited by the radius of the incircle and conditions involving the prism height. After calculating, I find the maximum volume is 192√3, which depends on certain planar coordinates.**Examining the prism's volume** I'm analyzing the volume of a prism with a designated lateral face. By orienting the prism's axis vertically and measuring its height, I find that the vertices form a rectangle on the sphere's surface. The sphere's center coordinates in relation to this face matter. The derived relationship gives me an area constraint based on a radius squared. I eventually calculate the maximum volume to be 192√3 when optimizing the height, using derivatives to reach this conclusion. It's fascinating to see how these geometric relationships come together!**Analyzing geometric properties** I'm considering a rectangle with all four corners resting on a sphere. In this case, the sphere's center projects onto a plane that's equidistant from these corners, identifying a unique center point for the rectangle. There’s a potential issue regarding whether the base of a quadrilateral prism is convex, since standard definitions require this. Also, I should prepare to write the proof of the area bound in Portuguese and derive the maximum quadrilateral area within a circle or disk.**Calculating derivatives and geometry** I'm working through the derivative of 144d - 4d³, which simplifies to 144 - 12d² = 0, leading me to d² = 12. Then I find F = 4 * 2√3 * (24) = 192√3 — that's good. Now I'll consider Case 2. If the designated face is lateral, I need to set the face plane along the prism axis and evaluate coordinates. The center projection of the sphere relates to the rectangular center at z = 0, ensuring the sphere's center is aligned correctly. The area optimization function appears manageable, and I see h ≤ 12.**Understanding the attainability cube** I'm exploring the attainability of a cube with sides measuring 4√3 cm. The spatial diagonal, given by a√3, equals 12 cm, representing the diameter of a sphere that completely contains the cube. This means the cube is perfectly centered inside the sphere with its vertices touching the sphere's surface. I need to confirm the dimensions, including the height of 4√3 cm and base properties. It's important to clarify these in a structured way, potentially for an exam, while avoiding any overcomplications. I also have to keep in mind the unique properties of the rectangle when projecting the sphere center onto its plane.**Reasoning about prism faces** I'm considering the properties of a quadrilateral prism. Any chosen face must be either the base or a lateral face, and both upper bounds will remain the same. I wonder if it's possible to have a lateral face that isn't a rectangle in a right prism. Since the lateral edges are perpendicular to the base, the lateral face would actually form a parallelogram, and if one edge aligns normally with the base edge, it turns out it has to be a rectangle. So, yes, it seems clear that every lateral face is indeed rectangular.**Considering face projection geometry** I'm thinking about ensuring that the center coordinate of the face is accurate, and it seems I won't need to go beyond that. However, I find it important to prove that projecting the sphere's center onto the face plane actually yields a rectangle center using all four corners. It feels like a vital step in solidifying my understanding of the relationship between these elements in the geometry. Let’s make sure everything is aligned properly!**Delivering geometric insights** I’m thinking about finding the equidistant point in a plane, specifically within a rectangle. The circumcenter occurs at the intersection of the diagonals, which seems correct! The vertices on the sphere's surface ensure the radius of our intersection circle aligns with the rectangle's circumradius, centering perfectly. In Case 1, the face could act as a base. The maximum area for a quadrilateral in a disk is bound by its diagonal lengths, keeping all calculations consistent. Everything checks out, so I'm ready to share!

I'm considering that this is a right quadrangular prism with one face's four vertices on the sphere, meaning that face is a cyclic quadrilateral inscribed in a circle, and the entire prism must fit within the sphere. If that face is a base, the other base's position is constrained by the sphere's geometry — I'm working out that the maximum height occurs when both bases end up inscribed on the sphere symmetrically, giving h ≤ 2d where d is the base plane's distance from center. Now I need to maximize the area of a cyclic quadrilateral for a given circle radius. Solving this, the optimal cyclic quadrilateral is a square, and optimizing d gives d = 2√3, producing a volume of 192√3 — which matches the familiar result that a cube maximizes box volume inscribed in a sphere, with edge length 4√3. That confirms the cube is optimal for this base-face case, so now I want to check whether having a lateral face inscribed in the sphere instead could yield an even larger volume. By symmetry the optimal configuration centers the prism at z0 = -h/2, giving each base vertex radius r with r² = 36 - h²/4, and the largest quadrilateral inscribed in that circle is a square of area 2r². Maximizing volume 2r²h over h gives h = 4√3, yielding the same maximum volume 192√3 as the cube — confirming the cube is indeed optimal among right quadrangular prisms. Using the diagonal bound, a convex quadrilateral inscribed in a disk of radius r has area at most 2r² since both diagonals can't exceed the diameter, giving equality for the inscribed square. This yields volume bound V ≤ 72h - h³/2, maximized at h = 4√3 giving V = 192√3, with equality when the base is centered symmetrically.

I need to figure out which face of the prism has its four vertices on the sphere - could be a base or a lateral face - and think about how this constraint interacts with maximizing the prism's volume. If that face is a base, I set up the base as a cyclic quadrilateral of circumradius r at distance d from the sphere's center, with r²+d²=36, then check how far the prism can extend before the opposite base's vertices exceed the sphere. This pushes toward the case where all eight vertices end up on the sphere, making it essentially an inscribed rectangular box, maximized as a cube with volume 192√3 — but that conflicts with the problem stating only four vertices lie on the sphere, so I suspect I'm misreading the setup and need to reconsider what configuration is actually intended. Working through the bound: for a prism at height slab, the base disc radius shrinks with offset from center, giving area ≤ 2ρ² for a quadrilateral, and optimizing height h yields exactly h=4√3, ρ²=24, matching the cube's 192√3 volume. So the cube remains optimal even under this general prism constraint, confirming the answer is 192√3 cm³ with edge 4√3 cm.

I'm thinking about a right rectangular prism with a square base of side a and height h, where one face's four vertices lie on the sphere's surface. I need to consider two cases: either that face is the square base, or it's a rectangular lateral face, since "one of the faces" could mean either. Maximizing the volume for the base-on-sphere case leads to a surprising result — the optimal shape turns out to be a cube with side 4√3, giving volume 192√3, since the sphere's diameter equals the cube's space diagonal. Now I'm turning to the second case where a lateral face (the a×h rectangle) sits on a great circle of the sphere instead, which changes how the rest of the prism extends from that face. Actually this simplifies to the diagonal condition 2a²+h²≤144, so I want to maximize a²h under that constraint. Substituting x=a², I'm maximizing xh subject to 2x+h²=144, solving for x in terms of h. Taking the derivative sets 144-3h²=0, giving h=4√3 and a²=48, which is just the cube case again with all vertices on the sphere — so this confirms the same optimum, giving V=192√3 cm³ with a=h=4√3.

The problem is in Portuguese. I will translate it. A questão envolve um prisma quadrangular reto inscrito numa esfera de raio 6 cm. Quatro vértices de uma face pertencem à superfície da esfera. O objetivo é encontrar o volume máximo e as dimensões correspondentes. Estou considerando se a base do prisma é um quadrilátero genérico ou um retângulo. A hipótese de base quadrada não parece indicada, pois o problema não especifica "base quadrada". Para maximizar o volume, a configuração ótima provavelmente será um paralelepípedo retângulo. Estou considerando se a face com os quatro vértices na esfera é uma base ou uma face lateral. A hipótese de base retangular parece mais adequada para modelar o problema. A expressão "dimensões da base" sugere um retângulo com comprimento e largura. Vou modelar o prisma como um paralelepípedo retângulo. A esfera não é necessariamente a circunferência do prisma. Apenas os quatro vértices de uma face estão na superfície; os demais podem estar dentro da esfera. A face com os quatro vértices na esfera é uma base do prisma. Os vértices formam um círculo na interseção entre o plano da face e a esfera. A base é um quadrilátero inscrito em círculo. O volume máximo provavelmente ocorre quando a base é um quadrado, pois entre todos os quadriláteros inscritos na mesma circunferência, o quadrado tem a maior área. Vou modelar o problema com coordenadas. O centro da esfera é O e o raio é 6 cm. A face com os quatro vértices na superfície está a uma distância d do centro. O círculo de interseção tem raio r = sqrt(36 - d²). Como o prisma é reto, a altura é perpendicular ao plano dessa face. Reavaliei se a face com os quatro vértices na esfera é uma base ou uma face lateral. Para prisma reto, as faces laterais só são retângulos se a base for retangular. Isso reforça que o prisma deve ter base retangular. O volume é a área da base multiplicada pela altura. A face com os quatro vértices na esfera pode ser uma base ou uma face lateral, então devo analisar ambos os casos. O problema menciona "uma das faces", então devo verificar ambos os casos para encontrar o volume máximo. A restrição é que o prisma inteiro deve estar contido na esfera. Isso parece um problema de otimização pesado para uma fase do IME, mas não se trata de um cubo. Questiono se o volume máximo ocorre quando todos os vértices do prisma tocam a esfera. Para um paralelepípedo retângulo inscrito, o cubo é o caso ótimo. Mas aqui apenas uma face tem seus quatro vértices na superfície. O volume máximo de um paralelepípedo retângulo inscrito na esfera ocorre quando todos os oito vértices tocam a superfície. Isso reforça que o cubo é a configuração ótima. O volume máximo é o do cubo inscrito. A diagonal do cubo é 12 cm, então a aresta mede 4√3 cm e o volume é 192√3 cm³. Estou analisando o caso em que a face com os quatro vértices na esfera é uma face lateral. A base é um quadrilátero cíclico com raio de circunferência r = sqrt(36 - d²), onde d é a distância do centro da esfera ao plano da base. A altura h do prisma é perpendicular a esse plano, então a outra base fica a uma distância |d ± h| do centro. Para que o prisma inteiro esteja contido na esfera, todos os vértices da segunda base devem estar dentro dela. Estou modelando o prisma com coordenadas. O plano da base fica a uma distância p do centro da esfera, e o círculo de interseção tem raio r = sqrt(36 - p²). Os vértices da base estão a uma distância r do eixo perpendicular ao plano. A outra base está deslocada verticalmente por h, então cada vértice superior fica a uma distância p ± h do centro. A análise das coordenadas mostra que, para a face superior ficar dentro da esfera, a base deve estar no lado oposto ao centro. Isso impõe p ≤ −h/2 < 0. A área da base quadrilátera inscrita no círculo de raio r = sqrt(36 − q²) é máxima quando a base é um quadrado, com área 2r². O volume do prisma é essa área multiplicada pela altura h. Para maximizar o volume, devo maximizar a área da base, que é 2(R² − q²), e minimizar |q|. A restrição q ≤ −h/2 limita a posição da base. Como R é fixo, preciso verificar se há mais restrições ativas para encontrar o valor ótimo. A área da base quadrilátera máxima é 2(R² − h²/4). O volume do prisma é V = 2R²h − (1/2)h³. Esse valor ocorre quando q = −h/2, o que implica que os vértices superiores também ficam na superfície da esfera. Derivei a função volume V(h) = 2R²h − (1/2)h³. A altura ótima é h = (2/√3)R. O raio do círculo de interseção é r = √(2/3)R. A área da base quadrilátera máxima é 4/3R². O volume máximo é (8√3 / 9)R³. Para R = 6 cm, o resultado é 192√3 cm³. Esse valor coincide com o volume do cubo inscrito na esfera. Agora questiono se uma face lateral pode produzir volume maior. A base quadrilátera de área máxima já foi otimizada, então o cubo parece ser o máximo absoluto. Questiono se um prisma com base quadrilátera não retangular pode ter volume maior que o cubo. A hipótese de base irregular ou rombo é considerada, mas o prisma reto impõe restrições geométricas que dificultam essa possibilidade. Notei que qualquer retângulo inscrito numa circunferência tem diagonais como diâmetros. Isso confirma que os quatro vértices coplanares na esfera formam um retângulo. Agora estou definindo coordenadas para analisar volumes possíveis com a face lateral. Estou modelando o prisma com altura h ao longo do eixo z e base no plano xy. O volume é a área da base multiplicada pela altura, com os oito vértices restritos à esfera de raio R. A otimização deve encontrar candidatos onde pelo menos uma face tenha seus quatro vértices na superfície. Reavaliei a interpretação do problema. A expressão "prisma quadrangular reto" pode ser ambígua, pois o termo "quadrangular" não implica necessariamente retângulo. A parte (b) menciona "a altura e as dimensões da base", o que sugere que as medidas da base devem ser determinadas como parte da solução. A parte (b) menciona "as dimensões da base", o que sugere duas medidas distintas. Isso reforça que a base é retangular e não quadrada. O cubo, com altura igual ao lado da base, continua sendo a configuração ótima entre todos os paralelepípedos retângulos inscritos na esfera. A resposta da parte (a) é 192√3 cm³. A parte (b) exige a altura e as dimensões da base, que são 4√3 cm cada, formando um cubo. Não existe paralelepípedo retângulo inscrito na esfera com volume maior que o do cubo. A restrição de volume para um paralelepípedo retângulo inscrito na esfera é x² + y² + z² ≤ 144. Pelo teorema da média aritmética e geométrica, o produto xyz é máximo quando x = y = z = 4√3. Isso confirma que o cubo é o único prisma retangular com volume igual a 192√3 cm³. Estou verificando a geometria da face lateral. Os quatro vértices do retângulo vertical devem estar no círculo de interseção do plano com a esfera. Isso impõe que as coordenadas y dos dois vértices inferiores sejam opostas, e os superiores também. A base do prisma reto deve ser retangular. A simetria vertical dos vértices da face lateral impõe h = −2z_b, ou seja, z_top = −z_b. Os quatro pontos ficam simétricos em relação ao plano xy. A equação (l/2)² + (h/2)² + d² = R² confirma que a face lateral é um retângulo inscrito no círculo de interseção do plano com a esfera. Questiono se a face lateral pode existir fora do posicionamento simétrico em relação ao centro da esfera. A análise anterior assumiu que os vértices inferiores e superiores se distribuem simetricamente em relação ao plano xy, o que impõe que o centro da esfera fique exatamente no meio da altura do prisma. A simetria vertical dos vértices da face lateral é matematicamente obrigatória. Para que todos os quatro pontos de um retângulo vertical estejam na esfera, suas coordenadas z devem ser espelhadas em relação ao plano equatorial perpendicular. Isso fixa o centro da esfera exatamente no meio da altura do prisma, impedindo qualquer deslocamento vertical. A base quadrilátera deve estar contida no círculo de raio ρ_max = √(R² − h²/4) no plano z = ±h/2. Isso restringe as coordenadas x e y dos vértices da base. A face lateral fixa exige apenas que seus quatro vértices estejam na esfera. Os demais vértices do prisma podem ficar dentro do disco de interseção. Isso permite variar h e d para maximizar o volume. A área da base quadrilátera depende de h, então preciso encontrar o máximo possível para cada altura. A equação δ² + (l/2)² + (h/2)² = R² confirma que qualquer corda do círculo de interseção, conectando dois vértices na superfície da esfera, produz um retângulo lateral cujos quatro vértices ficam na esfera. A altura h permanece fixa. Agora estou testando se o centro da esfera pode estar deslocado verticalmente. Mesmo sem assumir simetria, a condição de que os quatro vértices do retângulo lateral estejam na superfície impõe x1² + y1² = x2² + y2². Isso restringe os pontos inferiores e superiores a terem a mesma projeção no plano xy, mantendo a altura fixa. Relei o enunciado original. O prisma é completamente contido na esfera, e apenas os quatro vértices de uma face pertencem à superfície. Isso reforça que a base quadrilátera deve estar no círculo de interseção do plano com a esfera. Questiono se a área máxima de um quadrilátero inscrito no círculo de raio ρ é realmente 2ρ². Suspeito de uma confusão na diagonal do quadrado. Confirmei que a diagonal do quadrado inscrito no círculo de raio ρ é 2ρ, logo o lado vale ρ√2 e a área é 2ρ². Um retângulo alongado teria área menor, e qualquer quadrilátero irregular também não poderia superar essa área máxima. Questiono se a hipótese de que apenas os quatro vértices de uma face estejam na superfície é suficiente para garantir o volume máximo. A interpretação do termo "completamente contido" permanece ambígua. Notei que a menção à "prova IME 2027" pode ser fictícia ou futura, já que as provas do IME são anuais. Estou verificando se a base pode estar posicionada fora do centro da esfera. Se a base estiver a uma distância q do centro, a altura h deve ser negativa para que os vértices superiores permaneçam dentro da esfera. Isso reforça que o centro da esfera deve ficar exatamente no meio da altura do prisma. Estou verificando se o centro da esfera pode estar deslocado verticalmente. Se a base estiver a uma distância q do centro, a altura h deve satisfazer h ≤ 2q para que os vértices superiores permaneçam dentro da esfera. Isso reforça que o centro deve ficar exatamente no meio da altura do prisma. A altura do prisma é 4√3 cm. O lado da base quadrada também é 4√3 cm. A diagonal do quadrado inscrito no círculo de interseção vale 2√24 = 4√6. Isso confirma que a configuração é um cubo. A diagonal da face lateral do cubo é 4√6. A distância do centro da esfera até o centro dessa face é 2√3. O quadrado da soma dessas distâncias é 36, igual ao quadrado do raio. Isso confirma que os vértices ficam na superfície da esfera. A derivada da função volume V(h) = 72h − 1,5h³ anula em h = 4√3. Isso reforça que o cubo é o único prisma retangular com volume máximo. Vou apresentar a solução em português com desenvolvimento passo a passo, conforme solicitado. A resposta final será apresentada ao final. Questiono se paralelepípedos arbitrários poderiam ter volume maior que o cubo. Apenas paralelepípedos retangulares possuem esfera circunscrita, pois só eles têm diagonal constante. Questiono se um prisma reto com base quadrilátera não retangular é um paralelepípedo. A definição exige bases paralelogramas. A definição de paralelepípedo exige bases paralelogramas. Um prisma com base quadrilátera genérica não é um paralelepípedo, mas ainda assim pode ter volume máximo igual ao cubo. A função volume V(h) = 72h − 0,5h³ atinge o máximo em h = 4√3. Valores de altura menores ou maiores produzem volumes inferiores. Isso confirma que o cubo é o único prisma retangular com volume máximo. Agora estruturo a prova em casos. O primeiro caso examina quando a face com vértices na esfera é uma base do prisma. No primeiro caso, a face com vértices na esfera é uma base do prisma. A distância do centro da esfera ao plano da base é |q|, e a altura h deve ser tal que a outra base fique dentro da esfera. O volume é V(h) = 2(R² − h²/4)h. Para maximizar a área da base quadrilátera inscrita no círculo de raio r = √(R² − h²/4), usamos o quadrado inscrito de área 2r². Derivando a função volume V(h) = 2(R² − h²/4)h e anulando o resultado, encontro h = 4√3 cm. Com R = 6 cm, o raio do círculo de interseção é √24 = 2√6 cm, e a área da base quadrilátera inscrita vale 48 cm². O lado do quadrado é 4√3 cm, confirmando que a configuração é um cubo com volume de 192√3 cm³. A segunda derivada é negativa, validando que se trata de um máximo absoluto. Agora analiso o caso em que a face com vértices na superfície da esfera é uma face lateral. No caso de a face com vértices na esfera ser uma face lateral, a área máxima da base quadrilátera inscrita no disco de interseção continua sendo 2ρ². Isso reproduz a mesma função volume V(h) = 72h − 0,5h³, cujo máximo absoluto ocorre em h = 4√3 cm. Portanto, o volume máximo é 192√3 cm³, atingido apenas quando o prisma é um cubo de aresta 4√3 cm. A diagonal espacial de qualquer paralelepípedo retangular inscrito na esfera é limitada pelo diâmetro da esfera. Isso impõe a² + b² + h² ≤ 144, reforçando que o volume máximo absoluto é 192√3 cm³. A condição de que os vértices de uma face estejam na superfície não altera esse limite. Questiono se um quadrilátero de área maior que 48 cm² pode existir em uma seção transversal da esfera. A área máxima de uma seção circular é πR² ≈ 113 cm², então áreas intermediárias são possíveis. Isso sugere que o volume máximo pode não ser apenas o cubo. A área de cada base quadrilátera fica limitada por 2(R² − z²), onde z é a distância do centro da esfera ao plano da base. Como a face com vértices na superfície exige que uma dessas áreas atinja o limite máximo, a configuração que maximiza o volume permanece sendo o cubo. A área de cada base fica limitada pelo círculo de interseção mais restritivo. Para maximizar o volume, as bases devem ficar simétricas em relação ao centro da esfera. Posições assimétricas reduzem a área da base e, portanto, o volume. O volume máximo absoluto de um prisma quadrangular reto inscrito na esfera é 192√3 cm³, atingido apenas pelo cubo. A condição de que os vértices de uma face estejam na superfície não altera esse resultado. Ainda resta a dúvida sobre por que o problema inclui essa restrição. A condição dos vértices na superfície pode ser uma pista para inscrever a face no círculo de interseção. A frase "completamente contido" também pode indicar que a esfera é o sólido, não apenas a superfície. A esfera S é o sólido de raio 6, e a superfície é a fronteira. O cubo está completamente contido nesse sólido. A resposta final é 192√3 cm³. Questiono se o problema pode ser mais simples do que parece. Questiono se o problema exige que todos os oito vértices do prisma estejam na superfície da esfera. Se sim, a configuração forçaria uma posição simétrica entre as bases, restringindo o volume máximo. A condição de que apenas quatro vértices estejam na superfície da esfera não altera o resultado. Prisma com apenas quatro vértices na fronteira teria área menor e volume inferior ao cubo. O volume máximo absoluto permanece 192√3 cm³, atingido apenas pelo cubo. A área da base fica limitada por 2(R² − max(d_bottom², d_top²)). A simetria entre as bases em relação ao centro da esfera minimiza esse máximo. Isso reforça que o volume máximo ocorre quando as bases estão simétricas.

We need answer in Portuguese. Need solve. Let's restate problem. "Considere um prisma quadrangular reto, completamente contido em uma esfera S de raio 6 cm. Sabe-se que os quatro vértices de uma das faces do prisma pertencem à superfície da esfera S. Determine: (a) o volume máximo que esse prisma pode assumir; (b) a altura e as dimensões da base do prisma correspondentes a esse volume máximo." Need understand geometry. Right quadrangular prism (rectangular box?) "prisma quadrangular reto" means right prism with quadrangular base, likely rectangular base? In Portuguese geometry, "prisma quadrangular reto" often means right prism with rectangular base? It could be any quadrilateral base? "quadrangular" maybe quadrilateral, but usually "paralelepípedo reto-retângulo" rectangular parallelepiped? Here "prisma quadrangular reto" maybe base is a quadrilateral, not necessarily rectangle? But completely contained in sphere, one face's four vertices belong to sphere surface. A face is a rectangle? For a right prism, base is congruent quadrilateral; if one face (base) has all vertices on sphere surface, then that face is cyclic quadrilateral inscribed in circle (intersection plane sphere). Since sphere contains prism and face vertices on sphere. Need maximum volume over right prisms with rectangular bases? The face that belongs to sphere could be base or lateral face. Since right prism: bases are congruent parallel polygons; lateral faces rectangles. "uma das faces" could be base or lateral face. But all four vertices of a face on sphere => the face is inscribed in sphere's great/small circle? Let's analyze. A right quadrangular prism has two congruent quadrilateral bases and four rectangular lateral faces. If a lateral face's four vertices belong to sphere surface: Its plane intersects sphere in a circle; the face is a rectangle inscribed in that circle. The prism lies inside sphere. Need maximize volume. If a base's four vertices belong to sphere: base quadrilateral inscribed in a circle (intersection of base plane with sphere). Prism extends perpendicular to base plane one side? Since completely contained in sphere, both bases inside sphere; top base vertices maybe not on sphere. Height limited by sphere. Need maximize volume. But phrase: "Sabe-se que os quatro vértices de uma das faces do prisma pertencem à superfície da esfera S." One face (could be base or lateral) has four vertices on sphere. Because prism completely contained in sphere, and one face's vertices on sphere. Inscribed prism in sphere? Typically an inscribed prism in sphere means all vertices on sphere; but here only one face. Since prism has 8 vertices; if all vertices on sphere then prism inscribed in sphere. But only one face. We need determine max volume. There may be multiple configurations: if a lateral face has vertices on sphere, then maybe the opposite lateral face inside; height limited? Let's examine. A right prism with rectangular base perhaps? A "prisma quadrangular reto" has base a quadrilateral; if not specified rectangle? In many Brazilian problems, "prisma quadrangular reto" means base is a rectangle? Actually "prisma quadrangular" = quadrangular prism, base is quadrilateral; "reto" = lateral edges perpendicular to bases. Could be regular quadrangular prism = square base (cubo/paralelepípedo). But not necessarily regular. We need maximize volume for any right quadrangular prism contained in sphere, with four vertices of one face on sphere. Need infer likely answer: Maybe maximum volume is 144√3? Let's solve. Let's set up. Sphere radius R=6 centered O. Consider a right prism. Let one base be a quadrilateral Q inscribed in a circle (intersection of plane Π with sphere radius r ≤ R). Since Q's four vertices on sphere surface, they lie on circle C = S ∩ Π. The base Q is a cyclic quadrilateral inscribed in circle radius r. The prism height h extends perpendicular to Π, either entirely on one side of Π (if chosen face base) or maybe if lateral face then different. Volume = area(Q) * h. Since prism completely in sphere: all points of prism at distance ≤ R from O. The base plane Π at distance d from O (0 ≤ d ≤ R). The circle radius r = sqrt(R^2 - d^2). The prism extends from Π along normal direction by h (one side) to plane Π' parallel at distance h. Since prism contained in sphere, the farthest cross-section? Actually every point of prism including two bases and volume must be inside sphere. If base Q is inscribed in circle in Π, and prism extends height h perpendicular to Π. For any point in prism, coordinate (x,y,z) with base point (x,y) ∈ Q and z∈[0,h] along normal. Its distance squared = r_xy^2 + (d+z)^2? Need coordinate center: Let Π: z=d? Let normal direction from center to plane; if plane at distance d, choose z-axis perpendicular, plane z=d (center at O=(0,0,0)). Then point in base has coordinates (x,y,d), x^2+y^2=r^2? Wait base vertices on sphere: x^2+y^2+d^2=R^2 => x^2+y^2 = R^2-d^2 = r^2. Points in base Q lie within circle x^2+y^2≤r^2. Prism extends along z direction: if Q is bottom base at z=d, top base at z=d+h (assuming extends away from center? If d≥0). Then points have x^2+y^2 + (d+z)^2 ≤ R^2 for all z∈[0,h], (x,y)∈Q. Since x^2+y^2≤r^2 = R^2-d^2 at z=0 and points with x^2+y^2 can equal r^2 at vertices but for interior less. To ensure top z=d+h inside sphere: max x^2+y^2 in Q? The farthest point from z-axis among Q may be r if Q contains circle boundary? For cyclic quadrilateral inscribed in circle radius r, it may occupy all around? A quadrilateral cannot have points on circle with full radial extent? It has 4 vertices on circle; the maximum x^2+y^2 for convex hull is r^2 (at vertices), but area interior max radius maybe r if vertices? The farthest point from center in convex hull of vertices lies on boundary maybe vertex? Since convex hull of points on circle has maximum radius r at vertices? Actually any point in convex hull of points on circle has norm ≤ max vertex norm? The disk convex hull? The maximum of convex function over convex hull occurs at extreme points? Norm squared convex, maximum over convex compact occurs at extreme points (vertices), so ≤ r^2. But if a vertex is at radius r, maximum r^2. So to contain top plane entirely in sphere, need r^2 + (d+h)^2 ≤ R^2. Because point of top base with radius r (vertex on circle) has distance sqrt(r^2+(d+h)^2). Since Q's vertices on circle, yes the top base (translation) has vertices distance sqrt(r^2+(d+h)^2) from O? Wait top base vertices coordinates same x,y as bottom base, z=d+h, so distance squared = x^2+y^2+(d+h)^2 = r^2+(d+h)^2. It must ≤ R^2 => (d+h)^2 ≤ R^2 - r^2 = d^2 => d+h ≤ d (assuming d≥0) impossible unless h=0. Contradiction? Wait if prism extends from plane z=d away from center (positive z), top base at z=d+h, distance to center increases if h>0, d>0; top vertices distance >R for h>0. So prism cannot have base face on sphere and extend away from center; it must extend toward center? Let's reconsider. If one face's four vertices belong to sphere, but prism is contained in sphere. If the face is on sphere at distance d from center, and prism extends inward (toward center) perhaps, then top base at z=d-h (if h ≤ d) closer to center, distance squared = x^2+y^2+(d-h)^2 = r^2+(d-h)^2 = R^2 - d^2 + (d-h)^2 ≤ R^2? For h>0, yes less. So prism must extend from the spherical face toward the center, not away. Because outward would leave sphere. Thus the face with vertices on sphere is an outer face, and prism lies on the other side of it (inside). So if one base is inscribed in sphere's surface, the prism extends inward (height toward center). The opposite base is inside sphere, not on sphere necessarily. Then height h ≤ d (distance from center to that face plane) if plane at z=d and extends to z=0 (center) or beyond? Wait if d is distance from center O to plane Π, and prism extends inward: plane Π at z=d, other base at z=d-h. Need contained: max distance from O for points in top base? Top base at z=d-h, with radial coordinate ≤ r (since same Q), so distance squared ≤ r^2 + (d-h)^2. Since r^2=R^2-d^2. This ≤ R^2 iff (d-h)^2 ≤ d^2 => |d-h| ≤ d => 0 ≤ h ≤ 2d (assuming h≥0, d≥0). So h can be up to 2d: top base plane could cross center (z=0) if h=d; if h>d, top plane at z=d-h negative (other side of center), still distance squared ≤ R^2? If h=2d, top at z=-d (same distance but other side); vertices distance squared = r^2 + (-d)^2 = R^2, so top vertices also on sphere. So prism symmetric about center? Then both bases on sphere. If h<2d, top inside; if h=2d both bases on sphere (prism inscribed). This makes sense: for a prism with one base inscribed in a small circle away from center, extending inward can reach opposite side; max height limited by sphere on opposite side: h ≤ 2d? Actually for any z between d-h and d, coordinate distance squared ≤ r^2 + z^2 where |z| ≤ max(d, |d-h|) maybe. Need r^2 + z_max^2 ≤ R^2 = r^2 + d^2 => |z| ≤ d. So z in [-d, d]? If prism from z=d-h to z=d; all points z between d-h and d must satisfy |z|≤ d (since max |z| = max(|d|, |d-h|) if h≤2d and d≥0). If h>2d, top z< -d, then |z|>d -> violation. Thus h≤2d. So given face plane at distance d, prism height h ∈ [0, 2d] for containment (assuming prism extends inward). Volume V = A * h, with A = area of cyclic quadrilateral Q in circle radius r = sqrt(R^2-d^2). For fixed d, A max? Q can be any quadrilateral inscribed in circle radius r. Max area of quadrilateral inscribed in circle radius r is square? Among all quadrilaterals inscribed in circle, max area is square? I think yes: among all n-gons inscribed in circle, regular n-gon maximizes area. For quadrilateral, square area = side √2 r? Actually square inscribed in circle radius r: side r√2, area = 2r^2. Max area of quadrilateral inscribed in circle = 2r^2? Let's verify: For any quadrilateral with sides a,b,c,d and semiperimeter s, area² ≤ (s-a)(s-b)(s-c)(s-d), equality cyclic (Brahmagupta), for given sum? But inscribed in circle radius r constraints. For fixed circumradius r, max area among cyclic quadrilaterals maybe square? Yes by symmetry and convexity, regular n-gon. So max A = 2r^2 = 2(R^2 - d^2). Then volume V(d,h) ≤ 2(R^2 - d^2) * h, with 0≤d≤R, 0≤h≤2d. For fixed d, max at h=2d (provided d≤R). Then V_max over d: 2(R^2 - d^2)*2d = 4d(R^2 - d^2). Max over d∈[0,R]. f(d)=4d(R^2 - d^2)=4R^2 d - 4d^3. f'(d)=4R^2 -12d^2=0 => d=R/√3. f''=-24d <0 (for d>0), max. f_max = 4*(R/√3)*(R^2 - R^2/3)=4*(R/√3)*(2R^2/3)=8R^3/(3√3) = 8√3 R^3 /9? R=6 => 8*(216)/(3√3)=1728/(3√3)=576/√3=192√3? Let's compute: 576/√3 = 576√3/3 = 192√3 cm³. Alternatively 8R^3/(3√3) with R=6: 8*216/(3√3)=1728/(3√3)=576/√3=192√3. So volume max = 192√3 cm³. Then corresponding height h=2d = 2R/√3 = 12/√3 = 4√3 cm. Base dimensions: Q is square inscribed in circle radius r = sqrt(R^2 - d^2)=sqrt(36 - 36/3)=sqrt(24)=2√6 cm. So base is square of side? Inscribed square in circle radius r: diagonal = 2r = 4√6; side = diagonal/√2 = 4√6/√2 = 4√3? Wait 4√6 / √2 = 4√3? Since √6/√2=√3. So side = 4√3 cm. Area = (4√3)^2 = 48 cm²? Or 2r^2=2*24=48. Yes. Volume = 48 * 4√3 = 192√3. Good. But is this the only interpretation? Need consider if the face with four vertices on sphere is a lateral face rather than base. Let's analyze. A right quadrangular prism has lateral faces rectangles. If four vertices of a lateral face belong to sphere surface, then that lateral face is a rectangle whose plane intersects sphere in a circle; its four vertices lie on that circle (since all four on sphere). The prism also has another lateral face parallel (opposite) and two bases perpendicular to lateral face? Let denote prism: base is quadrilateral Q in plane Π1 (bottom) and congruent top in parallel plane Π2, with lateral edges perpendicular to bases. A lateral face is a rectangle formed by one side of base and corresponding side of top. Its four vertices: two bottom base vertices (on one side of Q) and two top base vertices (corresponding side translated). They all lie on sphere. So the sphere contains a rectangle face (lateral). The prism cross-section perpendicular to that face? Need maximize volume. Let coordinate system. Let the lateral face lie in plane, say plane x=0? Its four vertices on sphere. Since face is rectangle, its sides: one vertical edge? Actually lateral face's vertices: bottom edge = side of base (length a, lying in base plane), top edge parallel congruent in top base plane; lateral edges perpendicular to base are vertical (height h). The lateral face has dimensions: base side length a (horizontal in base plane) and height h (vertical). All four vertices on sphere. Thus the rectangle of dimensions a × h is inscribed in a circle of radius r (intersection of sphere with that plane), so its diagonal a² + h² = (2r)^2 = 4r²? But r ≤ R, and plane at distance d from center. The rectangle must be cyclic with circumradius r. But how does prism orientation relate to sphere center? Since it's a right prism, if lateral face is on sphere, the prism extends inward perpendicular to that lateral face (toward interior). The base plane is perpendicular to lateral face? Need set coordinates. Let's model general prism. A right prism with rectangular base? Wait if lateral face is rectangle, then its sides: one side is base edge (horizontal), other is lateral edge (vertical). The base quadrilateral Q must have at least one side equal to the horizontal dimension a of this lateral face, and base plane contains that side. The lateral face plane is perpendicular to base plane? In a right prism, lateral faces are perpendicular to base planes? Actually in a right prism, lateral edges are perpendicular to base planes, so lateral faces are rectangles; each lateral face is perpendicular to base plane. Yes. So the lateral face plane is perpendicular to base plane. The four vertices on sphere: two adjacent vertices of base Q and their translated counterparts. This means the base side segment (length a) lies in base plane, and the lateral edge (height h) lies perpendicular to base plane; the rectangle belongs to a plane that contains base edge and is perpendicular to base plane. Its four vertices all on sphere. Need maximize volume V = area(Q) * h. Given one lateral face rectangle with sides a (base edge) and h (height) is inscribed in a circle of radius r (intersection sphere with lateral face plane), so a² + h² = (2r)^2 = 4(R² - d²), where d is distance from center O to lateral face plane. Also prism contained in sphere imposes constraints on other vertices and volume. We need maximize over choices of prism: base quadrilateral Q with one side a, height h, and sphere relationship. The lateral face vertices on sphere means the rectangle's four vertices are on sphere. But the prism may extend inward relative to that lateral face; its opposite lateral face and rest of prism inside sphere. The base plane contains side a and is perpendicular to lateral face plane. The top base is translation by h along normal to base plane (which lies in lateral face plane direction? Actually lateral edge direction is perpendicular to base plane, and lies in the lateral face plane? Wait lateral face plane contains base edge direction u (in base plane) and lateral edge direction v perpendicular to base plane. So v is normal to base plane. The lateral face plane is spanned by u and v. The opposite lateral face is parallel shifted? The base is a quadrilateral in plane spanned by u and w (some other direction in base plane). The prism extends along v. The sphere contains prism. We can choose coordinates: Let base plane be z=0 (but lateral face? Maybe easier). Let lateral face plane be x=0? Actually base plane perpendicular to lateral face plane, so set lateral face plane as y=0? Let's set: lateral face plane Π_L: y=0. It contains base edge along x-axis and lateral edge along z-axis. The base plane is perpendicular to Π_L, so base plane is z=0? Wait if lateral face contains z-axis (vertical), then base plane must be perpendicular to z-axis? Let's define: lateral edge direction = z-axis (height h along z). Base plane = xy-plane (z=0). Then lateral face plane containing base edge and z-axis: base edge lies in xy-plane, say along x-axis; and z-axis. So plane y=0 (x-z plane). Yes. Base is quadrilateral in xy-plane, with one side along x-axis from (0,0,0) to (a,0,0). The lateral face at y=0 contains this side and vertical edges to (0,0,h) and (a,0,h). The four vertices of this lateral face: (0,0,0), (a,0,0), (0,0,h), (a,0,h). They all lie on sphere S of radius R=6. The sphere center O has coordinates? Also sphere contains entire prism. Need find position of O relative to prism. The plane y=0 intersects sphere in a circle (radius r ≤ R) containing rectangle vertices. Distance from O to plane y=0 is d ≤ R. The circle radius r = sqrt(R² - d²). The rectangle has diagonal L = sqrt(a² + h²) = 2r (since inscribed in circle? Wait rectangle's circumcircle is intersection of sphere with plane y=0; its center is projection of O onto plane y=0; radius r. Since rectangle cyclic, diagonal equals diameter 2r). So a² + h² = 4(R² - d²). Also the sphere center O may be at distance d from plane y=0, but its projection onto plane y=0 is center of rectangle? The circle center is projection of O. For rectangle circumcenter, it must coincide with circle center. Thus projection of O onto plane y=0 is the rectangle center (a/2, 0, h/2). Because rectangle vertices symmetric. So O coordinates: projection along normal to plane y=0 (i.e., y-direction) by distance d: O = (a/2, ±d?, h/2). Since plane y=0; distance d means O_y = ±d. But choice sign affects prism containment. Also prism lies on one side of lateral face? Since lateral face vertices on sphere, to contain prism, the prism must extend inward from that face. The lateral face is on boundary; prism lies on one side (positive y perhaps). The sphere center should be on same side as prism? If O_y = +d, then sphere extends around center; the prism (if placed in y>0) maybe inside if d sufficiently large? Let's analyze. But the prism has base in z=0 plane: quadrilateral Q with vertices (0,0,0), (a,0,0), and two other vertices (x1,y1,0), (x2,y2,0) forming a quadrilateral. Top base at z=h translated. The lateral face y=0 is one side. The prism interior (for convex prism) is on one side of each lateral face. Since base Q in z=0; if Q lies in y≥0? The side along x-axis is boundary y=0; the rest of Q likely y>0 (assuming convex quadrilateral with that side as edge). So prism lies in y≥0 half-space (because base Q in y≥0 and top translation). Thus prism is in y≥0. Then for sphere to contain it, sphere center should perhaps be in y≥0 too? If O_y = -d (plane y=0 at distance d but center opposite side), then sphere extends into y>0 from -d to -d+R etc. The prism starts at y=0 and extends y>0; possible if d≤R and prism inside sphere? Need check. But if O_y=-d, the circle radius r same; rectangle vertices on sphere. The sphere on y>0 side is limited: max y = -d + R (since center y=-d radius R). Need prism y max ≤ -d + R. But prism extends unbounded in y? Actually base Q has finite y extent; need contain. The lateral face y=0 is closest to center? If center at -d, distance d; sphere on opposite side extends to y=0? The point on sphere furthest y is -d + R = R - d. So prism can extend up to y=R-d. Fine. If center at +d, sphere in y>0 extends from 0? Actually sphere centered y=+d includes y=0 plane? Circle; prism lies y≥0; sphere may contain up to y=d+R, plenty. But prism must be within sphere; both possible? However because prism contains the lateral face on sphere and extends inward; "inward" relative to sphere means toward center. If center at y=+d, the prism extending into y>0 goes away from center (since center at +d, interior of sphere includes y<d? Actually sphere center at y=d; points with y>d are outside? Wait sphere is all points distance ≤R from (a/2,d,h/2). So y can be d ± something; y=0 is distance d from center, boundary. y>0 up to d+R? If prism extends y>0, it might go toward or away? At y=0, on sphere; increasing y initially moves closer to center? If center y=d>0, then y>0 but <d moves closer; y>d moves beyond center. But prism height? Hmm. But containment and maximizing force maybe choose orientation where center lies on same side as prism? Let's derive. Given prism with base Q in z=0, lateral face y=0 has vertices on sphere. The prism lies in half-space y≥0 (assuming Q interior y>0). For any point in prism, coordinates (x,y,z) with y≥0, z∈[0,h], and (x,y) in base Q. Sphere center O=(a/2, s*d?, h/2) with s=±1? Actually distance from O to plane y=0 is |O_y| = d. The rectangle center is (a/2, 0, h/2). So O = (a/2, σ d, h/2), with σ=±1. The sphere contains prism iff for all prism points, (x - a/2)^2 + (y - σ d)^2 + (z - h/2)^2 ≤ R². Since equality for lateral face vertices; for other points maybe inequality. The maximal horizontal extent of base Q? Q is a quadrilateral with one side from (0,0) to (a,0) along x-axis, lying in y≥0. To maximize volume, q should likely be a square? But constrained by base side a and lateral face? Let's consider general base. Given base Q with side a along x-axis and height h. The prism lateral face y=0 has four vertices on sphere. But the opposite lateral face y=? The base Q may have other side(s) not necessarily symmetric. We need maximize area(Q) * h subject to sphere containment and a² + h² = 4(R² - d²), d≤R, and center distance. Maybe maximum occurs when base is square? Need prove. Since base is quadrilateral with one side a, the maximum area for given a under constraints maybe? Let's analyze. Base Q in xy-plane, y≥0, with one edge from (0,0) to (a,0). The other vertices P=(u,v), Q_=(w,v?) no, quadrilateral order: (0,0), (a,0), (x1,y1), (x2,y2) with y_i≥0. Area = (1/2)|...|. For fixed a, max area of quadrilateral with one side a and lying in y≥0 within some region maybe? But containment in sphere imposes region for base: For z=0, points (x,y,0) in prism. Sphere inequality: (x - a/2)^2 + (y - σ d)^2 + (h/2)^2 ≤ R². So base Q must lie inside ellipse/circle? Rearr: (x - a/2)^2 + (y - σ d)^2 ≤ R² - h²/4 =: ρ². But from a² + h² = 4(R² - d²) => R² - h²/4 = R² - (4(R²-d²) - a²)/4? Wait h² = 4(R²-d²) - a². Then R² - h²/4 = R² - (4(R²-d²)-a²)/4 = R² - (R²-d²) + a²/4 = d² + a²/4. So constraint for z=0: (x - a/2)^2 + (y - σ d)^2 ≤ a²/4 + d². Since x∈[0,a]? Actually base includes x from 0 to a maybe? The side length a; other vertices could have x outside? But convex quadrilateral with side a likely x coordinates between? Not necessarily. But for max area likely within rectangle? Let's see. The circle center (a/2, σ d) radius sqrt(a²/4 + d²). This is exactly the sphere cross-section at z=0 (the plane through base). At z=0, sphere intersection is a circle with radius sqrt(R² - h²/4)? Wait sphere center O at (a/2,σd,h/2), radius R; plane z=0 distance h/2 from center, so intersection circle radius sqrt(R² - h²/4)=sqrt(a²/4 + d²). Yes. So base Q must be contained in disk D0: (x - a/2)^2 + (y - σ d)^2 ≤ a²/4 + d². Also y≥0? But disk may extend y<0; but since Q y≥0 maybe if σ? Also lateral face y=0 is boundary; disk intersects line y=0 at x=0 and x=a? Let's check: y=0: (x-a/2)^2 + d² ≤ a²/4 + d² => (x-a/2)^2 ≤ a²/4 => 0≤x≤a. Good, so base edge from 0 to a lies on circle boundary (since equality). The disk is outside? For y>0, if σ=+1 center y=d>0; if σ=-1 center y=-d<0 (disk offset). Both possible. Q must be convex quadrilateral with edge endpoints on circle? Actually endpoints (0,0),(a,0) on boundary. Other vertices must lie in disk and in y≥0? But convex hull with edge lies in y≥0 if vertices have y≥0. The disk may include negative y but irrelevant. Now area of quadrilateral with one side a inside a disk. For fixed disk (given a,d) and side endpoints on boundary, max area of quadrilateral with that side? The maximum area convex polygon with n sides inscribed in a circle? But here quadrilateral with one side fixed as chord of circle, and other vertices inside disk (not necessarily on boundary). To maximize area, likely other vertices on boundary? The area of quadrilateral = area triangle (0,0),(a,0),P + triangle (0,0),(a,0),R? Actually let order: O=(0,0), A=(a,0), B, C. Area = 1/2 |OA x OB + ...|. For fixed OA, area = 1/2 a (y_B + y_C?) depending orientation. If B and C on same side y≥0, area = 1/2 a (y_B + y_C)? Let's compute: Polygon O,A,B,C. Shoelace: area = 1/2 | x_O y_A - y_O x_A + x_A y_B - y_A x_B + x_B y_C - y_B x_C + x_C y_O - y_C x_O | = 1/2 |0 + a y_B - 0 + x_B y_C - y_B x_C +0| = 1/2 | a y_B + x_B y_C - y_B x_C |. If B,C ordered, area = 1/2 [a y_B + x_B y_C - y_B x_C]? For convex with O,A,B,C counterclockwise, yes area = 1/2 (a y_B + x_B y_C - y_B x_C). Not simply sum y. But maybe max area for fixed chord and disk occurs when B,C on circle? By convexity, maximum area of convex polygon with fixed vertices? For polygon with one edge fixed and other vertices in convex set, maximum area occurs when all other vertices on boundary of convex set? I think yes: area as function of vertex is linear, maximum over convex compact at extreme points; but with multiple vertices need each vertex extreme? The feasible set for each vertex independent disk, so area linear in B and C? Area = 1/2 (a y_B + det(B,C))? det(B,C)=x_B y_C - y_B x_C is bilinear, not linear in B alone. So not trivial. But likely symmetric: B and C are mirror? Max area might be square? Let's examine. If we choose B and C on circle, quadrilateral inscribed in circle with side fixed? Actually endpoints O,A on circle; B,C also on circle? If B,C on boundary, quadrilateral inscribed in same circle. Among quadrilaterals inscribed in a given circle with one side fixed a, max area maybe the one symmetric? Let's derive. Alternatively, use geometry: For fixed side a and disk radius ρ = sqrt(a²/4+d²) centered at midpoint M=(a/2, σ d). The disk is not necessarily centered on side; its center offset σd perpendicular to side. The maximum area of quadrilateral with base a and vertices in disk likely occurs by choosing B,C on the two arcs symmetric? Since disk is symmetric with respect to perpendicular bisector of chord OA? The disk center lies on perpendicular bisector? M=(a/2, σd) lies on perpendicular bisector x=a/2, yes. The whole problem symmetric about line x=a/2. The maximum area should be symmetric: B and C symmetric with respect to x=a/2, and likely both on the far side of disk from chord? Let's set coordinate relative to midpoint: Let u axis along side OA, v axis perpendicular (y direction). Disk center at (0, σd) in (u,v) coordinates, radius ρ. Chord endpoints at u=±a/2, v=0. The disk includes points satisfying u² + (v - σd)² ≤ ρ² = a²/4 + d², v≥0 maybe (if Q in v≥0). The maximum area of quadrilateral with base a and two other vertices B,C in v≥0. By symmetry and convexity, optimal B and C symmetric: B=(0? no) maybe B=(u,v), C=(-u,v) or (u,v) with two vertices? Wait quadrilateral order O,A,B,C; if symmetric about v-axis, B and C mirror: B=(u,v), C=(-u,v)? But order O( -a/2,0), A(a/2,0), B(u,v), C(-u,v) would self intersect? Let's set O left, A right, B upper right, C upper left. Area =? Shoelace: area = 1/2 [a*v + u*v - v*(-u)? Let's compute: O=(-a/2,0), A=(a/2,0), B=(u,v), C=(-u,v). Area = 1/2 | x_O y_A - y_O x_A + x_A y_B - y_A x_B + x_B y_C - y_B x_C + x_C y_O - y_C x_O | =1/2 |0 + (a/2)v - 0 + u v - v(-u) + 0 - 0| =1/2 [ (a/2)v + 2uv ]? Wait (a/2)v + u*v - v*(-u) = (a/2)v + uv + uv = (a/2+2u)v. But area should be v*(a +? maybe). Let's use formula: Area = (1/2)[ (a/2)*v + u*v - v*(-u) ] = (1/2)[ (a/2)v + 2uv ] = v(a/4 + u). Hmm if v>0. But for u positive/negative? If B and C above, area = average height times base? The polygon O-A-B-C area = area triangle OAB + triangle OBC? Actually diagonal OB: area OAB = 1/2 * a * v? Coordinates O(-a/2,0), A(a/2,0), B(u,v): area = 1/2 |det(A-O, B-O)| = 1/2 * a * v (since base a horizontal). Triangle OBC? Area = 1/2 |det(B-O, C-O)| = 1/2 |(u+a/2, v) x (-u+a/2, v)| = 1/2 |(u+a/2)v - v(-u+a/2)| = 1/2 |v(u+a/2 + u - a/2)| = 1/2 |2uv| = |uv|. So total area = (1/2) a v + |u| v? If u≥0, area = v(a/2 + u). But if u can be negative, |u|. However with B upper right (u>0), C upper left (-u? Actually C=(-u,v) where u>0? Then C x negative; O at -a/2, A at a/2; order O,A,B,C: if B=(u,v) with u maybe between? C=(-u,v) with u>0 -> C left of center. Then polygon O-A-B-C: O left, A right, B right, C left -> convex if u? It is convex if v>0 and maybe u< a/2? But B could be left of A? Need. Area formula with u>0, C=(-u,v) gives triangle OBC area =? O=(-a/2,0), B=(u,v), C=(-u,v). det(B-O,C-O)= (u+a/2, v) x (-u+a/2, v)= (u+a/2)v - v(-u+a/2)= v(u+a/2 + u - a/2)=2uv => area=v u. Total = a v/2 + uv = v(a/2+u). This can exceed a v? If u large, but containment restricts u≤? Since B,C in disk: u² + (v-σd)² ≤ ρ². Also if B,C on circle maybe. But maybe optimal quadrilateral is not symmetric with B,C same v? Could be one vertex on disk boundary and other? Let's derive rigorously. Maybe there is another known approach: The prism with one lateral face inscribed in sphere? Perhaps maximum volume occurs when prism is actually a cube? Let's test. Suppose all 8 vertices of prism on sphere? Then prism inscribed in sphere. For a right prism with rectangular base? Max volume of inscribed rectangular box in sphere is cube? Actually max volume of rectangular parallelepiped inscribed in sphere radius R is cube with side 2R/√3? Volume = (2R/√3)^3 = 8R^3/(3√3) = same as our earlier? Wait cube side 2R/√3 = 12/√3=4√3, volume = (4√3)^3 = 64*3√3? No (4√3)^3 = 64*3√3? Actually (4√3)^2=48, times 4√3=192√3. Yes same volume 192√3. Interesting! Our earlier solution for base inscribed gave same volume. But that was a prism with one base inscribed, height 4√3 and square base side 4√3, so it's a cube (all dimensions 4√3) with all 8 vertices on sphere. Indeed if d=R/√3, r=2√6, square side 4√3, height h=2d=4√3. So prism is a cube inscribed in sphere. So answer likely cube side 4√3, volume 192√3. But need be careful: Could the face with vertices on sphere be lateral face of a rectangular box not cube? Maybe maximum also same cube? Let's check if volume could be larger with lateral face configuration. Since cube is global max among rectangular boxes inscribed in sphere? For a right prism with any quadrilateral base, max volume might be cube? Let's verify. General problem: Right quadrangular prism contained in sphere, with one face's vertices on sphere. The cube's all vertices on sphere. Is it possible to have a prism with larger volume than inscribed cube? The maximum volume of any convex body contained in sphere? Sphere volume ~904.78; cube volume ~332.55, less. But maybe non-cube prism could have larger? I recall maximum volume rectangular box in sphere is cube (symmetry). For general prism (not necessarily rectangular base) maybe? The problem statement likely expects use of AM-GM or Lagrange to find cube. Let's solve using coordinate and optimization; maybe simpler. Let's consider general right prism with base a quadrilateral Q (not necessarily square). But one face's four vertices on sphere. Because prism is "completely contained" in sphere. Could the face be a base as I first assumed? In many prism problems, "uma das faces" maybe base? Since a prism has bases and lateral faces; but the phrase "quatro vértices de uma das faces" could be any. However if it were a base, the prism would have one base on sphere and the other possibly inside. My earlier volume max 192√3 with cube. If it were a lateral face, maybe same cube but with lateral face on sphere? Let's see cube has all faces; any face's four vertices on sphere, including lateral. So if the condition allows lateral face, cube is still candidate. Max volume maybe same? Need prove. Let's formulate general optimization for any right prism with one face (base or lateral) inscribed in sphere. Since a cube satisfies both (all faces), and its volume 192√3 likely maximum. Need show no larger prism. Alternative approach: Any right prism contained in sphere of radius R has volume ≤ volume of inscribed cube? Is that true? For arbitrary prism? There is theorem: Among all parallelepipeds inscribed in sphere, cube maximizes volume. But for general prism with quadrilateral base maybe? I think yes? Actually a prism can have base with more area than square for same circumradius? Max area quadrilateral inscribed in circle is square (2r²), so for given height and base circle radius r, square base max. Then volume ≤ 2r² h. For sphere containment, need relation between r,h,d. My earlier derived max 192√3. That assumed base face on sphere and prism extends inward. If lateral face on sphere, we can perhaps derive similar bound. Let's solve lateral face case to ensure no larger volume. Suppose a lateral face rectangle is inscribed in circle radius r in plane Π at distance d from O. Its dimensions a (base edge) and h (height), with a²+h²=4r²=4(R²-d²). The prism volume V = A_base * h, where base Q has one side a and lies in plane perpendicular to lateral face. Need bound A_base in terms of a and sphere containment. Maybe use similar "face" argument: The prism can be seen as having a face (lateral) inscribed in sphere. The volume maybe maximized when base Q is also square and h=a? Let's optimize. Let's set coordinates as before: Lateral face y=0, vertices (0,0,0),(a,0,0),(0,0,h),(a,0,h). Sphere center O=(a/2, σd, h/2), with d² = R² - (a²+h²)/4, d≥0, σ=±1. Base plane z=0 contains side (0,0)-(a,0). Base Q is quadrilateral in z=0, y≥0, with edge along x from 0 to a. The sphere at z=0 is disk D: (x-a/2)^2 + (y-σd)^2 ≤ R² - h²/4 = ρ² = a²/4 + d²? Let's recompute: R² - h²/4 = R² - [4(R²-d²)-a²]/4 = d² + a²/4. Yes. So base Q must be contained in disk D of radius ρ = sqrt(d² + a²/4) centered at (a/2, σd), with edge endpoints on boundary (since y=0, x=0,a satisfy equality). Also Q must lie in y≥0? Since prism same side as base; if disk center y=σd. If σ=+1, disk centered above side (positive y); if σ=-1, center below side (negative y). Since Q y≥0 and edge on y=0, for σ=-1 the disk extends from y=-d-ρ to -d+ρ = -d+sqrt(d²+a²/4). Is this positive? sqrt(d²+a²/4) > d, so upper y positive. So disk includes y>0 region. For σ=+1, disk includes y from d-ρ to d+ρ; d-ρ negative? ρ>d, so includes y=0. Both possible. But containment of prism might force σ? Because sphere must contain entire prism including z=h top? Check top base z=h: sphere cross-section disk at z=h has same radius ρ? Since sphere center z=h/2, plane z=h distance h/2, same radius sqrt(R² - h²/4)=ρ. Top base Q_h (translation of Q by h in z) must lie in disk D_h identical to D in xy coordinates (since translation along z). So Q must be subset of D. Same constraint for both bases and all intermediate, since for z∈[0,h], cross-section disk radius sqrt(R² - (z-h/2)^2) ≥? At z=0,h radius ρ; at z=h/2 radius R. So if Q⊂D0 then maybe all points? Since D_z contains D0? At z=0 radius ρ; at z=h/2 radius R > ρ; so yes D0⊂D_z. Thus base constraint suffices. So lateral face case reduces to: choose a>0, h>0, d≥0 with a²+h² ≤? Actually a²+h² = 4(R²-d²) ≤ 4R², so d² = R² - (a²+h²)/4 ≥0 => a²+h² ≤4R². Also sphere containment: Q in disk D of radius ρ=sqrt(d²+a²/4). Need maximize V = area(Q)*h subject to Q convex quadrilateral with edge a and Q⊂D, 0≤? a,h,d relation. Now for fixed a,d (hence disk D), what's max area of quadrilateral with one side a inside disk D? Then optimize over a,h,d. This is a 2D geometric optimization. Let's solve maximum area of quadrilateral with one side a and contained in a disk of radius ρ = sqrt(d² + a²/4), where disk center lies on perpendicular bisector of side at distance d from side (signed). The disk is symmetric about perpendicular bisector. The side endpoints are on circle boundary. We can parameterize quadrilateral with base OA length a and other vertices B,C in disk, y≥0. By symmetry, maximum area likely occurs when B,C lie on circle and symmetric about perpendicular bisector? Need prove or at least derive. Let's attempt optimization. Set coordinates centered at midpoint of base: O=(-a/2,0), A=(a/2,0). Disk center M=(0, σd) with radius ρ, where ρ² = a²/4 + d². Let σ = +1 or -1. But if σ=-1, M below side, disk mostly? Actually d maybe? Because ρ>d, the circle intersects both sides. Since Q y≥0, but disk may have center below; still max area might use σ=+1? Let's see later. The disk equation: x² + (y - σd)² ≤ ρ² = a²/4 + d². Since side endpoints satisfy x=±a/2,y=0: a²/4 + d² = ρ² yes. So side is a chord of circle at distance |σd| from center? If σ=+1, chord at distance d below center (since center above chord), and chord length a = 2√(ρ² - d²) = 2*(a/2)=a. Good. If σ=-1, center below chord, chord also at distance d from center (above center). But disk is mirrored; since Q y≥0 maybe using upper part. But if σ=-1, disk is centered below; the disk may extend less above? The maximum y-coordinate of disk is -d + ρ = ρ - d. Since ρ² = d² + a²/4 => ρ - d = a²/(4(ρ+d))? Positive. The area with σ=-1 might be smaller because disk "wasted" below side. So likely σ=+1 (sphere center on same side as prism), allowing larger disk above side. Let's assume σ=+1 for max; later justify. So disk D: x² + (y-d)² ≤ a²/4 + d², y≥0, with center above side. The chord OA at bottom. We need max area quadrilateral O,A,B,C with B,C∈D, y≥0. Since area function: Area = area of triangle OAB + triangle OAC? Actually quadrilateral O-A-B-C can be triangulated by diagonal OB or AC. For convex with O,A,B,C in order, area = area(△OAB) + area(△OBC). = (1/2) a y_B + (1/2)|det(B-O,C-O)? Wait triangle OBC area = 1/2 |det(B-O, C-O)|. But if C is after B, maybe area = 1/2 [a y_B + det(B,C)?]. Use shoelace: Area = 1/2 | x_A y_B + x_B y_C - y_B x_C |? Actually with O=(-a/2,0), A=(a/2,0): Area = 1/2 [ (a/2) y_B + x_B y_C - y_B x_C +? Wait compute: Σ x_i y_{i+1}: O: (-a/2)*0=0; A: (a/2)*y_B = a y_B/2; B: x_B y_C; C: x_C*0=0. Σ y_i x_{i+1}: O:0; A:0*x_B=0; B: y_B x_C; C: y_C*(-a/2)= -a y_C/2. Area = 1/2 [a y_B/2 + x_B y_C - y_B x_C + a y_C/2]? Wait include? Let's do correctly: Shoelace: Sum1 = x_O*y_A + x_A*y_B + x_B*y_C + x_C*y_O = (-a/2)*0 + (a/2)*y_B + x_B*y_C + x_C*0 = (a/2)y_B + x_B y_C. Sum2 = y_O*x_A + y_A*x_B + y_B*x_C + y_C*x_O = 0 + 0 + y_B x_C + y_C*(-a/2) = y_B x_C - (a/2)y_C. Area = 1/2 | Sum1 - Sum2 | = 1/2 | (a/2)y_B + x_B y_C - y_B x_C + (a/2)y_C | = 1/2 [ (a/2)(y_B+y_C) + x_B y_C - y_B x_C ]. But determinant (B-O, C-O) = (x_B+a/2, y_B) x (x_C+a/2, y_C) = (x_B+a/2)y_C - y_B(x_C+a/2) = x_B y_C + (a/2)y_C - y_B x_C - (a/2)y_B. Then Area = 1/2 [ (a/2)(y_B+y_C) + x_B y_C - y_B x_C ] = 1/2 [ det(B-O,C-O) + a y_B ]? Since det + a y_B = x_B y_C + (a/2)y_C - y_B x_C - (a/2)y_B + a y_B = x_B y_C - y_B x_C + (a/2)y_C + (a/2)y_B. Yes. So area = 1/2 [det(B-O,C-O) + a y_B] if O,A,B,C order? Hmm. Alternatively area of quadrilateral = area triangle OAB + triangle OBC (using diagonal OB) = 1/2 a y_B + 1/2 |det(B-O,C-O)|. But if C and A are on opposite sides of line OB? For convex O-A-B-C, diagonal OB lies inside; triangle OBC area = 1/2 |det(B-O,C-O)|. Since O,A,B,C convex, C and A on same side? Actually line OB: A and C lie on same side? In convex quadrilateral O-A-B-C, diagonal OB splits into triangles OAB and OBC; C lies opposite side of OB from A? Let's test square: O(-1,0), A(1,0), B(1,1), C(-1,1). Det(B-O,C-O) = (2,1)x(0,1)=2*1 -1*0=2 positive; area OBC=1; area OAB=1/2*2*1=1; total 2. Yes. So area = 1/2 a y_B + 1/2 det(B-O,C-O) if det positive. Good. Thus Area = 1/2 [ a y_B + det(B-O,C-O) ]. Given B,C in disk, maximize F = a y_B + det(B-O,C-O) = a y_B + (x_B+a/2)y_C - y_B(x_C+a/2) = a y_B + x_B y_C + a y_C/2 - y_B x_C - a y_B/2 = a y_B/2 + a y_C/2 + x_B y_C - y_B x_C = (a/2)(y_B+y_C) + x_B y_C - y_B x_C. For fixed x_B,y_B, F is linear in x_C,y_C: F = y_C(x_B + a/2) + y_B(-x_C - a/2?) Wait F = (a/2)(y_B+y_C)+ x_B y_C - y_B x_C = y_B*(a/2) + y_C*(a/2 + x_B) - y_B x_C. Not purely linear because -y_B x_C term. But if B fixed, optimize over C in disk to maximize linear function L(C)= y_C*(a/2 + x_B) - y_B x_C + const. The maximum of a linear functional over disk occurs at boundary; specifically C on circle in direction of gradient (a/2+x_B, -y_B). So C on circle (not interior). Similarly for B for fixed C, F linear? F as function of B: (a/2)y_B + x_B y_C - y_B x_C = y_B(a/2 - x_C) + x_B y_C. Linear in B (x_B,y_B) for fixed C. Max over disk at boundary. Thus both B and C on circle. Good. So B,C on circle D boundary. The maximum area quadrilateral with inscribed in circle? Actually with B,C on same circle and base chord OA on circle, the quadrilateral O,A,B,C is cyclic? Since all four vertices on same circle? But B,C on circle D, O,A also on D boundary (yes side endpoints). So quadrilateral is cyclic (inscribed in circle D). Max area among cyclic quadrilaterals with fixed chord OA? Since all vertices on circle, area of cyclic quadrilateral is maximized when it is regular? But with one side fixed, not all sides equal necessarily. For a given circumcircle and one chord fixed, the maximum area quadrilateral maybe is isosceles trapezoid? Let's optimize over B,C on circle. Parameterize circle D: radius ρ, center (0,d) in our coordinates (where σ=+1). Points on circle: O=(-a/2,0), A=(a/2,0) correspond to central angles? Let's use polar center M=(0,d). Let ρ=√(a²/4+d²). The chord OA has distance d from center; central angle 2α where sin α = (a/2)/ρ, cos α = d/ρ? Actually half-angle α with sin α = (a/2)/ρ, cos α = d/ρ. The chord endpoints at angles π+α? Let's set center M. Vector to O=(-a/2, -d), angle θ_O = π + α? Since sin? Coordinates relative M: (-a/2, -d). Angle from positive x: cos=-a/(2ρ), sin=-d/ρ. Let α such that cos α = a/(2ρ), sin α = d/ρ? Then θ_O = π + α? Because cos(π+α)=-cosα=-a/(2ρ), sin(π+α)=-sinα=-d/ρ. Yes. Vector to A=(a/2,-d): angle θ_A = 2π - α? cos= a/(2ρ), sin=-d/ρ. Usually θ_A = -α (or 2π-α). So minor arc OA has central angle 2α? From θ_A=-α to θ_O=π+α? Actually difference clockwise? The chord endpoints separated by minor arc? The shorter arc between O and A goes through bottom of circle (y minimum). Its central angle = 2α? Since θ_A=-α, θ_O=π+α; difference = π+2α > π, so minor arc is other way: from θ_O=π+α to θ_A=2π-α difference = π-2α? Wait compute: θ_A=2π-α (or -α). Difference θ_O - θ_A = (π+α) - (-α)=π+2α > π; other difference = 2π - (π+2α)=π-2α < π if α<π/2. So minor arc OA has central angle π-2α? Let's verify α small (d small): chord near diameter? If d=0 (center on chord), α=0, chord diameter, endpoints opposite, central angle π. Formula π-2α=π yes. If d large, α near? a small, chord near top? α? sinα=a/(2ρ) small, cosα=d/ρ near1, α small; minor arc near bottom? Hmm maybe not needed. Max area quadrilateral with vertices O,A,B,C on circle. Since area depends on arcs. There is formula for cyclic quadrilateral area = (1/2) sin(arc AB)*... Not simple. Maybe use symmetry: The maximum area should be symmetric about perpendicular bisector of OA: B and C mirror? But with two additional vertices, for fixed chord, max area of cyclic quadrilateral likely when the two other vertices are symmetric and lie on the arc opposite the chord? Let's test. If B,C chosen on major arc opposite chord (the upper arc), with B,C symmetric: B=(u,v), C=(-u,v) with u≥0? Then quadrilateral O-A-B-C is isosceles trapezoid? Actually O(-a/2,0), A(a/2,0), B=(u,v), C=(-u,v). This is symmetric. Area = (a + 2u?) wait area = (a + 2u)/2 * v? Actually trapezoid bases lengths a (bottom) and 2u (top? distance between B,C = 2u) if B,C have same y, area = (a + 2u) v /2 = (a/2 + u)v. This matches earlier formula with symmetric B,C above (if u>0). But our earlier formula for symmetric B=(u,v), C=(-u,v): area = v(a/2 + u). Yes. If B,C on circle, can express v in terms of u: circle x²+(y-d)²=ρ² => u²+(v-d)²=ρ² = a²/4 + d² => v = d + sqrt(ρ² - u² - d²?) Wait ρ² = a²/4 + d², so u² + (v-d)² = a²/4 + d² => (v-d)² = a²/4 + d² - u² = ρ² - u². Since v likely above? d positive, v could be d + sqrt(ρ²-u²) or d - sqrt(...). To maximize v? For area = v(a/2 + u), likely v positive, choose upper branch v = d + sqrt(ρ² - u²). (If d - sqrt could be negative for large u, but area maybe smaller). Domain u∈[?]. Also v≥0 => sqrt(ρ²-u²) ≥ -d always if ρ≥d; yes. Then area A(u) = (a/2 + u)(d + sqrt(ρ² - u²)), with ρ²=a²/4+d². Max over u∈[0,?]. Since symmetric, u maybe up to ρ? But if u too large, v negative? Upper branch v≥0 always? d -? no, upper always ≥d≥0. But B,C maybe on upper arc. Also need quadrilateral convex and O,A,B,C order; u can be from 0 to a/2? Actually if u > a/2, C x=-u < -a/2, O may not be leftmost? But polygon order O(-a/2,0), A(a/2,0), B(u,v), C(-u,v): if u>a/2, then B x > A x, still convex? O left, A mid, B right, C left; yes possible (kite-like). But if u extremely large, v maybe small? However u max? v real requires u≤ρ. Let's optimize. Alternatively, there is known theorem: Among all quadrilaterals inscribed in a circle with one side fixed, the maximum area is achieved by an isosceles trapezoid? Let's verify via calculus. Let's maximize A(u) = (a/2+u)(d + √(ρ²-u²)). Let ρ² = a²/4 + d². Derivative: A' = (d + s) + (a/2+u)( -u/s ) where s=√(ρ²-u²). Set A'=0: (d+s) - u(a/2+u)/s =0 => (d+s)s = u(a/2+u) => d s + s² = u a/2 + u² => d s + (ρ² - u²) = u a/2 + u² => d s + ρ² - u² - u a/2 - u² =0 => d s + ρ² - u a/2 - 2u² =0. ρ² = a²/4 + d². So d s + a²/4 + d² - u a/2 - 2u² =0. This is messy. Maybe max at u=a/2? Symmetric square? If u=a/2, B=(a/2,v), C=(-a/2,v), rectangle with width a? Then v = d + √(ρ² - a²/4)= d + √(d²)= d+d=2d. Area = (a/2 + a/2)*2d = 2ad. Also other square? If base Q is square? Our earlier cube had a=4√3, h=4√3, d=R/√3=2√3? Wait for cube lateral face: a=4√3, h=4√3, R=6. a²+h²=96+48? Actually a²+h²=48+48=96, 4(R²-d²)=4(36-d²)=96 => d²=12 => d=2√3. Radius ρ=√(a²/4+d²)=√(12+12)=√24=2√6. Area base Q must be square? For cube, base Q is square side a=4√3, area 48. In lateral face coordinate, base Q with edge a=4√3, h=4√3; base Q is square? Actually cube: base is square side 4√3; one side a=4√3; other sides also 4√3. The base Q lies in z=0, contained in disk D radius ρ=2√6 ≈4.899 centered at (a/2=3.464, d=3.464)?? Wait a=4√3≈6.928, half=3.464, d=2√3≈3.464, ρ²=a²/4+d²=12+12=24. Base square vertices: (0,0), (a,0), (a,a), (0,a). Does this lie in disk? Check vertex (a,a): (a/2)^2+(a-d)^2? (3.464)^2+(6.928-3.464)^2=12+12=24 yes. So B=(a,a) and C=(0,a)? But our symmetric form with u? The square has other vertices (a,a) and (0,a), not symmetric B=(u,v), C=(-u,v) with same v? Actually if base square, other vertices are (a,a) and (0,a); not symmetric about x=a/2? Wait disk center x=a/2; square vertices: (0,0), (a,0), (a,a), (0,a). This is symmetric: reflect x=a/2 maps (a,a) to (0,a). So B=(a,a), C=(0,a). In our coordinates O=(-a/2,0), A=(a/2,0), B=(a/2? no). Let's shift x by -a/2: O=(-a/2,0), A=(a/2,0); original (a,a) -> (a/2, a); original (0,a)->(-a/2, a). So B=(a/2, a), C=(-a/2, a). That's u=a/2, v=a. Yes symmetric. Area = (a/2 + a/2)*a = a² =48. In our formula with a=4√3, u=a/2=2√3≈3.464, v=a=4√3≈6.928; but circle upper branch v = d + √(ρ²-u²) = 2√3 + √(24-12)=2√3+√12=2√3+2√3=4√3 yes. Good. Is this max? Let's check derivative at u=a/2 maybe. A=(a/2+u)(d+s). At u=a/2, s=√(ρ² - a²/4)=d, so A=(a)(2d)=2ad. Derivative? d=2√3, a=4√3? A' = (d+s) + (a/2+u)(-u/s) = (2d) + (a)(-a/(2d))? u/s=a/(2d). =2d - a²/(2d). Set =0 => 4d² = a². Here d=2√3, a=4√3: 4*12=48, a²=48 yes. So derivative zero. Second derivative? likely max if? So max at u=a/2 =? if 4d² = a²? But our a,d relation from a²+h² = 4(R²-d²) and h? For cube h=a? Actually a=h=4√3; h =? In lateral face, h = 4√3. Then a²+h² = a²+a²=2a² =4(R²-d²). With a=4√3, 2*48=96=4(36-d²)=>d²=12. So yes 4d²=48=a². Thus cube satisfies u=a/2 stationary. Is it global max? Let's examine. But maybe for fixed a,d (with a² ≠4d²), max area might be at different u. Need maximize V = A_max(a,d)*h with a²+h² =4(R²-d²). We can compute A_max. Let's derive general A_max for quadrilateral in disk with chord a. Maybe formula: A_max = ? If 4d² = a²? Hmm. Let's solve optimization more generally. Option: Use known result: For a given circle radius ρ and chord length a, the maximum area of a quadrilateral inscribed in the circle with one side a is (a/2)√(4ρ² - a²)? Let's test. Square case ρ=√24≈4.899, a=6.928: (a/2)*√(4ρ² - a²) = (3.464)*√(96-48)=3.464*√48=3.464*6.928=24? But area square 48. Not. Maybe formula area max = a√(ρ² - a²/4)? For square: 6.928*√(24-12)=6.928*√12=6.928*3.464=24? No. Hmm. Let's derive A_max for cyclic quadrilateral with one chord fixed. Perhaps area = (1/2) d1 d2 sinθ? For cyclic quadrilateral, area = 1/2 (ab+cd) sin? Actually. Let's parameterize using angles. Let circle radius ρ. Let chord OA subtend central angle 2φ? Previously minor arc? Let's define half-angle θ such that chord a = 2ρ sin θ? Wait if chord from angles ±θ around bottom? Let's choose coordinate with center M and chord horizontal at distance d above/below. Let the chord endpoints have angles? Better: Let M origin, circle radius ρ. Let chord OA at distance d from center, chord length a = 2√(ρ²-d²). Let's set chord vertical? Let half-angle α with sin α = a/(2ρ), cos α = d/ρ. The chord endpoints relative to M: O=(-a/2, -d), A=(a/2, -d) in previous. In polar, angles π+α and 2π-α? As above. Let other vertices B,C on circle at angles β, γ. The area of cyclic quadrilateral O,A,B,C = (1/2)ρ² [ sin(θ_A-θ_O) + sin(θ_B-θ_A) + sin(θ_C-θ_B) + sin(θ_O-θ_C) ]? Actually area = (1/2)ρ² Σ sin(Δθ_i) for vertices in order. Simpler: area = (1/2)ρ² [ sin(arc OA) + sin(arc AB) + sin(arc BC) + sin(arc CO) ] if arcs <π? But with signs. Maybe use formula area = (1/2)(ac? no). Let's try another approach. Since disk D is convex and side endpoints on boundary, the maximum area quadrilateral with one side a might be achieved when B and C are the points on the circle diametrically opposite to the midpoint of arcs? For fixed chord, the area as function of B,C may be maximized by choosing B,C at the top of circle? Let's test. If B and C both equal top point of circle (0, d+ρ) -> degenerate triangle area = a(d+ρ)/2? Actually quadrilateral degenerates to triangle OAB twice? Area = a v/2? Not max likely. Let's perform optimization for general a,d. Coordinates centered: O=(-a/2,0), A=(a/2,0), disk D: x²+(y-d)²≤ρ², ρ²=a²/4+d², d≥0. Max area F = (a/2)(y_B+y_C)+x_B y_C - y_B x_C. Since B,C on boundary, use angle parameter. Let M=(0,d), radius ρ. Let point on circle: (ρ cosθ, d + ρ sinθ). Then relative coordinates. O angle? O=(-a/2,0) = (ρ cosθ_O, d + ρ sinθ_O) => sinθ_O = -d/ρ, cosθ_O=-a/(2ρ). Let θ_O = π+α? As before sinα=d/ρ, cosα=a/(2ρ). So θ_O = π+α? cosπ+α=-cosα=-a/(2ρ), sinπ+α=-sinα=-d/ρ. Yes. A=(a/2,0): θ_A=2π-α? cos=cosα=a/(2ρ), sin=-sinα=-d/ρ => θ_A=-α (or 2π-α). Let B=(ρ cosβ, d+ρ sinβ), C=(ρ cosγ, d+ρ sinγ). Assume order O,A,B,C counterclockwise around circle? The vertices on circle; for convex quadrilateral, arcs should be in order. The arc from O to A via bottom? Central angle difference from θ_O=π+α to θ_A=2π-α (CCW) is π-2α? Wait if α small, yes. Then B,C lie on remaining major arc? Actually quadrilateral O,A,B,C with O,A near bottom, B,C on upper arc. Order: O (π+α), A(2π-α), then B,C on upper arc from 2π-α to π+α+2π? Hmm. Let's parameterize B,C on the major arc opposite chord, i.e., θ ∈ [2π-α, 2π+π+α]? More simply, choose B,C on upper arc between angles -α and π+α (the long way). Let's set B at angle β ∈ ( -α, π+α)? Wait A at -α, O at π+α (same as -π+α? Since π+α equivalent to -π+α mod 2π). The arc from A(-α) through 0, π/2, π to O(π+α) has central angle π+2α? Actually difference (π+α) - (-α)=π+2α >π, so it's major arc. The minor arc is the other way from O(π+α) through 3π/2 to A(2π-α) or -α. For quadrilateral O,A,B,C, to be convex with O,A consecutive, B,C likely on the major arc (upper) so order O -> A via minor? Wait order around circle for convex quadrilateral: O, A, B, C should be in cyclic order (either clockwise or counterclockwise). Starting at O, going clockwise (decreasing angle) maybe: O (π+α) -> lower? Let's use counterclockwise from O: O angle π+α; next A angle 2π-α (in CCW from π+α to 2π-α = π-2α) then B,C in (2π-α, 2π+π+α? Actually after A=2π-α, CCW to next O+2π =3π+α? The arc length = (3π+α) - (2π-α)=π+2α. So B,C lie in this arc: angles θ ∈ [2π-α, 3π+α] (mod). This corresponds to upper major arc. Let β,γ with -α ≤ β ≤ γ ≤ π+α? Since 2π-α equivalent to -α; 3π+α equivalent to π+α. So yes B,C angles β,γ ∈ [-α, π+α] with β≤γ. Then area formula maybe. Let's derive area using shoelace in terms of θ. Let vertices P0=O(π+α), P1=A(-α), P2=B(β), P3=C(γ) with -α ≤ β ≤ γ ≤ π+α (but β,γ in this interval). The area = (1/2)ρ² [ sin(θ1-θ0) + sin(θ2-θ1) + sin(θ3-θ2) + sin(θ0+2π-θ3) ]? Actually sum of sin of arc differences. Here θ0=π+α, θ1=2π-α (or -α+2π), θ2=β? Wait β is in [-α, π+α] not adjusted. Let's keep angles unwrapped: θ0=π+α, θ1=2π-α, θ2=2π+β? Since β in [-α,π+α] => 2π+β in [2π-α, 3π+α]. θ3=2π+γ. Then differences: θ1-θ0 = (2π-α)-(π+α)=π-2α; θ2-θ1=2π+β - (2π-α)=β+α; θ3-θ2=γ-β; θ0+2π - θ3 = (3π+α) - (2π+γ)=π+α-γ. All positive if β,γ in range. Area = (1/2)ρ² [ sin(π-2α) + sin(β+α) + sin(γ-β) + sin(π+α-γ) ]. sin(π-2α)=sin2α=2 sinα cosα = 2*(a/(2ρ))*(d/ρ)= a d / ρ². sin(π+α-γ)= sin(π - (γ-α))? = sin(γ-α)? Since sin(π - x)=sin x, x=γ-α? Actually π+α-γ = π - (γ-α). So sin = sin(γ-α). Good (if γ-α ∈ [0,π]? γ≤π+α => γ-α≤π; yes). Thus Area = (1/2)ρ² [ sin2α + sin(β+α) + sin(γ-β) + sin(γ-α) ]. Since sin2α =? Also a=2ρ sinα, d=ρ cosα. ρ² sin2α = 2ρ² sinα cosα = 2ρ²*(a/(2ρ))*(d/ρ)=a d. Good. Now maximize S = sin(β+α) + sin(γ-β) + sin(γ-α) over -α ≤ β ≤ γ ≤ π+α. This is 2D trig optimization independent of ρ,a,d except α. Let x = β+α, y = γ-β, z = γ-α? Note x+y = γ+α? Wait x=β+α; y=γ-β; then x+y = γ+α. z=γ-α = x+y -2α? Actually γ-α = (β+α)+(γ-β)-2α = x+y-2α. Also constraints: β≥-α => x≥0; γ≤π+α => γ-α ≤ π => z≤π; β≤γ => y≥0; γ-β = y ≤? γ≤π+α and β≥-α => y≤π+2α? But also γ-α≤π => x+y-2α≤π => x+y≤π+2α. Also β+α = x ≤? β≤γ => x ≤ γ+α. Hmm. Maybe use variables u=β+α ∈ [0, ?], v=γ-α ∈ [0, π], with β≤γ => u - α ≤ v + α => u - v ≤ 2α? Actually β = u-α, γ=v+α; β≤γ => u-α ≤ v+α => u - v ≤ 2α. Also γ-β = (v+α)-(u-α)=v-u+2α ≥0 => v-u+2α ≥0 => u - v ≤ 2α same. So constraints: 0≤u≤? β≤? β≤π+α? x=β+α≤π+2α? But γ≤π+α => v≤π; u=β+α ≤ γ+α = v+2α ≤ π+2α. So u∈[0, π+2α], v∈[0,π], u-v≤2α (i.e., v ≥ u-2α). Also γ-β = v-u+2α ≥0 same. Objective S = sin u + sin(v-u+2α) + sin v? Wait sin(β+α)=sin u; sin(γ-β)=sin(v-u+2α); sin(γ-α)=sin v. Yes S = sin u + sin(v-u+2α) + sin v. Need maximize over u,v with 0≤u≤π+2α, 0≤v≤π, v ≥ u-2α (if u>2α). Also β≤γ ensures v-u+2α≥0. This is more complex. But likely maximum at u=π/2? v=π? Let's test symmetric case? In symmetric B,C with β=-? Actually if B,C symmetric about y-axis (x=0), their angles? B=(u,v) coordinates. Let's relate. Alternatively, use known result: For fixed chord, max area quadrilateral inscribed in circle occurs when the opposite arc is divided into three equal parts? Hmm. Let's test numeric for cube: a=6.928, d=3.464, ρ=4.899, α? sinα=a/(2ρ)=6.928/(9.798)=0.7071 => α=π/4 (45°). Then ρ²=24, d=ρ cosα=4.899*0.707=3.464 yes. Chord central angle 2α? Actually chord length 2ρ sinα =2*4.899*0.707=6.928. Minor arc below central angle 2α=π/2? Wait chord endpoints O,A at angles π+α=225° and 2π-α=315°; minor arc between them (through bottom) is 90° = 2α? Yes. In cube symmetric, B,C at (a/2,a)= (3.464,6.928), (-3.464,6.928). Angles? B: cosβ=0.707? x=3.464=ρ cosβ => cosβ=0.707, y=d+ρ sinβ=6.928 => sinβ=(6.928-3.464)/4.899=3.464/4.899=0.707 => β=π/4 (45°). C: x=-3.464 => cosγ=-0.707, sinγ=0.707 => γ=3π/4 (135°). So β=45°, γ=135°. Compute S = sin(β+α)=sin(45+45)=sin90=1; sin(γ-β)=sin90=1; sin(γ-α)=sin90=1; sum=3. Area = (1/2)ρ² [sin2α +3] = 0.5*24 [1+3]=48. Good. Can S exceed 3? Each sine ≤1, sum ≤3, achieved when all arguments = π/2 mod? Conditions: β+α = π/2 => β=π/2 - α = 45°; γ-β=π/2; γ-α=π/2 => γ=π/2+α=135°. This gives S=3. So global max S=3, achieved by unique β,γ. Therefore A_max = (1/2)ρ² [sin2α + 3] = (1/2)ρ² [2 sinα cosα + 3] =? Since sin2α = a d / ρ², so A_max = (1/2)(a d + 3ρ²) = (1/2)(a d + 3(a²/4 + d²)) = (3a²/8) + (1/2)ad + (3/2)d². Let's verify with cube: a=4√3≈6.928, d=2√3≈3.464: 3a²/8 = 3*48/8=18; (1/2)ad = 0.5*6.928*3.464=12; (3/2)d²=1.5*12=18; total 48. Yes. But is this valid for any a,d? We assumed max S=3 requires β=π/2 - α, γ=π/2 + α. Need ensure β,γ in feasible interval [-α, π+α] and β≤γ. β=π/2 - α ≥ -α? Yes if π/2≥0. β≤π+α? π/2 - α ≤ π+α => π/2 ≤ π+2α true. γ=π/2+α ≤ π+α? Yes. So feasible. Also need β+? The vertices order O,A,B,C: β=π/2 - α, γ=π/2+α. For α∈? If α>π/2? But chord length <2ρ, α∈(0,π/2) because cosα=d/ρ>0. So β>0, γ<π. Good. So max area formula valid for all 0<α<π/2, i.e., d>0. If d=0 (α=π/2? chord diameter, sinα=1 cosα=0; formula works? Then β=π/2-π/2=0, γ=π/2+π/2=π, S=sin0+sinπ+sinπ=0? Wait sin(β+α)=sinπ/2=1; sin(γ-β)=sinπ=0; sin(γ-α)=sinπ/2=1; S=2 not 3? Let's check d=0: chord diameter, points O,A opposite, α=π/2. Our formula A_max=(1/2)ρ²[sinπ +? sin2α=0 + S]. S max? Let's solve S=sin u + sin(v-u+π) + sin v = sin u - sin(v-u) + sin v? Actually sin(v-u+π)= -sin(v-u). With constraints u∈[0,2π?], v∈[0,π], v≥u-π. Max? For d=0, ρ=a/2, disk includes chord as diameter. Quadrilateral with diameter as side? Max area? If base is diameter, any quadrilateral inscribed in circle with one side diameter is right? The maximum area of quadrilateral with one side diameter a and other vertices on semicircle? Probably area = (1/2) a * (height1+height2)? Let's test square? If a=diameter, side a cannot be square? Inscribed quadrilateral with one side diameter: other vertices on semicircle; max area maybe when other vertices split semicircle into three equal? Let's numeric: a=2, ρ=1, d=0. Max area maybe √3? Our formula with α=π/2: A_max=(1/2)*1[0+ S_max]. Need S_max? Let's optimize S=sin u - sin(v-u)+ sin v. Choose u? If u=π/2, v=π? S=1 - sin(π/2)+0=0? Not. Let's use known max area of cyclic quadrilateral with one side diameter? The other two vertices on circle; area = (a/2)(y_B + y_C?) Hmm. For d=0, disk center on chord; upper semicircle y≥0. Max area likely = (a/2)*? Maybe choose B,C both at top? Degenerate. Let's solve later; not critical. But for our problem with R=6 and d>0 likely fine. However to be rigorous, need handle d=0 separately; but maximum likely at d>0. So A_max formula seems plausible. Now we have lateral face configuration. Let a>0, h>0, d≥0 satisfy a² + h² = 4(R² - d²), so d² = R² - (a²+h²)/4 ≥0 => a²+h² ≤4R². Also d≥0. The base area A_max(a,d) = (3a²/8) + (1/2) a d + (3/2)d². Volume V = A_max * h. Need maximize over a,h,d≥0 with a²+h² = 4(R²-d²) (or d² = R² - (a²+h²)/4). We can express d = sqrt(R² - (a²+h²)/4) if a²+h² ≤4R². Need maximize. Alternatively, maybe our earlier base-face configuration gave volume 192√3. Need compare lateral-face max. Let's optimize V_lateral. Let R=6. Let a,h>0, d≥0. Constraint a²+h²+4d² = 4R² =144. Let define s = a²+h² ≤144, d² = (144 - s)/4 = 36 - s/4, d = sqrt(36 - s/4). A_max = (3a²/8) + (1/2) a sqrt(36 - s/4) + (3/2)(36 - s/4) = (3a² - (3/2)s? Wait (3/2)d² = (3/2)(36 - s/4)=54 - 3s/8. Since s=a²+h², (3/2)d² =54 - (3/8)a² - (3/8)h². Then A_max = (3a²/8) + (1/2) a d + 54 - (3/8)a² - (3/8)h² = 54 + (1/2) a d - (3/8)h². Nice! Because a² term cancels. So A_max = 54 - (3/8)h² + (1/2) a d. Volume V = h [54 - (3/8)h² + (1/2)a d]. Also d = sqrt(36 - (a²+h²)/4). This is a function of a,h. Need maximize V(a,h) = 54h - (3/8)h³ + (1/2) a h sqrt(36 - (a²+h²)/4). Domain a>0, h>0, a²+h² ≤144. Check cube: a=h=4√3≈6.928. s=96, d=sqrt(36-24)=√12=2√3≈3.464. V = 54*6.928 - (3/8)*(6.928)^3 + 0.5*6.928*6.928*3.464 = 374.1 - (0.375*332.55?) Wait h³=332.55? (6.928)^3=332.55; *0.375=124.7; first=374.1-124.7=249.4; second=0.5*48*3.464=83.14; total=332.55 =192√3 yes. Could there be larger? Let's optimize. V(a,h) = h [54 - 3h²/8 + (a/2) d]. For fixed h, maximize over a>0 with a² ≤144 - h². Let g(a)= (a/2) sqrt(36 - (a²+h²)/4) = (a/2) sqrt((144 - h² - a²)/4)? Since 36 - h²/4 - a²/4 = (144 - h² - a²)/4, sqrt = (1/2) sqrt(144 - h² - a²). Thus g(a)= (a/4) sqrt(144 - h² - a²). Let K=144 - h² >0. g(a)= (a/4) sqrt(K - a²), 0<a<√K. Max of g over a: derivative: g = 1/4 * a (K - a²)^{1/2}; g' = 1/4 [ (K - a²)^{1/2} + a * (1/2)(K-a²)^{-1/2}(-2a) ] = 1/4 [ sqrt(K-a²) - a² / sqrt(K-a²) ] = 1/4 [ (K - a² - a²)/sqrt(K-a²) ] = 1/4 (K - 2a²)/sqrt(K-a²). Set =0 => a² = K/2 => a = √(K/2). Max g = (1/4)*√(K/2)*√(K - K/2)= (1/4)*√(K/2)*√(K/2)= K/8? Compute: sqrt(K/2)*sqrt(K/2)=K/2; /4 = K/8. So max g = K/8 = (144 - h²)/8. Achieved a² = K/2 = (144 - h²)/2 = 72 - h²/2. Also need a²+h² = (72 - h²/2)+h² =72 + h²/2 ≤144 => h²≤144 => h≤12. Good. Thus for fixed h, max over a of V is h [54 - 3h²/8 + (144 - h²)/8] = h [54 + (144 - h²)/8 - 3h²/8] = h [54 + 144/8 - h²/8 - 3h²/8] = h [54 + 18 - 4h²/8] = h [72 - h²/2] = 72h - (1/2)h³. Now maximize over h∈ (0,12] (since a>0, h<12; h=0 volume 0). F(h)=72h - h³/2. F'(h)=72 - (3/2)h² =0 => (3/2)h²=72 => h²=48 => h=4√3 (≈6.928). Yes cube. F''=-3h<0 => max. F(4√3)=72*(4√3) - (1/2)*(4√3)^3 = 288√3 - (1/2)*(64*3√3?) (4√3)^3 = 64*3√3? Actually (4√3)^3 = 64 * (√3)^3? Wait (4√3)^3 = 64*3√3? Compute: (4√3)^2=48, times 4√3 =192√3. Then half =96√3. 288√3 - 96√3 =192√3. Yes same. So lateral face configuration also max 192√3, achieved same cube (a=h=4√3, d=2√3? Wait our a here is base edge, h is height, d same? Yes cube). Thus both interpretations yield same maximum and cube. Good. Therefore answer likely: max volume = 192√3 cm³; height = 4√3 cm; base dimensions: square of side 4√3 cm. Need present solution. Need craft final response in Portuguese, item by item. Should show development. Let's plan proof concise. We need solve as IME candidate. Could use simpler approach: Since prism is right and one face's vertices on sphere. We can assume the face with vertices on sphere is a base? But problem says "uma das faces" not necessarily base. To be safe, we can argue: The maximum volume occurs when the prism is a cube inscribed in sphere, with all vertices on sphere, hence any face works. But need derive not assume. Maybe present a rigorous but concise solution using optimization. Since final answer desired oververbosity 5, not too detailed but enough. We can do: Let R=6. Consider the prism. Let the face whose vertices lie on sphere be a base (if lateral, analogous; or treat both). Actually maybe simpler: Any maximum prism can be oriented so that the face on sphere is a base, because if it's a lateral face, we can slice? Hmm not trivial. Need avoid invalid assumption. Alternative: Use general result: Among all right quadrangular prisms contained in a sphere, with one face's four vertices on sphere, the maximum volume is achieved by a cube inscribed in the sphere. Then prove via considering the face as a rectangle? But base may not be rectangle? "Prisma quadrangular" maybe base is quadrilateral not necessarily rectangle; but a cube base is square. Need justify max base area. Maybe present solution via two cases: Case 1: The face is a base. Let base be a quadrilateral inscribed in a circle of radius r at distance d from center. Area base ≤ 2r² (max cyclic quadrilateral area; achieved by square). Let h be height. Sphere containment forces h ≤ 2d (or h=2d for max). Then V ≤ 2(R²-d²)*2d = 4d(R²-d²) ≤ 192√3. This is elegant. Case 2: The face is a lateral face. Need show similar bound. But maybe problem intended "face" could be any, but many solutions may assume base? Let's see. The phrase "quatro vértices de uma das faces" could be base or lateral. In IME, they expect rigorous? The condition "completamente contido" and "quatro vértices de uma das faces pertencem à superfície" in a prism: If the face is lateral, the base is not necessarily cyclic? However a lateral face is a rectangle; its four vertices on sphere. The prism might be non-square. But our optimization showed same max. But proving in exam may be longer. Maybe there is a simpler argument: Any right quadrangular prism contained in a sphere can be enclosed in a rectangular parallelepiped? Hmm. Let's examine if a non-cube prism with lateral face on sphere could have larger volume than 192√3? Our optimization suggests no. But need present proof maybe using inequalities. Maybe use known theorem: For any prism inscribed in sphere (at least one face on sphere), volume ≤ volume of inscribed cube. Could state and prove via area and height. Let's attempt a unified proof via " if one face is on sphere, the opposite face is at distance ..." Maybe use "enlarge" to a prism with all vertices on sphere? Given any right prism contained in sphere with one face on sphere, can we translate/expand to an inscribed prism (all vertices on sphere) with at least same volume? Not always? Let's see. If one face on sphere and prism contained, the opposite face interior. The volume could be increased by extending height until opposite face hits sphere? But that would require moving the first face? If fix first face on sphere, extending height inward until opposite face also touches sphere maybe; since sphere symmetric, if first face at distance d, extending h=2d makes opposite face also on sphere (same circle). Then volume = A*2d. Is this maximal for fixed d? Our base-face case yes. For lateral-face case, maybe similar: fix lateral face on sphere, extend prism inward until opposite lateral face touches sphere? The opposite lateral face is parallel at distance a? Actually base Q has other side parallel to the lateral face? The prism extends along lateral edge direction? Wait if lateral face on sphere, the prism's "height" h is along lateral edges; extending prism along that direction? The opposite lateral face is at distance a? The volume = base area * h? No h is lateral edge length; if increase h, the opposite lateral face moves outward? But if lateral face on sphere fixed, increasing h (lateral edge) moves the opposite lateral face away? Actually h is length of lateral edge; the opposite lateral face is parallel to the inscribed lateral face but at distance h along normal? Wait lateral face plane is y=0; prism extends in direction perpendicular to base plane (which lies within lateral face? Actually lateral edge direction v is perpendicular to base plane; it lies in the lateral face plane? No, lateral face plane contains v and base edge u. The opposite lateral face is a parallel plane shifted by some distance in direction w? The prism has two bases separated by h in v direction; but the lateral face itself is spanned by u and v; its opposite lateral face is parallel to it but shifted by the base width perpendicular to u in base plane? Hmm confusing. Let's stick to coordinate proof for final. But perhaps final answer can simply present solution assuming the face with vertices on sphere is a base, because "prisma quadrangular reto" usually base = face? Not exactly, but many vestibular problems assume all vertices of a prism inscribed? Here only one face. The examiner may expect considering base as the face? Let's read statement: "Sabe-se que os quatro vértices de uma das faces do prisma pertencem à superfície da esfera S." It doesn't say base. If it were lateral, the base could be any quadrilateral; complicated. The problem likely from IME and intended more elaborate? Need solve correctly. Let's search memory: There is an IME 2027? Maybe impossible due current date 2026? Actually current date 2026, IME 2027 maybe future? But question maybe from "IME 2027" simulated? Anyway. Let's attempt a robust solution. We can set up general prism in sphere and derive max via known maximum volume of inscribed prism? Let's examine if the maximum volume of any prism contained in sphere with one face on sphere is indeed cube; we can prove by noting the prism can be associated with a rectangular box? Hmm. A right prism with quadrilateral base. Its volume = area base * height. The base is contained in intersection of sphere with base plane? No, only one face vertices on sphere, not base necessarily. Alternative: Use central symmetry? The maximum volume of a prism contained in a sphere likely occurs when the prism is inscribed (all vertices on sphere) and is a cube. This can be justified by: given any prism contained in sphere, reflect it through center and take convex hull? The convex hull of prism and its central reflection is centrally symmetric prism? Volume may increase? The maximum centrally symmetric body in sphere? The cube? There is theorem: maximum volume parallelepiped inscribed in sphere is cube; least surface area? But general prism? Maybe use "if one face is inscribed, the prism can be extended symmetrically to a double prism (prism + reflected) which is centrally symmetric and has volume 2V, contained in sphere? Then by known max volume of centrally symmetric prism (or box) in sphere, double volume ≤ 2*192√3? Let's see: Reflect prism through sphere center. The union of original and reflected may not be convex; its convex hull maybe a larger centrally symmetric body containing two prisms. Volume of convex hull? Not simple. But we can perhaps prove: For any prism contained in sphere with one face on sphere, its volume ≤ volume of a cube. Consider its "midplane" maybe. Let's derive via coordinates more generally for any face. Actually our coordinate optimization for lateral face already proves. We can present a unified coordinate proof? Too long. Maybe final answer can include a note: "Para efeito de maximização, podemos supor que a face em questão é uma base; se fosse lateral, basta repetir o argumento com a área máxima de um retângulo inscrito..." But not detailed. Given exam response, maybe acceptable to solve assuming base is the face? But if grader knows, might expect both? Let's see problem statement: "prisma quadrangular reto" in Portuguese: a "prisma quadrangular" has base a quadrilateral; "reto" means lateral edges perpendicular. The face with four vertices on sphere could be any; but perhaps because it's "completamente contido", if a lateral face with vertices on sphere, then the two bases would intersect sphere? Not necessarily. But maybe one can show that in maximal configuration it must be a base? Let's test: Our lateral face cube has all faces including lateral on sphere, so both. But if not cube, maybe the face on sphere could be lateral; does our optimization yield same max but with non-square base? For fixed h=4√3 and a=4√3, cube. So final dimensions same. The proof may be simpler if we assume base; result same. Let's attempt a concise rigorous solution for base case and then add: "O caso em que a face é lateral é análogo e conduz ao mesmo resultado." Maybe enough? But user wants solution; maybe not require exhaustive. Let's craft final answer: 1. Let a face (base) of the prism be a quadrilateral ABCD inscribed in circle C = S ∩ π, where π is its plane at distance d from O (sphere center). Let r = √(R²-d²). The area of any quadrilateral inscribed in a circle of radius r is at most 2r² (e.g., by Brahmagupta or standard: max area cyclic quadrilateral is square). So base area ≤ 2(R²-d²). 2. Since prism contained in sphere, if its height is h and base plane is at distance d from O, then the opposite base must lie within sphere; the largest h is 2d (when opposite base also touches sphere). Thus h ≤ 2d. Then V ≤ 2(R²-d²)*2d = 4R²d - 4d³. Let f(d)=4d(R²-d²), 0≤d≤R. f'(d)=4(R²-3d²)=0 => d=R/√3. Max f = 4*(R/√3)*(2R²/3)=8R³/(3√3). With R=6: V_max = 8*216/(3√3)=1728/(3√3)=576/√3=192√3 cm³. 3. Equality conditions: base area max square, r²=R²-d²=36-12=24 => r=2√6; square side = r√2 =? Wait square inscribed in circle radius r: diagonal =2r=4√6; side = diagonal/√2 =4√3. Height h=2d=4√3. So prism is cube side 4√3. Need ensure "altura e dimensões da base": altura = 4√3 cm; base square side 4√3 cm. But wait: In step 2, if the face on sphere is a base, the height h extends inward; max h=2d because opposite base also on sphere? Let's explain: Choosing coordinate with base plane z=d, prism extends to z=d-h. Since contained, for all points of top base (same circle radius r) distance squared ≤R² => r²+(d-h)² ≤ R² = r²+d² => |d-h| ≤ d => h≤2d. The max occurs at h=2d (or h=0 trivial). Good. If h=2d, top base plane z=-d; top vertices also on sphere? Actually top vertices coordinates same (x,y) with x²+y²=r², z=-d => distance squared r²+d²=R², yes. So all eight vertices on sphere. Nice. Need include "a face could be lateral" note. Let's append: "Se a face em questão for lateral, o mesmo tipo de argumento (área máxima de um quadrilátero inscrito em círculo e limitação da altura) conduz à mesma desigualdade; portanto não há perda de generalidade." But is that true? Our lateral optimization was more complex. Maybe not acceptable. Let's see if we can present a short argument for lateral face without full optimization. Maybe for lateral face, we can use: The lateral face is a rectangle of sides x (base edge) and h (height), with x²+h² = (2r)² = 4(R²-d²). The base area A of prism with one side x is at most? We found A_max = 54 + ... Not simple. But maybe there is a simpler bound: Volume = A h ≤ ? We can bound A by (x+? )? Hmm. Alternative: Use known inequality: For a prism with one lateral face rectangle inscribed in circle, its volume ≤ 4d(R²-d²) as well? Let's test: Our lateral max volume also 192√3, but our derivation involved A_max expression. Could we derive a simple upper bound V ≤ 4d(R²-d²)? For cube: d=2√3, 4d(R²-d²)=4*3.464*24=332.55 = V. So yes. Is it generally true? Let's check with lateral configuration: We need show V ≤ 4d(R²-d²) for some relation? But d in lateral case is distance from sphere center to lateral face plane, not base plane. R²-d² = r² = (a²+h²)/4. So 4d(R²-d²)= d(a²+h²). For cube d=2√3, a=h=4√3, d(a²+h²)=3.464*(48+48)=332.55 yes. But is V ≤ d(a²+h²)? Our V = A h; A ≈? For cube A=48, h=6.928 =>332.55. d(a²+h²)=3.464*96=332.55. Maybe true? Let's test with a=2, h maybe? Our formula A_max = 54 -3h²/8 +0.5 a d; V = h A_max. Need check if V ≤ d(a²+h²)? Probably not always? Let's test arbitrary a=4,h=6,R=6. Then s=16+36=52, d=sqrt(36-13)=√23≈4.796. A_max=54 -3*36/8 +0.5*4*4.796 =54 -13.5 +9.592=50.092; V=300.55. d(a²+h²)=4.796*52=249.4, V > bound. So not true. So base-face bound cannot directly apply. Maybe final answer can state: "A condição de que os quatro vértices de uma face pertencem à esfera é satisfeita pela configuração de volume máximo encontrada (um cubo). Como o cubo é inscrito na esfera, ele é o prisma de maior volume possível entre todos os prismas contidos na esfera (fato conhecido/justificado pela simetria)." But "fato conhecido" not ideal. Given exam, perhaps they expect cube and proof via symmetry. Let's search memory: IME problem maybe asks "Determine volume máximo" and answer 192√3. Many solutions use "max volume of inscribed cube" maybe. Let's attempt a rigorous but concise proof for any prism using "if a face is on sphere, the prism can be placed inside a rectangular box with same or larger volume"? Hmm. Maybe use the following theorem: Among all prisms contained in a sphere, the maximum volume is attained by a cube inscribed in the sphere. This can be proven by noting that the volume of a prism is area base * height, and both base area and height are maximized simultaneously? But not trivial. Given time, final answer with base-case proof and a note that the maximum occurs when all vertices are on sphere (cube) likely accepted. Let's craft final in Portuguese. But let's be more precise: Since problem says "uma das faces", perhaps the intended solution is to realize that the face with vertices on sphere can be taken as a base, because if it were lateral, the prism would not be "completamente contido" in maximal configuration? Actually cube satisfies both. Maybe they assume "face" = base? Many authors: "prisma quadrangular" has bases and lateral faces; "uma das faces" could be any, but often "face" includes lateral. However the phrase "os quatro vértices de uma das faces" if lateral would be rectangle; if base would be quadrilateral. Maybe the problem statement may have a diagram? The image only text. It doesn't specify. But "quadrangular" could mean base is a quadrilateral, not necessarily rectangle; if lateral face is rectangle, the base could be any quadrilateral, making problem too hard. So likely they mean a base? But in Portuguese "face" of prism: "bases" and "faces laterais"; "face" generic. Hmm. Let's see if a lateral face being inscribed in sphere imposes that the prism is actually a cube? No, our optimization allowed non-cube with lower volume. So not impossible. Maybe we should present both cases in final? But that would be long. Desired oververbosity 5; can include a paragraph "A face considerada pode ser uma base (o caso lateral é inteiramente análogo e leva à mesma limitação; por brevidade, detalhamos o caso da base)." But "análogo" may not be accepted if not shown. Let's derive a unified elegant proof using "the projection of the prism onto the face plane"? Maybe. Let's attempt to prove general upper bound V ≤ 192√3 for any right prism with one face on sphere. Use geometric inequality: volume of any prism ≤ volume of largest rectangular parallelepiped inscribed in sphere? But prism base not necessarily rectangle; however any quadrilateral base can be transformed? Not obvious. Maybe use "Among all quadrilaterals inscribed in a circle, square maximizes area; among all rectangular faces (a×h) inscribed in a circle, square maximizes area? Actually for lateral face, a²+h²=4r²; the volume depends on base area. Maybe for fixed lateral face rectangle area (or perimeter), max volume when base is square? Could be. Let's consider optimizing over all possible prisms via Lagrange with general base polygon. But too much. Given final answer format likely just result, maybe proof can be sketchy. Let's produce final answer: - Start: "Seja O o centro da esfera, R=6. Considere a face cujos vértices estão na esfera. Se essa face é uma base, seja d a distância de O ao plano da base e r=√(R²-d²) o raio da circunferência interseção. A base é um quadrilátero inscrito nessa circunferência; sua área máxima é 2r². Seja h a altura do prisma. Como o prisma está contido na esfera, a base oposta deve satisfazer r²+(d-h)²≤R², logo h≤2d. Portanto V≤2(R²-d²)·2d =4d(R²-d²). Maximizando f(d)=4d(R²-d²) em [0,R], obtemos d=R/√3, Vmax=8R³/(3√3)=192√3 cm³. Em igualdade, a base é um quadrado de lado 4√3 cm e altura 4√3 cm." But this only covers base case. Add: "Observação: se a face inicial for lateral, o mesmo raciocínio, aplicado ao retângulo inscrito na circunferência da face, conduz à mesma desigualdade final (pois o volume também é limitado por 4d(R²-d²)?)." But is it true? We saw not dimensionally with d of lateral. Let's not claim. Maybe we can state "A configuração de volume máximo encontrada é um cubo inscrito na esfera; como qualquer prisma contido na esfera tem volume no máximo o de um cubo inscrito (resultado clássico de otimização), o resultado segue." This is a known fact? Is it true? For rectangular box yes. For general prism maybe not? Let's test possible counterexample: A very flat prism with large base? The max area of quadrilateral inscribed in sphere cross-section maybe? Suppose base circle radius r=R (d=0), base max square area 2R²=72; height h=0? If h=2d=0, volume 0. If base at d=R/2, r²=27, area max 54, h=2d=R=6, volume 324? Wait R=6, d=3, h=6, volume=54*6=324 >192√3? Let's compute: R=6, d=3, r²=36-9=27, base max area=2r²=54, h≤2d=6 => V=324. But is such prism contained in sphere? Base at distance 3 from center, circle radius √27≈5.196; top base at z=-3 (h=2d), distance from center same 3; vertices distance sqrt(r²+9)=sqrt(27+9)=6=R, so both bases on sphere. Prism height 6, base square area 54, volume 324. But is this prism contained? Yes, all vertices on sphere? Both bases square inscribed in small circles at distance 3; prism is a right prism with square base oriented? It would have all 8 vertices on sphere? Wait top base vertices maybe not corresponding? They are on sphere (same circle at z=-3). But the prism lies between z=3 and z=-3; all points inside? The maximum distance from center for points in prism occurs at vertices? For convex prism, distance squared = x²+y²+z²; x²+y² maximum r²=27 at vertices of base (since square inscribed); |z| maximum max(3,3)=3? Actually z ranges -3 to 3, so |z|≤3, so distance²≤27+9=36=R². So yes contained, with all 8 vertices on sphere? Wait top base vertices have same x,y as bottom? If prism is right, top base is translation by vector (0,0,-6) from bottom? Then top base vertices have same x,y as bottom, z=-3, so distance²=r²+9=36, yes. They all lie on sphere. So the prism is inscribed in sphere (all 8 vertices on sphere) but not a cube: base square side? r=√27, side = r√2=√54=3√6? Diagonal=2r=2√27=6√3; side= (6√3)/√2=3√6 ≈7.348. Height=6. Volume = (3√6)² *6 =54*6=324 cm³. This is larger than 192√3≈332? Wait 192√3 ≈332.55, so 324 is less. Okay. But our f(d) max at d=R/√3≈3.464 gave 332.55 >324. So okay. Could there be a prism with base at d=3, h=6, volume 324 <332.55. So base case max fine. For lateral face, maybe maximum also cube. So final okay. Let's produce final answer with derivation for base case and note that the maximum configuration found is a cube; since any prism contained in sphere has volume ≤ that of inscribed cube (can be justified by the fact that the cube is the solution of the same optimization when the face is any face; the other case is analogous). Maybe acceptable. Let's craft final in Portuguese, with equations. Potential final: ``` Seja R=6. Considere a face cujos quatro vértices estão em S. Inicialmente tome-a como uma base do prisma (se for lateral, o argumento é análogo). Seja d a distância do centro O da esfera ao plano dessa face e r=√(36-d²) o raio da circunferência interseção. A base é um quadrilátero convexo inscrito nessa circunferência; sua área é máxima quando é um quadrado, valendo 2r²=2(36-d²). Se h é a altura do prisma, a base oposta está no plano z=d-h (adotando o plano da face em z=d e o centro em z=0). Como o prisma está contido na esfera, para os vértices da base oposta (mesma circunferência) devemos ter r²+(d-h)² ≤ R² = r²+d², logo |d-h|≤d, isto é, 0≤h≤2d. Assim, V ≤ 2(36-d²) * 2d = 4d(36-d²). A função f(d)=4d(36-d²), 0≤d≤6, tem máximo quando f'(d)=4(36-3d²)=0 ⇒ d=√12=2√3. Logo Vmax = 4*(2√3)*(36-12)=8√3*24=192√3 cm³. Na igualdade: h=2d=4√3 cm e a base é um quadrado de área 2r²=2(36-12)=48 cm², logo lado = √48=4√3 cm. ``` Need compute f max correctly: f(2√3)=4*2√3*(36-12)=8√3*24=192√3 yes. Then item b: height = 4√3 cm; base = square of side 4√3 cm. Add note: "Como a solução ótima é um cubo (todas as arestas 4√3), se a face considerada for lateral o resultado é o mesmo." Good. Let's finalize.

**Exploring Maximum Volume** I'm currently analyzing the geometric constraints presented. My focus is on determining how the dimensions of a quadrangular prism, inscribed within a sphere of radius 6 cm, impact its volume. Specifically, I'm investigating the scenario where one face's vertices lie on the sphere's surface to find the maximum possible volume. **Investigating Prism Shape** I'm clarifying the definition of a "right quadrangular prism" in this context. I need to determine if "quadrangular" implies a square base, or if any quadrilateral is permissible, especially given the constraint of one face having vertices on the sphere's surface. This distinction is crucial for proceeding. **Defining Prism Vertices** I'm clarifying the vertex placement. While symmetry suggests a rectangular base for maximum volume, the wording "completely contained" and only one face's vertices necessarily on the sphere's surface implies the other four vertices are simply *within* or on the sphere, not necessarily all on the surface. **Refining Prism Constraints** I'm re-evaluating the core constraint that only one face's vertices lie on the sphere's surface. If the prism is fully contained and a specific face's vertices are on the sphere, this implies a deviation from the maximally inscribed cube. I'm questioning if this constraint inherently limits the volume compared to a fully inscribed prism. **Clarifying Volume Logic** My thoughts are now focused on the implication that any prism contained within the sphere will have a volume less than or equal to the maximum inscribed prism. I'm reconsidering the cube inscribed in the sphere, as all its vertices are on the surface, which means any face's vertices are also on the sphere. This makes me question the unique constraint in the problem statement. **Analyzing Face Vertex Condition** I'm reconsidering the cube inscribed within the sphere. It fulfills the condition that all four vertices of any given face lie on the sphere's surface. This makes me question if the problem implies a scenario beyond simply the maximal inscribed cube. **Investigating Examiner Intent** I'm considering why the examiner would phrase the problem this way. My current hypothesis is they envisioned a specific configuration where a face's vertices lie on the sphere, and the prism extends from this face. I'm now exploring how far this prism can extend while remaining within the sphere. **Analyzing Vertex Location** I'm processing the implications of a face's vertices being precisely on the sphere. This means those four points lie on a circle within a specific plane. For a right prism, the opposite face's vertices share x-y coordinates but are at a different z-level. Their distance to the sphere's center must also be less than or equal to the radius. **Calculating Prism Height** I'm analyzing the case where one face's vertices lie on the sphere. For a right prism, this implies the opposite vertices must be within or on the sphere, leading to a maximum height condition that forces all vertices onto the sphere, effectively making it a fully inscribed prism. **Deriving Maximum Height** I'm focusing on the constraint of one face's vertices being on the sphere. This leads to a specific height calculation where all vertices must lie on the sphere's surface for maximum volume, effectively making it a fully inscribed prism. **Revising Volume Assumptions** I'm now realizing that a square base assumption is too restrictive. The problem implies a rectangular base for maximum volume, where all eight vertices must lie on the sphere's surface, making it a fully inscribed prism, not just a specific face. **Revising Geometric Focus** I'm now shifting my analysis. The crucial detail isn't a specific face being on the sphere, but rather the implication that only *one* face's vertices are necessarily on the sphere, suggesting the prism is not a fully inscribed cube. My current focus is on how this partial constraint affects the possible dimensions and volume. **Exploring Prism Orientation** I am now considering how the orientation of the prism within the sphere impacts the condition of one face's vertices lying on the surface. My thoughts are on the implications of this face being a base versus a lateral face. **Confirming Sphere Inclusion** I'm confirming that a prism is fully contained within a sphere if and only if all eight of its vertices lie on or within the sphere. This fundamental geometric principle means the maximum volume prism within the sphere will indeed have all its vertices on the spherical surface. **Revising Maximum Volume Logic** I'm realizing that the maximum volume for any quadrangular prism inscribed within a sphere is achieved when it's a cube. The condition of only one face's vertices being on the sphere's surface is actually satisfied by the fully inscribed cube. Therefore, the maximum volume is indeed that of the inscribed cube. **Confirming Inscribed Cube** My analysis has confirmed that for any quadrangular prism inscribed within a sphere, the maximum volume is achieved by a cube. The constraint of one face's vertices on the sphere is inherently satisfied by a fully inscribed cube. **Questioning Problem Phrasing** I am questioning the specific phrasing regarding one face's vertices lying on the sphere. If a prism is fully contained, the maximal volume is achieved by a cube, where all vertices are on the sphere, thus satisfying this condition. I am trying to understand the examiner's intent behind this specific wording. **Questioning The Setup** I am now questioning the interpretation of "completely contained" and the implication that the prism might not be fully inscribed. The phrasing about one face's vertices on the sphere's surface is leading me to reconsider how the prism might be positioned, and if it could extend beyond a hemisphere. **Revising Question Origin** I'm now shifting focus from purely geometric interpretation to the context of the problem's origin. I've identified stylistic elements in the provided text strongly suggesting it's from a specific IME exam, likely a recent one. This understanding may inform the intent behind the phrasing. **Investigating Examiner Intent** I am now considering why the examiner specifically mentioned only one face's vertices on the sphere. My hypothesis is they intended for the prism not to be fully inscribed, and I'm exploring how this partial constraint impacts the maximum possible volume. **Confirming Max Volume** My thoughts have solidified: the maximum volume is indeed achieved by the fully inscribed cube. This shape inherently satisfies the condition that one face's vertices lie on the sphere's surface, and no other prism contained within the sphere can exceed this volume. **Exploring Centered Prisms** I'm reconsidering the standard approach. My initial thoughts focused on prisms centered at the origin, leading to the inscribed cube. However, the prompt's phrasing about one face's vertices on the sphere suggests a different configuration. I am now analyzing prisms that are not necessarily centered. **Analyzing Off-Center Prisms** My thinking has shifted to the implications of a prism not being centered within the sphere. If one face has all its vertices on the sphere's surface, those vertices must lie on a circle. I'm now calculating the maximum area for such a face, assuming it's a square inscribed within that circle. **Investigating Face Geometry** I'm focusing on the geometry of the face with vertices on the sphere. If these four vertices lie on the sphere, they form a circle. For maximum area, this face must be a square inscribed in that circle. I'm calculating the dimensions of this square. **Calculating Face Area** My current thoughts are on determining the area of the square face whose vertices lie on the sphere's surface. This square is inscribed within a circle on the sphere's surface. I'm calculating the side length of this square to be $R\sqrt{2}$. **Considering Prism Height** With the base face determined, I'm now analyzing the height of the prism. Since it's a right prism, the opposite face will be parallel. I need to determine the maximum possible height while ensuring the entire prism remains within the sphere. **Determining Prism Position** I'm exploring the positioning of the prism. The face on the sphere's surface is at a specific height. The opposite face's vertices must be within or on the sphere. I'm calculating the maximum height based on this constraint. **Revising Volume Calculations** My thoughts are now on calculating the volume of this off-center prism. I have the dimensions of the square base and the derived maximum height. I'm combining these to determine the prism's volume under this new assumption. **Comparing Volumes** I am now comparing the volume of this off-center prism with the volume of the inscribed cube. My goal is to ascertain if this alternative configuration yields a larger volume, thereby addressing the examiner's specific wording. **Revising Geometric Configuration** I'm reassessing the prism's orientation. My previous calculations assumed the face with vertices on the sphere was a base. I'm now considering the possibility that this face is lateral, and the prism's orientation is not as initially assumed, potentially altering the volume calculation. **Questioning Base Shape** I am questioning the assumption that the base must be rectangular. While the lateral faces of a right prism are rectangular, the base can be any quadrilateral. My current focus is on whether a non-rectangular base could yield a larger volume, or if the maximum area for vertices on a circle still implies a square. **Confirming Axial Symmetry** My current thoughts confirm that the axis of the prism must pass through the sphere's center. This is due to the perpendicular relationship between the base and the direction connecting the sphere's center to the center of the circular section formed by the face on the sphere's surface. This validates my prior assumption of exact symmetry.

**Analyzing Geometric Constraints** I am currently focused on understanding the geometric relationship between the quadrangular prism and the sphere. My immediate task is to fully grasp how the prism is inscribed within the sphere and the specific condition that its vertices lie on the sphere's surface. This will be crucial for formulating the subsequent steps. **Clarifying Geometric Possibilities** I've been digging into the problem's phrasing in Portuguese. Specifically, the mention of "uma das faces do prisma" (one of the prism's faces) lying on the sphere's surface is key. My current thought is that this face must be one of the bases, as the lateral faces are rectangles and thus always inscribable in a circle. This has significant implications for the geometry. **Confirming Optimal Configuration** My latest thinking confirms that for maximum volume, a right quadrangular prism inside a sphere will indeed have all eight vertices on the sphere. This implies both bases are cyclic quadrilaterals inscribed within circles on the sphere. The optimal shape for these bases, to maximize area, is a square. **Investigating Ambiguity** I'm now revisiting the problem's wording concerning "uma das faces do prisma." The phrasing suggests not all vertices are necessarily on the sphere, which contradicts my previous assumptions of a fully inscribed prism. I need to explore how a face lying on the sphere impacts the prism's configuration and volume calculation. **Considering Lateral Faces** My latest thought is that if a lateral face of the prism is on the sphere, its vertices are always concyclic. This means the other two vertices of that base, and their counterparts, only need to be inside or on the sphere, not necessarily on the surface itself. This opens up new configuration possibilities. **Refining Containment Conditions** I've realized a critical detail: for a right prism, the base quadrilateral $Q$ must be contained within a circle whose radius depends on the sphere's radius and the z-coordinates of the bases. This implies $Q$ must fit within a disk of radius $r = \sqrt{R^2 - \max(z_1^2, z_2^2)}$. **Deriving Prism Dimensions** My current focus is on understanding how the condition of a base face's vertices lying on the sphere dictates the prism's dimensions. For a base at $z = z_1$, all four vertices must lie on the sphere's intersection circle. This implies the other base at $z = z_2$ must be contained within the sphere. For maximum height, I'm exploring the case where the bases are equidistant from the sphere's center. **Exploring Lateral Face Configurations** I'm now considering the scenario where a lateral face lies on the sphere. This implies its vertices are concyclic, and crucially, that the two bases must be equidistant from the sphere's center, meaning $z_1 = -z_2$. The bases are then contained within a disk of radius $r = \sqrt{R^2 - (h/2)^2}$. **Analyzing Face-Sphere Intersection** I'm currently trying to understand why the problem statement specifically mentions one face's vertices lying on the sphere's surface. It seems redundant if the goal is maximum volume, as an inscribed prism with all vertices on the sphere would naturally satisfy this. I'm exploring if this detail implies a specific configuration or is meant to guide the solver towards a particular proof. **Considering Base Planes** I am now considering the implications if a base plane passes through the sphere's center. My current thought is that this configuration is unlikely to maximize prism volume, as it limits the achievable height. I'm focusing on how the condition of four vertices of a face on the sphere’s surface guides the placement of these base planes. **Proving Square Base** My current thinking is that the phrasing "as dimensões da base" suggests the base might not be a square initially, and I need to prove it is. This means the student must demonstrate that the optimal base shape for maximum volume within a sphere is indeed a square, not just assume it. **Exploring Base Geometry** I am now focusing on the area of a general quadrilateral inscribed in a circle of radius $r$. My current thought is that by using central angles and the sine function, I can demonstrate that the maximum area for such a quadrilateral occurs when it is a square. **Proving Square Base** My current thinking is that the phrasing "as dimensões da base" suggests the base might not be a square initially, and I need to prove it is. This means I must demonstrate that the optimal base shape for maximum volume within a sphere is indeed a square, not just assume it. The diagonal proof is particularly elegant for this. **Maximizing Prism Area** My current thoughts are focused on the area of a general quadrilateral inscribed within a circle of radius $r$. I'm using central angles and the sine function to show that the maximum area for such a quadrilateral is achieved when it's a square. This seems crucial for the overall volume maximization. **Deriving Optimal Height** I've successfully derived the optimal height for maximum prism volume within the sphere. Both calculus and AM-GM inequality confirm that $h = \frac{2\sqrt{3}}{3}R$. For $R=6$, this yields $h = 4\sqrt{3}$ cm. **Confirming Cube Shape** My latest calculations confirm that the prism of maximum volume inscribed in the sphere is indeed a cube. The side length of this cube is $4\sqrt{3}$ cm, and its space diagonal matches the sphere's diameter, as expected. **Revising Volume Calculations** I'm re-evaluating my volume calculations, specifically the arithmetic for the cube and the derivative method. I've identified an error in my previous AM-GM application and a mismatch in the derivative result. My current focus is on reconciling these discrepancies and confirming the correct maximum volume. **Revising Containment Interpretation** I'm reassessing the exact meaning of "completely contained within a sphere" versus "inscribed." My current thought is that the phrasing about one face having vertices on the sphere is key, possibly implying not all vertices are necessarily on the sphere's surface, which requires deeper analysis. **Exploring Base Configurations** My latest thinking is that the problem statement implies the base vertices lie on a *spherical section*, not necessarily the sphere itself. I'm now considering how placing a face on the sphere's surface constrains the prism's dimensions and how this differs from a fully inscribed prism. **Confirming Maximization Strategy** I've realized that the problem's conditions strongly suggest the maximum volume occurs when all vertices of the prism lie on the sphere. My current focus is rigorously demonstrating this for a right quadrangular prism, showing that both bases must be equidistant from the sphere's center and their vertices must lie on the sphere. This will set up the optimal configuration for volume calculation. **Exploring Diagonal Lengths** I am now examining the lengths of the diagonals of the quadrilateral base. My current thinking is to use these diagonal lengths and the angle between them to express the area of the base, which is a crucial step towards maximizing the prism's volume. **Maximizing Quadrilateral Area** I'm analyzing how to maximize the area of a quadrilateral inscribed within a circle. My current focus is on using the lengths and the angle between its diagonals to show that the maximum area is achieved when the quadrilateral is a square. This directly impacts the base of the prism.