# Answer: 128

## Solution:

Let our cubes be $A B C D A_{1} B_{1} C_{1} D_{1}$ and $A B C D A_{2} B_{2} C_{2} D_{2}$ with a common face $A B C D$. Let the triangular section of the first cube be $K L M$, where point $K$ lies on $A A_{1}$, point $L$ on $A B$, and point $M$ on $A D$. One side of the quadrilateral section of the second cube is the segment $L M$. The other two sides are the extensions of segments $K L$ and $K M$ to the faces of the second cube, which we will call segments $L P$ and $M Q$. For the section to be a quadrilateral, points $P$ and $Q$ must lie on the same face of the second cube, which can only be the face $A_{2} B_{2} C_{5} D_{2}$.

Thus, the quadrilateral section of the second cube is a trapezoid $L M Q P$. Finding its maximum area is equivalent to finding the maximum area of triangle $K P Q$, which is similar to triangle $K L M$. Let the similarity coefficient be $k=\frac{K P}{K L}$. Then $k^{2}=\frac{S_{K P Q}}{S_{K L M}}=\frac{S_{K P Q}}{S}$. Therefore, our problem is equivalent to finding the maximum similarity coefficient.

On the other hand, by Thales' theorem, $k=\frac{K P}{K L}=\frac{K A_{2}}{K A}=\frac{K A+A A_{2}}{K A}=1+\frac{A A_{2}}{K A}$. Thus, the similarity coefficient is greater the smaller $K A$ is, which means our task is to minimize $K A$, or equivalently, to minimize $K A_{2}$.

Suppose we have a triangle with vertices located on three edges of the cube, at distances $x, y$, and $z$. Let's find the formula for the area of this triangle. This can be done in different ways, for example, using the vector product, or by calculating the area of the tetrahedron formed by the vertices of the triangle and the vertex of the cube in two ways, but we will compute this area using Heron's formula, knowing the sides of the triangle: $a=\sqrt{x^{2}+y^{2}}, b=\sqrt{x^{2}+z^{2}}$, and $c=\sqrt{y^{2}+z^{2}}$.

$$
\begin{aligned}
& S=\frac{\sqrt{(a+b+c)(a+b-c)(c+(b-a))(c-(b-a))}}{4}=\frac{\sqrt{\left((a+b)^{2}-c^{2}\right)\left(c^{2}-(b-a)^{2}\right)}}{4}= \\
= & \frac{\sqrt{\left(a^{2}+b^{2}+c^{2}-2 a b\right)\left(c^{2}-a^{2}-b^{2}+c^{2}+2 a b\right)}}{4}=\underbrace{\frac{\sqrt{4 a^{2} b^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}}}{4}}_{\text {This formula is also sometimes called Heron's formula }}=
\end{aligned}
$$

$$
\begin{gathered}
=\frac{\sqrt{4\left(x^{2}+y^{2}\right)\left(x^{2}+z^{2}\right)-\left(\left(x^{2}+y^{2}\right)+\left(x^{2}+z^{2}\right)-\left(y^{2}+z^{2}\right)\right)^{2}}}{4}=\frac{\sqrt{4 x^{4}+4 x^{2} y^{2}+4 x^{2} z^{2}+4 y^{2} z^{2}-\left(2 x^{2}\right)^{2}}}{4}= \\
=\frac{\sqrt{x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}}}{2}
\end{gathered}
$$

Let's look at this formula for triangle $K P Q$ and segments $x=A_{2} P, y=A_{2} Q, z=A_{2} K$. On the one hand, we need to minimize $z$, and on the other hand, maximize the area. Clearly, for this, $x$ and $y$ should be as large as possible, that is, equal to the edge $\ell$.

As we know, $k=\frac{K A+A A_{2}}{K A}=\frac{K A+\ell}{K A}$, so $K A \cdot k=K A+\ell$, from which $K A=\frac{\ell}{k-1}$, $K A_{2}=K A+\ell=\frac{k \ell}{k-1}$.

Substituting these values into the formula, we get:

$S_{K P Q}=\frac{1}{2} \sqrt{\ell^{4}+\ell^{2} \cdot\left(\frac{k \ell}{k-1}\right)^{2}+\ell^{2} \cdot\left(\frac{k \ell}{k-1}\right)^{2}}=\frac{\ell^{2}}{2(k-1)} \sqrt{(k-1)^{2}+2 k^{2}}$.

Accordingly, $S=\frac{S_{K P Q}}{k^{2}}=\frac{\ell^{2} \sqrt{(k-1)^{2}+2 k^{2}}}{2 k^{2}(k-1)}$, from which $\frac{4 S^{2}}{\ell^{4}}=\frac{(k-1)^{2}+2 k^{2}}{k^{4}(k-1)^{2}}=\frac{1}{k^{4}}+\frac{2}{k^{2}(k-1)^{2}}$. The right-hand side of this equation decreases for $k>1$, so the equation has no more than one solution. The specific solution is often easily found from this equation, as it is usually an integer.

For example, in the first variant, we get the equation $\frac{1}{k^{4}}+\frac{2}{k^{2}(k-1)^{2}}=\frac{4 \cdot 16^{2}}{12^{4} \cdot \frac{8}{11}}=\frac{11}{2 \cdot 3^{4}}$, from which it is immediately desirable to check $k=3$, which turns out to be correct.

The answer is the difference in the areas of the two triangles and is equal to $\left(k^{2}-1\right) S$.

| $S$ | 16 |
| :---: | :---: |
| $\ell$ | $12 \sqrt[4]{\frac{8}{11}}$ |
| $k$ | 3 |
| Answer | 128 |
# Problem 7. (4 points)

Two cubes with an edge of $12 \sqrt[4]{\frac{8}{11}}$ share a common face. A section of one of these cubes by a certain plane is a triangle with an area of 16. The section of the other by the same plane is a quadrilateral. What is the maximum value that its area can take?