Factorize the Box

Relevant wiki: Sophie Germain Identity

The base area of the cuboid = $\sqrt{((x-y)^2 + y^2)((x+y)^2 + y^2)}$

By using Sophie Germain Identity , we can apply:

$x^4 + 4y^4 = ((x-y)^2 + y^2)((x+y)^2 + y^2)$

Hence, the base area = $\sqrt{x^4 + 4y^4}$

Now since the triangle $ABC$ is a right triangle, and let $BD$ be the red height, $\dfrac{x-y}{y} = \dfrac{y}{x+y}$ because of similarity.

Then, $x^2 - y^2 = y^2$ .

$x^2 = 2y^2$ or $x = y\sqrt{2}$ .

Therefore, the base area = $\sqrt{x^4 + 4y^4} = \sqrt{4y^4 + 4y^4} = (2\sqrt{2})y^2$ .

Then, the cuboid's volume = $(2\sqrt{2})y^{2}x = 4y^3 = 108$ .

As a result, $y^3 = 27$ . $y = \boxed{3}$ .

I ended up solving it without applying the identity in question. I was actually confused the entire time as I was progressing and just waiting to see when I would need the identity but I never did. Anyways, let's start. Let's say that the length of the segment AB is a and that the length of the segment BC is b. Then because the diagonal is 2x we know that $a^2 + b^2 = 4x^2$ .

Now focusing on the triangle ABC we find that $y^2 + (x-y)^2 = a^2$ and $y^2 + (x+y)^2 = b^2$ . Now comes the crucial step. Let's add both sides to get $y^2 + (x-y)^2 + y^2 + (x+y)^2 = a^2 + b^2$ . We can now simplify the left side of this equality using simple algebra to get $4y^2 + 2x^2$ . For the right side, we already know that it is equal to $4x^2$ so we have that $4y^2 + 2x^2 = 4x^2$ which yields $2y^2 = x^2$ and therefore $y = \frac{x}{\sqrt{2}}$ . This is half the puzzle. Now to conclude I will rewrite the volume of the cube. We know that the volume of the cube is $xab$ but this is simply x times the area of the top face. And the top face is simply two times the area of the triangle ABC. So let's compute the area of the triangle ABC: $\frac{2xy}{2} = xy = x \frac{x}{\sqrt{2}} = \frac{x^2}{\sqrt{2}}$ . And the top face, which is two times this, is equal to $\sqrt{2}x^2$ . Therefore the volume of the cube is $\sqrt{2}x^3$ and to conclude we have the equality $\sqrt{2}x^3 = 108$ which implies $x^3 = \frac{108}{\sqrt{2}}$ .

If we now cube the expression we had for y we get that $y^3 = \frac{x^3}{2 \sqrt{2}}$ . And plugging our value for x gives us $y^3 = \frac{108}{4} = 27$ and thus $y = 3$ .

Using the 2nd Euclid's Theorem (x-y)(x+y)=y^2 Thus x=y sqrt2 Volume=x 2x y sqrt2 Substituting 108=4y^3--->y=3

The area of the triangle is $\frac{1}{2}×2x×y=xy$ .

Note that the cuboid has a base twice this area $2xy$ , height $x$ and hence volume $V=2x^2y$ .

Now study the triangle a bit closer.

Let's say D is the point where the perpendicular meets the line AC, and E is the midpoint of AC. Drawing a circle with center E and radius x, B must lie on the circle, since ABC is a right angled triangle. The right angled triangle EDB has two sides y and hypothenuse x so that $x= y\sqrt{2}$ , and we can write $V=108=2(y\sqrt{2})^2y=4y^3 \Rightarrow y=\sqrt[3]{\frac{108}{4}}=\boxed{3}$ .