2011 AMC 12B Problems/Problem 18

Forgive me if this isn't a great solution, it's my first one.

We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths $\frac{1}{2}, 1$ and $\frac{\sqrt{3}}{2}$ .

Next, take a cross-section of the pyramid, forming a triangle with the top of the triangle and the midpoints of two opposite sides of the square base.

This triangle is isosceles with a base of 1 and two sides of length $\frac{\sqrt{3}}{2}$ .

The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides $\frac{1}{2}, \frac{\sqrt2}{2}$ and $\frac{\sqrt{3}}{2}$ .

The cube, touching all four triangular faces, will form a similar pyramid which sits on top of the cube. If the cube has side length $x$ , the pyramid has side length $\frac{x\sqrt{2}}{2}$ .

Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid.

$x$ + $\frac{x\sqrt{2}}{2}$ = $\frac{\sqrt2}{2}$ .

$x\left(1+\frac{\sqrt{2}}{2} \right)$ = $\frac{\sqrt{2}}{2}$

$x\left(2+\sqrt{2}\right)$ = $\sqrt{2}$

$x$ = $\frac{\sqrt{2}}{2+\sqrt{2}}$ $\cdot$ $\frac{2-\sqrt{2}}{2-\sqrt{2}}$ = $\frac{2\sqrt{2}-2}{4-2}$ = $\sqrt{2}-1}$ (Error compiling LaTeX. Unknown error_msg) = side length of cube.

$\left(\sqrt{2}-1}\right)^3$ (Error compiling LaTeX. Unknown error_msg) = $(\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3$ = $2\sqrt{2} - 6 +3\sqrt{2} - 1$ = $5\sqrt{2} - 7$

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2011 AMC 12B Problems/Problem 18