2007 AMC 12A Problems/Problem 20

From AoPSWiki

Problem

Corners are sliced off a unit cube so that the six faces each become regular octagons. What is the total volume of the removed tetrahedra?

$\mathrm{(A)}\ \frac{5\sqrt{2}-7}{3}\qquad \mathrm{(B)}\ \frac{10-7\sqrt{2}}{3}\qquad \mathrm{(C)}\ \frac{3-2\sqrt{2}}{3}\qquad \mathrm{(D)}\ \frac{8\sqrt{2}-11}{3}\qquad \mathrm{(E)}\ \frac{6-4\sqrt{2}}{3}$

Solution

Since the sides of a regular polygon are equal in length, we can call each side $x$ . Examine one edge of the unit cube: each contains two slanted diagonal edges of an octagon and one straight edge. The diagonal edges form $45-45-90 \triangle$ right triangles, making the distance on the edge of the cube $\frac{x}{\sqrt{2}}$ . Thus, $2 \cdot \frac{x}{\sqrt{2}} + x = 1$ , and $x = \frac{1}{\sqrt{2} + 1} \cdot \left(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right) = \sqrt{2} - 1$ .

Each of the cut off corners is a pyramid, whose volume can be calculated by $V = \frac 13 Bh$ . Use the base as one of the three congruent isosceles triangles, with the height being one of the edges of the pyramid that sits on the edges of the cube. The height is $\frac x{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$ . The base is a $45-45-90 \triangle$ with leg of length $1 - \frac{1}{\sqrt{2}}$ , making its area $\frac 12\left(1 - \frac 1{\sqrt{2}}\right)^2 = \frac{3 - 2\sqrt{2}}4$ . Plugging this in, we get that the area of one of the tetrahedra is $\frac 13 \left(1 - \frac{1}{\sqrt{2}}\right)\left(\frac{3 - 2\sqrt{2}}4\right) = \frac{10 - 7\sqrt{2}}{24}$ . Since there are 8 removed corners, we get an answer of $\frac{10 - 7\sqrt{2}}{3} \Longrightarrow \mathrm{B}$

2007 AMC 12A (Problems)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Personal tools

Navigation

Search

Toolbox

2007 AMC 12A Problems/Problem 20

From AoPSWiki

Problem

Solution

See also