1983 AIME Problems/Problem 11

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Problem

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

size(180);import three; pathpen = black+linewidth(0.65); pointpen = black;currentprojection = perspective(30,-20,10);real s =...

Solution 1

First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\sqrt{2}$ . We apply the Pythagorean Theorem to find that the height is equal to $6$ .

size(180);import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);cu...

Next, we complete the figure into a triangular prism, and find the area, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$ .

Now, we subtract off the two extra pyramids that we included, whose combined area is $2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$ .

Thus, our answer is $432-144=\boxed{288}$ .

Solution 2

Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . now, we have a regular tetrahedron $EFGH$ , which has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$ . Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S^3}{12}$ , where S is the side length of the tetrahedron, the volume of our original solid is:

$V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}$

1983 AIME (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1983 AIME Problems/Problem 11

From AoPSWiki

Contents

Problem

Solution 1

Solution 2

See also