1983 AIME Problems/Problem 11

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?

$[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label("A",A,W); label("B",B,S); label("C",C,SE); label("D",D,NE); label("E",E,N); label("F",F,N); [/asy]$

Solutions

Solution 1

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

$[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label("A",A,S); label("B",B,S); label("C",C,S); label("D",D,NE); label("E",E,N); label("F",F,N); label("$12\sqrt{2}$",(E+F)/2,N); label("$6\sqrt{2}$",(A+B)/2,S); label("6",(3*s/2,s/2,3),ENE); [/asy]$

Next, we complete t he figure into a triangular prism, and find its volume, 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 volume 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

$[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label("A",A,(-1,-1,0)); label("B",B,( 2,-1,0)); label("C",C,( 1, 1,0)); label("D",D,(-1, 1,0)); label("E",E,(0,0,1)); label("F",F,(0,0,1)); label("G",G,(0,0,-1)); label("H",H,(0,0,-1)); [/asy]$

Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . Now, we have a regular tetrahedron $EFGH$ , which by symmetry 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:

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

Solution 3

We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$ ; thus, we will integrate with respect to height from $0$ to $6$ , noting that each cross section of height $dh$ is a rectangle. The volume is then $\int_0^h(wl) \ \text{d}h$ , where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\sqrt{2}-\sqrt{2}h$ since it decreases linearly with respect to $h$ , and $l=6\sqrt{2}+\sqrt{2}h$ since it similarly increases linearly with respect to $h$ . Now we solve: $\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}$ .

Solution 4

Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$ , and that little portion that hangs out has a length of $3\sqrt2$ . This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$ , and a height of $3\sqrt{2}$ . Since there are two of these, we can compute the sum of the volumes of these two to be $72$ . Now we are left with a triangular prism with a base of dimensions $3\sqrt6, 3\sqrt6, 3\sqrt2$ and a height of $6\sqrt2$ . We can compute the volume of this to be 216, and thus our answer is $\boxed{288}$ .

pi_is_3.141

Solution 5

From solution 1, the height of the solid is 6. Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this rectangular prism is ( 6*sqrt(2) )^2 * 6 * 1/2 = 216. Notice that the solid is symetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a tetrahedron. This tetrahedron has sidelength of 6sqrt(2) and since the formula for the volume of a tetrahedron is s^3/6sqrt(2) the volume of this tetrahedron is 72. 216 + 72 = \boxed{288}.From solution 1, the height of the solid is 6. Construct a triangular prism with base $ ABCD $, and with the height of the solid. The volume of this rectangular prism is \[ \left( 6\sqrt{2} \right)^2 \times 6 \times \frac{1}{2} = 216. \] Notice that the solid is symmetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a tetrahedron. This tetrahedron has a side length of $ 6\sqrt{2} $ and since the formula for the volume of a tetrahedron is \[ \frac{s^3}{6\sqrt{2}}, \] the volume of this tetrahedron is \[ \frac{(6\sqrt{2})^3}{6\sqrt{2}} = 72. \] Thus, the total volume is \[ 216 + 72 = \boxed{288}. \]

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

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