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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 triangular prism is $( 6\sqrt{2} )^2 \cdot 6 \cdot 1/2 = 216$ 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 regular tetrahedron. This tetrahedron has sidelength of $6\sqrt{2}$ and since the formula for the volume of a regular tetrahedron is $\frac{s^3}{6\sqrt{2}}$ the volume of this tetrahedron is $72$. $216 + 72 = \boxed{288}$.

~skibidi solver

See Also

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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