1983 AIME Problems/Problem 11
Contents
Problem
The solid shown has a square base of side length . The upper edge is parallel to the base and has length
. All other edges have length
. Given that
, 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); label("B",B); label("C",C); label("D",D); label("E",E,N); label("F",F,N); [/asy]](http://latex.artofproblemsolving.com/1/b/f/1bf705e2ad6de8631e5bcc2d287c3fe2ffa8c74e.png)
Solution 1
First, we find the height of the figure by drawing a perpendicular from the midpoint of to
. The hypotenuse of the triangle is the median of equilateral triangle
one of the legs is
. We apply the Pythagorean Theorem to find that the height is equal to
.
![[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); label("B",B); label("C",C); label("D",D); label("E",E,N); label("F",F,N); label("$12\sqrt{2}$",(E+F)/2,N); label("$6\sqrt{2}$",(A+B)/2); label("6",(3*s/2,s/2,3),ENE); [/asy]](http://latex.artofproblemsolving.com/6/7/b/67be1761d8d554b812d62abea793e1f2413f27d9.png)
Next, we complete the figure into a triangular prism, and find the area, which is .
Now, we subtract off the two extra pyramids that we included, whose combined area is .
Thus, our answer is .
Solution 2
Extend and
to meet at
, and
and
to meet at
. now, we have a regular tetrahedron
, which has twice the volume of our original solid. This tetrahedron has side length
. Using the formula for the volume of a regular tetrahedron, which is
, where S is the side length of the tetrahedron, the volume of our original solid is:
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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