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,W); label("B",B,S); label("C",C,SE); label("D",D,NE); label("E",E,N); label("F",F,N); [/asy]](http://latex.artofproblemsolving.com/2/1/9/2192c6ce4b71d13bb3e1bc8bdfb68314b8de73f3.png)
Solutions
Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of to
. The hypotenuse of the triangle formed is the median of equilateral triangle
, and one of the legs is
. We apply the Pythagorean Theorem to deduce that the height is
.
![[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]](http://latex.artofproblemsolving.com/a/0/4/a0406c7f42ae451f7ee423c91bd5b18d4cd6b556.png)
Next, we complete t he figure into a triangular prism, and find its volume, which is .
Now, we subtract off the two extra pyramids that we included, whose combined volume is .
Thus, our answer is .
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]](http://latex.artofproblemsolving.com/c/0/d/c0daaf96bd5d22d9e77dbca583857221ea1bdb77.png)
Extend and
to meet at
, and
and
to meet at
. Now, we have a regular tetrahedron
, which by symmetry 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:
.
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 ; thus, we will integrate with respect to height from
to
, noting that each cross section of height
is a rectangle. The volume is then
, where
is the width of the rectangle and
is the length. We can express
in terms of
as
since it decreases linearly with respect to
, and
since it similarly increases linearly with respect to
. Now we solve:
.
Solution 4
Draw an altitude from a vertex of the square base to the top edge. By using triangle ratios, we obtain that the altitude has a length of
, and that little portion that hangs out has a length of
. This is a triangular pyramid with a base of
, and a height of
. Since there are two of these, we can compute the sum of the volumes of these two to be
. Now we are left with a triangular prism with a base of dimensions
and a height of
. We can compute the volume of this to be 216, and thus our answer is
.
pi_is_3.141
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 |