Difference between revisions of "1983 AIME Problems/Problem 11"

Revision as of 15:59, 26 March 2011

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?

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

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);
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);
D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); 
D(B--Ba--Ca--C,dashed+d);D(A--Aa--Da--D,dashed+d);D(E--(E.x,E.y,0),dashed+l);D(F--(F.x,F.y,0),dashed+l);
D(Aa--E--Da,dashed+d); D(Ba--F--Ca,dashed+d);
MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);MP("12\sqrt{2}",(E+F)/2,N);MP("6\sqrt{2}",(A+B)/2);MP("6",(3*s/2,s/2,3),ENE);
 (Error making remote request. Unknown error_msg)

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 • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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Difference between revisions of "1983 AIME Problems/Problem 11"

Revision as of 15:59, 26 March 2011

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 7: / Line 7: @@
 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);
-D(A--B--C--D--A--E--D); D(B--F--C); D(E--F);
+draw(A--B--C--D--A--E--D);
-MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);
+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></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms -->
 == Solution 1 ==
 First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of <math>AD</math> to <math>EF</math>. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] <math>ADE</math> one of the legs is <math>3\sqrt{2}</math>. We apply the [[Pythagorean Theorem]] to find that the height is equal to <math>6</math>.