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

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MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);
 
MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);
 
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms -->
 
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms -->
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== Solution ==
+
== 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>.
 
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>.
 
<center><asy>
 
<center><asy>
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Thus, our answer is <math>432-144=\boxed{288}</math>.
 
Thus, our answer is <math>432-144=\boxed{288}</math>.
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 +
== Solution 2 ==
 +
Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. now, we have a regular tetrahedron <math>EFGH</math>, which has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is:
 +
 +
<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math>
  
 
== See also ==
 
== See also ==

Revision as of 22:21, 13 March 2009

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? 

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);
D(A--B--C--D--A--E--D); D(B--F--C); D(E--F);
MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("E",E,N);MP("F",F,N);
 (Error making remote request. Unknown error_msg)

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:

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

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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