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

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</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 1 ==
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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 ==
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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:
 
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>
 
<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math>
  
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== See also ==
+
== See Also ==
 
{{AIME box|year=1983|num-b=10|num-a=12}}
 
{{AIME box|year=1983|num-b=10|num-a=12}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 06:02, 16 April 2012

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

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 $6$.

[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]

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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