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Difference between revisions of "1984 AIME Problems/Problem 9"
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== Solution ==
 
== Solution ==
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[[Image:1984_AIME-9.png]]
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<!--<center><asy>
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import three; pointpen=black;pathpen=black;
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triple A=(0,0,0),B=(3,0,0),C=(5,2.2,0),D=(1.5,4,4);
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currentprojection=perspective(1,-1,1,Z,B);
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D(A--B--C--A--D--B--D--C);
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MP("A",A);MP("B",B);MP("C",C);MP("D",D);
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</asy></center>-->
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[[Image:1984_AIME-9.png|center]]
  
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Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. The height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron, so <math>h = \frac{1}{2} 8 = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020</math>.
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Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. The height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron, so <math>h = \frac{1}{2} 8 = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = \boxed{020}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=8|num-a=10}}
 
{{AIME box|year=1984|num-b=8|num-a=10}}
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* [[AIME Problems and Solutions]]
 
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* [[American Invitational Mathematics Examination]]
 
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* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 18:03, 9 April 2008

Problem

In tetrahedron $ABCD$, edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$.

Solution

1984 AIME-9.png

Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. The height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron, so $h = \frac{1}{2} 8 = 4$. The volume of the tetrahedron is thus $\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = \boxed{020}$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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