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Difference between revisions of "1984 AIME Problems/Problem 9"

m (meh 3D asy is difficult)
(replace with 3d asymptote)
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== Solution ==
 
== Solution ==
−
<!--<center><asy>
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<span style="font-size:50%">For non-asymptote version of image, see [[:Image:1984_AIME-9.png]].</span><center><asy>
−
import three; pointpen=black;pathpen=black;
+
size(200);
−
triple A=(0,0,0),B=(3,0,0),C=(5,2.2,0),D=(1.5,4,4);
+
import three; pointpen=black;pathpen=black+linewidth(0.6); pen small = fontsize(10);
−
currentprojection=perspective(1,-1,1,Z,B);
+
triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0);  
−
D(A--B--C--A--D--B--D--C);
+
currentprojection=perspective(16,-10,8);
−
MP("A",A);MP("B",B);MP("C",C);MP("D",D);
 
−
</asy></center>-->
 
−
[[Image:1984_AIME-9.png|center]]
 
  
−
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>.
+
/* draw pyramid - other lines + angles */
 +
D(A--B--C--A--D--B--D--C);
 +
D(D--Da--Db--cycle);D(rightanglemark(D,Da,Db));D(rightanglemark(A,Db,D));D(anglemark(Da,Db,D,12));
 +
 
 +
/* labeling points */
 +
MP("A",A);MP("B",B);MP("C",C);MP("D",D,N);MP("30^{\circ}",Db+(0,.35,0.08),NE,small);
 +
MP("3",(A+B)/2); MP("15\mathrm{cm}^2",(Db+C)/2+(0,-0.5,-0.1),NE,small); MP("12\mathrm{cm}^2",(A+D)/2,NW,small);
 +
</asy></center>
 +
 
 +
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 ==

Revision as of 20:06, 24 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

For non-asymptote version of image, see Image:1984_AIME-9.png.

size(200);
import three; pointpen=black;pathpen=black+linewidth(0.6); pen small = fontsize(10);
triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0); 
currentprojection=perspective(16,-10,8);

/* draw pyramid - other lines + angles */
D(A--B--C--A--D--B--D--C); 
D(D--Da--Db--cycle);D(rightanglemark(D,Da,Db));D(rightanglemark(A,Db,D));D(anglemark(Da,Db,D,12));

/* labeling points */
MP("A",A);MP("B",B);MP("C",C);MP("D",D,N);MP("30^{\circ}",Db+(0,.35,0.08),NE,small);
MP("3",(A+B)/2); MP("15\mathrm{cm}^2",(Db+C)/2+(0,-0.5,-0.1),NE,small); MP("12\mathrm{cm}^2",(A+D)/2,NW,small);
 (Error making remote request. Unknown error_msg)

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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