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

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In [[tetrahedron]] <math>ABCD</math>, [[edge]] <math>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>.
 
In [[tetrahedron]] <math>ABCD</math>, [[edge]] <math>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>.
  
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== Solution ==
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== Solution 1==
 
<center><asy>
 
<center><asy>
 
/* modified version of olympiad modules */
 
/* modified version of olympiad modules */
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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>. Because the problem does not specify, we may assume both <math>ABC</math> and <math>ABD</math> to be isosceles triangles. Thus, 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} \cdot15 \cdot 4 = \boxed{020}</math>.
 
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>. Because the problem does not specify, we may assume both <math>ABC</math> and <math>ABD</math> to be isosceles triangles. Thus, 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} \cdot15 \cdot 4 = \boxed{020}</math>.
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== Solution 2 (Rigorous)==
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It is clear that <math>DX=8</math> and <math>CX=10</math> where <math>X</math> is the foot of the perpendicular from <math>D</math> and <math>C</math> to side <math>AB</math>. Thus <math>[DXC]=\frac{ab\sin{c}}{2}=20=5*h \rightarrow h = 4</math> where h is the height of the tetrahedron from <math>D</math>. Hence, the volume of the tetrahedron is <math>bh/3=15*4/3=\boxed{020}</math>
  
 
== Solution 2 Kinda Sketchy==
 
== Solution 2 Kinda Sketchy==

Revision as of 15:04, 18 June 2018

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 1

[asy] /* modified version of olympiad modules */ import three; real markscalefactor = 0.03; path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 anglemark(triple A, triple B, triple C, real t=8 ... real[] s) { triple M,N,P[],Q[]; path3 mark; int n=s.length; M=t*markscalefactor*unit(A-B)+B; N=t*markscalefactor*unit(C-B)+B; for (int i=0; i<n; ++i) { P[i]=s[i]*markscalefactor*unit(A-B)+B; Q[i]=s[i]*markscalefactor*unit(C-B)+B; } mark=arc(B,M,N); for (int i=0; i<n; ++i) { if (i%2==0) { mark=mark--reverse(arc(B,P[i],Q[i])); } else { mark=mark--arc(B,P[i],Q[i]); } } if (n%2==0 && n!=0) mark=(mark--B--P[n-1]); else if (n!=0) mark=(mark--B--Q[n-1]); else mark=(mark--B--cycle); return mark; } size(200); import three; defaultpen(black+linewidth(0.7)); 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(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight); draw(surface(A--B--D--cycle),rgb(0.7,0.6,0.6),nolight); /* draw pyramid - other lines + angles */ draw(A--B--C--A--D--B--D--C); draw(D--Da--Db--cycle); draw(rightanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15)); /* labeling points */ label("$A$",A,SW);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$30^{\circ}$",Db+(0,.35,0.08),(1.5,1.2),small); label("$3$",(A+B)/2,S); label("$15\mathrm{cm}^2$",(Db+C)/2+(0,-0.5,-0.1),NE,small); label("$12\mathrm{cm}^2$",(A+D)/2,NW,small); [/asy]

Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, 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 $\frac{1}{3}Bh = \frac{1}{3} \cdot15 \cdot 4 = \boxed{020}$.

Solution 2 (Rigorous)

It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$. Thus $[DXC]=\frac{ab\sin{c}}{2}=20=5*h \rightarrow h = 4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $bh/3=15*4/3=\boxed{020}$

Solution 2 Kinda Sketchy

Make faces $ABC$ and $ABD$ right triangles. This makes everything a lot easier. Then do everything in solution 1.

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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