1984 AIME Problems/Problem 9

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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.

/* modified version of olympiad modules */
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);
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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 (ProblemsAnswer KeyResources)
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