1984 AIME Problems/Problem 9

Revision as of 10:57, 9 August 2019 by Anstar (talk | contribs) (Solution 3 Kinda Sketchy)


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 $bh/3=15*4/3=\boxed{020}$ ~ ShreyJ

Solution 3 (Sketchy)

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

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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