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1984 AIME Problems/Problem 9

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 $[DXC]=\frac{ab\sin{c}}{2}=20=5 \cdot h \rightarrow h = 4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $\frac{bh}{3}=15\cdot \frac{4}{3}=\boxed{020}$ ~ Mathommill


(Note this actually isn't rigorous because they never proved that the height from $D$ to $XC$ is the altitude of the tetrahedron.

Solution 3 (Sketchy)

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

Solution 4 (coord/vector bash)

We can use 3D coordinates.

Let $A = (0, 0, 0)$ and $B = (3, 0, 0).$ WLOG, let $D = \left(\frac{3}{2}, 8, 0\right)$, because the area of $\Delta{ABD} = 12$ and the tetrahedron area won't change if we put it somewhere else with $y=8.$

To find $C$, we can again let the $x$-coordinate be $\frac{3}{2}$ for simplicity. Note that $C$ is $10$ units away from $AB$ because the area of $\Delta{ABC}$ is $15$. Since the angle between $ABD$ and $ABC$ is $30^\circ$, we can form a 30-60-90 triangle between $A$, $B$, and an altitude dropped from $C$ onto face $ABD$. Since $10$ is the hypotenuse, we get $5\sqrt{3}$ and $5$ as legs. Then $y=5\sqrt{3}$ and $z=5$, so $C = \left(\frac{3}{2}, 5\sqrt{3}, 5\right).$

(I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.)

Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula $\frac{1}{3}Bh.$ Letting $\Delta{ABC}$ be the base we have $B = 15$ (from the problem statement). We need to find the distance between $D$ and $ABC$, and to do this, we should find the projection of $D$ onto face $ABC$.

Note that we can simplify this to projecting $D$ onto $\mathbf{\overrightarrow{C}}.$ This is because we know the projection will have the same $x$-coordinate as $D$ and $C$, as both are $\frac{3}{2}.$ Now we find $\text{proj}_{\mathbf{\overrightarrow{D}}} \mathbf{\overrightarrow{C}}$, or plugging in our coordinates, $\text{proj}_{\langle\frac{3}{2}, 5\sqrt{3}, 5\rangle} \left\langle\frac{3}{2}, 8, 0\right\rangle$.

Let the $x$-coordinates for both be $0$ for simplicity, because we can always add a $\frac{3}{2}$ at the end. Using the projection formula, we get \[\langle 0, 6, 2\sqrt{3}\rangle.\]

Finally, we calculate the distance between $\left(\frac{3}{2}, 6, 2\sqrt{3}\right)$ and $D$ to be $4$. So the height is $4$, and plugging into our tetrahedron formula we get \[\frac{1}{3}\cdot 15\cdot 4 = \boxed{20}.\]

-PureSwag

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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