# 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

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 $D = (\frac{3}{2}, 8, 0)$, 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 $C = (\frac{3}{2}, 5\sqrt{3}, 5).$

(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, $\text{proj}_{<\frac{3}{2}, 5\sqrt{3}, 5>} <\frac{3}{2}, 8, 0>$.

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 $$<0, 6, 2\sqrt{3}>.$$

Finally, we calculate the distance between $(\frac{3}{2}, 6, 2\sqrt{3})$ 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