# 1999 AIME Problems/Problem 15

## Problem

Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?

## Solution $[asy]defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("$$A$$",A,SW);label("$$B$$",B,NW);label("$$C$$",C,SE); label("$$D$$",foot(A,B,C),NE);label("$$E$$",foot(B,A,C),SW);label("$$F$$",foot(C,A,B),NW);label("$$P$$",P,NW);label("$$Q$$",Q,NE);label("$$R$$",R,SE);[/asy]$ $[asy]import three; defaultpen(linewidth(0.6)); currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); [/asy]$

As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$. The crux of this problem is the following lemma.

Lemma: The point $O$ is the orthocenter of $\triangle ABC$.

Proof. Observe that $$OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;$$ the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$. Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$. Analogously, $O$ lies on the $B$-altitude and $C$-altitude of $\triangle ABC$, and so $O$ is, indeed, the orthocenter of $\triangle ABC$.

To find the coordinates of $O$, we need to find the intersection point of altitudes $BE$ and $AD$. The equation of $BE$ is simply $x=16$. $AD$ is perpendicular to line $BC$, so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$. $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$, therefore $y=\frac{3}{4} x$. These two lines intersect at $(16,12)$, so that's the base of the height of the tetrahedron.

Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$. From the Pythagorean Theorem, $h=\sqrt{BS^2-SO^2}$. However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$.

The area of the base is $102$, so the volume is $\frac{102*12}{3}=\boxed{408}$.

## Alternate Solution 1

Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$, $(8, 12, 0)$, and $(25, 12, 0)$. We can compute the area of this triangle as $102$. Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$. Clearly, the height of the pyramid is $z$. The desired volume is thus $\frac{102z}{3} = 34z$.

We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$, $VP = PB$, and $VQ = QC$. We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$. The desired volume is thus $34 \times 12 = \boxed{408}$.

## Alternate Solution 2

The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\sqrt{13},15,17$) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras) $p^2+q^2=4^2\cdot{13}$ $q^2+r^2=15^2$ $r^2+p^2=17^2$

to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$

Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is $\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$

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