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1999 AIME Problems/Problem 15

Revision as of 21:22, 15 March 2010 by Kubluck (talk | contribs) (→‎Alternate Solution)

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]

Let $D$, $E$, $F$ be the feet of the altitudes to sides $BC$, $CA$, $AB$, respectively, of $\triangle ABC$. The base of the tetrahedron is the orthocenter $O$ of the large triangle, so we just need to find that, then it's easy from there.

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

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 thus arrive at a fairly simple system of equations, yielding $z = 12$. The desired volume is thus $34 \times 12 = \boxed{408}$.

See also

1999 AIME (Problems • Answer Key • Resources)
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