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

Let $D$ , $E$ , $F$ be the foots of the altitudes of 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 wish to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is just $x=16$ . $AD$ is perpendicular to the 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 $y=\dfrac{3}{4} x$ . These two equations 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 $h=12$ .

And the area of the base is 104, so the volume is

$\dfrac{104*12}{3}=\boxed{408}$

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Final Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1999 AIME Problems/Problem 15

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