1999 AIME Problems/Problem 15

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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

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);d...

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...

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 $104$ , so the volume is $\frac{104*12}{3}=\boxed{408}$ .

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 $34 \times 12 = \boxed{408}$ .

1999 AIME (Problems • Resources)
Preceded by Problem 14	Followed by Last 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

From AoPSWiki

Contents

Problem

Solution

Alternate Solution

See also