1999 AIME Problems/Problem 15

Revision as of 11:21, 15 October 2007 by 1=2 (talk | contribs)

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


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


The base of the height of the tetrahedron is the orthocenter of the big triangle, so we just need to find that, then it's easy from there.


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


The equations of those two heights are $x=16$, and $y=\dfrac{3}{4} x$. They intersect at $(16,12)$, so that's the base of the height of the tetrahedron. From the pythagorean theorem,

$h=\sqrt{(\dfrac{8\sqrt{13}}{2})^2-8^2}=12$

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

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

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Final Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions