Difference between revisions of "1999 AIME Problems/Problem 15"

Revision as of 17:32, 15 October 2007

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

$[asy] size(100); draw((0,0)--(34,0)--(16,24)--(0,0)); draw((17,0)--(25,12)--(8,12)--(17,0)); [/asy]$

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

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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,

$h=\sqrt{\left(\frac{8\sqrt{13}}{2}\right)^2-8^2}=12$

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

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

1999 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1999 AIME Problems/Problem 15"

Revision as of 17:32, 15 October 2007

Problem

Solution

See also

@@ Line 5: / Line 5: @@
 <asy>
 size(100);
-draw((0,0)--(34,0)--(16,24));
+draw((0,0)--(34,0)--(16,24)--(0,0));
-draw((17,0)--(25,12)--(8,12));
+draw((17,0)--(25,12)--(8,12)--(17,0));
 </asy>