Difference between revisions of "1999 AIME Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | {{ | + | {{image}} |
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| + | 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. | ||
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| + | {{image}} | ||
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| + | The equations of those two heights are <math>x=16</math>, and <math>y=\dfrac{3}{4} x</math>. They intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron. From the pythagorean theorem, | ||
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| + | <math>h=\sqrt{(\dfrac{8\sqrt{13}}{2})^2-8^2}=12</math> | ||
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| + | And the area of the base is 104, so the volume is | ||
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| + | <math>\dfrac{104*12}{3}=\boxed{408}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1999|num-b=14|after=Final Question}} | {{AIME box|year=1999|num-b=14|after=Final Question}} | ||
Revision as of 10:21, 15 October 2007
Problem
Consider the paper triangle whose vertices are 
and 
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 
, and 
. They intersect at 
, so that's the base of the height of the tetrahedron. From the pythagorean theorem,
And the area of the base is 104, so the volume is
See also
| 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 | ||
| All AIME Problems and Solutions | ||
