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

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== Solution ==
 
== Solution ==
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<asy>
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<center>[[Image:AIME_1999_Solution_15_1.png]][[Image:AIME_1999_Solution_15_2.png]]</center>
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size(100);
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Let <math>D</math>, <math>E</math>, <math>F</math> be the foots of the altitudes of sides <math>BC</math>, <math>CA</math>, <math>AB</math>, respectively, of <math>\triangle ABC</math>.
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draw((0,0)--(34,0)--(16,24)--(0,0));
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The base of the [[tetrahedron]] is the [[orthocenter]] <math>O</math> of the large triangle, so we just need to find that, then it's easy from there.
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draw((17,0)--(25,12)--(8,12)--(17,0));
 
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</asy>
 
  
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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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To find the coordinates of <math>O</math>, we wish to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is just <math>x=16</math> . <math>AD</math> is perpendicular to the line <math>BC</math>, so the slope of <math>AD</math> is equal to the negative reciprocal of the slope of <math>BC</math>. <math>BC</math> has slope <math>\frac{24-0}{16-34}=-\frac{4}{3}</math>, therefore <math>y=\dfrac{3}{4} x</math>. These two equations intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron.  
  
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{{image}}
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Let <math>S</math> be the foot of altitude <math>BS</math> in <math>\triangle BPQ</math>. From the pythagorean theorem, <math>h=\sqrt{BS^2-SO^2}</math>. However, since <math>S</math> and <math>O</math> are, by coincidence, the same point, <math>SO=0</math> and <math>h=12</math>.
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<!-- this can't be easily made from asymptote, so I guess an upload would be better -->
 
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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{\left(\frac{8\sqrt{13}}{2}\right)^2-8^2}=12</math>
 
  
 
And the area of the base is 104, so the volume is
 
And the area of the base is 104, so the volume is

Revision as of 19:32, 3 November 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

AIME 1999 Solution 15 1.pngAIME 1999 Solution 15 2.png

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

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

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
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