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

m (Proofreading)
(→‎Solution: subst asy code by minsoens)
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== Solution ==
 
== Solution ==
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<center>[[Image:AIME_1999_Solution_15_1.png]][[Image:AIME_1999_Solution_15_2.png]]</center>
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<center><asy>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);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);</asy><asy>import three; defaultpen(linewidth(0.6));
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currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); </asy></center><!-- Asymptote renderings of Image:AIME_1999_Solution_15_1.png, Image:AIME_1999_Solution_15_2.png, by Minsoens -->
 
Let <math>D</math>, <math>E</math>, <math>F</math> be the feet of the altitudes to sides <math>BC</math>, <math>CA</math>, <math>AB</math>, respectively, of <math>\triangle ABC</math>.
 
Let <math>D</math>, <math>E</math>, <math>F</math> be the feet of the altitudes to sides <math>BC</math>, <math>CA</math>, <math>AB</math>, respectively, of <math>\triangle ABC</math>.
 
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.
 
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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To find the coordinates of <math>O</math>, we need to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is simply <math>x=16</math>. <math>AD</math> is [[perpendicular]] to 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=\frac{3}{4} x</math>. These two lines intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron.  
 
To find the coordinates of <math>O</math>, we need to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is simply <math>x=16</math>. <math>AD</math> is [[perpendicular]] to 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=\frac{3}{4} x</math>. These two lines intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron.  
  
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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>.
+
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>.
  
 
The area of the base is <math>104</math>, so the volume is <math>\frac{104*12}{3}=\boxed{408}</math>.
 
The area of the base is <math>104</math>, so the volume is <math>\frac{104*12}{3}=\boxed{408}</math>.

Revision as of 11:29, 26 April 2008

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]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);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);[/asy][asy]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), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); [/asy]

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 $\frac{104*12}{3}=\boxed{408}$.

See also

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