Difference between revisions of "1999 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Consider the paper triangle whose vertices are <math>(0,0), (34,0),</math> and <math>(16,24).</math> The vertices of its midpoint triangle are the | + | Consider the paper triangle whose vertices are <math>(0,0), (34,0),</math> and <math>(16,24).</math> The vertices of its midpoint triangle are the [[midpoint]]s 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 == | == Solution == | ||
Line 16: | Line 16: | ||
== See also == | == See also == | ||
{{AIME box|year=1999|num-b=14|after=Last Question}} | {{AIME box|year=1999|num-b=14|after=Last Question}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 11:30, 26 April 2008
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
![[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]](http://latex.artofproblemsolving.com/8/b/c/8bca170a0544db80c5bdf360300276df0efcb5e9.png)
![[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]](http://latex.artofproblemsolving.com/0/3/3/0334c5d9ccf52482e6d919c1ccf99cd120d73464.png)
Let ,
,
be the feet of the altitudes to sides
,
,
, respectively, of
.
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.
To find the coordinates of , we need to find the intersection point of altitudes
and
. The equation of
is simply
.
is perpendicular to line
, so the slope of
is equal to the negative reciprocal of the slope of
.
has slope
, therefore
. These two lines intersect at
, so that's the base of the height of the tetrahedron.
Let be the foot of altitude
in
. From the Pythagorean Theorem,
. However, since
and
are, by coincidence, the same point,
and
.
The area of the base is , so the volume is
.
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 |