Difference between revisions of "1999 AIME Problems/Problem 15"
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== Solution == | == Solution == | ||
− | <center> | + | <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(" |
+ | 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. | ||
− | Let <math>S</math> be the foot of altitude <math>BS</math> in <math>\triangle BPQ</math>. From the | + | 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 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
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 |