Difference between revisions of "2002 AIME I Problems/Problem 15"
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Polyhedron <math>ABCDEFG</math> has six faces. Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math> The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math> The distance from <math>E</math> to face <math>ABCD</math> is 12. Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> | Polyhedron <math>ABCDEFG</math> has six faces. Face <math>ABCD</math> is a square with <math>AB = 12;</math> face <math>ABFG</math> is a trapezoid with <math>\overline{AB}</math> parallel to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math> The other three faces are <math>ADEG, BCEF,</math> and <math>EFG.</math> The distance from <math>E</math> to face <math>ABCD</math> is 12. Given that <math>EG^2 = p - q\sqrt {r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> | ||
− | == Solution == | + | == Solution 1 == |
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*One may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>. | *One may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=12</math>. Since <math>G</math> lies on this plane, we must have <math>2\cdot3+b=12</math>, so <math>b=6</math>. Therefore, <math>a=-6\pm\sqrt{55-6^2}=-6\pm\sqrt{19}</math>. So <math>G(-6\pm\sqrt{19},-3,6)</math>. | ||
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+ | ==Solution 2== | ||
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+ | We let <math>A</math> be the origin, or <math>(0,0,0)</math> and <math>B = (0,0,12)</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2002|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:06, 15 February 2017
Contents
Problem
Polyhedron has six faces. Face is a square with face is a trapezoid with parallel to and and face has The other three faces are and The distance from to face is 12. Given that where and are positive integers and is not divisible by the square of any prime, find
Solution 1
Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: , , and . Since is an isosceles trapezoid and is an isosceles triangle, we have symmetry about the -plane.
Therefore, the -component of is 0. We are given that the component is 12, and it lies over the square, so we must have so (the other solution, does not lie over the square). Now let and , so is parallel to . We must have , so .
The last piece of information we have is that (and its reflection, ) are faces of the polyhedron, so they must all lie in the same plane. Since we have , , and , we can derive this plane.* Let be the extension of the intersection of the lines containing . It follows that the projection of onto the plane must coincide with the , where is the projection of onto the plane . by a ratio of , so the distance from to the plane is and by the similarity, the distance from to the plane is . The altitude from to has height . By similarity, the x-coordinate of is . Then .
Now that we have located , we can calculate : Taking the negative root because the answer form asks for it, we get , and .
- One may also do this by vectors; , so the plane is . Since lies on this plane, we must have , so . Therefore, . So .
Solution 2
We let be the origin, or and
See also
2002 AIME I (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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