2002 AIME I Problems/Problem 15
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 . Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, , so , where a and b are variables.
We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have and . Therefore, the x coordinate of is , so .
We also know that and are coplanar, so they all lie on the plane . Since is on it, then . Also, since is contained, then . Finally, since is on the plane, then . Therefore, . Since , then , or . Therefore, the two permissible values of are . The only one that satisfies the conditions of the problem is , from which the answer is .
Solution 3 (minimal coordinates, Apollonius)
Denote the foot of the altitude from to be . Let the projection of onto be . We seek . Let . Then we get . Because the diagram is symmetrical, . So, . We find .
Extend and to meet the plane . Since is a quadrilateral and all on a plane, then the extension of and will meet the lines and , respectively. Call these intersections and . Let .
Using the Law of Cosines on gives . Using Law of Cosines on gives the equation . Now, using Apollonius' Theorem on the same triangle gives . Equating the two gives . Solving gives us .
Finally, plugging into either expression for gives . Since and is parallel to , by the midpoint theorem, .
Then .
-RyanZhu817
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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