2000 AIME II Problems/Problem 12
Problem
The points , and lie on the surface of a sphere with center and radius . It is given that , , , and that the distance from to is , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Let be the foot of the perpendicular from to the plane of . By the Pythagorean Theorem on triangles , and we get:
It follows that , so is the circumcenter of .
By Heron's Formula the area of is (alternatively, a triangle may be split into and right triangles):
From , we know that the circumradius of is:
Thus by the Pythagorean Theorem again,
So the final answer is .
Solution 2 (Vectors)
We know the radii to ,, and form a triangular pyramid . We know the lengths of the edges . First we can break up into its two component right triangles and . Let the axis be perpendicular to the base and axis run along , and occupy the other dimension, with the origin as . We look at vectors and . Since is isoceles we know the vertex is equidistant from and , hence it is units along the axis. Hence for vector , in form it is where is the height (answer) and is the component of the vertex along the axis. Now on vector , since is along , and it is along axis, it is . We know both vector magnitudes are . Solving for yields , so Answer = .
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
this is highly trivial for an AIME #12