2011 AIME I Problems/Problem 8
Problem
In triangle , , , and . Points and are on with on , points and are on with on , and points and are on with on . In addition, the points are positioned so that , , and . Right angle folds are then made along , , and . The resulting figure is placed on a level floor to make a table with triangular legs. Let be the maximum possible height of a table constructed from triangle whose top is parallel to the floor. Then can be written in the form , where and are relatively prime positive integers and is a positive integer that is not divisible by the square of any prime. Find .
Solution
Note that the area is given by Heron's formula and it is . Let denote the length of the altitude dropped from vertex i. It follows that $h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}}, h_a = \frac{40\sqrt{221}}{23}$ (Error compiling LaTeX. ! Extra }, or forgotten $.). From similar triangles we can see that . We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields .
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.