2011 AIME I Problems/Problem 8
Contents
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 1
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 . 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 .
Solution 2
As from above, we can see that the length of the altitude from A is the longest. Thus the highest table is formed when X and Y meet up. Let the distance of this point from C be x, then the distance from B will be 23 - x. Let h be the height of the table. From similar triangles, we have where A is the area of triangle ABC. Similarly, . Therefore, and hence .
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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