Difference between revisions of "2011 AIME I Problems/Problem 8"
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dot("$Y$",Y[1],dir(250)); | dot("$Y$",Y[1],dir(250)); | ||
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+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=Asvy1s-6Rx0 | ||
==Solution 1== | ==Solution 1== |
Revision as of 23:29, 4 March 2021
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 .
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=Asvy1s-6Rx0
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 B be , making the distance from C . Let be the height of the table. From similar triangles, we have where A is the area of ABC. Similarly, . Therefore, and hence .
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 |
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