Difference between revisions of "2009 AIME I Problems/Problem 12"
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==Solution 3== | ==Solution 3== | ||
Revision as of 04:26, 30 September 2022
Solution 3
As in Solution , let and be the intersections of with and respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let . Let . Let . The semi-perimeter of is . Since the lengths of the sides of are , and , the square of its area by Heron's formula is .
The radius of is . Therefore . As is the in-circle of , the area of is also , and so the square area is .
Therefore Dividing both sides by we get: and so . The semi-perimeter of is therefore and the whole perimeter is . Now , so the ratio of the perimeter of to the hypotenuse is and our answer is
and .
See also
2009 AIME I (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.