2002 AMC 10A Problems/Problem 23
Problem 23
Points and lie on a line, in that order, with and . Point is not on the line, and . The perimeter of is twice the perimeter of . Find .
Solution
First, we draw an altitude to from . Let it intersect at . As is isosceles, we immediately get , so the altitude is . Now, let . Using the Pythagorean Theorem on , we find . From symmetry, as well. Now, we use the fact that the perimeter of is twice the perimeter of .
We have so . Squaring both sides, we have which nicely rearranges into . Hence, AB is 9 so our answer is .
Simpler Solution
Let be the foot of the altitude from to Then because is isosceles. By the Pythagorean triple the altitude is Since is the only primitive Pythagorean triple with leg we test Since this works, giving us
~dolphin7
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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