Difference between revisions of "2002 AMC 10A Problems/Problem 23"
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Revision as of 11:13, 4 July 2013
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 BC from E.Let it intersect at M. As triangle BEC is isosceles, we immediately get MB=MC=6, so the altitude is 8. Now, let . Using the Pythagorean Theorem on triangle EMA, 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
.
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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