Difference between revisions of "2002 AMC 10A Problems/Problem 23"
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Revision as of 10:13, 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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