Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Solution == | == Solution == | ||
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Let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have | Let <math>O</math> be the incenter. Because <math>MO \parallel BC</math> and <math>BO</math> is the angle bisector, we have | ||
Revision as of 10:44, 17 February 2014
Problem
Triangle has side-lengths
and
The line through the incenter of
parallel to
intersects
at
and $\overbar{AC}$ (Error compiling LaTeX. Unknown error_msg) at
What is the perimeter of
Solution
Let be the incenter. Because
and
is the angle bisector, we have
It then follows due to alternate interior angles and base angles of isosceles triangles that . Similarly,
. The perimeter of
then becomes
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.