Difference between revisions of "2011 AMC 12A Problems/Problem 13"
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== Problem == | == Problem == | ||
− | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\ | + | Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> parallel to <math>\overline{BC}</math> intersects <math>\overline{AB}</math> at <math>M</math> and <math>\overline{AC}</math> at <math>N.</math> What is the perimeter of <math>\triangle AMN?</math> |
<math> | <math> |
Revision as of 20:46, 13 March 2015
Problem
Triangle has side-lengths and The line through the incenter of parallel to intersects at and 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.