Difference between revisions of "2011 AIME I Problems/Problem 4"
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Extend <math>{CM}</math> and <math>{CN}</math> such that they intersects lines <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. | Extend <math>{CM}</math> and <math>{CN}</math> such that they intersects lines <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. | ||
Since <math>{BM}</math> is the angle bisector of angle B,and <math>{CM}</math> is perpendicular to <math>{BM}</math> ,so , <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math> .For the same reason,<math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>. | Since <math>{BM}</math> is the angle bisector of angle B,and <math>{CM}</math> is perpendicular to <math>{BM}</math> ,so , <math>BP=BC=120</math>, M is the midpoint of <math>{CP}</math> .For the same reason,<math>AQ=AC=117</math>,N is the midpoint of <math>{CQ}</math>. | ||
− | Hence<math>MN=\frac{PQ}{2}.But < | + | Hence<math>MN=\frac{PQ}{2}</math>.But <math>PQ=BP+AQ-AB=120+117-125=112,so</math>MN=\boxed{56}$. |
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=3|num-a=5}} | {{AIME box|year=2011|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:28, 7 November 2015
Problem 4
In triangle , , and . The angle bisector of angle intersects at point , and the angle bisector of angle intersects at point . Let and be the feet of the perpendiculars from to and , respectively. Find .
Solution
Extend and such that they intersects lines at points and , respectively. Since is the angle bisector of angle B,and is perpendicular to ,so , , M is the midpoint of .For the same reason,,N is the midpoint of . Hence.But MN=\boxed{56}$.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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