Difference between revisions of "2011 AIME I Problems/Problem 4"
m (→Solution 2) |
m (→Solution 1) |
||
Line 5: | Line 5: | ||
== Solution 1 == | == Solution 1 == | ||
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. | Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <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 <math>B</math>, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>, and <math>N</math> is the midpoint of <math>{CQ}</math>. |
Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>. | Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>. | ||
Latest revision as of 19:27, 10 November 2019
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 1
Extend and such that they intersect line at points and , respectively. Since is the angle bisector of angle , and is perpendicular to , so , and is the midpoint of . For the same reason, , and is the midpoint of . Hence . But , so .
Solution 2
[There seem to be some mislabeled points going on here but the idea is sound.] Let be the incenter of . Now, since and , we have is a cyclic quadrilateral. Consequently, . Since , we have that . Letting be the point of contact of the incircle of with side , we have thus
Solution 3 (Bash)
Project onto and as and . and are both in-radii of so we get right triangles with legs (the in-radius length) and . Since is the hypotenuse for the 4 triangles ( and ), are con-cyclic on a circle we shall denote as which is also the circumcircle of and . To find , we can use the Law of Cosines on where is the center of . Now, the circumradius can be found with Pythagorean Theorem with or : . To find , we can use the formula and by Heron's, . To find , we can find since . . Thus, and since , we have . Plugging this into our Law of Cosines formula gives . To find , we use LoC on . Our formula now becomes . After simplifying, we get .
--lucasxia01
Solution 4
Because , is cyclic.
Ptolemy on CMIN:
by angle addition formula.
.
Let be where the incircle touches , then . , for a final answer of .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.