Difference between revisions of "2011 AIME I Problems/Problem 4"
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Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>. | Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>. | ||
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>. | <math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>. | ||
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+ | == Solution 5 (Very fast) == | ||
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+ | Extend <math>CM</math> and <math>CN</math> to meet <math>AB</math> at <math>X</math> and <math>Y.</math> Note that <math>CM=MX</math> and <math>CN = NY</math> so <math>MN = \displaystyle\frac{XY}{2}.</math> To find <math>XY</math> we just do a system of equations and we find <math>XY=112</math> so <math>XY = \boxed{056}.</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:07, 27 September 2020
Contents
Problem
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 . , so .
Solution 2
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 (LoC) 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 .
Solution 5 (Very fast)
Extend and to meet at and Note that and so To find we just do a system of equations and we find so
Video Solution
https://www.youtube.com/watch?v=vkniYGN45F4
~Shreyas S
Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s
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.