Difference between revisions of "2011 AIME I Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
− | Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp MC</math> and <math>IN \perp NC</math>, we have <math>CMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = CI</math>. Since <math>\sin \angle MIN = \sin (90^\circ - \tfrac 12 \angle BAC) = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math>. Thus, <math>MN=\frac{117+120-125}{2}=\boxed{056}</math> | + | Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp MC</math> and <math>IN \perp NC</math>, we have <math>CMIN</math> is a cyclic quadrilateral. Consequently (by the Law of Sines), <math>\frac{MN}{\sin \angle MIN} = 2R = CI</math> (note <math>R</math> is referring to the radius of <math>CMIN</math>'s circle and not the incircle). Since <math>\sin \angle MIN = \sin (90^\circ - \tfrac 12 \angle BAC) = \cos \angle IAK</math>, we have that <math>MN = AI \cdot \cos \angle IAK</math>. Letting <math>X</math> be the point of contact of the incircle of <math>ABC</math> with side <math>AC</math>, we have <math>AX=MN</math>. Thus, <math>MN=\frac{117+120-125}{2}=\boxed{056}</math> |
== Solution 3 (Bash) == | == Solution 3 (Bash) == |
Revision as of 14:15, 22 October 2022
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 , must be an isoceles triangle, so , and is the midpoint of . For the same reason, , and is the midpoint of . Hence . Since so .
Solution 2
Let be the incenter of . Now, since and , we have is a cyclic quadrilateral. Consequently (by the Law of Sines), (note is referring to the radius of 's circle and not the incircle). 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 .
Video Solution
https://www.youtube.com/watch?v=yIUBhWiJ4Dk ~Mathematical Dexterity
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.