Difference between revisions of "2011 AIME I Problems/Problem 4"
(→Solution: Nitpicking. Also previous post was done as it is clear that the assertion of OQ being a midline and MQC, NOC being isosceles was not addressed as some have asked why the facts were true) |
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− | == Problem | + | == Problem == |
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | ||
+ | == 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. | ||
+ | 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>. <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>. | ||
− | == Solution == | + | == 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 - \frac{\angle BAC}{2}) = \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) == | ||
+ | Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines (LoC) formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>. | ||
− | + | --lucasxia01 | |
− | + | == Solution 4== | |
− | + | Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. | |
− | + | Ptolemy on CMIN: | |
+ | <math>CN*MI+CM*IN=CI*MN</math> | ||
+ | <math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math> | ||
− | + | <math>MN = CI \sin \angle MCN</math> by angle addition formula. | |
− | + | <math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=vkniYGN45F4 | ||
+ | |||
+ | ~Shreyas S | ||
+ | |||
+ | Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s | ||
== 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:26, 14 July 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 .
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 |
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