Difference between revisions of "2011 AIME I Problems/Problem 4"

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== Problem 4 ==
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== 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 ==  
 
== 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 <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>.
 
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>. <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>IM \perp LC</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>
+
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) ==  
 
== 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 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>.
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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
 
--lucasxia01
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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>.
 +
 +
==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}}
 
{{MAA Notice}}

Revision as of 20:26, 14 July 2020

Problem

In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.

Solution 1

Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively. Since ${BM}$ is the angle bisector of angle $B$, and ${CM}$ is perpendicular to ${BM}$, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, and $N$ is the midpoint of ${CQ}$. Hence $MN=\frac{PQ}{2}$. $PQ=BP+AQ-AB=120+117-125=112$, so $MN=\boxed{056}$.

Solution 2

Let $I$ be the incenter of $ABC$. Now, since $IM \perp MC$ and $IN \perp NC$, we have $CMIN$ is a cyclic quadrilateral. Consequently, $\frac{MN}{\sin \angle MIN} = 2R = CI$. Since $\sin \angle MIN = \sin (90 - \frac{\angle BAC}{2}) = \cos \angle IAK$, we have that $MN = AI \cdot \cos \angle IAK$. Letting $X$ be the point of contact of the incircle of $ABC$ with side $AC$, we have $AX=MN$. Thus, $MN=\frac{117+120-125}{2}=\boxed{056}$

Solution 3 (Bash)

Project $I$ onto $AC$ and $BC$ as $D$ and $E$. $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$. Since $IC$ is the hypotenuse for the 4 triangles ($\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$. To find $MN$, we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$. Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$: $r^2 + 56^2 = (2R)^2$. To find $r$, we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$. To find $\angle MCN$, we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$. $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$. Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$, we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$. Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$. To find $\cos{\angle C}$, we use LoC on $\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$. Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$. After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$.

--lucasxia01

Solution 4

Because $\angle CMI = \angle CNI = 90$, $CMIN$ is cyclic.

Ptolemy on CMIN:

$CN*MI+CM*IN=CI*MN$

$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN$

$MN = CI \sin \angle MCN$ by angle addition formula.

$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$.

Let $H$ be where the incircle touches $BC$, then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$. $a=120, b=117, c=125$, for a final answer of $\boxed{056}$.

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 (ProblemsAnswer KeyResources)
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