Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(soln 5)
 
(25 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
== Problem 4 ==
+
== 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>.
+
<asy>
−
Hence <math>MN=\frac{PQ}{2}</math>. But <math>PQ=BP+AQ-AB=120+117-125=112</math>, so <math>MN=\boxed{056}</math>.
+
defaultpen(fontsize(10)+0.8); size(200);
 +
pen p=fontsize(9)+linewidth(3);
 +
pair A,B,C,D,K,L,M,N,P,Q;
 +
A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L);
 +
draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5);
 +
dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p);
 +
label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10));
 +
</asy>
 +
Since <math>{BM}</math> is the angle bisector of angle <math>B</math> and <math>{CM}</math> is perpendicular to <math>{BM}</math>, <math>\triangle BCP</math> must be an isoceles triangle, 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=\tfrac 12 PQ</math>. Since <cmath>PQ=BP+AQ-AB=120+117-125=112,</cmath> so <math>MN=\boxed{056}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
−
[There seem to be some mislabeled points going on here but the idea is sound.]
+
Let <math>I</math> be the incenter of <math>ABC</math>. Since <math>I</math> lies on <math>BM</math> and <math>AN</math>, <math>IM \perp MC</math> and <math>IN \perp NC</math>, so <math>\angle IMC + \angle INC = 180^\circ</math>. This means that <math>CMIN</math> is a cyclic quadrilateral. By the Law of Sines, <math>\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI</math>, where <math>R</math> is the radius of the circumcircle of <math>CMIN</math>. Since <math>\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI</math>, we have that <math>MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI</math>. Letting <math>H</math> be the point of contact of the incircle of <math>ABC</math> with side <math>BC</math>, we have <math>MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH</math>. Thus, <math>MN = s - AB = \frac{117+120-125}{2}=\boxed{056}</math>.
−
Let <math>I</math> be the incenter of <math>ABC</math>. Now, since <math>AM \perp LC</math> and <math>AN \perp KB</math>, we have <math>AMIN</math> is a cyclic quadrilateral. Consequently, <math>\frac{MN}{\sin \angle MIN} = 2R = AI</math>. Since <math>\angle MIN = 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>.
+
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
Line 21: Line 28:
 
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic.  
 
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic.  
  
−
Ptolemy on CMIN:
+
Applying Ptolemy's theorem on CMIN:
  
−
<math>CN*MI+CM*IN=CI*MN</math>
+
<math>CN \cdot MI+CM \cdot IN=CI \cdot MN</math>
  
−
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math>
+
<math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI \cdot MN</math>
  
−
<math>MN = CI \sin \angle MCN</math> by angle addition formula.
+
<math>MN = CI \sin \angle MCN</math> by sine angle addition formula.
  
 
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>.  
 
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>.  
Line 33: Line 40:
 
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>.
 +
 +
Note: This is similar to Solution 2 after the first four lines
 +
 +
==Solution 5 (Trig Bash)==
 +
Applying [[Ptolemy's Theorem]] on the cyclic quadrilateral <math>MINC</math>, we find that
 +
 +
<math>MI\cdot CN + IN\cdot MC = MN\cdot IC</math>.
 +
 +
<math>\angle CIN=\frac{\alpha+\gamma}{2}</math> and <math>\angle MIC=\frac{\beta+\gamma}{2}</math> by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that
 +
 +
<math>MI=IC\cdot\cos\left(\frac{\beta+\gamma}{2}\right),</math> <math>MC=IC\cdot\sin\left(\frac{\beta+\gamma}{2}\right),</math> <math>IN=IC\cdot\cos\left(\frac{\alpha+\gamma}{2}\right),</math> <math>CN=IC\cdot\sin\left(\frac{\alpha+\gamma}{2}\right).</math>
 +
 +
Plugging in the values and simplifying results in <math>MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)</math> by the angle-addition identity <math>\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)</math>.
 +
 +
Before we continue, we would like to simplify the value in the sine function. We see that <math>\frac{\alpha+\beta+2\gamma}{2}=\frac{\gamma}{2}+\frac{\alpha+\beta+\gamma}{2}=\frac{\gamma}{2}+90</math>. Using the fact that <math>\cos(A)=\sin(90-A)</math> results in
 +
 +
<math>\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)=\sin\left(90+\frac{\gamma}{2}\right)=\sin\left(90-(-\frac{\gamma}{2})\right)=\cos\left(-\frac{\gamma}{2}\right)=\cos\left(\frac{\gamma}{2}\right).</math>
 +
 +
How do we simplify <math>IC</math>? Well, we can perform the [[Law of Sines]] on triangle <math>AIC</math>. This results in:
 +
 +
<math>\frac{AC}{\sin(\angle AIC)}=\frac{IC}{\sin\left(\frac{\alpha}{2}\right)}</math>
 +
 +
The value of <math>\angle AIC</math> is <math>\frac{\alpha+2\beta+\gamma}{2}</math> by the Exterior Angle Theorem on <math>\triangle ABI</math>, so the value of <math>\sin(\angle AIC)</math> is equivalent to the value of <math>\cos\left(\frac{\beta}{2}\right)</math> by a similar argument as above. Then rearranging yields <math>IC = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}</math>.
 +
 +
Going back to the previous formula <math>MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)</math> and substituting values yields:
 +
 +
<math>MN = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}</math>.
 +
 +
Finally, using the formulae <math>\sin\left(\frac{\alpha}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{bc}}</math> and <math>\cos\left(\frac{\alpha}{2}\right)=\sqrt{\frac{s(s-a)}{bc}}</math> (where <math>s</math> is half the perimeter of the triangle), we reach our final value:
 +
 +
<math>MN = b\cdot\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}\cdot\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{s(s-b)}{ac}}}</math>
 +
 +
<math>=\frac{\sqrt{s(s-b)(s-c)^2}}{\sqrt{s(s-b)}}</math>
 +
 +
<math>=\sqrt{(s-c)^2}</math>
 +
 +
<math>=s-c</math>
 +
 +
<math>=181-125</math>
 +
 +
<math>=\boxed{056}.</math>
 +
 +
==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 ==
 
== 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}}

Latest revision as of 11:08, 15 July 2024

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. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); [/asy] Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$, $\triangle BCP$ must be an isoceles triangle, 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 $MN=\tfrac 12 PQ$. Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$.

Solution 2

Let $I$ be the incenter of $ABC$. Since $I$ lies on $BM$ and $AN$, $IM \perp MC$ and $IN \perp NC$, so $\angle IMC + \angle INC = 180^\circ$. This means that $CMIN$ is a cyclic quadrilateral. By the Law of Sines, $\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI$, where $R$ is the radius of the circumcircle of $CMIN$. Since $\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI$, we have that $MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI$. Letting $H$ be the point of contact of the incircle of $ABC$ with side $BC$, we have $MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH$. Thus, $MN = s - AB = \frac{117+120-125}{2}=\boxed{056}$.

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.

Applying Ptolemy's theorem on CMIN:

$CN \cdot MI+CM \cdot IN=CI \cdot MN$

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

$MN = CI \sin \angle MCN$ by sine 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 $\boxed{056}$.

Note: This is similar to Solution 2 after the first four lines

Solution 5 (Trig Bash)

Applying Ptolemy's Theorem on the cyclic quadrilateral $MINC$, we find that

$MI\cdot CN + IN\cdot MC = MN\cdot IC$.

$\angle CIN=\frac{\alpha+\gamma}{2}$ and $\angle MIC=\frac{\beta+\gamma}{2}$ by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that

$MI=IC\cdot\cos\left(\frac{\beta+\gamma}{2}\right),$ $MC=IC\cdot\sin\left(\frac{\beta+\gamma}{2}\right),$ $IN=IC\cdot\cos\left(\frac{\alpha+\gamma}{2}\right),$ $CN=IC\cdot\sin\left(\frac{\alpha+\gamma}{2}\right).$

Plugging in the values and simplifying results in $MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)$ by the angle-addition identity $\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$.

Before we continue, we would like to simplify the value in the sine function. We see that $\frac{\alpha+\beta+2\gamma}{2}=\frac{\gamma}{2}+\frac{\alpha+\beta+\gamma}{2}=\frac{\gamma}{2}+90$. Using the fact that $\cos(A)=\sin(90-A)$ results in

$\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)=\sin\left(90+\frac{\gamma}{2}\right)=\sin\left(90-(-\frac{\gamma}{2})\right)=\cos\left(-\frac{\gamma}{2}\right)=\cos\left(\frac{\gamma}{2}\right).$

How do we simplify $IC$? Well, we can perform the Law of Sines on triangle $AIC$. This results in:

$\frac{AC}{\sin(\angle AIC)}=\frac{IC}{\sin\left(\frac{\alpha}{2}\right)}$

The value of $\angle AIC$ is $\frac{\alpha+2\beta+\gamma}{2}$ by the Exterior Angle Theorem on $\triangle ABI$, so the value of $\sin(\angle AIC)$ is equivalent to the value of $\cos\left(\frac{\beta}{2}\right)$ by a similar argument as above. Then rearranging yields $IC = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}$.

Going back to the previous formula $MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)$ and substituting values yields:

$MN = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}$.

Finally, using the formulae $\sin\left(\frac{\alpha}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos\left(\frac{\alpha}{2}\right)=\sqrt{\frac{s(s-a)}{bc}}$ (where $s$ is half the perimeter of the triangle), we reach our final value:

$MN = b\cdot\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}\cdot\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{s(s-b)}{ac}}}$

$=\frac{\sqrt{s(s-b)(s-c)^2}}{\sqrt{s(s-b)}}$

$=\sqrt{(s-c)^2}$

$=s-c$

$=181-125$

$=\boxed{056}.$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&oldid=222615"