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"

(→‎Solution)
(Added another solution)
Line 8: Line 8:
 
Hence<math>MN=\frac{PQ}{2}</math>.But <math>PQ=BP+AQ-AB=120+117-125=112</math>,so<math>MN=\boxed{56}</math>.
 
Hence<math>MN=\frac{PQ}{2}</math>.But <math>PQ=BP+AQ-AB=120+117-125=112</math>,so<math>MN=\boxed{56}</math>.
  
−
== Solution 2 (Bash) ==  
+
== Solution 2 ==
 +
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{56}</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 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>. <math>\square</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 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>. <math>\square</math>
  

Revision as of 00:54, 30 December 2017

Problem 4

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 intersects lines ${AB}$ at points $P$ and $Q$, respectively. Since ${BM}$ is the angle bisector of angle B,and ${CM}$ is perpendicular to ${BM}$ ,so , $BP=BC=120$, M is the midpoint of ${CP}$ .For the same reason,$AQ=AC=117$,N is the midpoint of ${CQ}$. Hence$MN=\frac{PQ}{2}$.But $PQ=BP+AQ-AB=120+117-125=112$,so$MN=\boxed{56}$.

Solution 2

Let $I$ be the incenter of $ABC$. Now, since $AM \perp LC$ and $AN \perp KB$, we have $AMIN$ is a cyclic quadrilateral. Consequently, $\frac{MN}{\sin \angle MIN} = 2R = AI$. Since $\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK$, 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 $MN=\frac{117+120-125}{2}=\boxed{56}$

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 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 $MN^2 = 3136 \implies MN = \boxed{056}$. $\square$

--lucasxia01

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=89271"