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Difference between revisions of "2003 AIME I Problems/Problem 10"

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By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math>
 
By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math>
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=== Solution 6 (Very simple) ===
 
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Draw point K so that MK is parallel to
 
  
 
== See also ==
 
== See also ==

Revision as of 16:31, 21 April 2019

Problem

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

[asy] pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw following */ pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); [/asy]


Solutions

[asy] pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw following */ pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); [/asy]

Solution 1

[asy] pointpen = black; pathpen = black+linewidth(0.7); size(220); /* We will WLOG AB = 2 to draw following */ pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23))), N=(2-M.x,M.y); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); D(C--D(MP("N",N))--B--N--M,linetype("6 6")+linewidth(0.7)); [/asy]

Take point $N$ inside $\triangle ABC$ such that $\angle CBN = 7^\circ$ and $\angle BCN = 23^\circ$.

$\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ$. Also, since $\triangle AMC$ and $\triangle BNC$ are congruent (by ASA), $CM = CN$. Hence $\triangle CMN$ is an equilateral triangle, so $\angle CNM = 60^\circ$.

Then $\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ$. We now see that $\triangle MNB$ and $\triangle CNB$ are congruent. Therefore, $CB = MB$, so $\angle CMB = \angle MCB = \boxed{83^\circ}$.

Solution 2

From the givens, we have the following angle measures: $m\angle AMC = 150^\circ$, $m\angle MCB = 83^\circ$. If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$. Then apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get

\[\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin (97^\circ - \theta)}\]

Clearing denominators, evaluating $\sin 150^\circ = \frac 12$ and applying one of our trigonometric identities to the result gives

\[\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta\]

and multiplying through by 2 and applying the double angle formula gives

\[\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta\]

and so $\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta$; since $0^\circ < \theta < 180^\circ$, we must have $\theta = 83^\circ$, so the answer is $\boxed{83}$.

Solution 3

Without loss of generality, let $AC = BC = 1$. Then, using the Law of Sines in triangle $AMC$, we get $\frac {1}{\sin 150} = \frac {MC}{\sin 7}$, and using the sine addition formula to evaluate $\sin 150 = \sin (90 + 60)$, we get $MC = 2 \sin 7$.

Then, using the Law of Cosines in triangle $MCB$, we get $MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1$, since $\cos 83 = \sin 7$. So triangle $MCB$ is isosceles, and $\angle CMB = \boxed{83}$.

Solution 4

Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros

First, take point $E$ outside of $\triangle{ABC}$ so that $\triangle{CEB}$ is equilateral. Then, connect $A$, $C$, and $M$ to $E$. Also, let $ME$ intersect $AB$ at $F$. $\angle{MCE} = 83^\circ - 60^\circ = 23^\circ$, $CE = AB$, and (trivially) $CM = CM$, so $\triangle{MCE} \cong \triangle{MCA}$ by SAS congruence. Also, $\angle{CMA} = \angle{CME} = 150^\circ$, so $\angle{AME} = 60^\circ$, and $AM = ME$, making $\triangle{AME}$ also equilateral. (it is isosceles with a $60^\circ$ angle) $\triangle{MAF} \cong \triangle{EAF}$ by SAS ($MA = AE$, $AF = AF$, and $m\angle{MAF} = m\angle{EAF} = 30^\circ$), and $\triangle{MAB} \cong \triangle{EAB}$ by SAS ($MA = AE$, $AB = AB$, and $m\angle{MAB} = m\angle{EAB} = 30^\circ$). Thus, $\triangle{BME}$ is isosceles, with $m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ$. Also, $\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ$, so $\angle{CME} = 150^\circ - 67^\circ = \boxed{83^\circ}$.

Solution 5 (Ceva)

Noticing that we have three concurrent cevians, we apply Ceva's theorem:

\[(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM)\] \[(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)\]

using the fact that $\sin 83 = \cos 7$ and $(\sin 7)(\cos 7) = 1/2 (\sin 14)$ we have:

\[(\sin 23)(\sin x) = (\sin 14)(\sin 37-x)\]

By inspection, $x=14^\circ$ works, so the answer is $180-83-14= \boxed{083}$

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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