2003 AIME I Problems/Problem 10

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 $\angle CMB.$

Solution

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

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 $\angle CMB = \angle MCB = \boxed{083^\circ}$ .

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 $\boxed{083}$ .

Solution 3

Without loss of generality, let $AC = BC = 1$ . Then, using 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 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 $\angle CMB = \boxed{083}$ .

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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