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

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[[Triangle]] <math> ABC </math> is [[isosceles triangle | isosceles]] with <math> AC = BC </math> and <math> \angle ACB = 106^\circ. </math> Point <math> M </math> is in the interior of the triangle so that <math> \angle MAC = 7^\circ </math> and <math> \angle MCA = 23^\circ. </math> Find the number of degrees in <math> \angle CMB. </math>
 
[[Triangle]] <math> ABC </math> is [[isosceles triangle | isosceles]] with <math> AC = BC </math> and <math> \angle ACB = 106^\circ. </math> Point <math> M </math> is in the interior of the triangle so that <math> \angle MAC = 7^\circ </math> and <math> \angle MCA = 23^\circ. </math> Find the number of degrees in <math> \angle CMB. </math>
  
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__TOC__
 
== Solution ==
 
== Solution ==
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[[Image:2003_I_AIME-10.png]]
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<center><asy>
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pointpen = black; pathpen = black+linewidth(0.7); size(220);
  
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From the givens, we have the following [[angle]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>.  If we define <math>m\angle CMB = \theta</math> then we also have <math>m\angle CBM = 97^\circ - \theta</math>.  Then Apply the [[Law of Sines]] to triangles <math>\triangle AMC</math> and <math>\triangle BMC</math> to get
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/* We will WLOG AB = 2 to draw following */
  
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<math>\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}</math>
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pair A=(0,0), B=(2,0), C=(1,Tan(37)), M=IP(A--(2Cos(30),2Sin(30)),B--B+(-2,2Tan(23)));
 +
 
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D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);
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</asy></center>
 +
 
 +
=== Solution 1 ===
 +
<center><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));
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</asy></center>
 +
 
 +
Take point <math>N</math> inside <math>\triangle ABC</math> such that <math>\angle CBN = 7^\circ</math> and <math>\angle BCN = 23^\circ</math>.
 +
 
 +
<math>\angle MCN = 106^\circ - 2\cdot 23^\circ = 60^\circ</math>. Also, since <math>\triangle AMC</math> and <math>\triangle BNC</math> are congruent (by ASA), <math>CM = CN</math>. Hence <math>\triangle CMN</math> is an [[equilateral triangle]], so <math>\angle CNM = 60^\circ</math>.
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Then <math>\angle MNB = 360^\circ - \angle CNM - \angle CNB = 360^\circ - 60^\circ - 150^\circ = 150^\circ</math>. We now see that <math>\triangle MNB</math> and <math>\triangle CNB</math> are congruent. Therefore, <math>CB = MB</math>, so <math>\angle CMB = \angle MCB = \boxed{083^\circ}</math>.
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 +
=== Solution 2 ===
 +
From the givens, we have the following [[angle]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>.  If we define <math>m\angle CMB = \theta</math> then we also have <math>m\angle CBM = 97^\circ - \theta</math>.  Then apply the [[Law of Sines]] to triangles <math>\triangle AMC</math> and <math>\triangle BMC</math> to get
 +
 
 +
<cmath>\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}</cmath>
  
 
Clearing [[denominator]]s, evaluating <math>\sin 150^\circ = \frac 12</math> and applying one of our [[trigonometric identities]] to the result gives
 
Clearing [[denominator]]s, evaluating <math>\sin 150^\circ = \frac 12</math> and applying one of our [[trigonometric identities]] to the result gives
  
−
<math>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</math>
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<cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath>
  
 
and multiplying through by 2 and applying the [[double angle formula]] gives
 
and multiplying through by 2 and applying the [[double angle formula]] gives
  
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<math>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</math>
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<cmath>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</cmath>
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 +
and so <math>\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta \Longleftrightarrow \tan 7^{\circ} = \cot \theta</math>; since <math>0^\circ < \theta < 180^\circ</math>, we must have <math>\theta = 83^\circ</math>, so the answer is <math>083</math>.
  
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and so <math>\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta</math>
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=== Solution 3 ===
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[[Without loss of generality]], let <math>AC = BC = 1</math>.  Then, using [[Law of Sines]] in triangle <math>AMC</math>, we get <math>\frac {1}{\sin 150} = \frac {MC}{\sin 7}</math>, and using the sine addition formula to evaluate <math>\sin 150 = \sin (90 + 60)</math>, we get <math>MC = 2 \sin 7</math>. 
  
−
and, since <math>0^\circ < \theta < 180^\circ</math>, we must have <math>\theta = 83^\circ</math>, so the answer is <math>083</math>.
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Then, using [[Law of Cosines]] in triangle <math>MCB</math>, we get <math>MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1</math>, since <math>\cos 83 = \sin 7</math>. So triangle <math>MCB</math> is isosceles, and <math>\angle CMB = \boxed{083}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:32, 10 June 2008

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 $083$.

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 $\angle CMB = \boxed{083}$.

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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