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

Latest revision as of 01:08, 30 September 2024

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 formulas 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 $180-83-14= \boxed{083}$

Solution 6

Let $\angle{APC} = \theta^{\circ}$ Using sine rule on $\triangle{APB}, \triangle{APC}$ , letting $AP=d$ we get : $\frac{d}{1} = \frac{\sin{7^{\circ}}}{\sin{150^{\circ}}} = 2\sin{7^{\circ}}= \frac{\sin{14^{\circ}}}{\cos{7^{\circ}}}= \frac{\sin{14^{\circ}}}{\sin{83^{\circ}}}= \frac{\sin{(97-\theta)^{\circ}}}{\sin{\theta^{\circ}}}$ Simplifying, we get that $\cos{(14-\theta)^{\circ}}-\cos{(14+\theta)^{\circ}}=\cos{(14-\theta)^{\circ}}-\cos{(180-\theta)^{\circ}},$ from where $\cos{(14-\theta)^{\circ}}=\cos{(180-\theta)^{\circ}}$ Simplifying more, we get that $\sin{97^{\circ}} \cdot \sin{(\theta-83)^{\circ}} = 0$ , so $\theta = 83^{\circ}$ NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities ~Prabh1512

Solution 7

Because protractors are allowed on the AIME, it is practical to solve this problem by carefully drawing the picture and measuring the angle.

@@ Line 1: / Line 1: @@
 == Problem ==
 [[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>
+<center><asy>
+pointpen = black; pathpen = black+linewidth(0.7); size(220);
-== Solution ==
+/* We will WLOG AB = 2 to draw following */
-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
-<math>\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}</math>
+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></center>
+__TOC__
+== Solutions ==
+<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)));
+D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);
+</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));
+</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>.
+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{83^\circ}</math>.
+=== 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
-<math>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</math>
+<cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath>
+and multiplying through by 2 and applying the [https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Double-angle_identities| double angle formulas] gives
+<cmath>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</cmath>
+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>\boxed{83}</math>.
+=== Solution 3 ===
+[[Without loss of generality]], let <math>AC = BC = 1</math>.  Then, using the [[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>.
+Then, using the [[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{83}</math>.
+=== 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 <math>E</math> outside of <math>\triangle{ABC}</math> so that <math>\triangle{CEB}</math> is equilateral. Then, connect <math>A</math>, <math>C</math>, and <math>M</math> to <math>E</math>. Also, let <math>ME</math> intersect <math>AB</math> at <math>F</math>. <math>\angle{MCE} = 83^\circ - 60^\circ = 23^\circ</math>, <math>CE = AB</math>, and (trivially) <math>CM = CM</math>, so <math>\triangle{MCE} \cong
+\triangle{MCA}</math> by SAS congruence. Also, <math>\angle{CMA} = \angle{CME} = 150^\circ</math>, so <math>\angle{AME} = 60^\circ</math>, and <math>AM = ME</math>,
+making <math>\triangle{AME}</math> also equilateral. (it is isosceles with a <math>60^\circ</math> angle) <math>\triangle{MAF} \cong \triangle{EAF}</math> by SAS (<math>MA = AE</math>,
+<math>AF = AF</math>, and <math>m\angle{MAF} = m\angle{EAF} = 30^\circ</math>), and <math>\triangle{MAB} \cong \triangle{EAB}</math> by SAS (<math>MA = AE</math>, <math>AB = AB</math>, and
+<math>m\angle{MAB} = m\angle{EAB} = 30^\circ</math>). Thus, <math>\triangle{BME}</math> is isosceles, with <math>m\angle{BME} = m\angle{BEM} = 60^\circ + 7^\circ = 67^\circ</math>. Also, <math>\angle{EMB} + \angle {CMB} = \angle{CME} = 150^\circ</math>, so <math>\angle{CME} = 150^\circ - 67^\circ  = \boxed{83^\circ}</math>.
+=== Solution 5 (Ceva) ===
+Noticing that we have three concurrent cevians, we apply Ceva's theorem:
+<cmath>(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\sin \angle BCM) </cmath>
+<cmath>(\sin 23)(\sin 30)(\sin x) = (\sin 7)(\sin 37-x)(\sin 83)</cmath>
+using the fact that <math>\sin 83 = \cos 7</math> and <math>(\sin 7)(\cos 7) = 1/2 (\sin 14)</math> we have:
+<cmath> (\sin 23)(\sin x) = (\sin 14)(\sin 37-x)</cmath>
-and multiplying through by 2 and applying the [[double angle formula]] gives
+By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math>
+=== Solution 6 ===
-<math>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</math>
+Let <math>\angle{APC} = \theta^{\circ}</math>
+Using sine rule on <math>\triangle{APB}, \triangle{APC}</math>, letting <math>AP=d</math> we get :
+<math>\frac{d}{1} = \frac{\sin{7^{\circ}}}{\sin{150^{\circ}}} = 2\sin{7^{\circ}}= \frac{\sin{14^{\circ}}}{\cos{7^{\circ}}}= \frac{\sin{14^{\circ}}}{\sin{83^{\circ}}}= \frac{\sin{(97-\theta)^{\circ}}}{\sin{\theta^{\circ}}}</math>
+Simplifying, we get that
+<math>\cos{(14-\theta)^{\circ}}-\cos{(14+\theta)^{\circ}}=\cos{(14-\theta)^{\circ}}-\cos{(180-\theta)^{\circ}},</math> from where <math>\cos{(14-\theta)^{\circ}}=\cos{(180-\theta)^{\circ}}</math>
+Simplifying more, we get that <math>\sin{97^{\circ}} \cdot \sin{(\theta-83)^{\circ}} = 0</math>, so <math>\theta = 83^{\circ}</math>
+NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities
+~Prabh1512
-and so <math>\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta</math>
+=== Solution 7 ===
-and, since <math>0^\circ < \theta < 180^\circ</math>, we must have <math>\theta = 83^\circ</math>, so the answer is <math>083</math>.
+Because protractors are allowed on the AIME, it is practical to solve this problem by carefully drawing the picture and measuring the angle.
 == See also ==
@@ Line 24: / Line 102: @@
 [[Category:Intermediate Geometry Problems]]
 [[Category:Intermediate Trigonometry Problems]]
+{{MAA Notice}}

2003 AIME I (Problems • Answer Key • Resources)
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