Difference between revisions of "2003 AIME I Problems/Problem 10"
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== Problem == | == 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> | [[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); | ||
+ | |||
+ | /* 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> | ||
+ | |||
__TOC__ | __TOC__ | ||
− | == | + | == Solutions == |
<center><asy> | <center><asy> | ||
pointpen = black; pathpen = black+linewidth(0.7); size(220); | pointpen = black; pathpen = black+linewidth(0.7); size(220); | ||
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<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>. | <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{ | + | 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 === | === 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 | 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> | + | <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 | ||
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<cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath> | <cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath> | ||
− | and multiplying through by 2 and applying the [ | + | 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> | <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> | + | 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 === | === Solution 3 === | ||
− | [[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>. | + | [[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> | ||
+ | |||
+ | By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math> | ||
+ | === Solution 6 === | ||
+ | |||
+ | 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 | ||
+ | |||
+ | === 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. | |
== See also == | == See also == |
Latest revision as of 02:08, 30 September 2024
Problem
Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees in
Contents
Solutions
Solution 1
Take point inside such that and .
. Also, since and are congruent (by ASA), . Hence is an equilateral triangle, so .
Then . We now see that and are congruent. Therefore, , so .
Solution 2
From the givens, we have the following angle measures: , . If we define then we also have . Then apply the Law of Sines to triangles and to get
Clearing denominators, evaluating and applying one of our trigonometric identities to the result gives
and multiplying through by 2 and applying the double angle formulas gives
and so ; since , we must have , so the answer is .
Solution 3
Without loss of generality, let . Then, using the Law of Sines in triangle , we get , and using the sine addition formula to evaluate , we get .
Then, using the Law of Cosines in triangle , we get , since . So triangle is isosceles, and .
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 outside of so that is equilateral. Then, connect , , and to . Also, let intersect at . , , and (trivially) , so by SAS congruence. Also, , so , and , making also equilateral. (it is isosceles with a angle) by SAS (, , and ), and by SAS (, , and ). Thus, is isosceles, with . Also, , so .
Solution 5 (Ceva)
Noticing that we have three concurrent cevians, we apply Ceva's theorem:
using the fact that and we have:
By inspection, works, so the answer is
Solution 6
Let Using sine rule on , letting we get : Simplifying, we get that from where Simplifying more, we get that , so 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.
See also
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 | ||
All AIME Problems and Solutions |
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