Difference between revisions of "2003 AIME I Problems/Problem 10"
(→Solution 4) |
m (→Solution 4) |
||
Line 62: | Line 62: | ||
=== Solution 4 === | === Solution 4 === | ||
− | + | Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. | |
+ | |||
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>, | 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>. | 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>. |
Revision as of 13:22, 20 May 2018
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 formula 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.
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.