Difference between revisions of "2003 AIME I Problems/Problem 10"
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=== Solution 4 === | === Solution 4 === | ||
− | + | Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. | |
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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 14:22, 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
![[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]](http://latex.artofproblemsolving.com/0/4/c/04c30cea6b00089941864b9928804588638a9952.png)
Contents
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]](http://latex.artofproblemsolving.com/0/4/c/04c30cea6b00089941864b9928804588638a9952.png)
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]](http://latex.artofproblemsolving.com/e/c/8/ec8d127e80906c0e7baf73d163450eef066397cc.png)
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 |
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