Difference between revisions of "1996 USAMO Problems/Problem 5"
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Let <math>ABC</math> be a triangle, and <math>M</math> an interior point such that <math>\angle MAB=10^\circ </math>, <math>\angle MBA=20^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles. | Let <math>ABC</math> be a triangle, and <math>M</math> an interior point such that <math>\angle MAB=10^\circ </math>, <math>\angle MBA=20^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles. | ||
− | ==Solution== | + | ==Solution 1== |
Clearly, <math>\angle AMB = 150^\circ</math> and <math>\angle AMC = 110^\circ</math>. Now by the Law of Sines on triangles <math>ABM</math> and <math>ACM</math>, we have <cmath>\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}</cmath> and <cmath>\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.</cmath> Combining these equations gives us <cmath>\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.</cmath> Without loss of generality, let <math>AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}</math> and <math>AC = \sin 20^\circ \sin 110^\circ</math>. Then by the Law of Cosines, we have | Clearly, <math>\angle AMB = 150^\circ</math> and <math>\angle AMC = 110^\circ</math>. Now by the Law of Sines on triangles <math>ABM</math> and <math>ACM</math>, we have <cmath>\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}</cmath> and <cmath>\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.</cmath> Combining these equations gives us <cmath>\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.</cmath> Without loss of generality, let <math>AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}</math> and <math>AC = \sin 20^\circ \sin 110^\circ</math>. Then by the Law of Cosines, we have | ||
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Thus, <math>AB = BC</math>, our desired conclusion. | Thus, <math>AB = BC</math>, our desired conclusion. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <center> | ||
+ | <asy> | ||
+ | |||
+ | pair A,B,C,M; | ||
+ | A=(0,0); | ||
+ | B=(1,2); | ||
+ | C=(2,0); | ||
+ | M=(0.8,1.1); | ||
+ | |||
+ | draw(A--B); | ||
+ | draw(B--C); | ||
+ | draw(C--A); | ||
+ | draw(A--M); | ||
+ | draw(B--M); | ||
+ | draw(C--M); | ||
+ | |||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </asy> | ||
+ | </center> | ||
+ | |||
+ | By the law of sines, <math>\frac{BM}{sin(10^\circ)}=\frac{AM}{sin(20^\circ)}</math> and <math>\frac{CM}{sin(40^\circ)}=\frac{AM}{sin(30^\circ)}</math>, so <math>\frac{BM}{CM}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>. | ||
+ | |||
+ | Let <math>\angle MBC=x</math>. Then, <math>\angle MCB=80^\circ-x</math>. By the law of sines, <math>\frac{BM}{CM}=\frac{sin(80^\circ-x)}{sin(x)}</math>. | ||
+ | |||
+ | So, we have <math>\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>. | ||
+ | |||
+ | First, let's focus on <math>sin(20^\circ)sin(40^\circ)</math>. By the identities <math>sin(A-B)sin(A+B)=cos^2(B)-cos^2(A)</math> and <math>cos(3A)=4cos^3(A)-3cos(A)</math>, we have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | sin(20^\circ)sin(40^\circ)&=sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)\ | ||
+ | &=cos^2(10^\circ)-cos^2(30^\circ)\ | ||
+ | &=cos^2(10^\circ)-\frac{3}{4}\ | ||
+ | &=\frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}\ | ||
+ | &=\frac{cos(30^\circ)}{4cos(10^\circ)} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Substituting back in to the original equality, and using the identity <math>sin(2A)=2sin(A)cos(A)</math> and the facts that <math>sin(30^\circ)=1/2</math> and <math>cos(30^\circ)=sin(60^\circ)</math>, we have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\ | ||
+ | &=\frac{4cos(10^\circ)sin(10^\circ)sin(30^\circ)}{cos(30^\circ)}\ | ||
+ | &=\frac{2sin(20^\circ)sin(30^\circ)}{cos(30^\circ)}\ | ||
+ | &=\frac{sin(20^\circ)}{sin(60^\circ)} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, <math>sin(80^\circ-x)sin(60^\circ)=sin(20^\circ)sin(x)</math>. Then, using the identity <math>sin(A)sin(B)=\frac{1}{2}(cos(A-B)-cos(A+B))</math>, | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)\ | ||
+ | \frac{1}{2}(cos(20^\circ-x)-cos(140^\circ-x))&=\frac{1}{2}(cos(20^\circ-x)-cos(20^\circ+x))\ | ||
+ | cos(140^\circ-x)&=cos(20^\circ+x) | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | The only acute angle satisfying this equality is <math>x=60^\circ</math>. Therefore, <math>\angle ACB=80^\circ-x+30^\circ=50^\circ</math> and <math>\angle BAC=10^\circ+40^\circ=50^\circ</math>. Thus, <math>\triangle ABC</math> is isosceles. | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:34, 11 June 2014
Problem
Let be a triangle, and an interior point such that , , and . Prove that the triangle is isosceles.
Solution 1
Clearly, and . Now by the Law of Sines on triangles and , we have and Combining these equations gives us Without loss of generality, let and . Then by the Law of Cosines, we have
Thus, , our desired conclusion.
Solution 2
By the law of sines, and , so .
Let . Then, . By the law of sines, .
So, we have .
First, let's focus on . By the identities and , we have
Substituting back in to the original equality, and using the identity and the facts that and , we have
Therefore, . Then, using the identity ,
The only acute angle satisfying this equality is . Therefore, and . Thus, is isosceles.
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