Difference between revisions of "1996 USAMO Problems/Problem 5"
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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>. | 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>. | ||
− | + | Combining, we have <math>\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>. From here, we can use the given trigonometric identities at each step: | |
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<cmath> | <cmath> | ||
− | \begin{ | + | \begin{array}[t]{llr} |
\frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\ | \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\ | ||
− | &=\frac{ | + | sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=sin(10^\circ)sin(30^\circ)sin(x)\ |
− | &=\frac{ | + | sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(30^\circ)=1/2]\ |
− | &=\frac{sin( | + | sin(80^\circ-x)sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)\ |
− | + | sin(80^\circ-x)(cos^2(10^\circ)-cos^2(30^\circ))&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(A-B)sin(A+B)=cos^2 B-cos^2 A]\ | |
− | + | sin(80^\circ-x)(cos^2(10^\circ)-\frac{3}{4})&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(30^\circ)=\frac{\sqrt{3}}{2}]\ | |
− | + | sin(80^\circ-x) \frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)\ | |
− | + | sin(80^\circ-x) \frac{cos(30^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(3A)=4cos^3 A-3cos A]\ | |
− | + | sin(80^\circ-x)cos(30^\circ)&=2sin(10^\circ)cos(10^\circ)sin(x)\ | |
− | + | sin(80^\circ-x)cos(30^\circ)&=sin(20^\circ)sin(x)&[sin(2A)=2sin A cos A ]\\ | |
− | \ | + | sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)&[cos(30^\circ)=sin(60^\circ)]\ |
− | 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))&[sin A sin B=\frac{1}{2}(cos(A-B)-cos(A+B))]\ |
− | \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) | cos(140^\circ-x)&=cos(20^\circ+x) | ||
− | \end{ | + | \end{array} |
</cmath> | </cmath> | ||
Revision as of 10:20, 12 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, .
Combining, we have . From here, we can use the given trigonometric identities at each step:
The only acute angle satisfying this equality is . Therefore, and . Thus, is isosceles.
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