1996 USAMO Problems/Problem 5
Problem
Let be a triangle, and an interior point such that , , and . Prove that the triangle is isosceles.
Solution
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:
\begin{equations*}[t]{llr} \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\\[10] sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=sin(10^\circ)sin(30^\circ)sin(x)\\[10] sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(30^\circ)=1/2]\\[10] sin(80^\circ-x)sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)\\[10] 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]\\[10] 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}]\\[10] sin(80^\circ-x) \frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)\\[10] 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]\\[10] sin(80^\circ-x)cos(30^\circ)&=2sin(10^\circ)cos(10^\circ)sin(x)\\[10] sin(80^\circ-x)cos(30^\circ)&=sin(20^\circ)sin(x)&[sin(2A)=2sin A cos A ]\\[10] sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)&[cos(30^\circ)=sin(60^\circ)]\\[10] \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))]\\[10] cos(140^\circ-x)&=cos(20^\circ+x) \end{equations*} (Error compiling LaTeX. ! LaTeX Error: Environment equations* undefined.)
The only acute angle satisfying this equality is . Therefore, and . Thus, is isosceles.
Solution 3
If then by Angle Sum in a Triangle we have . By Trig Ceva we have Because is monotonic increasing over , there is only one solution to the equation. We claim it is , which will make isosceles with .
Notice that as desired.
See Also
1996 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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