Difference between revisions of "1996 USAMO Problems/Problem 5"
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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. | 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. | ||
+ | ==Solution 3== | ||
+ | If <math>\angle{MBC} = x</math> then by Angle Sum in a Triangle we have <math>\angle{MCB} = 80^\circ - x</math>. By Trig Ceva we have | ||
+ | <cmath>\sin 10^\circ \sin x \sin 30^\circ = \sin (80^\circ - x) \sin 40^\circ \sin 20^\circ.</cmath> | ||
+ | Because <math>\dfrac{\sin x}{\sin (80^\circ - x)}</math> is monotonic increasing over <math>(0, \dfrac{\pi}{2})</math>, there is only one solution <math>0 \le x \le \dfrac{\pi}{2}</math> to the equation. We claim it is <math>x = 60^\circ</math>, which will make <math>ABC</math> isosceles with <math>\angle{A} = \angle{C}</math>. | ||
+ | |||
+ | Notice that | ||
+ | <cmath>\sin 20^\circ \sin 20^\circ \sin 40^\circ = 2 \sin 10^\circ \cos 10^\circ \sin 20^\circ \sin 40^\circ</cmath> | ||
+ | <cmath>= \sin 10^\circ (\sin 10^\circ + \frac{1}{2}) \sin 40^\circ</cmath> | ||
+ | <cmath>= \sin 10^\circ (\frac{1}{2} \sin 40^\circ + \frac{1}{2} (\cos 30^\circ - \cos 50^\circ))</cmath> | ||
+ | <cmath>= \sin 10^\circ \frac{1}{2} \cos 30^\circ</cmath> | ||
+ | <cmath>= \sin 10^\circ \sin 30^\circ \sin 60^\circ,</cmath> | ||
+ | as desired. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:36, 13 February 2015
Contents
[hide]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:
\begin{array}[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{array} (Error compiling LaTeX. Unknown error_msg)
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.
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