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1996 USAMO Problems/Problem 5

Revision as of 10:20, 12 June 2014 by Mjhassan (talk | contribs) (→‎Solution 2)

Problem

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$ , $\angle MAC= 40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

Solution 1

Clearly, $\angle AMB = 150^\circ$ and $\angle AMC = 110^\circ$. Now by the Law of Sines on triangles $ABM$ and $ACM$, we have \[\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}\] and \[\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.\] Combining these equations gives us \[\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.\] Without loss of generality, let $AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}$ and $AC = \sin 20^\circ \sin 110^\circ$. Then by the Law of Cosines, we have

\begin{align*} BC^2 &= AB^2 + AC^2 - 2(AB)(BC)\cos\angle BAC\\ &= \frac{1}{16} + \sin^2 20^\circ\sin^2 110^\circ - 2\left(\frac{1}{4}\right)\sin 20^\circ\sin 110^\circ\cos 50^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \frac{1}{2}\sin 20^\circ\sin 110^\circ\sin 40^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \sin 20^\circ\sin 110^\circ\sin 20^\circ\cos 20^\circ \\ &= \frac{1}{16} \end{align*}

Thus, $AB = BC$, our desired conclusion.

Solution 2

[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("\(A\)",A,SW); label("\(B\)",B,N); label("\(C\)",C,SE); label("\(M\)",M,NE); [/asy]

By the law of sines, $\frac{BM}{sin(10^\circ)}=\frac{AM}{sin(20^\circ)}$ and $\frac{CM}{sin(40^\circ)}=\frac{AM}{sin(30^\circ)}$, so $\frac{BM}{CM}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}$.

Let $\angle MBC=x$. Then, $\angle MCB=80^\circ-x$. By the law of sines, $\frac{BM}{CM}=\frac{sin(80^\circ-x)}{sin(x)}$.

Combining, we have $\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}$. From here, we can use the given trigonometric identities at each step:

\[\begin{array}[t]{llr} \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\\ sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=sin(10^\circ)sin(30^\circ)sin(x)\\ sin(80^\circ-x)sin(20^\circ)sin(40^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(30^\circ)=1/2]\\ sin(80^\circ-x)sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)&=\frac{1}{2}sin(10^\circ)sin(x)\\ sin(80^\circ-x)(cos^2(10^\circ)-cos^2(30^\circ))&=\frac{1}{2}sin(10^\circ)sin(x)&[sin(A-B)sin(A+B)=cos^2 B-cos^2 A]\\ sin(80^\circ-x)(cos^2(10^\circ)-\frac{3}{4})&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(30^\circ)=\frac{\sqrt{3}}{2}]\\ sin(80^\circ-x) \frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)\\ sin(80^\circ-x) \frac{cos(30^\circ)}{4cos(10^\circ)}&=\frac{1}{2}sin(10^\circ)sin(x)&[cos(3A)=4cos^3 A-3cos A]\\ sin(80^\circ-x)cos(30^\circ)&=2sin(10^\circ)cos(10^\circ)sin(x)\\ sin(80^\circ-x)cos(30^\circ)&=sin(20^\circ)sin(x)&[sin(2A)=2sin A cos A ]\\ sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)&[cos(30^\circ)=sin(60^\circ)]\\ \frac{1}{2}(cos(20^\circ-x)-cos(140^\circ-x))&=\frac{1}{2}(cos(20^\circ-x)-cos(20^\circ+x))&[sin A sin B=\frac{1}{2}(cos(A-B)-cos(A+B))]\\ cos(140^\circ-x)&=cos(20^\circ+x) \end{array}\]

The only acute angle satisfying this equality is $x=60^\circ$. Therefore, $\angle ACB=80^\circ-x+30^\circ=50^\circ$ and $\angle BAC=10^\circ+40^\circ=50^\circ$. Thus, $\triangle ABC$ is isosceles.


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