1996 USAMO Problems/Problem 5

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, $\frac{BM}{CM}=\frac{sin(80^\circ-x)}{sin(x)}$ .

So, we have $\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}$ .

First, let's focus on $sin(20^\circ)sin(40^\circ)$ . By the identities $sin(A-B)sin(A+B)=cos^2(B)-cos^2(A)$ and $cos(3A)=4cos^3(A)-3cos(A)$ , we have

$\begin{align*} sin(20^\circ)sin(40^\circ)&=sin(30^\circ-10^\circ)sin(30^\circ+10^\circ)\\ &=cos^2(10^\circ)-cos^2(30^\circ)\\ &=cos^2(10^\circ)-\frac{3}{4}\\ &=\frac{4cos^3(10^\circ)-3cos(10^\circ)}{4cos(10^\circ)}\\ &=\frac{cos(30^\circ)}{4cos(10^\circ)} \end{align*}$

Substituting back in to the original equality, and using the identity $sin(2A)=2sin(A)cos(A)$ and the facts that $sin(30^\circ)=1/2$ and $cos(30^\circ)=sin(60^\circ)$ , we have

$\begin{align*} \frac{sin(80^\circ-x)}{sin(x)}&=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}\\ &=\frac{4cos(10^\circ)sin(10^\circ)sin(30^\circ)}{cos(30^\circ)}\\ &=\frac{2sin(20^\circ)sin(30^\circ)}{cos(30^\circ)}\\ &=\frac{sin(20^\circ)}{sin(60^\circ)} \end{align*}$

Therefore, $sin(80^\circ-x)sin(60^\circ)=sin(20^\circ)sin(x)$ . Then, using the identity $sin(A)sin(B)=\frac{1}{2}(cos(A-B)-cos(A+B))$ ,

$\begin{align*} sin(80^\circ-x)sin(60^\circ)&=sin(20^\circ)sin(x)\\ \frac{1}{2}(cos(20^\circ-x)-cos(140^\circ-x))&=\frac{1}{2}(cos(20^\circ-x)-cos(20^\circ+x))\\ cos(140^\circ-x)&=cos(20^\circ+x) \end{align*}$

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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1996 USAMO Problems/Problem 5

Problem

Solution 1

Solution 2