1996 USAMO Problems/Problem 5


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:

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) \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)&=sin(20^\circ)sin(x)&[sin(2A)=2sin A cos A ]\\[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]
\end{equations*} (Error compiling LaTeX. Unknown error_msg)

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.

Solution 3

If $\angle{MBC} = x$ then by Angle Sum in a Triangle we have $\angle{MCB} = 80^\circ - x$. By Trig Ceva we have \[\sin 10^\circ \sin x \sin 30^\circ = \sin (80^\circ - x) \sin 40^\circ \sin 20^\circ.\] Because $\dfrac{\sin x}{\sin (80^\circ - x)}$ is monotonic increasing over $(0, \dfrac{\pi}{2})$, there is only one solution $0 \le x \le \dfrac{\pi}{2}$ to the equation. We claim it is $x = 60^\circ$, which will make $ABC$ isosceles with $\angle{A} = \angle{C}$.

Notice that \[\sin 20^\circ \sin 20^\circ \sin 40^\circ = 2 \sin 10^\circ \cos 10^\circ \sin 20^\circ \sin 40^\circ\] \[= \sin 10^\circ (\sin 10^\circ + \frac{1}{2}) \sin 40^\circ\] \[= \sin 10^\circ (\frac{1}{2} \sin 40^\circ + \frac{1}{2} (\cos 30^\circ - \cos 50^\circ))\] \[= \sin 10^\circ \frac{1}{2} \cos 30^\circ\] \[= \sin 10^\circ \sin 30^\circ \sin 60^\circ,\] as desired.

See Also

1996 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

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