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Difference between revisions of "1996 USAMO Problems/Problem 5"
(→‎Solution)
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Let <math>ABC</math> be a triangle, and <math>M</math> an interior point such that <math>\angle MAB=10^\circ </math>, <math>\angle MBA=20^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles.
 
Let <math>ABC</math> be a triangle, and <math>M</math> an interior point such that <math>\angle MAB=10^\circ </math>, <math>\angle MBA=20^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles.
  
−
==Solution==
+
==Solution 1==
  
 
Clearly, <math>\angle AMB = 150^\circ</math> and <math>\angle AMC = 110^\circ</math>. Now by the Law of Sines on triangles <math>ABM</math> and <math>ACM</math>, we have <cmath>\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}</cmath> and <cmath>\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.</cmath> Combining these equations gives us <cmath>\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.</cmath> Without loss of generality, let <math>AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}</math> and <math>AC = \sin 20^\circ \sin 110^\circ</math>. Then by the Law of Cosines, we have
 
Clearly, <math>\angle AMB = 150^\circ</math> and <math>\angle AMC = 110^\circ</math>. Now by the Law of Sines on triangles <math>ABM</math> and <math>ACM</math>, we have <cmath>\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}</cmath> and <cmath>\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.</cmath> Combining these equations gives us <cmath>\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.</cmath> Without loss of generality, let <math>AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}</math> and <math>AC = \sin 20^\circ \sin 110^\circ</math>. Then by the Law of Cosines, we have
Line 18: Line 18:
  
 
Thus, <math>AB = BC</math>, our desired conclusion.
 
Thus, <math>AB = BC</math>, our desired conclusion.
 +
 +
==Solution 2==
 +
 +
<center>
 +
<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>
 +
</center>
 +
 +
By the law of sines, <math>\frac{BM}{sin(10^\circ)}=\frac{AM}{sin(20^\circ)}</math> and <math>\frac{CM}{sin(40^\circ)}=\frac{AM}{sin(30^\circ)}</math>, so <math>\frac{BM}{CM}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>.
 +
 +
Let <math>\angle MBC=x</math>. Then, <math>\angle MCB=80^\circ-x</math>. By the law of sines, <math>\frac{BM}{CM}=\frac{sin(80^\circ-x)}{sin(x)}</math>.
 +
 +
So, we have <math>\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>.
 +
 +
First, let's focus on <math>sin(20^\circ)sin(40^\circ)</math>. By the identities <math>sin(A-B)sin(A+B)=cos^2(B)-cos^2(A)</math> and <math>cos(3A)=4cos^3(A)-3cos(A)</math>, we have
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Substituting back in to the original equality, and using the identity <math>sin(2A)=2sin(A)cos(A)</math> and the facts that <math>sin(30^\circ)=1/2</math> and <math>cos(30^\circ)=sin(60^\circ)</math>, we have
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore, <math>sin(80^\circ-x)sin(60^\circ)=sin(20^\circ)sin(x)</math>. Then, using the identity <math>sin(A)sin(B)=\frac{1}{2}(cos(A-B)-cos(A+B))</math>,
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
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.
 +
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:34, 11 June 2014

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.


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