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Difference between revisions of "1996 USAMO Problems/Problem 5"
(Undo revision 156978 by Edud looc (talk))
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'''Problem:'''
+
==Problem==
  
−
----
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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=10^\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
  
 +
<cmath>
 +
\begin{align*}
 +
BC^2 &= AB^2 + AC^2 - 2(AB)(BC)\cos\angle BAC\\
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&= \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 \\
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&= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \sin 20^\circ\sin 110^\circ\sin 20^\circ\cos 20^\circ \\
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&= \frac{1}{16}
 +
\end{align*}
 +
</cmath>
  
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Thus, <math>AB = BC</math>, our desired conclusion.
  
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'''Solution:'''
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===Solution 2===
−
----
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 +
<center>
 +
<asy>
 +
 
 +
pair A,B,C,M;
 +
A=(0,0);
 +
B=(1,2);
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C=(2,0);
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M=(0.8,1.1);
 +
 
 +
draw(A--B);
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draw(B--C);
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draw(C--A);
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draw(A--M);
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draw(B--M);
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draw(C--M);
 +
 
 +
label("\(A\)",A,SW);
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label("\(B\)",B,N);
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label("\(C\)",C,SE);
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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>.
 +
 
 +
Combining, we have <math>\frac{sin(80^\circ-x)}{sin(x)}=\frac{sin(10^\circ)sin(30^\circ)}{sin(20^\circ)sin(40^\circ)}</math>. From here, we can use the given trigonometric identities at each step:
 +
 
 +
<cmath>
 +
\begin{equations*}[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{equations*}
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</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.
 +
 
 +
===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.
 +
 
 +
== See Also ==
 +
{{USAMO newbox|year=1996|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 15:35, 28 June 2021

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

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{equations*}[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{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 (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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

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