1996 USAMO Problems/Problem 5: Difference between revisions

Revision as of 12:36, 13 February 2015

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)}\\[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{array} (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.

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 71: / Line 71: @@
 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.
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