Difference between revisions of "1996 USAMO Problems/Problem 5"
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==Solution== | ==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 | ||
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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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </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*} | ||
+ | </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 be a triangle, and an interior point such that , , and . Prove that the triangle is isosceles.
Solution
Solution 1
Clearly, and . Now by the Law of Sines on triangles and , we have and Combining these equations gives us Without loss of generality, let and . Then by the Law of Cosines, we have
Thus, , our desired conclusion.
Solution 2
By the law of sines, and , so .
Let . Then, . By the law of sines, .
Combining, we have . 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 . Therefore, and . Thus, is isosceles.
Solution 3
If then by Angle Sum in a Triangle we have . By Trig Ceva we have Because is monotonic increasing over , there is only one solution to the equation. We claim it is , which will make isosceles with .
Notice that 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.