Difference between revisions of "1996 USAMO Problems/Problem 5"
(→Solution 2) |
(Tag: Undo) |
||
(4 intermediate revisions by 3 users not shown) | |||
Line 3: | Line 3: | ||
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 1== | + | ==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 19: | Line 20: | ||
Thus, <math>AB = BC</math>, our desired conclusion. | Thus, <math>AB = BC</math>, our desired conclusion. | ||
− | ==Solution 2== | + | ===Solution 2=== |
<center> | <center> | ||
Line 52: | Line 53: | ||
<cmath> | <cmath> | ||
− | \begin{ | + | \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] | \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)&=sin(10^\circ)sin(30^\circ)sin(x)\[10] | ||
Line 66: | Line 67: | ||
\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] | \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) | cos(140^\circ-x)&=cos(20^\circ+x) | ||
− | \end{ | + | \end{equations*} |
</cmath> | </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. | 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== | + | ===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 | 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> | <cmath>\sin 10^\circ \sin x \sin 30^\circ = \sin (80^\circ - x) \sin 40^\circ \sin 20^\circ.</cmath> | ||
Line 84: | Line 85: | ||
as desired. | 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.