Difference between revisions of "1989 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
− | Let <math> | + | Let <math>a</math>, <math>b</math>, <math>c</math> be the three sides of a [[triangle]], and let <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math>, be the angles opposite them. If <math>a^2+b^2=1989c^2</math>, find |
<center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | <center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | We can draw the altitude h to c, to get two right | + | === Solution 1 === |
+ | We can draw the [[altitude]] <math>h</math> to <math>c</math>, to get two [[right triangle]]s. <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the [[cotangent]]. From the definition of area, <math>h=\frac{2A}{c}</math>, so <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math>. | ||
− | + | Now we evaluate the numerator: | |
− | From the | + | <cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath> |
+ | |||
+ | From the [[Law of Cosines]] and the sine area formula, | ||
+ | |||
+ | <cmath>\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ | ||
+ | \sin{\gamma}&= \frac{2A}{ab}\\ | ||
+ | \cot{\gamma}&= \frac{\cos \gamma}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*}</cmath> | ||
+ | |||
+ | Then <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | <cmath>\begin{align*} | ||
+ | \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | By the Law of Cosines, | ||
+ | |||
+ | <cmath>a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2</cmath> | ||
+ | |||
+ | Now | ||
+ | |||
+ | <cmath>\begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}} = \frac {\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}} = \frac {ab}{c^2}\cos{\gamma} = \frac {ab}{c^2} \cdot \frac {994c^2}{ab}\\ &= \boxed{994}\end{align*}</cmath> | ||
+ | |||
+ | |||
+ | === Solution 3=== | ||
+ | |||
+ | Use Law of cosines to give us <math>c^2=a^2+b^2-2ab\cos(\gamma)</math> or therefore <math>\cos(\gamma)=\frac{994c^2}{ab}</math>. Next, we are going to put all the sin's in term of <math>\sin(a)</math>. We get <math>\sin(\gamma)=\frac{c\sin(a)}{a}</math>. Therefore, we get <math>\cot(\gamma)=\frac{994c}{b\sin a}</math>. | ||
+ | |||
+ | Next, use Law of Cosines to give us <math>b^2=a^2+c^2-2ac\cos(\beta)</math>. Therefore, <math>\cos(\beta)=\frac{a^2-994c^2}{ac}</math>. Also, <math>\sin(\beta)=\frac{b\sin(a)}{a}</math>. Hence, <math>\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}</math>. | ||
+ | |||
+ | Lastly, <math>\cos(\alpha)=\frac{b^2-994c^2}{bc}</math>. Therefore, we get <math>\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}</math>. | ||
+ | |||
+ | Now, <math>\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}</math>. After using <math>a^2+b^2=1989c^2</math>, we get <math>\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}</math>. | ||
− | |||
− | + | === Solution 4=== | |
− | |||
− | <math>\ | + | Let <math>\gamma</math> be <math>(180-\alpha-\beta)</math> |
− | <math>\cot{\ | + | <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}</math> |
− | + | WLOG, assume that <math>a</math> and <math>c</math> are legs of right triangle <math>abc</math> with <math>\beta = 90^o</math> and <math>c=1</math> | |
− | <math>\ | + | By Pythagorean theorem, we have <math>b^2=a^2+1</math>, and the given <math>a^2+b^2=1989</math>. Solving the equations gives us <math>a=\sqrt{994}</math> and <math>b=\sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>. |
+ | We see that our derived equation equals to <math>\tan^2 \alpha</math> as <math>\tan \beta</math> approaches infinity. | ||
+ | Evaluating <math>\tan^2 \alpha</math>, we get <math>\boxed{994}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1989|num-b=9|num-a=11}} | {{AIME box|year=1989|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:55, 11 August 2017
Problem
Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Contents
Solution
Solution 1
We can draw the altitude to , to get two right triangles. , from the definition of the cotangent. From the definition of area, , so .
Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
Then .
Solution 2
By the Law of Cosines,
Now
Solution 3
Use Law of cosines to give us or therefore . Next, we are going to put all the sin's in term of . We get . Therefore, we get .
Next, use Law of Cosines to give us . Therefore, . Also, . Hence, .
Lastly, . Therefore, we get .
Now, . After using , we get .
Solution 4
Let be
WLOG, assume that and are legs of right triangle with and
By Pythagorean theorem, we have , and the given . Solving the equations gives us and . We see that , and .
We see that our derived equation equals to as approaches infinity. Evaluating , we get .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.