Difference between revisions of "1989 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
− | 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=1989^{}_{}c^2</math>, find | + | 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=1989^{}_{}c^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> | ||
== Solution == | == Solution == | ||
− | We can draw the altitude h to c, to get two right | + | 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 therefore <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math>. |
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− | <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the cotangent. | ||
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− | From the definition of area, <math>h=\frac{2A}{c}</math>, so therefore <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math> | ||
Now we evaluate the numerator: | Now we evaluate the numerator: | ||
− | < | + | <cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath> |
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− | <math> | + | From the [[Law of Cosines]] (<math>R</math> is the [[circumradius]]), |
− | + | <cmath>\begin{eqnarray*}\cos{\gamma}&=&\frac{1988c^2}{2ab}\ | |
− | + | \sin{\gamma}&=&\frac{c}{2R}\ | |
− | + | \cot{\gamma}&=&\frac{1988cR}{ab}\end{eqnarray*}</cmath> | |
+ | Since <math>R=\frac{abc}{4A}</math>, <math>\cot{\gamma}=\frac{1988c^2}{4A}</math>. <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\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}} | ||
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+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 13:04, 25 November 2007
Problem
Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Solution
We can draw the altitude to , to get two right triangles. , from the definition of the cotangent. From the definition of area, , so therefore .
Now we evaluate the numerator:
From the Law of Cosines ( is the circumradius),
Since , .
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 |