Difference between revisions of "1989 AIME Problems/Problem 10"

Revision as of 18:57, 11 November 2007

Problem

Let $a_{}^{}$ , $b_{}^{}$ , $c_{}^{}$ be the three sides of a triangle, and let $\alpha_{}^{}$ , $\beta_{}^{}$ , $\gamma_{}^{}$ , be the angles opposite them. If $a^2+b^2=1989^{}_{}c^2$ , find

$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$

Solution

We can draw the altitude h to c, to get two right triangles.

$\cot{\alpha}+\cot{\beta}=\frac{c}{h}$ , from the definition of the cotangent.

From the definition of area, $h=\frac{2A}{c}$ , so therefore $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}$

Now we evaluate the numerator:

$\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}$ .

$\cos{\gamma}=\frac{1988c^2}{2ab}$ , from the Law of Cosines

$\sin{\gamma}=\frac{c}{2R}$ , where R is the circumradius.

$\cot{\gamma}=\frac{1988cR}{ab}$

Since $R=\frac{abc}{4A}$ , $\cot{\gamma}=\frac{1988c^2}{4A}$

$\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=994$

@@ Line 4: / Line 4: @@
 == Solution ==
-{{solution}}
+We can draw the altitude h to c, to get two right triangles.
+<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>
+Now we evaluate the numerator:
+<math>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</math>.
+<math>\cos{\gamma}=\frac{1988c^2}{2ab}</math>, from the [[Law of Cosines]]
+<math>\sin{\gamma}=\frac{c}{2R}</math>, where R is the circumradius.
+<math>\cot{\gamma}=\frac{1988cR}{ab}</math>
+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}=994</math>
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Difference between revisions of "1989 AIME Problems/Problem 10"

Revision as of 18:57, 11 November 2007

Problem

Solution

See also