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

m (→‎See also)
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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 triangles.
+
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>.
−
 
 
−
<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:
 
Now we evaluate the numerator:
  
−
<math>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</math>.
+
<cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath>
−
 
 
−
<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.
 
  
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<math>\cot{\gamma}=\frac{1988cR}{ab}</math>
+
From the [[Law of Cosines]] (<math>R</math> is the [[circumradius]]),
  
−
Since <math>R=\frac{abc}{4A}</math>, <math>\cot{\gamma}=\frac{1988c^2}{4A}</math>
+
<cmath>\begin{eqnarray*}\cos{\gamma}&=&\frac{1988c^2}{2ab}\\
−
 
+
\sin{\gamma}&=&\frac{c}{2R}\\
−
<math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=994</math>
+
\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}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:Intermediate Trigonometry Problems]]

Revision as of 14:04, 25 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:

\[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\]

From the Law of Cosines ($R$ is the circumradius),

\begin{eqnarray*}\cos{\gamma}&=&\frac{1988c^2}{2ab}\\ \sin{\gamma}&=&\frac{c}{2R}\\ \cot{\gamma}&=&\frac{1988cR}{ab}\end{eqnarray*}

Since $R=\frac{abc}{4A}$, $\cot{\gamma}=\frac{1988c^2}{4A}$. $\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}$

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
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