Difference between revisions of "1985 AIME Problems/Problem 9"

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In a [[circle]], [[parallel]] [[chord]]s of [[length]]s 2, 3, and 4 determine [[central angle]]s of <math>\alpha</math>, <math>\beta</math>, and <math>\alpha + \beta</math> [[radian]]s, respectively, where <math>\alpha + \beta < \pi</math>. If <math>\cos \alpha</math>, which is a [[positive]] [[rational number]], is expressed as a [[fraction]] in lowest terms, what is the sum of its [[numerator]] and [[denominator]]?
 
In a [[circle]], [[parallel]] [[chord]]s of [[length]]s 2, 3, and 4 determine [[central angle]]s of <math>\alpha</math>, <math>\beta</math>, and <math>\alpha + \beta</math> [[radian]]s, respectively, where <math>\alpha + \beta < \pi</math>. If <math>\cos \alpha</math>, which is a [[positive]] [[rational number]], is expressed as a [[fraction]] in lowest terms, what is the sum of its [[numerator]] and [[denominator]]?
 
== Solution ==
 
== Solution ==
{{solution}}
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All chords of a given length in a given circle subtend the same [[arc]] and therefore the same central angle.  Thus, by the given, we can re-arrange our chords into a [[triangle]] with the circle as its [[circumcircle]]. 
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{{image}}
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This triangle has [[semiperimeter]] <math>\frac{2 + 3 + 4}{2}</math> so by [[Heron's formula]] it has [[area]] <math>K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}</math>.  The area of a given triangle with sides of length <math>a, b, c</math> and circumradius of length <math>R</math> is also given by the formula <math>K = \frac{abc}{4R}</math>, so <math>\frac6R = \frac{3}{4}\sqrt{15}</math> and <math>R = \frac8{\sqrt{15}}</math>.
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Now, consider the triangle formed by two radii and the chord of length 2.  This [[isosceles triangle]] has vertex [[angle]] <math>\alpha</math>, so by the [[Law of Cosines]],
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<math>2^2 = R^2 + R^2 - 2R^2\cos \alpha</math>
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and so <math>\cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}</math> and the answer is <math>17 + 32 = 049</math>.
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== See also ==
 
== See also ==
 
* [[1985 AIME Problems/Problem 8 | Previous problem]]
 
* [[1985 AIME Problems/Problem 8 | Previous problem]]

Revision as of 13:13, 1 December 2006

Problem

In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution

All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle.


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This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$. The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$, so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$.

Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle $\alpha$, so by the Law of Cosines,

$2^2 = R^2 + R^2 - 2R^2\cos \alpha$

and so $\cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}$ and the answer is $17 + 32 = 049$.

See also