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 == | ||
| − | {{ | + | 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 12:13, 1 December 2006
Problem
In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of 
, 
, and 
radians, respectively, where 
. If 
, 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.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
This triangle has semiperimeter 
so by Heron's formula it has area 
. The area of a given triangle with sides of length 
and circumradius of length 
is also given by the formula 
, so 
and 
.
Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle , so by the Law of Cosines,
and so 
and the answer is 
.
