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

m (imgs)
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== Problem ==
 
== Problem ==
−
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 lengths 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 == <!-- Images obsoleted Image:1985_AIME-9.png, Image:1985_AIME-9a.png by asymptote -->
−
[[Image:1985_AIME-9.png]]
+
<center><asy>
 +
size(200);
 +
pointpen = black; pathpen = black + linewidth(0.8);
 +
real r = 8/15^0.5, a = 57.91, b = 93.135;
 +
pair O = (0,0), A = r*expi(pi/3);
 +
D(CR(O,r));
 +
D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle);
 +
D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle);
 +
D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle);
 +
MP("2",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE);
 +
MP("3",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE);
 +
MP("4",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE);
 +
D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5));
 +
label("\(\alpha+\beta\)",(0.08,0.08),NE,fontsize(8));
 +
</asy></center>
  
 
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]].   
 
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]].   
−
 
+
<center><asy>
−
[[Image:1985_AIME-9a.png]]
+
size(200);
 +
pointpen = black; pathpen = black + linewidth(0.8);
 +
real r = 8/15^0.5, a = 57.91, b = 93.135;
 +
pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A;
 +
D(CR(O,r));
 +
D(O--A1--A2--cycle);
 +
D(O--A2--A3--cycle);
 +
D(O--A1--A3--cycle);
 +
MP("2",(A1+A2)/2,NE);
 +
MP("3",(A2+A3)/2,E);
 +
MP("4",(A1+A3)/2,E);
 +
D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5));
 +
label("\(\alpha\)",(0.07,0.16),NE,fontsize(8));
 +
label("\(\beta\)",(0.12,-0.16),NE,fontsize(8));
 +
</asy></center>
  
 
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>.
 
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>.
  
−
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]],
+
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]],
  
−
<math>2^2 = R^2 + R^2 - 2R^2\cos \alpha</math>
+
<cmath>2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}</cmath>
−
 
+
and the answer is <math>17 + 32 = \boxed{049}</math>.
−
and so <math>\cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}</math> and the answer is <math>17 + 32 = 049</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1985|num-b=8|num-a=10}}
 
{{AIME box|year=1985|num-b=8|num-a=10}}
−
* [[AIME Problems and Solutions]]
 
−
* [[American Invitational Mathematics Examination]]
 
−
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
−
 
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 18:51, 9 April 2008

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

[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP("2",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE); MP("3",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE); MP("4",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE); D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5)); label("\(\alpha+\beta\)",(0.08,0.08),NE,fontsize(8)); [/asy]

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.

[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP("2",(A1+A2)/2,NE); MP("3",(A2+A3)/2,E); MP("4",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); label("\(\alpha\)",(0.07,0.16),NE,fontsize(8)); label("\(\beta\)",(0.12,-0.16),NE,fontsize(8)); [/asy]

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,

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

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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