Difference between revisions of "2000 AMC 12 Problems/Problem 24"

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== Problem ==
 
== Problem ==
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[[Image:2000_12_AMC-24.png|right]]
 
If circular [[arc]]s <math>AC</math> and <math>BC</math> have [[center]]s at <math>B</math> and <math>A</math>, respectively, then there exists a [[circle]] [[tangent (geometry)|tangent]] to both <math>\stackrel{\frown}{AC}</math> and <math>\stackrel{\frown}{BC}</math>, and to <math>\overline{AB}</math>. If the length of <math>\stackrel{\frown}{BC}</math> is <math>12</math>, then the [[circumference]] of the circle is  
 
If circular [[arc]]s <math>AC</math> and <math>BC</math> have [[center]]s at <math>B</math> and <math>A</math>, respectively, then there exists a [[circle]] [[tangent (geometry)|tangent]] to both <math>\stackrel{\frown}{AC}</math> and <math>\stackrel{\frown}{BC}</math>, and to <math>\overline{AB}</math>. If the length of <math>\stackrel{\frown}{BC}</math> is <math>12</math>, then the [[circumference]] of the circle is  
  
 
<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math>
 
<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math>
  
[[Image:2000_12_AMC-24.png]]
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== Solution ==
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[[Image:2000_12_AMC-24a.png|left]]
  
== Solution ==
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Since <math>AB,BC,AC</math> are all [[radius|radii]], it follows that <math>\triangle ABC</math> is an [[equilateral triangle]].
[[Image:2000_12_AMC-24a.png]]
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Draw the circle with center <math>A</math> and radius <math>\overline{AB}</math>. Then let <math>D</math> be the point of tangency of the two circles, and <math>E</math> be the intersection of the smaller circle and <math>\overline{AD}</math>. Let <math>F</math> be the intersection of the smaller circle and <math>\overline{AB}</math>. Also define the radii <math>r_1 = AB, r_2 = \frac{DE}{2}</math> (note that <math>DE</math> is a diameter of the smaller circle, as <math>D</math> is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and <math>D</math>).
  
Since <math>AB,BC,AC</math> are all [[radius|radii]], it follows that <math>\triangle ABC</math> is an [[equilateral triangle]]. Draw the circle with center <math>A</math> and radius <math>\overline{AB}</math>. Then let <math>D</math> be the point of tangency of the two circles, and <math>E</math> be the intersection of the smaller circle and <math>\overline{AD}</math>. Let <math>F</math> be the intersection of the smaller circle and <math>\overline{AB}</math>. Also define the radii <math>r_1 = AB, r_2 = \frac{DE}{2}</math> (note that <math>DE</math> is a diameter of the smaller circle, as <math>D</math> is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and <math>D</math>).
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By the [[Power of a Point Theorem]],  
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<cmath>AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) \cdot AD.</cmath>
  
By the [[Power of a Point Theorem]], <math>AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) AD</math>. Since <math>AD = r_1</math>, then <math>\frac{r_1^2}{4} = r_1 (r_1 - 2r_2) \Longrightarrow r_2 = \frac{3r_1}{8}</math>. Since <math>ABC</math> is equilateral, <math>\angle BAC = 60^{\circ}</math>, and so <math>\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi\r_1 \Longrightarrow r_1 = \frac{36}{\pi}</math>. Thus <math>r_2 = \frac{27}{2}\pi</math> and the circumference of the circle is <math>27\ \mathrm{(D)}</math>.
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Since <math>AD = r_1</math>, then <math>\frac{r_1^2}{4} = r_1 (r_1 - 2r_2) \Longrightarrow r_2 = \frac{3r_1}{8}</math>. Since <math>ABC</math> is equilateral, <math>\angle BAC = 60^{\circ}</math>, and so <math>\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi r_1 \Longrightarrow r_1 = \frac{36}{\pi}</math>. Thus <math>r_2 = \frac{27}{2}\pi</math> and the circumference of the circle is <math>27\ \mathrm{(D)}</math>.
 
   
 
   
 
== See also ==
 
== See also ==

Revision as of 11:45, 5 January 2008

Problem

2000 12 AMC-24.png

If circular arcs $AC$ and $BC$ have centers at $B$ and $A$, respectively, then there exists a circle tangent to both $\stackrel{\frown}{AC}$ and $\stackrel{\frown}{BC}$, and to $\overline{AB}$. If the length of $\stackrel{\frown}{BC}$ is $12$, then the circumference of the circle is

$\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28$

Solution

2000 12 AMC-24a.png

Since $AB,BC,AC$ are all radii, it follows that $\triangle ABC$ is an equilateral triangle.

Draw the circle with center $A$ and radius $\overline{AB}$. Then let $D$ be the point of tangency of the two circles, and $E$ be the intersection of the smaller circle and $\overline{AD}$. Let $F$ be the intersection of the smaller circle and $\overline{AB}$. Also define the radii $r_1 = AB, r_2 = \frac{DE}{2}$ (note that $DE$ is a diameter of the smaller circle, as $D$ is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and $D$).

By the Power of a Point Theorem, \[AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) \cdot AD.\]

Since $AD = r_1$, then $\frac{r_1^2}{4} = r_1 (r_1 - 2r_2) \Longrightarrow r_2 = \frac{3r_1}{8}$. Since $ABC$ is equilateral, $\angle BAC = 60^{\circ}$, and so $\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi r_1 \Longrightarrow r_1 = \frac{36}{\pi}$. Thus $r_2 = \frac{27}{2}\pi$ and the circumference of the circle is $27\ \mathrm{(D)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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