Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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== Problem == | == Problem == | ||
− | 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]] to both <math>AC</math> and <math>BC</math>, and to <math>\overline{AB}</math>. If the length of <math>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 == | == Solution == | ||
− | {{ | + | [[Image:2000_12_AMC-24a.png]] |
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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]]. 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]], <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>. | ||
== See also == | == See also == |
Revision as of 11:28, 5 January 2008
Problem
If circular arcs and have centers at and , respectively, then there exists a circle tangent to both and , and to . If the length of is , then the circumference of the circle is
Solution
Since are all radii, it follows that is an equilateral triangle. Draw the circle with center and radius . Then let be the point of tangency of the two circles, and be the intersection of the smaller circle and . Let be the intersection of the smaller circle and . Also define the radii (note that is a diameter of the smaller circle, as 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 ).
By the Power of a Point Theorem, . Since , then . Since is equilateral, , and so $\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi\r_1 \Longrightarrow r_1 = \frac{36}{\pi}$ (Error compiling LaTeX. ! Missing $ inserted.). Thus and the circumference of the circle is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
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 |