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

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== Problem ==
 
== Problem ==
 
[[Image:2000_12_AMC-24.png|right]]
 
[[Image:2000_12_AMC-24.png|right]]
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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>\overarc{AC}</math> and <math>\overarc{BC}</math>, and to <math>\overline{AB}</math>. If the length of <math>\overarc{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>

Revision as of 23:15, 12 December 2018

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 $\overarc{AC}$ and $\overarc{BC}$, and to $\overline{AB}$. If the length of $\overarc{BC}$ 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 1

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 $27\ \mathrm{(D)}$.

(Alternatively, the Pythagorean Theorem can also be used to find $r_2$ in terms of $r_1$. Notice that since AB is tangent to circle $O$, $\overline{OF}$ is perpendicular to $\overline{AF}$. Therefore,

\[AF^2 + OF^2 = AO^2\] \[\left(\frac {r_1}{2}\right)^2 + r_2^2 = (r_1 - r_2)^2\]

After simplification, $r_2 = \frac{3r_1}{8}$.

Solution 2 (Pythagorean Theorem)

First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$. $\frac {1}{6}({2}{\pi})AB=12 \implies AB=36/{\pi}$. Next, connect the center of the circle to side $AB$, and call this length $r$, and call the foot $M$. Since $ABC$ is equilateral, it follows that $MB=18/{\pi}$, and $OA$ (where O is the center of the circle) is $36/{\pi}-r$. By the pythagorean theorem, you get $r^2+(18/{\pi})^2=(36/{\pi}-r)^2 \implies r=27/2{\pi}$. Finally, we see that the circumference is $2{\pi}*27/2{\pi}=\boxed{(D)27}$.

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

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