2000 AMC 12 Problems/Problem 24


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

[asy] label("A", (0,0), W); label("B", (64,0), E); label("C", (32, 32*sqrt(3)), N); draw(arc((0,0),64,0,60)); draw(arc((64,0),64,120,180)); draw((0,0)--(64,0)); draw(circle((32, 24), 24)); [/asy]

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


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 $2{\pi}\cdot 27/2{\pi}=\boxed{(D)27}$.

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution


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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