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

Revision as of 09:28, 8 November 2021

Problem

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

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

Solutions

Solution 1

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

https://youtu.be/QyeaoEtgu-Y

https://youtu.be/NsQbhYfGh1Q?t=3466

~ pi_is_3.14

@@ Line 4: / Line 4: @@
 [[Image:2000_12_AMC-24.png]]
-<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math>
+<math>\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28</math>
 == Solutions ==

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

Page

Toolbox

Search