Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See also)
(→‎Solution)
Line 16: Line 16:
  
 
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>.
 
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>.
−
+
 
 +
(Alternatively, the [[Pythagorean Theorem]] can also be used to find <math>r_2</math> in terms of <math>r_1</math>. Notice that since AB is tangent to circle <math>O</math>, <math>\overline{OF}</math> is perpendicular to <math>\overline{AF}</math>. Therefore,
 +
 
 +
<cmath>AF^2 + OF^2 = AO^2</cmath>
 +
<cmath>(\frac {r_1}{2})^2 + r_2^2 = (r_1 - r_2)^2</cmath>
 +
 
 +
After simplification, <math>r_2 = \frac{3r_1}{8}</math>.)
 +
 
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}

Revision as of 02:51, 20 December 2014

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

(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,

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_24&oldid=66617"