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

Latest revision as of 12:04, 17 December 2024

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

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

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

Solutions

Solution (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 $A$ . $\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}$ . 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 $MB=\frac{18}{\pi}$ , and $OA$ (where $O$ is the center of the circle) is $\frac{36}{\pi}-r$ . By the Pythagorean Theorem, you get $r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}$ . Finally, we see that the circumference is $2{\pi}\cdot \frac{27}{2\pi}=\boxed{(D)27}$ .

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/QyeaoEtgu-Y

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.

@@ Line 15: / Line 15: @@
 == Solutions ==
-=== Solution 2 (Pythagorean Theorem) ===
+=== Solution (Pythagorean Theorem) ===
-First, note the triangle <math>ABC</math> is equilateral. Next, notice that since the arc <math>BC</math> has length 12, it follows that we can find the radius of the sector centered at <math>A</math>. <math>\frac {1}{6}({2}{\pi})AB=12 \implies AB=36/{\pi}</math>. Next, connect the center of the circle to side <math>AB</math>, and call this length <math>r</math>, and call the foot <math>M</math>. Since <math>ABC</math> is equilateral, it follows that <math>MB=18/{\pi}</math>, and <math>OA</math> (where O is the center of the circle) is <math>36/{\pi}-r</math>.  By the Pythagorean Theorem, you get <math>r^2+(18/{\pi})^2=(36/{\pi}-r)^2 \implies r=27/2{\pi}</math>. Finally, we see that the circumference is <math>2{\pi}\cdot 27/2{\pi}=\boxed{(D)27}</math>.
+First, note the triangle <math>ABC</math> is equilateral. Next, notice that since the arc <math>BC</math> has length 12, it follows that we can find the radius of the sector centered at <math>A</math>. <math>\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}</math>. Next, connect the center of the circle to side <math>AB</math>, and call this length <math>r</math>, and call the foot <math>M</math>. Since <math>ABC</math> is equilateral, it follows that <math>MB=\frac{18}{\pi}</math>, and <math>OA</math> (where <math>O</math> is the center of the circle) is <math>\frac{36}{\pi}-r</math>.  By the Pythagorean Theorem, you get <math>r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}</math>. Finally, we see that the circumference is <math>2{\pi}\cdot  \frac{27}{2\pi}=\boxed{(D)27}</math>.
 == Video Solution by OmegaLearn ==

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