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

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== Solutions ==
 
== Solutions ==
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=== Solution 2 (Pythagorean Theorem) ===
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=== 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=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>.
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== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 22:48, 20 August 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 $A$. $\frac {1}{6}({2}{\pi})AB=12 \implies AB=36/{\pi}$. 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

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

~ pi_is_3.14

Video Solution

https://youtu.be/QyeaoEtgu-Y

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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