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Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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== Problem ==
 
== Problem ==
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[[Image:2000_12_AMC-24.png|right]]
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If circular arcs <math>AC</math> and <math>BC</math> have centers at <math>B</math> and <math>A</math>, respectively, then there exists a circle tangent to both <math>\overarc {AC}</math> and <math>\overarc{BC}</math>, and to <math>\overline{AB}</math>. If the length of <math>\overarc{BC}</math> is <math>12</math>, then the circumference of the circle is
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If circular [[arc]]s <math>AC</math> and <math>BC</math> have [[center]]s at <math>B</math> and <math>A</math>, respectively, then there exists a [[circle]] [[tangent (geometry)|tangent]] to both <math>\stackrel{\frown}{AC}</math> and <math>\stackrel{\frown}{BC}</math>, and to <math>\overline{AB}</math>. If the length of <math>\stackrel{\frown}{BC}</math> is <math>12</math>, then the [[circumference]] of the circle is  
 
  
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<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math>
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<asy>
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label("A", (0,0), W);
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label("B", (64,0), E);
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label("C", (32, 32*sqrt(3)), N);
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draw(arc((0,0),64,0,60));
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draw(arc((64,0),64,120,180));
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draw((0,0)--(64,0));
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draw(circle((32, 24), 24));
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</asy>
  
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== Solution ==
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<math>\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28</math>
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[[Image:2000_12_AMC-24a.png|left]]
 
  
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Since <math>AB,BC,AC</math> are all [[radius|radii]], it follows that <math>\triangle ABC</math> is an [[equilateral triangle]].  
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== Solutions ==
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=== Solution 2 (Pythagorean Theorem) ===
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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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Draw the circle with center <math>A</math> and radius <math>\overline{AB}</math>. Then let <math>D</math> be the point of tangency of the two circles, and <math>E</math> be the intersection of the smaller circle and <math>\overline{AD}</math>. Let <math>F</math> be the intersection of the smaller circle and <math>\overline{AB}</math>. Also define the radii <math>r_1 = AB, r_2 = \frac{DE}{2}</math> (note that <math>DE</math> is a diameter of the smaller circle, as <math>D</math> 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 <math>D</math>).
 
  
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By the [[Power of a Point Theorem]],
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== Video Solution by OmegaLearn ==
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<cmath>AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) \cdot AD.</cmath>
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https://youtu.be/NsQbhYfGh1Q?t=3466
  
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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>.
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~ pi_is_3.14
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== See also ==
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== Video Solution ==
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https://youtu.be/QyeaoEtgu-Y
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== See Also ==
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}
 
{{AMC12 box|year=2000|num-b=23|num-a=25}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 18:59, 18 April 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 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

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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