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

Latest revision as of 19:59, 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

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 1: / Line 1: @@
 == Problem ==
-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]] to both <math>AC</math> and <math>BC</math>, and to <math>\overline{AB}</math>. If the length of <math>BC</math> is <math>12</math>, then the [[circumference]] of the circle is
+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
-<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math>
+<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>
-{{image}}
+<math>\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28</math>
-== Solution ==
-{{solution}}
+== Solutions ==
+=== Solution 2 (Pythagorean Theorem) ===
-== See also ==
+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>.
+== Video Solution by OmegaLearn ==
+https://youtu.be/NsQbhYfGh1Q?t=3466
+~ pi_is_3.14
+== Video Solution ==
+https://youtu.be/QyeaoEtgu-Y
+== See Also ==
 {{AMC12 box|year=2000|num-b=23|num-a=25}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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