Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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== Problem == | == Problem == | ||
− | + | 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 | |
− | If circular | ||
− | < | + | <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> | ||
− | + | <math>\textbf {(A)}\ 24 \qquad \textbf {(B)}\ 25 \qquad \textbf {(C)}\ 26 \qquad \textbf {(D)}\ 27 \qquad \textbf {(E)}\ 28</math> | |
− | |||
− | + | == Solutions == | |
+ | === 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>. | ||
− | + | == Video Solution by OmegaLearn == | |
+ | https://youtu.be/NsQbhYfGh1Q?t=3466 | ||
− | + | ~ pi_is_3.14 | |
− | |||
− | + | == Video Solution == | |
+ | https://youtu.be/QyeaoEtgu-Y | ||
− | + | == See Also == | |
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− | == See | ||
{{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]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:48, 20 August 2024
Contents
Problem
If circular arcs and have centers at and , respectively, then there exists a circle tangent to both and , and to . If the length of is , then the circumference of the circle is
Solutions
Solution (Pythagorean Theorem)
First, note the triangle is equilateral. Next, notice that since the arc has length 12, it follows that we can find the radius of the sector centered at . . Next, connect the center of the circle to side , and call this length , and call the foot . Since is equilateral, it follows that , and (where O is the center of the circle) is . By the Pythagorean Theorem, you get . Finally, we see that the circumference is .
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=3466
~ pi_is_3.14
Video Solution
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.