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 2 (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 | |
− | + | ||
− | == See | + | == Video Solution == |
+ | https://youtu.be/QyeaoEtgu-Y | ||
+ | |||
+ | == 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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:59, 18 April 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 2 (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.