Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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== Solutions == | == Solutions == | ||
− | === Solution | + | === 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 23:48, 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.