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 | + | |
+ | [[Image:2000_12_AMC-24.png]] | ||
<math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math> | <math>\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28</math> | ||
− | == Solution == | + | == Solutions == |
− | [[Image:2000_12_AMC-24a.png | + | === Solution 1 === |
+ | [[Image:2000_12_AMC-24a.png]] | ||
Since <math>AB,BC,AC</math> are all [[radius|radii]], it follows that <math>\triangle ABC</math> is an [[equilateral triangle]]. | Since <math>AB,BC,AC</math> are all [[radius|radii]], it follows that <math>\triangle ABC</math> is an [[equilateral triangle]]. | ||
Line 24: | Line 26: | ||
After simplification, <math>r_2 = \frac{3r_1}{8}</math>. | After simplification, <math>r_2 = \frac{3r_1}{8}</math>. | ||
− | == See | + | === 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 == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=3466 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == 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}} | {{MAA Notice}} |
Latest revision as of 04:38, 2 February 2021
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 1
Since are all radii, it follows that is an equilateral triangle.
Draw the circle with center and radius . Then let be the point of tangency of the two circles, and be the intersection of the smaller circle and . Let be the intersection of the smaller circle and . Also define the radii (note that is a diameter of the smaller circle, as 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 ).
By the Power of a Point Theorem,
Since , then . Since is equilateral, , and so . Thus and the circumference of the circle is .
(Alternatively, the Pythagorean Theorem can also be used to find in terms of . Notice that since AB is tangent to circle , is perpendicular to . Therefore,
After simplification, .
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
https://youtu.be/NsQbhYfGh1Q?t=3466
~ pi_is_3.14
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.