Difference between revisions of "2014 AMC 10B Problems/Problem 19"
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Two concentric circles have radii <math>1</math> and <math>2</math>. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle? | Two concentric circles have radii <math>1</math> and <math>2</math>. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle? | ||
− | + | <math> | |
+ | \textbf{(A)}\ \frac{1}{6}\qquad | ||
+ | \textbf{(B)}\ \frac{1}{4}\qquad | ||
+ | \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\qquad | ||
+ | \textbf{(D)}\ \frac{1}{3}\qquad | ||
+ | \textbf{(E)}\ \frac{1}{2}\qquad</math> | ||
+ | [[Category:Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
− | == | + | Let the center of the two circles be <math>O</math>. Now pick an arbitrary point <math>A</math> on the boundary of the circle with radius <math>2</math>. We want to find the range of possible places for the second point, <math>A'</math>, such that <math>AA'</math> passes through the circle of radius <math>1</math>. To do this, first draw the tangents from <math>A</math> to the circle of radius <math>1</math>. Let the intersection points of the tangents (when extended) with circle of radius <math>2</math> be <math>B</math> and <math>C</math>. Let <math>H</math> be the foot of the altitude from <math>O</math> to <math>\overline{BC}</math>. Then we have the following diagram. |
+ | |||
+ | <asy> | ||
+ | scale(200); | ||
+ | pair A,O,B,C,H; | ||
+ | A = (0,1); | ||
+ | O = (0,0); | ||
+ | B = (-.866,-.5); | ||
+ | C = (.866,-.5); | ||
+ | H = (0, -.5); | ||
+ | draw(A--C--cycle); | ||
+ | draw(A--O--cycle); | ||
+ | draw(O--C--cycle); | ||
+ | draw(O--H,dashed+linewidth(.7)); | ||
+ | draw(A--B--cycle); | ||
+ | draw(B--C--cycle); | ||
+ | draw(O--B--cycle); | ||
+ | dot("$A$",A,N); | ||
+ | dot("$O$",O,NW); | ||
+ | dot("$B$",B,W); | ||
+ | dot("$C$",C,E); | ||
+ | dot("$H$",H,S); | ||
+ | label("$2$",O--(-.7,-.385),N); | ||
+ | label("$1$",O--H,E); | ||
+ | draw(circle(O,.5)); | ||
+ | draw(circle(O,1)); | ||
+ | </asy> | ||
+ | |||
+ | We want to find <math>\angle BOC</math>, as the range of desired points <math>A'</math> is the set of points on minor arc <math>\overarc{BC}</math>. This is because <math>B</math> and <math>C</math> are part of the tangents, which "set the boundaries" for <math>A'</math>. Since <math>OH = 1</math> and <math>OB = 2</math> as shown in the diagram, <math>\triangle OHB</math> is a <math>30-60-90</math> triangle with <math>\angle BOH = 60^\circ</math>. Thus, <math>\angle BOC = 120^\circ</math>, and the probability <math>A'</math> lies on the minor arc <math>\overarc{BC}</math> is thus <math>\dfrac{120}{360} = \boxed{\textbf{(D)}\: \dfrac13}</math>. | ||
+ | |||
+ | ==Comment== | ||
+ | This problem is exactly the Bertrand paradox! I feel sorry for students who took 2014 AMC 10B :-( | ||
+ | The answers (E) 1/2, (D) 1/3, and (B) 1/4 are all correct, depending on what kind of uniform you choose. | ||
+ | (1) If you choose angle uniform, the probability is 1/3. | ||
+ | (2) If you choose the center of the chord is uniformly distributed in the big circle, the probability is 1/4. | ||
+ | (3) If you choose the center of the chord is uniformly distributed along a radius (i.e. 0 to 2 from the center of the big circle), the probability is 1/2. | ||
+ | |||
+ | -EUCLID_2003 | ||
+ | |||
{{AMC10 box|year=2014|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2014|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:27, 7 October 2024
Problem
Two concentric circles have radii and . Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
Solution
Let the center of the two circles be . Now pick an arbitrary point on the boundary of the circle with radius . We want to find the range of possible places for the second point, , such that passes through the circle of radius . To do this, first draw the tangents from to the circle of radius . Let the intersection points of the tangents (when extended) with circle of radius be and . Let be the foot of the altitude from to . Then we have the following diagram.
We want to find , as the range of desired points is the set of points on minor arc . This is because and are part of the tangents, which "set the boundaries" for . Since and as shown in the diagram, is a triangle with . Thus, , and the probability lies on the minor arc is thus .
Comment
This problem is exactly the Bertrand paradox! I feel sorry for students who took 2014 AMC 10B :-( The answers (E) 1/2, (D) 1/3, and (B) 1/4 are all correct, depending on what kind of uniform you choose. (1) If you choose angle uniform, the probability is 1/3. (2) If you choose the center of the chord is uniformly distributed in the big circle, the probability is 1/4. (3) If you choose the center of the chord is uniformly distributed along a radius (i.e. 0 to 2 from the center of the big circle), the probability is 1/2.
-EUCLID_2003
2014 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 20 | |
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