Difference between revisions of "2014 AMC 10B Problems/Problem 19"
Euclid 2003 (talk | contribs) (→See Also) |
Euclid 2003 (talk | contribs) (→Comment) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 43: | Line 43: | ||
==Comment== | ==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.
![[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]](http://latex.artofproblemsolving.com/b/7/c/b7caeda0d004ad915eefc557ca648dfe2ea51acd.png)
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) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
