Difference between revisions of "2014 AMC 10B Problems/Problem 19"
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| − | 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> | + | 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>. |
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==See Also== | ==See Also== | ||
{{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}} | ||
Revision as of 16:14, 9 February 2016
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 
.
See Also
| 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. 
