Difference between revisions of "2014 AMC 10B Problems/Problem 19"
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==Problem== | ==Problem== | ||
+ | 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>. | ||
==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}} |
Latest revision as of 19:44, 28 April 2021
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
.
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.