Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10B Problems/Problem 19"

(→‎Solution: fixed 120 degree sentence)
(Detailed Solution.)
Line 6: Line 6:
 
==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>
  
−
Pick any arbitrary point on the circle, and it is clear that the second point must lie on the arc of 120 degrees, on the opposite end of the circle from the point. Therefore the probability is <math>120/360=1/3</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>.
 +
 
 +
Solution and Diagram credited to happiface.
  
 
==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 03:03, 23 February 2014

Problem

Two concentric circles have radii $1$ and $2$. 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? 

$\textbf{(A) }\frac{1}{6}\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }\frac{2-\sqrt{2}}{2}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2}\qquad$

Solution

Let the center of the two circles be $O$. Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$. We want to find the range of possible places for the second point, $A'$, such that $AA'$ passes through the circle of radius $1$. To do this, first draw the tangents from $A$ to the circle of radius $1$. Let the intersection points of the tangents (when extended) with circle of radius $2$ be $B$ and $C$. Let $H$ be the foot of the altitude from $O$ to $\overline{BC}$. 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 $\angle BOC$, as the range of desired points $A'$ is the set of points on minor arc $\overarc{BC}$. This is because $B$ and $C$ are part of the tangents, which "set the boundaries" for $A'$. Since $OH = 1$ and $OB = 2$ as shown in the diagram, $\triangle OHB$ is a $30-60-90$ triangle with $\angle BOH = 60^\circ$. Thus, $\angle BOC = 120^\circ$, and the probability $A'$ lies on the minor arc $\overarc{BC}$ is thus $\dfrac{120}{360} = \boxed{\textbf{(D)}\: \dfrac13}}$ (Error compiling LaTeX. Unknown error_msg).

Solution and Diagram credited to happiface.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_19&oldid=60415"