Distance between Points

Geometry Level pending

Two circles of radius $1$ and $2$ share the same centre. A point is picked at random on each of the circles circumferences. What is the probability that the distance between the two points is less than $2$ ?

\frac{\tan^{-1}(\sqrt{10})}{2}

\tan^{-1}(\frac{\pi}{3})

\frac{\tan^{-1}(\sqrt{3})}{2\pi}

\frac{\tan^{-1}(3\sqrt{5})}{\pi}

\frac{\tan^{-1}(\sqrt{15})}{\pi}

1 solution

Max Patrick
Nov 19, 2019

Pick the point on the outer circle and rotate the entire set-up so that this point C is at the top, ie on the positive y-axis of the coordinate plane.

Draw a circle radius 2 about this point, this cutting the smaller circle at points A and B into a major and minor arc. Let B be in positive x.

Let Q be the angle between OB and the Y-axis.

Q/pi is the probability we need (a simplification of 2Q/2pi).

Notice that COB is an isosceles triangle, sides 2,2,1. Consider half of this, a right-angled triangle.

cos(Q) =0.5/2=1/4

The probability is therefore arccos(1/4)/pi which is equal to the correct answer.

Expressing it with arctangents of a square root seems too complicated to me!

Distance between Points

1 solution

0 pending reports