Game of probability In a game, played between $2$ players there is a circular field and one of the players is blindfolded, who stands in the center of the field. The other player stands at a fixed point on the circumference of the circular field. On the word GO, the blind-folded player starts running towards the edge of the field while the second player's aim is to run in and catch him before he moves out of the circle.
If the blind-folded player runs in a random direction with a constant speed $v$ while the second player runs towards the first player with a constant speed $m$ times $v$, what should be the value of $m$ such that the probability that the second player wins is $0.50$?
Sourece: https://erdos.sdslabs.co/problems/15
 A: The locus of a catch positions is a circle of Apollonius defined by the players initial positions and the ratio of their velocities. The probability of the second player win is a relative part of the 'game circle' not contained in the Apollonius' circle.
A: 0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle.
In that case the path of player 2 will be the hypotenuse of an isosceles right with side length $r$, the radius of the circle. the length of this path is $l=\sqrt 2 r$
since player 2 must cover $\sqrt 2 r $ in the same time as player 1 takes to cover the distance $r$, they must be moving $\sqrt 2$ times as quickly.
So $m=\sqrt 2$
A: 
Let at time t, the first player has moved a distance R1 while the second player who was situated at a point on the circumference at the start has caught the first player at the same time t.  Let the distance R2 be between his initial point on the circumference to the point where he caught the first player.  Let R be the radius of the circular field. Let the angle $\alpha$ be subtended by the  second player to the radius. As the player 1 moves towards the circle edge, R1,R2 and  $\alpha$ all change.  At any instantaneous point, we have a relation amongst R1,R2,R and $\alpha$ by cosine rule.
From cosine rule,
we get $ R_1^2 = R_2^2 + R^2 - 2R_2Rcos\alpha$
Taking the derivative, we get
$ 2R_1\frac{dR_1}{dt} = 2R_2\frac{dR_2}{dt} + 2R\frac{dR_2}{dt}cos\alpha -2RR_2sin\alpha \frac{d\alpha}{dt}$
$\frac{d\alpha}{dt} = \frac{mv}{R_2}$
$ R_1v = R_2mv + Rmv(cos\alpha - sin\alpha)$
$R_1 = mR_2 +Rm(cos\alpha - sin\alpha)$
$R_1 = m^2dR_1+ mR(cos\alpha - sin\alpha)$
$R_1(1-m^2) = mR(cos\alpha - sin\alpha)$
$R_1 = \frac{mR}{1-m^2} (cos\alpha - sin\alpha)$
Area of the Apollonius circle $= \int_{0}^{\pi}R_1dR_1d\alpha$
$=\int_{0}^{\pi}\int_{0}^{R}\frac{m^2R}{1-m^2)^2}(cos\alpha - sin\alpha)^2dRd\alpha$
$=\frac{m^2R^2}{2(1-m^2)^2} \int_{0}^{\pi}(cos^2\alpha+sin^2\alpha - 2sin\alpha cos\alpha)d\alpha$
$= \frac{m^2R^2}{2(1-m^2)^2}(\int_{0}^{\pi}(1-sin2\alpha)d\alpha$
$=\frac{m^2R^2}{2(1-m^2)}(\pi)$
Probability = Area of Apollonius Circle/ Area of the circular field ($\pi R^2$)
Required probability $= \frac{m^2}{2(1-m^2)^2} = \frac{1}{2}$
$\frac{m^2}{(1-m^2)^2} = 1$
Rearranging the terms you get $m^4-2m^2 +1 = m^2$
$m^4-3m^2+1 = 0$
Solving this you get $m = \dfrac{1+\sqrt{5}}{2}$
Thanks 
Satish
