A circle of radius 1 is drawn in the plane. Four non-overlapping circles each of radius 1, are drawn (externally) tangential to the original circle. An angle γ is chosen uniformly at random in the interval [0,2π). The probability that a half ray drawn from the centre of the original circle at an angle of γ intersects one of the other four circles can be expressed as ab, where a and b are coprime positive integers. What is the value of a+b?

Question

A circle of radius 1 is drawn in the plane. Four non-overlapping circles each of radius 1, are drawn (externally) tangential to the original circle. An angle γ is chosen uniformly at random in the interval [0,2π). The probability that a half ray drawn from the centre of the original circle at an angle of γ intersects one of the other four circles can be expressed as ab, where a and b are coprime positive integers. What is the value of a+b?

Answer 1

To find the probability that a half ray drawn from the center of the original circle intersects one of the other four circles, we need to consider the possible ranges of angles for which the intersection occurs.

Let's start by visualizing the scenario:

1. Draw a circle with a radius of 1 in the center.
2. Place four circles externally tangent to the original circle, each with a radius of 1.

Now, let's divide the problem into cases:

Case 1: The half ray intersects one of the four circles directly.
In this case, the angle γ must be within the range of ±π/2 from the tangent lines of each of the four external circles. Since each circle is tangent to the original circle, the angle range for each external circle is π/2 (90 degrees). Therefore, the total range of angles in this case is 4 * π/2 = 2π.

Case 2: The half ray intersects two of the four circles.
In this case, the angle γ must be within the range that intersects two tangent lines simultaneously. Since the tangent lines of each external circle form an angle of π/2 (90 degrees) with the tangent lines of the other two external circles, the allowed angle range for this case is π/2 (90 degrees) as well. Again, the total range of angles in this case is 4 * π/2 = 2π.

Case 3: The half ray intersects three or four of the four circles.
In these cases, the angle γ will overlap with the ranges considered in cases 1 and 2.

Now, to find the desired probability, we divide the sum of the angle ranges in cases 1 and 2 by the total angle range of 2π:

P = (2π + 2π) / 2π = 4/2 = 2/1

Therefore, the probability is 2/1, and the values of a and b are 2 and 1, respectively.

Finally, a + b = 2 + 1 = 3.