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 γis 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 a/b, where aand b are coprime positive integers. What is the value of a+b?
Details and assumptions
The half ray from the centre of the fifth circle at angle γ goes only in one direction, not both.
Well, a unit circle externally tangent to the central unit circle subtends a central angle of 60°.
Draw a diagram to see that the tangents from the center to the external circles form angles of 30° each.
So, if there are 4 external circles, they cover 240°.
Since there are 360° available, the chance that a central ray will intersect one of the external circles must then be 240/360 = 2/3
To solve this problem, we need to find the probability that a half ray drawn from the center of the original circle at a random angle intersects one of the other four circles.
Let's consider a fifth circle, tangent to the original circle externally. We want to determine the probability that a half ray drawn from the center of the original circle at an angle γ intersects this fifth circle. If we can find this probability, we can simplify the problem by considering the intersection or non-intersection with this fifth circle equivalent to the intersection or non-intersection with any of the other four circles.
To start solving the problem, we need to define some variables. Let the radius of the fifth circle be R. We can observe that the distance between the centers of the original circle and the fifth circle is 1 + R.
Now, suppose we draw a line segment from the center of the original circle to the point of intersection with the fifth circle, forming a right triangle. The base of this right triangle is the radius of the original circle, which is 1. The hypotenuse is the distance between the centers of the two circles, which is 1 + R.
Thus, according to the Pythagorean theorem, the length of the altitude of this right triangle is sqrt((1+R)^2 - 1^2). This altitude represents the maximum possible length of a half ray drawn from the center of the original circle such that it does not intersect the fifth circle.
Now, let's consider the numerator of our desired probability. We want to count the number of angles γ such that a half ray with angle γ will intersect the fifth circle. Since γ is chosen uniformly at random in the interval [0, 2π), we can treat γ as a random variable. To find the probability, we need to determine the length of the portion of the interval [0, 2π) for which these half rays intersect the fifth circle.
To find the length of this interval, we need to consider the angles at which the half rays completely miss the fifth circle. This corresponds to the length of the interval [0, 2π) minus the length of the interval for which the half rays hit the fifth circle.
The length of the interval for which the half rays intersect the fifth circle can be found by observing that the angle γ, denoting the position of the half ray, can take any value from 0 to 2π, minus twice the angle θ.
Now, θ can be found by considering the right triangle we formed earlier. The sine of θ can be calculated as the ratio of the altitude (sqrt((1+R)^2 - 1^2)) to the hypotenuse (1 + R). Therefore, we have sin(θ) = sqrt((1+R)^2 - 1^2) / (1 + R).
Next, we can calculate cos(θ) using the Pythagorean identity: cos^2(θ) + sin^2(θ) = 1. With this relationship, we find cos(θ) = sqrt(1 - sin^2(θ)).
Finally, the length of the interval for which the half rays intersect the fifth circle is given by 2(cos^(-1)(sqrt(1 - sin^2(θ)))).
Therefore, the desired probability is given by (2(cos^(-1)(sqrt(1 - sin^2(θ)))) / (2π).
To calculate the value of a/b, we need to simplify this expression and ensure that a and b are coprime integers.
Now, let's evaluate the expression and find the value of a and b to calculate their sum a+b.