Question: Two points on a circle of radius 1 are chosen at random. Find the probability that the distance between the two points is at most 1.

What I Tried:

First, I tried to made a sketch on desmos

I came up with the idea that the center of the blue circle can be shifted on the radius of the red circle (the red circle represents the original circle, and the blue circle represents the area of where a point could be at most 1 unit away from a selected point, which is the center of the blue circle).

I thought that I could find the average of the ratio of the area of intersection of the circles to the overall area when the center of the blue circle was on the edge of the red circle and 1 (because the red and blue circles would completely overlap).

I tried finding the ratio of the intersection of the circles to the overall area when the center of the blue circle was on the edge of the red circle, but now I'm stuck. I'm not even sure if this idea is logical. Can somebody please help? Any help would be appreciated

CheeseSteak1432 Apr 22, 2023

#1**+1 **

Let's think through this step-by-step:

We have a circle of radius 1

Two points are chosen at random on the circle

We want to find the probability that the distance between the two points is at most 1

To find the probability, we need to calculate:

The number of possibilities where the distance is at most 1

The total number of possibilities

Then divide the first by the second

To count the possibilities where the distance is at most 1:

** The two points can be anywhere on the circle, as long as they are at most 1 unit apart. This forms a "slice" of the circle with an central angle of up to pi radians (since the circle has circumference 2pi and radius 1).

** The area of a slice of a circle is (pi * r^2) * (theta / 360) where r is the radius and theta is the measure of the central angle in degrees. Here, r = 1 and theta = 180 degrees (since pi radians = 180 degrees), so the area of the slice is (pi * 1^2) * (180/360) = pi/2 square units.

** Since the points can be anywhere in this slice, the number of possibilities is (area of slice) = pi/2

The total number of possibilities is the area of the entire circle = pi.

**Therefore, the probability is (pi/2) / pi = 1/2**

Guest Apr 22, 2023

#2**+1 **

Thank you for your time and effort, but the website says that this is incorrect.

I have come up with the thought that if two points are chosen that are at most a distance of 1 unit apart from each other, the angle that they make with the center point of the circle is at most 60 degrees (equilateral triangle). This thought might be incorrect, but if it is correct it might have something to do with the answer.

CheeseSteak1432
Apr 22, 2023

#3**+1 **

Try this:

Let's call the two points A and B, and let O be the center of the circle.

Without loss of generality, we can assume that point A is located at the top of the circle. Then, the distance between point A and any other point B on the circle is given by the length of the arc subtending angle AOB.

If we fix point A and randomly choose a point B, the probability that the distance between A and B is at most 1 is the same as the probability that angle AOB is less than or equal to 60 degrees (since the length of an arc subtending an angle of 60 degrees on a unit circle is 1).

So, let's assume that point A is located at (1, 0) and let's consider the region of the circle to the right of A. The probability that point B falls in this region is 1/2. To find the probability that the angle AOB is less than or equal to 60 degrees, we need to find the length of the arc that subtends this angle and divide it by the total circumference of the circle.

The length of the arc subtending angle AOB is just the length of one-sixth of the circumference of the circle (since angle AOB is 60 degrees and there are 6 such angles in a full circle). The circumference of the circle is 2π, so the length of the arc is (1/6) * 2π = π/3. Therefore, the probability that the distance between the two points is at most 1 is:

(1/2) * (π/3) / (2π) = 1 / 12

**So the probability that the distance between the two points is at most 1 is 1/12.**

**P. S. I did it in another way and got: 3 / 4. But was less sure of that method!**

Guest Apr 22, 2023

#4**+1 **

The points can be chosen in 2π radians along the circle. The distance between two points on the circle is at most 1 if they are adjacent. There are 2 adjacent points on the circle, so the probability is 2/(2π)=1/π.

Guest Apr 23, 2023

#5**+1 **

Thanks for taking the time to answer my question, but I am slightly confused about the second parargaph. Firstly, did you mean that we should assume that A is located at (0, 1) instead of (1, 0) (this is assuming that the center of the circle is located at (0, 0))? Secondly, why don't we also consider the left side?

CheeseSteak1432
Apr 23, 2023

#6**+1 **

**Earlier errors are now fixed. I hope there are no more.**

Question: Two points on a circle of radius 1 are chosen at random. Find the probability that the distance between the two points is at most 1.

I have spent a long time on this, I think this answer is correct.

Orient your circle so that the interval joining the 2 points that you choose is above and parallel to the diameter.

The green semicircle is the top of the original circle.

If I draw 1 unit intervals paralel to the diameter how much of semicircle will not be included.

I needed a curve 1 unit to right of the original and the area in the bricked in area indicatesw the left over.

So the prob of the points being 1 unit or less apart is the clear green area over the whole green semicircle.

Area of Simicircle = \(\frac{\pi}{2}\)

**Area of the bricked sector** **sector** =

\(\frac{60}{360}*\pi\\ =\frac{\pi}{6}\\\)

**Area of bricked segment**

**\(=\frac{\pi}{6}-\frac{1}{2}*1*\sqrt{1^2-0.5^2}\\ =\frac{\pi}{6}-\frac{1}{2}*\frac{\sqrt{3}}{2}\\ =\frac{\pi}{6}-\frac{\sqrt{3}}{4}\\ =\frac{2\pi-3\sqrt3}{12}\)**

**Total bricked area **

\(=\frac{2\pi}{12}+\frac{2\pi}{12}-\frac{3\sqrt3}{12}\\ =\frac{4\pi-3\sqrt3}{12}\\ \)

**Area on non-bricked section**

\(\frac{6\pi}{12}-\frac{4\pi-3\sqrt3}{12}\\ =\frac{2\pi +3\sqrt3}{12}\\\)

Probability that the points will be less than or equal to 1 unit apart

\(=\frac{2\pi+3\sqrt3}{12}\div \frac{\pi}{2}\\ =\frac{3\sqrt3+2\pi}{6\pi} \\ \approx 61\%\)

Hopefully I have now fixed all the errors.

Melody Apr 23, 2023

#7**+1 **

Thank you for spending a long time on this explanation, but I am confused on how you found the bricked area. I see that you found the area of the sector, but what about the bricked area to the right of line segment DE?

CheeseSteak1432
Apr 24, 2023

#9

#10**+1 **

I actually found another solution on the website I got the problem from. Before I noticed that if two points are chosen that are at most a distance of 1 unit apart from each other, the angle that they make with the center point of the circle is at most 60 degrees (equilateral triangle). When I looked at the solution, I realized that that idea was actually a part of the solution. Let's say A is on the right side of the circle. That means point B could be chosen either above or below point A. This creates an arc of 120 degrees, and since there are 360 degrees in a circle, the correct solution is 120/360 = 1/3. Thank you all for giving your ideas!

CheeseSteak1432
Apr 28, 2023

#11**+1 **

Sorry Cheesesteak, but that answer is NOT correct. I am very pleased that you have responded though.

Take a point that is 0.3 units from the cente, now rotate that point 180 degrees to get the second point.

The two points are 0.6 units and 180 degrees apart.

You said the 2 points had to be less than 60 degrees apart. You have only found a small portion of the solutions

Melody
Apr 30, 2023