Unit Circle on Unit Square

Consider a unit square intersecting a unit circle at two of the square's vertices.

Note that the intersecting points and the center of the circle form an equilateral triangle. Now rotate the square around the inside of the circle, keeping one of the intersecting points in contact.

The center of the circle traces a 30 $^\circ$ arc with radius 1 on the square. If you keep on rotating the square around the inside of the circle, you get a shape consisting of 4 of these arcs. As long as the center of the circle is inside this shape or on its perimeter, the circle will completely cover the square.

Each quarter of this shape is a 30 $^\circ$ circular arc with two triangles cut from it.

The area of the entire shape is $\boxed{A=\frac{\pi}{3}+1-\sqrt{3} \approx 0.3151}.$

Let the square be centered at the origin. The arbitrary points $(x,y)$ chosen satisfies $|x|, |y| \leq \tfrac12$ .

In order to see if the square fits within the circle centered at $(x,y)$ , we only need to consider the vertices $(\pm\tfrac12, \pm\tfrac12)$ . Thus we need $(x \mp \tfrac12)^2 + (y \mp \tfrac12)^2 \leq 1^2,$ or indeed $(|x| + \tfrac12)^2 + (|y| + \tfrac12)^2 \leq 1.$

In an Excel spreadsheet I entered the formulas

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A1:  =RAND()/2
B1:  =RAND()/2
C1:  =IF((A1+0.5)^2+(B1+0.5)^2<=1,1,0)

and copied this about 25,000 times. The average in column C was $\boxed{0.315}$ .

Label the corners 1,2,3 and 4. Let $S_i$ be the set of points in the square no more than 1 unit away from corner $i$ . Then, since the area of the square is 1, $p$ is just equal to the area of $S_1 \cap S_2 \cap S_3 \cap S_4$ . See Andy Hayes' solution for visualization!

This area can be found with calculus, and is equal to $4\int_0^{\frac{\sqrt{3}}{2}-\frac{1}{2}}\sqrt {1-(x+\frac{1}{2})^2}-\frac{1}{2}\hspace{2mm}dx = \\4 (\frac{\pi}{12} + \frac{1}{4} - \frac{\sqrt{3}}{4}) \approx 0.315$

P=(area of square)/(area of the circle)

P=(1)^2/(pi*(1)^2)=1/pi=0.318 😊

Any point in the square which makes a circle of radius 1 must satisfy these conditions: 1. If a unit circle covers two vertices of the square it covers the line between them. 2. If the unit circle covers all the perimeter of the square then it covers the whole square. The only case in which the circle does not cover the square is if it does not cover the perimeter. We want to find the area of points in the square which when a point inside is a center of a circle of radius 1 it covers the square. Each vertex of the square must be covered. Hence, by condition 1 we only need to focus on the vertices and make sure all these are covered. Condition 2 lets us see we have covered the whole square.

To find all points in the square which will cover a particular vertex simply make this vertex a center of circle of radius one. This will cover all points in the square which which are a maximum of distance 1 away. All these points have the same relationship to the vertex meaning that when a circle is placed on these it covers the vertex. No other point in the square will do this. Applying this to every vertex will give this shape: The yellow area is the overlap of the 4 circles. All points in this yellow area can make a circle of radius 1 to cover all points in this circle. It is now required to calculate this yellow area. This is calculated by finding the area of this yellow shape (constructions drawn) This can be simply simplified that were each arc meets it makes a point which forms an equilateral triangle with two square vertices. This can be used to find the square length and the segment area.

The area of the yellow shape is: $\pi/3-\sqrt{3}+1$ Since probability is simply (area)/(area of square) = area/1 means that the probability is the area of the shape. To 3 decimals it is: 0.315

First we must find the Diagonal of a unit square. Using simple triangle identities we see that it's equal to $\sqrt{2}$ .

Now, for the unit circle to cover the square the radius of the circle needs exceed all distances from the origin of the circle to any one side of the square. Knowing that $\sqrt{2} > 1$ , we can see that there must be some region where this is not allowed. On the diagonal line that region would be equal to $\sqrt{2}-1$ . Turning the region into a two dimensional one all we have to do is square it and create a cube, whose area equals to $3-2\sqrt{2}$ . Because a square has two diagonal lines and each diagonal has two regions. We see that the probability of the unit circle failing to cover the square would be equal to: $\frac{4(3-2\sqrt{2})}{Area of Square}$ = $\frac{4(3-2\sqrt{2})}{1}$ = $0.6862...$

So $p$ = $1-0.6862...$ =0.3137... ~ 0.315

I wrote a Monte Carlo script to achieve this, by just looking at the distance of the four corners of the square from the random point within the square centered at the origin:

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from random import *
pout = 0
nsamp = 100000
for i in range(1,nsamp):
  x = random() - 0.5
  y = random() - 0.5 
  d1 = ( 0.5-x)**2 + ( 0.5-y)**2
  d2 = ( 0.5-x)**2 + (-0.5-y)**2
  d3 = (-0.5-x)**2 + ( 0.5-y)**2
  d4 = (-0.5-x)**2 + (-0.5-y)**2
  if (d1 > 1.0) or (d2 > 1.0) or (d3 > 1.0) or (d4 > 1.0):
    pout = pout + 1
print((nsamp - pout)/nsamp)