# Square Cross

Source: https://thealphajas.github.io/fun/probability/square-cross/
Topics: Uniform Distribution, Integration, Geometric Probability

## Problem

A square of side length $$20$$ is placed in front of you. You select a point uniformly at random from its interior. Then, independently, a circle of radius $$ R \sim \text{Unif}(0,10) $$ is formed around the selected point. What is the probability that this circle does not intersect the square at any point?

**Original Problem Link**: [Click here](https://www.quantguide.io/questions/square-cross)

## Solution

### Step 1 - Understanding the Core Idea  

Let’s build some intuition. Clearly, the best place to pick a point is the center of the square—a circle centered there can grow up to radius $$10$$ before touching an edge. Any point away from the center? Well, that’s where things get tricky—closer points to the boundary have a higher chance of their circle spilling out.

Try to visualize the locus of all points that are an equal distance from the closest edge. These points form a smaller concentric square, with perpendicular distance $$ x $$ from the center. Our goal is now clear:  

1. Find the probability of choosing a point at a given $$ x $$.  
2. Given $$ x $$, find the probability that a circle centered there does not intersect the square.  
3. Integrate over all possible $$ x $$.  

#### Visualizing the Locus  

The points that are an equal distance from the nearest edge form a square frame at distance $$ x $$. Here’s what it looks like:

![Locus of points at perpendicular distance x from the nearest edge of the square](/images/squarecross.png)

_(Yeah ik the diagram kinda sucks, but I believe in MS-Paint XD. Anyhow)_


Key observations:  

1. All points on the same locus are equivalent. The probability of a circle touching the square depends only on $$ x $$, not on the specific point along the locus.  
2. Probability of picking a point at distance $$ x $$:  
   - The area of the square frame at distance $$ x $$ is  
     $$
     4 \times (2x) \times dx = 8x \, dx
     $$
   - Since the total area of the square is $$400$$, the probability of selecting a point from this region is  
     $$
     \frac{8x \, dx}{400} = \frac{x \, dx}{50}
     $$

### Step 2 - Computing the Probability  

Now, assume a point at distance $$ x $$ has been chosen. What is the probability that a randomly chosen radius does not cause intersection?

- The favorable radii are those that do not exceed $$ x $$, i.e., $$ R \in [0, 10-x] $$
- Since $$ R \sim \text{Unif}(0,10) $$, the probability that $$ R \leq 10 - x $$ is simply $$\frac{10 - x}{10}$$

Thus, our probability function is now completely defined. To compute the final probability, we integrate:

$$
P(\text{No intersection}) = \int_0^{10} \frac{10-x}{10} \cdot \frac{x}{50} \, dx
$$

Expanding:

$$
P(\text{No intersection}) = \frac{1}{500} \int_0^{10} x(10-x) \, dx
$$

Breaking it down:

$$
\frac{1}{500} \int_0^{10} (10x - x^2) \, dx
$$

Computing the integral,

$$
\frac{1}{500} \left[ 5x^2 - \frac{x^3}{3} \right]_0^{10}
$$

$$
= \frac{1}{500} \left[ 500 - \frac{1000}{3} \right]
$$

$$
= \frac{1}{500} \times \frac{500}{3}
$$

$$
= \frac{1}{3}
$$

### Thus, the probability that the circle does not intersect the square is $$\frac{1}{3}$$