# Math Help - Square

1. ## Square

What is the probability that a point chosen at random within a square is closer to the centre than to the boundary?

2. Originally Posted by alexmahone
What is the probability that a point chosen at random within a square is closer to the centre than to the boundary?
This boils down to a problem in geometry .... Find the area of a square that is closer to the centre than the sides.

3. Originally Posted by mr fantastic
This boils down to a problem in geometry .... Find the area of a square that is closer to the centre than the sides.
You're probably referring to a square of side length $\frac{a}{2}$ located within the original square of side length $a$.

However, the corner of the inner square is $\frac{a}{2\sqrt{2}}$ units away from the centre whereas it is only $\frac{a}{4}$ units away from the boundary.

So I don't think the above approach is correct.

4. Originally Posted by alexmahone
You're probably referring to a square of side length $\frac{a}{2}$ located within the original square of side length $a$.

However, the corner of the inner square is $\frac{a}{2\sqrt{2}}$ units away from the centre whereas it is only $\frac{a}{4}$ units away from the boundary.

So I don't think the above approach is correct.
No I didn't mean that at all. Let me re-word it:

Find the area inside a square that is closer to the centre of the square than the sides of the square.

5. Originally Posted by mr fantastic
No I didn't mean that at all. Let me re-word it:

Find the area inside a square that is closer to the centre of the square than the sides of the square.
I get a "square" of length $\frac{a}{2}$ with circular corners of radius $\frac{a}{4}$. Is that right?

6. Originally Posted by alexmahone
I get a "square" of length $\frac{a}{2}$ with circular corners of radius $\frac{a}{4}$. Is that right?
No, that is not correct. Recall that a parabola is a set of points equally distance from a point and a line. So in this question the area is the intersection of the interiors of four parabolas. See the graph below.

7. Originally Posted by Plato
No, that is not correct. Recall that a parabola is a set of points equally distance from a point and a line. So in this question the area is the intersection of the interiors of four parabolas. See the graph below.

$\sqrt{x^2 + y^2} = 1 - y \Rightarrow \, ....$

$\sqrt{x^2 + y^2} = 1 - x \Rightarrow \, ....$

Since these parabolas are inverses of each other, they intersect on the line $y = x \, ....$

By symmetry, the required area is four times the required area in one quadrant of the square.

8. Originally Posted by mr fantastic
By symmetry, the required area is four times the required area in one quadrant of the square.
Or eight a fairly easy integral.