Square

**alexmahone** · January 23rd 2009, 01:58 AM

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

**mr fantastic** · January 23rd 2009, 02:22 AM

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.

**alexmahone** · January 23rd 2009, 03:02 AM

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.

**mr fantastic** · January 23rd 2009, 04:10 AM

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.

**alexmahone** · January 23rd 2009, 04:26 AM

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?

**Plato** · January 23rd 2009, 05:21 AM

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.

**mr fantastic** · January 23rd 2009, 12:37 PM

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.

I will expand slightly on what Plato has said. Think about this:

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

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

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.

**Plato** · January 23rd 2009, 12:59 PM

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.

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