Probability of hitting a target? 
A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge.

I have absolutely no idea how to approach the problem. Could someone get me started on the right path?
 A: A start: Draw the two diagonals of the square, and also the two lines that join midpoints of opposite sides. These $4$ lines divide the square into $8$ congruent right-angled isosceles triangles. By symmetry, it is enough to determine the probability that a randomly chosen point within a specific one of these triangles is closer to the centre of the square than it is to one of the sides of the square.
Assume (it makes no difference) that the original square has vertices $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$. Consider the triangle $T$ with vertices $(0,0)$, $(2,0)$, and $(2,2)$.  This triangle has area $2$. So for our probability, we will determine the area of the region $K$ in $T$ consisting of all points $(x,y)$ such that the distance from $(x,y)$ to the origin is $\le$ the distance from $(x,y)$ to the nearest side of the original square.  Then our required probability is the area of $K$ divided by $2$,
So let us determine the locus of all points $(x,y)$ in $T$ that are equidistant from the origin and the nearest side of the original square. 
The distance to the origin is $\sqrt{x^2+y^2}$, and the distance to the nearest side of the original square is $2-x$. So our locus is $\sqrt{x^2+y^2}=2-x$. Square and simplify. We get $x=\frac{1}{4}(4-y^2)$, part of a parabola. 
Our region $K$ is the region bounded by the $x$-axis, the parabola $x=\frac{1}{4}(4-y^2)$, and the line $y=x$.  We find the area of $K$ by integrating with respect to $y$.  
Setting up the integral is slightly tricky. We will be integrating from $y=0$ to where $x=\frac{1}{4}(4-y^2)$ meets $y=x$. That gives the equation $y=\frac{1}{4}(4-y^2)$, which has the solution $y=2\sqrt{2}-2$. Thus our required area is 
$$\int_0^{2\sqrt{2}-2} \left(\frac{1}{4}(4-y^2)-y\right)\,dy.$$
After calculating the area, don't forget to divide by $2$.
