Probability that a point is closer to a side than a diagonal So I have a rectangle in which a point is randomly chosen. One side is $a$ and the other is $b=a\sqrt3$. I am supposed to find the probability that a point is closer to a side than to the closest diagonal. 
I have found the probability that the point is closer to $a$ (0.71) and to $b$ (0.24). Now I was wondering how I can put these two probabilities together to form the asked-for probability. Thanks

I answered a very similar question involving a square instead of a $\sqrt3$ rectangle here.

Points in the shaded area are closer to a side than a diagonal. The rectangle divides into four triangles, each containing one shaded triangle; the apexes of the shaded triangles are the incentres of the larger triangles they lie in.
Assuming that the shorter side is 1, the shaded triangles on the east and west have altitude (from the rectangle sides) $\frac1{2\sqrt3}$, so have a combined area of $\frac1{2\sqrt3}$ also. Those on the north and south have altitude $\frac{\sqrt3}2\tan15^\circ=\sqrt3-\frac32$ and combined area $\sqrt3(\sqrt3-\frac32)=3-\frac{3\sqrt3}2$. The shaded area is therefore $3-\frac{3\sqrt3}2\frac1{2\sqrt3}=\frac{9-4\sqrt3}3$. The rectangle's area is of course $\sqrt3$, so its shaded proportion – and therefore the probability a point inside it is closer to a side than a diagonal – is
$$\frac{9-4\sqrt3}{3\sqrt3}=\sqrt3-\frac43=0.398717\dots$$
