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
 A: 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
$$\frac{9-4\sqrt3}{3\sqrt3}=\sqrt3-\frac43=0.398717\dots$$
A: The red areas are of points closer to a side of the rectangle. Its area is the sum of the triangles, two acutangle and two obtusangle. 
The two acutangle are each $1/3$ of the equilateral triangle having side $1$ and half diagonals. A diagonal is $\sqrt{3+1}=2$ so half diagonal is $1$ (that's why they are equilateral).
So the area of the two acutangle is $2\times \dfrac{1}{3} \cdot\dfrac{\sqrt 3}{4}=\dfrac{\sqrt 3}{6}$
As their sides are the angle bisector of the $30°$ angle the two obtusangle have the vertex that is the incentre (centre of the inscribed circle) of the bigger triangle formed by the diagonals whose area is $\dfrac{\sqrt 3}{4}$ and whose perimeter is $2+\sqrt 3$. 
The radius of the inscribed circle is $h=\dfrac{area}{half\,perimeter}=\dfrac{\frac{\sqrt 3}{4}}{\frac{2+\sqrt 3}{2}}=\dfrac{\sqrt{3}}{2 \left(2+\sqrt{3}\right)}$
The two obtusangle triangle have an area $\sqrt{3}\times \dfrac{\sqrt{3}}{2 \left(2+\sqrt{3}\right)}=\dfrac{3}{2 \left(2+\sqrt{3}\right)}$
Therefore the red area is $\dfrac{\sqrt 3}{6}+\dfrac{3}{2 \left(2+\sqrt{3}\right)}=\dfrac{1}{3} \left(9-4 \sqrt{3}\right)$
The area of the rectangle is $\sqrt 3$ so the probability of getting a point closer to the sides of the rectangle is $p=\dfrac{\frac{1}{3} \left(9-4 \sqrt{3}\right)}{\sqrt 3}=\dfrac{1}{3} \left(3 \sqrt{3}-4\right)\approx 0.3987$ 

