<s>[INST] 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
 [/INST] 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 $p=\dfrac{\frac{1}{3} \left(9-4 \sqrt{3}\right)}{\sqrt 3}=\dfrac{1}{3} \left(3 \sqrt{3}-4\right)\approx 0.3987$ 

</s>