[Geometry]
Problem: A point is selected at random from the interior of a right triangle with legs of length $2 \sqrt{3}$ and 4 . Let $p$ be the probability that the distance between the point and the nearest vertex is less than 2. Then $p$ can be written in the form $a+\sqrt{b} \pi$, where $a$ and $b$ are rational numbers. Compute $(a, b)$.

Solution: Label the triangle as $\triangle A B C$, with $A B=2 \sqrt{3}$ and $B C=4$. Let $D$ and $E$ lie on $\overline{A B}$ such that $D B=A E=2$. Let $F$ be the midpoint of $\overline{B C}$, so that $B F=F C=2$. Let $G$ and $H$ lie on $\overline{A C}$, with $A G=H C=2$. Now draw the arcs of radius 2 between $E$ and $G, D$ and $F$, and $F$ and $H$. Let the intersection of arc $D F$ and $\operatorname{arc} E G$ be $J$. Finally, let $M$ be the midpoint of $\overline{A B}$. The completed diagram is shown below.

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The region $R$ consisting of all points within $\triangle A B C$ that lie within 2 units of any vertex is the union of the three sectors $E A G, D B F$, and $F C H$. The angles of these sectors, being the angles $\angle A, \angle B$, and $\angle C$, sum to $180^{\circ}$, so the sum of their areas is $2 \pi$. Computing the area of $R$ requires subtracting the areas of all intersections of the three sectors that make up $R$.

The only sectors that intersect are $E A G$ and $D B F$. Half this area of intersection, the part above $\overline{M J}$, equals the difference between the areas of sector $D B J$ and of $\triangle M B J$. Triangle $M B J$ is a $1: \sqrt{3}: 2$ right triangle because $B M=\sqrt{3}$ and $B J=2$, so the area of $\triangle M B J$ is $\frac{\sqrt{3}}{2}$. Sector $D B J$ has area $\frac{1}{12}(4 \pi)=\frac{\pi}{3}$, because $\mathrm{m} \angle D B J=30^{\circ}$. Therefore the area of intersection of the sectors is $2\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)=\frac{2 \pi}{3}-\sqrt{3}$. Hence the total area of $R$ is $2 \pi-\left(\frac{2 \pi}{3}-\sqrt{3}\right)=\frac{4 \pi}{3}+\sqrt{3}$. The total area of $\triangle A B C$ is $4 \sqrt{3}$, therefore the desired probability is $\frac{\frac{4 \pi}{3}+\sqrt{3}}{4 \sqrt{3}}=\frac{\pi}{3 \sqrt{3}}+\frac{1}{4}$. Then $a=\frac{1}{4}$ and $b=\left(\frac{1}{3 \sqrt{3}}\right)^{2}=\frac{1}{27}$, hence the answer is $\left(\frac{1}{4}, \frac{1}{27}\right)$.

Final answer: $(\frac{1}{4}, \frac{1}{27})$