2. A point $P$ is chosen randomly inside the triangle with sides 13,20 , and 21 . Find the probability that the circle centered at $P$ with radius 1 will intersect at least one of the sides of the triangle.
2. A point $P$ is chosen randomly inside the triangle with sides 13,20 , and 21 . Find the probability that the circle centered at $P$ with radius 1 will intersect at least one of the sides of the triangle.
Answer: $75 / 196$
Solution 1: Let $A B C$ be a triangle with sides $B C=13, C A=20$ and $A B=21$ and let $S$ be the set of points $P$ such that the circle $\omega$ with radius 1 centered at $P$ intersects at least one of the sides of $A B C$. For a fixed side of $A B C$ (say $\ell$ ), $\omega$ intersects $\ell$ if and only if $P$ lies within one unit from $\ell$. This suggests we construct a triangle $A_{1} B_{1} C_{1}$ such that $A_{1} B_{1}\left\|A B, B_{1} C_{1}\right\| B C, C_{1} A_{1} \| C A$ and the corresponding parallel sides of $A_{1} B_{1} C_{1}$ and $A B C$ have distance 1. Thus, the set $S$ of such points $P$ forms a region $\mathcal{R}$ outside $A_{1} B_{1} C_{1}$ but inside $A B C$ and the probability is then the ratio of the areas of $\mathcal{R}$ and $A B C$.
Observe that $A B C$ and $A_{1} B_{1} C_{1}$ are similar and hence the ratio of their corresponding sides is constant, say $k>0$. Also, triangle $A B C$ is divided into four regions: the triangle $A_{1} B_{1} C_{1}$ and three trapezoids $A_{1} B_{1} B A, B_{1} C_{1} C B$ and $C_{1} A_{1} A C$. The region $\mathcal{R}$ then comprises these trapezoids. We use $[\mathcal{P}]$ to denote the area of region $\mathcal{P}$. Using Heron's formula with semiperimeter $s=27$, we see that $[A B C]=\sqrt{27(27-13)(27-20)(27-21)}=126$. As $A B C$ and $A_{1} B_{1} C_{1}$ are similar, $\left[A_{1} B_{1} C_{1}\right]:[A B C]=k^{2}$ and with $B_{1} C_{1}=13 k, C_{1} A_{1}=20 k, A_{1} B_{1}=21 k$, we obtain
$$
\begin{aligned}
{[A B C] } & =\left[A_{1} B_{1} C_{1}\right]+\left[A_{1} B_{1} B A\right]+\left[B_{1} C_{1} C B\right]+\left[C_{1} A_{1} A C\right] \\
126 & =126 k^{2}+\frac{1}{2}(21 k+21)+\frac{1}{2}(20 k+20)+\frac{1}{2}(13 k+13) \\
& =126 k^{2}+27 k+27
\end{aligned}
$$
so $126 k^{2}+27 k-99=9(k+1)(14 k-11)=0$ and $k=\frac{11}{14}$. Therefore, the probability is
$$
\frac{[\mathcal{R}]}{[A B C]}=1-\frac{\left[A_{1} B_{1} C_{1}\right]}{[A B C]}=1-k^{2}=1-\frac{121}{196}=\frac{75}{196} \text {. }
$$

Solution 2: The additional points in the figure below (not drawn to scale) are precisely what they appear to be.
We can determine the proportionality constant $k$ between $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ by determining $A_{1} B_{1}=A_{2} B_{2}=21-A A_{2}-B B_{2}$. Since $\triangle A A_{1} A_{2}$ and $\triangle A A_{1} A_{3}$ are congruent right triangles, then $A A_{1}$ bisects $\angle A$. Let $\alpha=\angle A_{1} A A_{2}$. Then $\tan \alpha=\frac{A_{1} A_{2}}{A A_{2}}=\frac{1}{A A_{2}}$ so $A A_{2}=\cot \alpha$.
From Heron's Formula, $[A B C]=126$. Since $126=\frac{1}{2}(21)(C D)$, then $C D=12$. By the Pythagorean Theorem, $A D=16$ and $B D=5$.
Let the bisector of $\angle A$ meet the altitude $C D$ at $E$. Thus $\frac{D E}{C E}=\frac{A D}{A C}=\frac{16}{20}=\frac{4}{5}$. Since $C E+D E=$ 12 , then $D E=\frac{16}{3}$. This implies $\tan \alpha=\frac{D E}{A D}=\frac{16 / 3}{16}=\frac{1}{3}$, and so $A A_{2}=\cot \alpha=3$.
Similarly, $B B_{1}$ bisects $\angle B$. If $\beta=\angle B_{1} B B_{2}$, then $B B_{2}=\cot \beta$. Extend $B B_{1}$ to meet $C D$ at $F$. Since $\frac{D F}{C F}=\frac{B D}{B C}=\frac{5}{13}$ and $C F+D F=12$, then $D F=\frac{10}{3}$. This implies $\tan \beta=\frac{D F}{B D}=\frac{10 / 3}{5}=\frac{2}{3}$, and so $B B_{2}=\cot \beta=1.5$.
Finally, $A_{2} B_{2}=21-3-1.5=16.5$. Thus, $k=\frac{A_{1} B_{1}}{A B}=\frac{16.5}{21}=\frac{11}{14}$, and so
$$
\frac{[A B C]-\left[A_{1} B_{1} C_{1}\right]}{[A B C]}=\frac{[A B C]-k^{2}[A B C]}{[A B C]}=1-k^{2}=\frac{75}{196}
$$