Solution:
Let $ABC$ be a triangle with sides $BC = 13$, $CA = 20$, and $AB = 21$ 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 $ABC$. For a fixed side of $ABC$ (say $\ell$), $\omega$ intersects $\ell$ if and only if $P$ lies within one unit from $\ell$. This suggests we construct a triangle $A_1B_1C_1$ such that $A_1B_1 \parallel AB$, $B_1C_1 \parallel BC$, $C_1A_1 \parallel CA$ and the corresponding parallel sides of $A_1B_1C_1$ and $ABC$ have distance $1$. Thus, the set $S$ of such points $P$ forms a region $\mathcal{R}$ outside $A_1B_1C_1$ but inside $ABC$ and the probability is then the ratio of the areas of $\mathcal{R}$ and $ABC$.

Observe that $ABC$ and $A_1B_1C_1$ are similar and hence the ratio of their corresponding sides is constant, say $k > 0$. Also, triangle $ABC$ is divided into four regions: the triangle $A_1B_1C_1$ and three trapezoids $A_1B_1BA$, $B_1C_1CB$, and $C_1A_1AC$. 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 $[ABC] = \sqrt{27(27-13)(27-20)(27-21)} = 126$. As $ABC$ and $A_1B_1C_1$ are similar, $[A_1B_1C_1] : [ABC] = k^2$ and with $B_1C_1 = 13k$, $C_1A_1 = 20k$, $A_1B_1 = 21k$, we obtain
$$\begin{aligned} [ABC] & = [A_1B_1C_1] + [A_1B_1BA] + [B_1C_1CB] + [C_1A_1AC] \\ 126 & = 126k^2 + \frac{1}{2}(21k + 21) + \frac{1}{2}(20k + 20) + \frac{1}{2}(13k + 13) \\ & = 126k^2 + 27k + 27 \end{aligned}$$
so $126k^2 + 27k - 99 = 9(k+1)(14k-11) = 0$ and $k = \frac{11}{14}$. Therefore, the probability is
$$\frac{[\mathcal{R}]}{[ABC]} = 1 - \frac{[A_1B_1C_1]}{[ABC]} = 1 - k^2 = 1 - \frac{121}{196} = \frac{75}{196}$$

Solution:
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 $\triangle A_1B_1C_1$ and $\triangle ABC$ by determining $A_1B_1 = A_2B_2 = 21 - AA_2 - BB_2$. Since $\triangle AA_1A_2$ and $\triangle AA_1A_3$ are congruent right triangles, then $AA_1$ bisects $\angle A$. Let $\alpha = \angle A_1AA_2$. Then $\tan \alpha = \frac{A_1A_2}{AA_2} = \frac{1}{AA_2}$ so $AA_2 = \cot \alpha$.

From Heron's Formula, $[ABC] = 126$. Since $126 = \frac{1}{2}(21)(CD)$, then $CD = 12$. By the Pythagorean Theorem, $AD = 16$ and $BD = 5$.

Let the bisector of $\angle A$ meet the altitude $CD$ at $E$. Thus $\frac{DE}{CE} = \frac{AD}{AC} = \frac{16}{20} = \frac{4}{5}$. Since $CE + DE = 12$, then $DE = \frac{16}{3}$. This implies $\tan \alpha = \frac{DE}{AD} = \frac{16/3}{16} = \frac{1}{3}$, and so $AA_2 = \cot \alpha = 3$.

Similarly, $BB_1$ bisects $\angle B$. If $\beta = \angle B_1BB_2$, then $BB_2 = \cot \beta$. Extend $BB_1$ to meet $CD$ at $F$. Since $\frac{DF}{CF} = \frac{BD}{BC} = \frac{5}{13}$ and $CF + DF = 12$, then $DF = \frac{10}{3}$. This implies $\tan \beta = \frac{DF}{BD} = \frac{10/3}{5} = \frac{2}{3}$, and so $BB_2 = \cot \beta = 1.5$.

Finally, $A_2B_2 = 21 - 3 - 1.5 = 16.5$. Thus, $k = \frac{A_1B_1}{AB} = \frac{16.5}{21} = \frac{11}{14}$, and so
$$\frac{[ABC] - [A_1B_1C_1]}{[ABC]} = \frac{[ABC] - k^2[ABC]}{[ABC]} = 1 - k^2 = \frac{75}{196}$$