113. The base of the pyramid $A B C E H$ is a convex quadrilateral $A B C E$, which is divided into two equal-area triangles by the diagonal $B E$. The length of the edge $A B$ is 1, the lengths of the edges $B C$ and $C E$ are equal. The sum of the lengths of the edges $A H$ and $E H$ is $\sqrt{2}$. The volume of the pyramid is $1 / 6$. Find the radius of the sphere with the largest volume among all spheres that can fit inside the pyramid.
113. Let $|E A|=x$, the area of $\triangle E M A$ will be the largest if $|E H|=|H A|=\frac{\sqrt{2}}{2}$, and in this case, it is equal to $\frac{x}{2} \sqrt{\frac{1}{2}-\frac{x^{2}}{4}}$. The distance from $B$ to the plane $E A H$ is not greater than $|A B|=1$. Since $S_{A E B}=S_{E B C}$

$$
\begin{aligned}
\frac{1}{12}=\frac{1}{2} V_{A B C E H}= & V_{A B E H} \leqslant \frac{x}{12} \sqrt{2-x^{2}}= \\
& =\frac{1}{12} \sqrt{x^{2}\left(2-x^{2}\right)} \leqslant \frac{1}{24}\left[x^{2}+\left(2-x^{2}\right)\right]=\frac{1}{12}
\end{aligned}
$$

Thus, $x=1$, and the segment $A B$ is perpendicular to the plane $E A H$. $A B C E$ is a square with side length 1.

Consider two triangular prismatic surfaces: the first is formed by the planes $A B C E, A H E$, and $B C H$; the second by $A B C E$, $E C H$, and $A B H$. Clearly, the radius of the largest sphere that fits inside the pyramid $A B C E H$ is equal to the radius of the smallest of the spheres inscribed in these prisms. The radius of the sphere inscribed in each of these prisms is equal to the radius of the circle inscribed in the perpendicular section. The perpendicular section of the first prism is a right triangle with legs 1 and $1 / 2$, the radius of the circle inscribed in this triangle is $\frac{3-\sqrt{5}}{4}$. The perpendicular section of the second prism is the triangle $A H E$, the radius of the circle inscribed in it is $\frac{\sqrt{2}-1}{2}>\frac{3-\sqrt{5}}{4}$.

Answer: $\frac{3-\sqrt{5}}{4}$.