[ Perpendicular Planes ]

The base of the pyramid $A B C E H$ is a convex quadrilateral $A B C E$, which is divided by the diagonal $B E$ into two equal-area triangles. The length of edge $A B$ is 1, and the lengths of edges $B C$ and $C E$ are equal to each other. The sum of the lengths of 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 $A B C E H$.
Since the diagonal $B E$ divides the quadrilateral $A B C E$ into two equal-area triangles, the triangular pyramids $A B E H$ and $B C E H$ with a common vertex $H$ are equal in volume. Therefore, $V(A B E H)=1 / 12$.

Let $B K$ be the height of the triangular pyramid $A B E H$, drawn from the vertex $B$. Then $B K \leq A B=1$. Therefore,

$$
\begin{gathered}
1 / 12=V(A B E H)=\frac{1}{3} S(A E H) \cdot B K=\frac{1}{3} \frac{1}{2} A H \cdot H E \cdot \sin \angle A H E \cdot B K \leq \\
\leq \frac{1}{3} \frac{1}{2} A H \cdot H E \cdot 1 \cdot 1=\frac{1}{6} A H \cdot H E \leq \frac{1}{6}\left(\frac{1}{2}(A H+H E)\right)^{2}=\frac{1}{6}\left(\frac{1}{2} \sqrt{2}\right)^{2}=1 / 12
\end{gathered}
$$

This is only possible if

$$
A H=H E=\sqrt{2} / 2, \angle A H E=90^{\circ}, B K=A B=1,
$$

and the point $K$ coincides with the point $A$, i.e., $A B$ is perpendicular to the plane $A H E$. Therefore, $A B \perp A E$. Since the base plane passes through the perpendicular to the plane of the face $A H E$, these planes are perpendicular, so the height $H M$ of the right triangle $A H E$ is the height of the pyramid $A B C E H$.

From the isosceles right triangle $A H E$, we find that $A E=1, H M=1 / 2$. The isosceles triangles $A B E$ and $C B E$ with a common base $B E$ are equal in area, so they are congruent. Therefore, the quadrilateral $A B C E$ is a square with a side length of 1, and the height of the pyramid $A B C E H$ is $1 / 2$ and passes through the midpoint $M$ of the side $A E$ of the base.

Consider the section of the pyramid by a plane passing through the points $H, M$ and the midpoint $N$ of the edge $B C$. We obtain a right triangle $H M N$ with sides

$$
M N=1, H M=1 / 2, H N=\sqrt{5} / 2
$$

Let $O$ be the center of the circle inscribed in the triangle $H M N$, and $r$ be its radius. Then

$$
r=\frac{1}{2}(M N+H M-H N)=(3-\sqrt{5}) / 4
$$

We will prove that a sphere of radius $r$ with center at point $O$ fits inside the pyramid $A B C E H$. Since the center of the sphere lies in a plane perpendicular to the faces $A H E$ and $B H C$, it touches these faces. Therefore, it is sufficient to establish that the distances from point $O$ to the planes of the faces $A H B$ and $C H E$ are not less than $r$.

Draw a plane through point $O$ parallel to the face $A H E$. Let this plane intersect the edges $A B, B H, C H$ and $C E$ at points $P, Q, R$ and $S$ respectively. From the theorem on the intersection of two parallel planes by a third, it follows that $P Q R S$ is an isosceles trapezoid. Let $F$ be the foot of the perpendicular dropped from point $O$ to the lateral side $P Q$ of this trapezoid. Then $O F$ is perpendicular to the plane of the face $A H B (O F \perp P Q, O F \perp A B)$.

Let $L$ be the point of tangency of the sphere with the base plane of the pyramid. Then

$$
\begin{gathered}
A P=S E=L M=r=(3-\sqrt{5}) / 4 \\
P S=A E=1, \\
R S=P Q=A H \cdot P B / A B=(\sqrt{2} / 2) \cdot(1-r) / 1=\sqrt{2}(\sqrt{5}+1) / 8 \\
Q R=B C \cdot M L / M N=1 \cdot r / 1=r=(3-\sqrt{5}) / 4
\end{gathered}
$$

Let $T$ be the midpoint of $P Q$, and $G$ be the foot of the perpendicular dropped from point $Q$ to the base $P C$ of the trapezoid $P Q R S$. Then $O Q$ is half the length of the midline of the trapezoid, and the right triangles $O F T$ and $Q G P$ are similar. Therefore,

$$
\begin{gathered}
O T=\frac{1}{4}(P S+Q R)=\frac{1}{4}(1+(3-\sqrt{5}) / 4)=(7-\sqrt{5}) / 16 \\
Q G=2 r=(3-\sqrt{5}) / 2 \\
O F=Q G \cdot O T / P Q=((3-\sqrt{5}) / 2) \cdot((7-\sqrt{5}) / 16) /(\sqrt{2}(\sqrt{5}+1) / 8)= \\
=(9 \sqrt{5}-19) /(4 \sqrt{2})>(3-\sqrt{5}) / 4
\end{gathered}
$$

Similarly, we will prove that the distance from point $O$ to the plane of the face $C H E$ is also less than $r$. Thus, the radius of the largest sphere that fits inside the pyramid $A B C E H$ is $(3-\sqrt{5}) / 4$.