[ Volume of a Tetrahedron and Pyramid

Given a regular quadrilateral pyramid PABCD ( $P$ - vertex) with the side of the base $a$ and the lateral edge $a$. A sphere with center at point $O$ passes through point $A$ and touches the edges $P B$ and $P D$ at their midpoints. Find the volume of the pyramid $O P C D$.

#
Let $M$ and $N$ be the midpoints of edges $PB$ and $PD$ respectively, and $K$ be the center of the base $ABCD$. Since the given sphere touches the line $PB$ at point $M$, its center $O$ lies in the plane passing through point $M$ and perpendicular to the line $PB$. Similarly, we can prove that point $O$ lies in the plane passing through point $N$ and perpendicular to the line $PD$. These two planes intersect along the line $AC$. Therefore, point $O$ lies on the line $AC$. Let $KO = x$, $OA = OM = ON = R$. Suppose that point $O$ lies between points $A$ and $K$. By the Pythagorean theorem in the right triangle $KOB$, we find that

$$
OB^2 = KO^2 + KB^2 = x^2 + \frac{a^2}{2}
$$

Since $OM$ lies in the plane perpendicular to $PB$, $OM \perp BM$, so

$$
R^2 = OM^2 = OB^2 - BM^2 = x^2 + \frac{a^2}{2} - \frac{a^2}{4} = x^2 + \frac{a^2}{4}
$$

On the other hand,

$$
R = OA = AK - KO = \frac{a \sqrt{2}}{2} - x
$$

From the equation

$$
x^2 + \frac{a^2}{4} = \left(\frac{a \sqrt{2}}{2} - x\right)^2
$$

we find that $x = \frac{a \sqrt{2}}{8}$. If point $O$ lies on the segment $CK$, then reasoning similarly, we get that $x < 0$, which is impossible. Further, we find:

$$
\begin{gathered}
\frac{CO}{AC} = \frac{CK + KO}{AC} = \frac{\frac{a \sqrt{2}}{2} + \frac{a \sqrt{2}}{8}}{\Delta \sqrt{2}} = \frac{5}{8} \\
S_{\triangle OCD} = \frac{CO}{AC} S_{\triangle ADC} = \frac{5}{8} \cdot \frac{a^2}{2} = \frac{5a^2}{16}
\end{gathered}
$$

Therefore,

$$
V_{OPCD} = \frac{1}{3} S_{\triangle OCD} \cdot PK = \frac{1}{3} \cdot \frac{5a^2}{16} \cdot \frac{a \sqrt{2}}{2} = \frac{5a^3 \sqrt{2}}{96}
$$

## Answer

$\frac{55^3 \sqrt{2}}{96}$.