[ $\underline{\text { C }}$ е $\underline{p}$ а, inscribed in the pyramid.]

A sphere is inscribed in a regular quadrilateral pyramid, touching the base and all lateral faces. The sphere divides the height of the pyramid in the ratio $4: 5$, measured from the apex of the pyramid. Find the volume of the pyramid if the side of the base of the pyramid is $a$.
Let $b$ be the apothem of the given pyramid with vertex $P$, and $V$ be the volume of the pyramid. Since the pyramid is regular, the center of the inscribed sphere lies on the height, and the sphere touches the lateral faces of the pyramid at points lying on the apothems, and the base at its center $M$. Consider the section of the pyramid by a plane passing through the apothems $PA$ and $PB$ lying in two opposite lateral faces. We obtain an isosceles triangle $APB$ with sides $PA = PB = b$, $AB = a$, and an inscribed circle with center $O$ lying on the height $PM$. If the circle intersects the height $PM$ at a point $K$ different from $M$, then by the problem's condition, $PK / MK = 4 / 5$. Let $PK = 4x$. Then

$$
MK = 5x, OM = \frac{1}{2} MK = 5x / 2, OP = OK + PK = 5x / 2 + 4x = 13x / 2
$$

The center of the circle inscribed in an angle lies on the angle bisector, so $BO$ is the bisector of triangle $PBM$. By the property of the bisector of a triangle,

$$
(a / 2) / b = BM / BP = OM / OP = (5x / 2) / (13x / 2) = 5 / 13,
$$

from which we find that $b = 13a / 10$. By the Pythagorean theorem in the right triangle $PBM$, we find that

$$
PM = \sqrt{PB^2 - BM^2} = \sqrt{169a^2 / 100 - a^2 / 4} = 6a / 5
$$

Therefore,

$$
V = \frac{1}{3} AB^2 \cdot PM = \frac{1}{3} a^2 \cdot 6a / 5 = 2a^3 / 5
$$

## Answer

$$
2a^3 / 5
$$

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