The side faces of the pyramid are inclined to the base plane at an angle of 45 degrees. A ball is inscribed in this pyramid.

The side faces of the pyramid are inclined to the base plane at an angle of 45 degrees. A ball is inscribed in this pyramid. In what ratio, counting from the top, does the center of this ball divide the height of the pyramid?

Since the side faces of the pyramid are inclined to the base plane at one angle (45 degrees), the pyramid is correct.

The ball is inscribed in the pyramid (or the sphere is inscribed in the pyramid), which means that the ball (sphere) touches each face of the pyramid (in this case, it is the so-called H and so-called M).

The center of the ball in this case lies at the height of the pyramid (T.O).

The line segments connecting the center of the ball to the tangency points are perpendicular to the tangent planes. Their lengths are equal to the radius of the ball. Thus, OH = OM.

Three perpendicular theorem: If a straight line drawn on a plane through the base of an inclined one is perpendicular to its projection, then it is perpendicular to the most inclined one.

Let SН be the height of the pyramid.

If we draw HK perpendicular to BC and connect K to the vertex S. By the three-perpendicular theorem, SK will also be perpendicular to BC. Therefore, the angle HKS is the linear angle of the dihedral and is equal to 45 degrees.

Since DHKS is rectangular, and the HKS angle is 45 degrees, therefore, the HSC angle is also 45 degrees.

Consider DOMS:

DОМS is rectangular, since ОМ is the radius of the ball.

sin ОSМ = ОМ / SO

sin 45 = ОМ / SO = ОМ / SO

According to the condition of the problem, it is necessary to find the SO / OH ratio.

Since OH = OM, and OM / SO =, then SO / OH = / 1

Answer: SO / OH = / 1