In a circle with center $O$, a trapezoid $A B C D (B C \| A D)$ is inscribed. In this same circle, a diameter $C E$ and a chord $B E$ are drawn, intersecting $A D$ at point $F$. Point $H$ is the foot of the perpendicular dropped from point $F$ to $C E$, $S$ is the midpoint of segment $E O$, and $M$ is the midpoint of $B D$. It is known that the radius of the circle is $R$, and $C H=9 R / 8$. Find $S M$.

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Since $C E$ is the diameter, then $B E \perp A D$. Through point $M$, draw a line parallel to $B E$, and through point $S$, draw a line parallel to $A D$. Let $P$ be the midpoint of $A D$, $Q$ the intersection of the drawn lines, $N$ the intersection of lines $B E$ and $Q S$, and $K$ the intersection of lines $M Q$ and $A D$ (see figure).

Let $B C=2 a, A D=2 b, B E=2 c, E F=d$. Then $R^{2}-b^{2}=O P^{2}=O F^{2}-a^{2}, O F^{2}-R^{2} / 64=F H^{2}=d^{2}-49 R^{2} / 64$. Subtracting, we get $b^{2}+d^{2}=7 R^{2} / 4+a^{2}$.

Notice that $B F \cdot F P=A F \cdot F D$, or $(2 c-d) d=(b-a)(b+a)=b^{2}-a^{2}, 2 c d=b^{2}+d^{2}-a^{2}=7 R^{2} / 4, b^{2}+d^{2}+2 c d=$ $a^{2}+7 R^{2} / 2$.

Triangles $N E S$ and $B E C$ are similar with a ratio of $1 / 4$, so $S Q=N Q-N S=F K-N S=a+b / 2-a / 2=b / 2$, $M Q=M K+K Q=1 / 2 B F+F N=1 / 2(2 c-d)+(d-c / 2)=c+d / 2$.

Therefore, $S M^{2}=M Q^{2}+S Q^{2}=1 / 4\left(c^{2}+d^{2}+2 c d+b^{2}\right)=1 / 4\left(c^{2}+a^{2}+7 R^{2} / 2\right)=1 / 4\left(R^{2}+7 R^{2} / 2\right)=9 R^{2} / 8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-18.jpg?height=717&width=720&top_left_y=-1&top_left_x=672)

## Answer

$\frac{3 R}{2 \sqrt{2}}$