Use the remarkable property of a circle and the theorem about the product of segments of intersecting chords.

## Solution

Let the diameter $A C$ intersect the chord $P Q$ at point $N$. Denote $A M=x, P Q=4 y$. Then $B M=A B-A M=a-x$, and since $N$ is the midpoint of $P Q$, we have $Q N=2 y$, so $M N=Q M=y$.

By the Pythagorean theorem in the right triangle $A B C$, we find that

$$
B C=\sqrt{A C^{2}-A B^{2}}=\sqrt{4 R^{2}-a^{2}}
$$

From the similarity of the right triangles $A B C$ and $A N M$, we get that $\frac{M N}{A M}=\frac{B C}{A C}$, from which

$$
y=M N=\frac{A M \cdot B C}{A C}=\frac{x \sqrt{4 R^{2}-a^{2}}}{2 R}
$$

By the theorem about the product of segments of intersecting chords

$$
A M \cdot B M=Q M \cdot P M
$$

or

$$
x(a-x)=y \cdot 3 y=3 y^{2}=3\left(\frac{x \sqrt{4 R^{2}-a^{2}}}{2 R}\right)^{2}=\frac{3}{4} \cdot \frac{x^{2}\left(4 R^{2}-a^{2}\right)}{R^{2}}
$$

From the equation

$$
x(a-x)=\frac{3}{4} \cdot \frac{x^{2}\left(4 R^{2}-a^{2}\right)}{R^{2}}
$$

we find that

$$
A M=x=\frac{4 a R^{2}}{16 R^{2}-3 a^{2}}
$$

## Answer

$\frac{4 a R^{2}}{16 R^{2}-3 a^{2}}$.
[ Inscribed angle, based on the diameter

In a circle of radius $R$, a chord $A B$ and a diameter $A C$ are drawn. A chord $P Q$, perpendicular to the diameter $A C$, intersects the chord $A B$ at point $M$. It is known that $A B=a, P M: M Q=3$. Find $A M$.

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