Given point $P$, circle $k$, and the chord $AB$ passing through $P$, such that $PA = AB = 1$. The tangents drawn from $P$ to circle $k$ touch the circle at points $C$ and $D$, and the intersection of $AB$ and $CD$ is point $M$. What is the length of $PM$?
Solution. Let the center of circle $k$ be $O$. According to the known construction of tangents, the Thalész circle of $O P$ intersects circle $k$ at the points of tangency $C, D$. Let the midpoint of chord $A B$ be $F$. Then the segment $O F$ is perpendicular to the chord $A B$, $\angle O F P = 90^{\circ}$, so the above Thalész circle $k_{1}$ also passes through point $F$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_e1d7c46a0d034546277fg-1.jpg?height=393&width=553&top_left_y=321&top_left_x=775)

# Figure 1

According to the theorem of secant segments, in circle $k$, $M A \cdot M B = M C \cdot M D$, and in circle $k_{1}$, $M C \cdot M D = M F \cdot M P$. From these two equalities, we get $M A \cdot M B = M F \cdot M P$. Introducing the notation $x = M A$, we have $M B = 1 - x$, $M F = \frac{1}{2} - x$, and finally $M P = 1 + x$ (Figure 2). The equation is: $x(1 - x) = \left(\frac{1}{2} - x\right)(1 + x)$. The solution to this equation is $x = \frac{1}{3}$, so the distance $P M$ is $P A + A M = 1 + x = \frac{4}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e1d7c46a0d034546277fg-1.jpg?height=222&width=553&top_left_y=1147&top_left_x=772)

Figure 2