[ Pythagorean Theorem (direct and inverse). ]

On a straight line, points $A, B, C$, and $D$ are arranged such that $A B=B C=C D$. Segments $A B, B C$, and $C D$ serve as diameters of circles. From point $A$, a tangent $l$ is drawn to the circle with diameter $C D$. Find the ratio of the chords cut by the line $l$ on the circles with diameters $A B$ and $B C$.
Drop perpendiculars from the centers of the circles to the line $l$.

## Solution

Let $O_{1}, O_{2}, O_{3}$ be the centers of circles $\Omega_{1}, \Omega_{2}, \Omega_{3}$ with radii $R$ and diameters $A B, B C$, and $C D$ respectively; $A M$ and $N K$ be the chords cut by the line $l$ on $\Omega_{1}$ and $\Omega_{2}$; $L$ be the point of tangency of the line $l$ with $\Omega_{3}$; $E$ and $F$ be the projections of points $O_{1}$ and $O_{2}$ onto the line $l$.

Triangle $A O_{1} E$ is similar to triangle $A O_{3} L$ with a similarity ratio of $1 / 5$, and triangle $A O_{2} F$ is similar to triangle $A_{3} L$ with a similarity ratio of $3 / 5$, therefore

$O_{1} E = 1 / 5 O_{3} L = R / 5, O_{2} F = 3 / 5 O_{3} L = 3 R / 5$

From the right triangles $O_{1} E M$ and $O_{2} F K$, using the Pythagorean theorem, we find that

$E M = \frac{2 R \sqrt{6}}{5}, F K = \frac{4 R}{5}$

Therefore, $A M: N K = E M: F K = \sqrt{6}: 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-23.jpg?height=446&width=1238&top_left_y=1186&top_left_x=410)

## Answer

$\sqrt{6}: 2$.