4. Points $A, B, C, D, E$ are sequentially located on a line, such that $A B=4, B C=D E=2, C D=3$. Circles $\Omega$ and $\omega$, touching each other, are such that $\Omega$ passes through points $D$ and $E$, and $\omega$ passes through points $B$ and $C$. Find the radii of circles $\Omega$ and $\omega$, given that their centers and point $A$ lie on the same line.

“Fiztech-2016", mathematics, solutions for ticket 12

we substitute into the first equation, the period $2 k \pi$ can be omitted in this process.
Answer: $r=\frac{9 \sqrt{3}}{2 \sqrt{17}}, R=\frac{8 \sqrt{3}}{\sqrt{17}}$.

Solution. Let the centers of circles $\Omega$ and $\omega$ be $O$ and $Q$ respectively. Drop perpendiculars $OH$ and $QT$ from points $O$ and $Q$ to line $AB$, then points $H$ and $T$ are the midpoints of chords $DE$ and $BC$ respectively (a diameter perpendicular to a chord bisects the chord). Therefore, $AT=5, AH=10$.

Let $QT=x$. Then $QH=2x$ (since $QT$ is the midline of triangle $AOH$), $r=QB=\sqrt{QT^2+BT^2}=\sqrt{x^2+1}$, $R=OD=\sqrt{OH^2+DH^2}=\sqrt{4x^2+1}$. We can express the segment $OQ$ in two ways. On one hand, since the circles touch externally, the distance between their centers is the sum of the radii, i.e., $OQ=R+r$. On the other hand, from the right trapezoid $HTQO$, we get that $OQ^2=TH^2+(QT-OH)^2=25+x^2$.

Thus, $\sqrt{1+x^2}+\sqrt{4x^2+1}=\sqrt{x^2+25}$, from which $5x^2+2+2\sqrt{4x^4+5x^2+1}=x^2+25 \Leftrightarrow 2\sqrt{4x^4+5x^2+1}=23-4x^2$.

Under the condition $23-4x^2 \geq 0$, the last equation is equivalent to the following: $16x^4+20x^2+4=529-184x^2+16x^4$, $x^2=\frac{175}{68}$. Then we get that $r=\frac{9 \sqrt{3}}{2 \sqrt{17}}, R=\frac{8 \sqrt{3}}{\sqrt{17}}$.