This is possible if $r_1,r_2,r_3$ are three consequtive positive numbers.How to prove it for other numbers?

Take a point $C$ on $\omega_3$. We will consider a fourth circle, $\alpha$, which is homothetically similar to $\omega_1$ with ratio $\frac{1}{2}$ and with $C$ as the homothetic center. Let $\alpha \cap \omega_2 = B_1, B_2$ and let $B_1C \cap \omega_1 = A_1$, $B_2C \cap \omega_1 = A_2$. Then both $A_1B_1C$ and $A_2B_2C$ are the desired segment.

One of a bronze medalist of IMO 2018 from Bangladesh told that it is an IMO 3/6 difficulty problem. Is it really. I am not sure

I agree with you.It is very hard problem.We have to find a way of drawing which is suitable for all conditions.

But this was a problem of Junior category. Again nobody of our country,bangladesh ever solved IMO 3/6 problem in exam hall. So,....I am quite confused

Thanic vhai or Joydip vhai? Who said that it is so tough?

Thanic vai

@above and @2above are bengali also!! nice to know that!

thegreatp.d wrote: Thanic vai He is very experienced about this type of geometrical problem.. Aniv wrote: @above and @2above are bengali also!! nice to know that!

DeZade2002 wrote: Take a point $C$ on $\omega_3$. We will consider a fourth circle, $\alpha$, which is homothetically similar to $\omega_1$ with ratio $\frac{1}{2}$ and with $C$ as the homothetic center. Let $\alpha \cap \omega_2 = B_1, B_2$ and let $B_1C \cap \omega_1 = A_1$, $B_2C \cap \omega_1 = A_2$. Then both $A_1B_1C$ and $A_2B_2C$ are the desired segment. Without homothety? any idea?

This is actually an easy problem. let $w_1$ is outmost circle and $w_3$ is the innermost circle Let $B$ a point on $w_2$ take a half turn at $B$. Let this transformation that take $w_3$ at $w'$ now $w'$ and $w_1$ meet at $A$ . let the $line AB$ meet $w_3$ at C and D where $BC < BD$ now we take A,B,C three points on $w_1 ,w_2 ,w_3$ with $AB=BC$ and we are done