Concentric Balance


Given three concentric circles $\omega_1, \omega_2, \omega_3$ with radii $r_1, r_2, r_3$ such that $r_1 + r_3 \geq 2r_2$, construct a line that intersects $\omega_1, \omega_2, \omega_3$ at $A, B, C$, respectively, such that $AB = BC$.  


Source: BdMO 2019 National Junior P10


Proof Based Problems  


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Solution

Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Figure 01

Figure 01: dividing $r_1$ equally

Figure 02

Figure 02: Construction


In Figure 1:

We first draw $AB=r_1$ and draw it's midpoint.

$M$. so $AM=\frac{r_1}{2}$



Now in Figure 2:

We do construction. First we select a random point E on biggest circle.

So $AE=r_3$.


Now we draw it's midpoint $F$.

Now we draw a circle with center $F$ with radius $\frac{r_1}{2}$.Name the circle $w$.


We know, $r_1+r_3 \geq 2r_2$

So, $\frac{r_1}{2}+\frac{r_3}{2} \geq r_2$


So, $w$ must intersect the 2nd big circle.Let's say at $G$.


Now draw a line parallel  to $FG$ through $A$

Which intersects small circle at $H$.


Now $AF=FE$,$FG || AH$ and $AH=2FG$

So, $HG=GE$ and $H,G,E$ collinear.


Thus we do the construction.

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