**Solution to Problem 17**
We will apply barycentric coordinates on $\triangle ABC$. The calculations are quite routine and requires one major insight.

$A=(1:0:0),A_1=(0

c),A_2=(0:s-c:s-b)$

Let the circumcircle of $\triangle AA_1A_2$ have the equation $$-a^2yz-b^2zx-c^2cy+(ux+vy+wz)(x+y+z)=0$$

This circle passes through $A$ making $u=0$. Plugging in the coordinates of $A_1$ and $A_2$ yields the system $$vb+wc=\dfrac{a^2bc}{b+c}$$

$$v(s-c)+w(s-b)=a(s-c)(s-b)$$

Solving this turns out to be surprisingly neat, contrary to what it looks like now. Alternatively, here's a trick you can use with symmetric linear equations that works sometimes in bary. The first equation suggests that both $v$ and $w$ have a $b+c$ in the denominator. The symmetric coefficients $b$ and $c$ suggest the numerators of $v$ and $w$ should have $c$ and $b$. Similarly, the symmetric coefficients in the second equation suggest the numerators also have $s-b$ and $s-c$ respectively. Putting this in the equations suggest that a factor of $a$ is missing from both, so we add that as well. The final solutions are $v=\dfrac{ac(s-b)}{b+c},w=\dfrac{ab(s-c)}{b+c}$ which indeed work.

Now finding the equation of $BB_1B_2$ simply becomes a matter of cyclically rotating $a,b,c$ in the above expressions. To find their radical axis, we simply subtract the two to get $$ac(s-b)(c+a)y-bc(s-a)(b+c)x+ab(s-c)(a-b)z=0$$

The other radical axes can be gained by cyclically rotating $a,b,c$ alongside $x,y,z$.

Let us stop here. Looks like the equations are getting out of hand. We need to show that these lines intersect on $OI$. It basically suffices to show that the equation of $OI$ is linearly dependent on the other three. Let's calculate $OI$ and hope we find something.

$$\begin{vmatrix}

x&a^2S_A& a \\

y&b^2S_B& b \\

z&c^2S_C& c

\end{vmatrix}=0$$

which turns into $$xbc(b-c)(s-a)+yca(c-a)(s-b)+zab(a-b)(s-c)=0$$

Do the coefficients look familiar? They seem to match up with the one term from each equation of the radical axis. Here's the equation of the radical axis again.

$$ac(s-b)(c+a)y-bc(s-a)(b+c)x+ab(s-c)(a-b)z=0$$

We do the only thing we can to get that term and eliminate the rest. Add the three radical equations up. The $-bc(s-a)(b+c)$ of this equation and $+bc(s-a)(b+c)$ from another equation will cancel each other out, and so the only terms left will be the ones in our equation of $OI$. We are done!!