Circumcenter and 3 circumcircles

Given an acute triangle $ABC$ and its circumcircle is centered at $O$ and has radius $R$.

$AO$ meets the circumcircle of $\triangle OBC$ at $X$, $BO$ meets the circumcircle of $\triangle OAC$ at $Y$, and $CO$ meets the circumcircle of $\triangle OAB$ at $Z$.

Show that $$AX + BY + CZ \geq 9R$$

Determine when equality occurs.

Solution

Let $AX$ meets $BC$ at $D$. And let $OO'$ be the diameter of the circumcircle of $\triangle OBC$, and let $M$ be the intersection of $OO'$ and $BC$.

$$\frac{OX}{OO'} = \frac{OM}{OD}$$ $$OX.OD = OO'.OM = OC^2 = R^2$$ $$AX = AO + OX = R + \frac{R^2}{OD} = \frac{R(R + OD)}{OD} = R\frac{AD}{OD}$$

If we define $E$ as the intersection of $BY$ and $AC$, and $F$ as the intersection of $CZ$ and $AB$, then similarly: $$BY = R \frac{BE}{OE}$$ $$CZ = R \frac{CF}{OF}$$

So: $$AX + BY + CZ = R(\frac{AD}{OD} + \frac{BE}{OE} + \frac{CF}{OF})$$ $$\geq \frac{9R}{\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF}}$$ But $OD / AD = A_{\triangle OBC} / A_{\triangle ABC}$ and so on, so: $$\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF} = 1$$ $$AX + BY + CZ \geq 9R$$

and equality happens when the area of $\triangle OBC$ is exactly $\triangle ABC / 3$. And this happens only when $O$ is the centroid of $\triangle ABC$, which means $ABC$ is an equilateral.