IMO Shortlist

$AZ$, $AI$ and $AY$ divide $\angle BAC$ into four equal angles; denote them by $\alpha$. In the same way we have four equal angles $\beta$ at $B$ and four equal angles $\gamma$ at $C$. Obviously $\alpha + \beta + \gamma = \frac{180^\circ}{4} = 45^\circ$; and $0^\circ < \alpha, \beta, \gamma < 45^\circ$.

Easy calculations in various triangles yield $\angle BIC = 180^\circ - 2\beta - 2\gamma = 180^\circ - (90^\circ - 2\alpha) = 90^\circ + 2\alpha$, hence (for $X$ is the incenter of triangle $BCI$, so $IX$ bisects $\angle BIC$) we have $\angle XIC = \angle BIX = \frac{1}{2} \angle BIC = 45^\circ + \alpha$ and with similar arguments $\angle CIY = \angle YIA = 45^\circ + \beta$ and $\angle AIZ = \angle ZIB = 45^\circ + \gamma$.

Furthermore, we have $\angle XIY = \angle XIC + \angle CIY = (45^\circ + \alpha) + (45^\circ + \beta) = 135^\circ - \gamma$, $\angle YIZ = 135^\circ - \alpha$, and $\angle ZIX = 135^\circ - \beta$.

Now we calculate the lengths of $IX$, $IY$ and $IZ$ in terms of $\alpha$, $\beta$ and $\gamma$. The perpendicular from $I$ on $CX$ has length $IX \cdot \sin \angle CXI = IX \cdot \sin (90^\circ + \beta) = IX \cdot \cos \beta$. But $CI$ bisects $\angle YCX$, so the perpendicular from $I$ on $CY$ has the same length, and we conclude
$$IX \cdot \cos \beta = IY \cdot \cos \alpha$$
To make calculations easier we choose a length unit that makes $IX = \cos \alpha$. Then $IY = \cos \beta$ and with similar arguments $IZ = \cos \gamma$.

Since $XYZ$ is equilateral we have $ZX = ZY$. The law of cosines in triangles $XYI$, $YZI$ yields
$$\begin{aligned} & ZX^2 = ZY^2 \\ \Longrightarrow & IZ^2 + IX^2 - 2 \cdot IZ \cdot IX \cdot \cos \angle ZIX = IZ^2 + IY^2 - 2 \cdot IZ \cdot IY \cdot \cos \angle YIZ \\ \Longrightarrow & IX^2 - IY^2 = 2 \cdot IZ \cdot (IX \cdot \cos \angle ZIX - IY \cdot \cos \angle YIZ) \\ \Longrightarrow & \underbrace{\cos^2 \alpha - \cos^2 \beta}_{\text{L.H.S.}} = \underbrace{2 \cdot \cos \gamma \cdot (\cos \alpha \cdot \cos (135^\circ - \beta) - \cos \beta \cdot \cos (135^\circ - \alpha))}_{\text{R.H.S.}}. \end{aligned}$$
A transformation of the left-hand side (L.H.S.) yields
$$\begin{aligned} \text{L.H.S.} & = \cos^2 \alpha \cdot (\sin^2 \beta + \cos^2 \beta) - \cos^2 \beta \cdot (\sin^2 \alpha + \cos^2 \alpha) \\ & = \cos^2 \alpha \cdot \sin^2 \beta - \cos^2 \beta \cdot \sin^2 \alpha \end{aligned}$$
$$\begin{aligned} & = (\cos \alpha \cdot \sin \beta + \cos \beta \cdot \sin \alpha) \cdot (\cos \alpha \cdot \sin \beta - \cos \beta \cdot \sin \alpha) \\ & = \sin (\beta + \alpha) \cdot \sin (\beta - \alpha) = \sin (45^\circ - \gamma) \cdot \sin (\beta - \alpha) \end{aligned}$$
whereas a transformation of the right-hand side (R.H.S.) leads to
$$\begin{aligned} \text{R.H.S.} & = 2 \cdot \cos \gamma \cdot (\cos \alpha \cdot (-\cos (45^\circ + \beta)) - \cos \beta \cdot (-\cos (45^\circ + \alpha))) \\ & = 2 \cdot \frac{\sqrt{2}}{2} \cdot \cos \gamma \cdot (\cos \alpha \cdot (\sin \beta - \cos \beta) + \cos \beta \cdot (\cos \alpha - \sin \alpha)) \\ & = \sqrt{2} \cdot \cos \gamma \cdot (\cos \alpha \cdot \sin \beta - \cos \beta \cdot \sin \alpha) \\ & = \sqrt{2} \cdot \cos \gamma \cdot \sin (\beta - \alpha) \end{aligned}$$
Equating L.H.S. and R.H.S. we obtain
$$\begin{aligned} & \sin (45^\circ - \gamma) \cdot \sin (\beta - \alpha) = \sqrt{2} \cdot \cos \gamma \cdot \sin (\beta - \alpha) \\ \Longrightarrow & \sin (\beta - \alpha) \cdot (\sqrt{2} \cdot \cos \gamma - \sin (45^\circ - \gamma)) = 0 \\ \Longrightarrow & \alpha = \beta \text{ or } \sqrt{2} \cdot \cos \gamma = \sin (45^\circ - \gamma). \end{aligned}$$
But $\gamma < 45^\circ$; so $\sqrt{2} \cdot \cos \gamma > \cos \gamma > \cos 45^\circ = \sin 45^\circ > \sin (45^\circ - \gamma)$. This leaves $\alpha = \beta$.

With similar reasoning we have $\alpha = \gamma$, which means triangle $ABC$ must be equilateral.