IMO Problem 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 $$\begin{array}{rl}& Z{X}^2=Z{Y}^2\\ ⟹& I{Z}^2+I{X}^2-2\cdot IZ\cdot IX\cdot \cos \angle ZIX=I{Z}^2+I{Y}^2-2\cdot IZ\cdot IY\cdot \cos \angle YIZ\\ ⟹& I{X}^2-I{Y}^2=2\cdot IZ\cdot (IX\cdot \cos \angle ZIX-IY\cdot \cos \angle YIZ)\\ ⟹& \underset{\text{L.H.S.}}{\underbrace{{\cos }^2\alpha -{\cos }^2\beta }}=\underset{\text{R.H.S.}}{\underbrace{2\cdot \cos \gamma \cdot (\cos \alpha \cdot \cos (135^{\circ }-\beta )-\cos \beta \cdot \cos (135^{\circ }-\alpha ))}}.\end{array}$$ A transformation of the left-hand side (L.H.S.) yields $$\begin{array}{rl}\text{L.H.S.}& ={\cos }^2\alpha \cdot ({\sin }^2\beta +{\cos }^2\beta )-{\cos }^2\beta \cdot ({\sin }^2\alpha +{\cos }^2\alpha )\\ & ={\cos }^2\alpha \cdot {\sin }^2\beta -{\cos }^2\beta \cdot {\sin }^2\alpha \end{array}$$ $$\begin{array}{rl}& =(\cos \alpha \cdot \sin \beta +\cos \beta \cdot \sin \alpha )\cdot (\cos \alpha \cdot \sin \beta -\cos \beta \cdot \sin \alpha )\\ & =\sin (\beta +\alpha )\cdot \sin (\beta -\alpha )=\sin (45^{\circ }-\gamma )\cdot \sin (\beta -\alpha )\end{array}$$ whereas a transformation of the right-hand side (R.H.S.) leads to $$\begin{array}{rl}\text{R.H.S.}& =2\cdot \cos \gamma \cdot (\cos \alpha \cdot (-\cos (45^{\circ }+\beta ))-\cos \beta \cdot (-\cos (45^{\circ }+\alpha )))\\ & =2\cdot \frac{\sqrt{2}}{2}\cdot \cos \gamma \cdot (\cos \alpha \cdot (\sin \beta -\cos \beta )+\cos \beta \cdot (\cos \alpha -\sin \alpha ))\\ & =\sqrt{2}\cdot \cos \gamma \cdot (\cos \alpha \cdot \sin \beta -\cos \beta \cdot \sin \alpha )\\ & =\sqrt{2}\cdot \cos \gamma \cdot \sin (\beta -\alpha )\end{array}$$ Equating L.H.S. and R.H.S. we obtain $$\begin{array}{rl}& \sin (45^{\circ }-\gamma )\cdot \sin (\beta -\alpha )=\sqrt{2}\cdot \cos \gamma \cdot \sin (\beta -\alpha )\\ ⟹& \sin (\beta -\alpha )\cdot (\sqrt{2}\cdot \cos \gamma -\sin (45^{\circ }-\gamma ))=0\\ ⟹& \alpha =\beta \text{ or }\sqrt{2}\cdot \cos \gamma =\sin (45^{\circ }-\gamma ).\end{array}$$ 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.