Korean Mathematical Olympiad

Suppose $△ACE$ is equilateral, then $△BDF$ is always equilateral, because three triangles $ACE$, $CDE$, $EFA$ are identical and shape of the hexagon $ABCDEF$ is invariable under the $120^{\circ }$ rotation around the center of $△ACE$. That is, since $B\to D\to F$ is obtained from the $120^{\circ }$ rotation, $△BDF$ is an equilateral triangle.

We now show that if three similar triangles $ABC$, $CDE$, $EFA$ are not of a isosceles triangle whose two same angles are $30^{\circ }$, the converse is true. First, let $\angle BAC=\angle DCE=\angle FEA=\alpha$, $\angle BCA=\angle DEC=\angle FAE=\beta$. Suppose that $△BDF$ is equilateral. Let $$\begin{array}{rl}\frac{AB}{AC}& =\frac{CD}{CE}=\frac{EF}{EA}=s\\ \frac{BC}{AC}& =\frac{DE}{CE}=\frac{FA}{EA}=t,\end{array}$$ $AC=a$, $CE=b$, $EA=c$. For $△ABF$, we have $$\begin{array}{rl}B{F}^2& =A{B}^2+F{A}^2-2AB\cdot FA\cdot \cos (\angle BAF)\\ & =(sa{)}^2+(tc{)}^2-2stac\cdot \cos (\alpha +\beta +A)\\ & =s^2{a}^2+t^2{c}^2-2stac[\cos (\alpha +\beta )\cos A-\sin (\alpha +\beta )\sin A].\end{array}$$

Similarly we have, $$B{D}^2=s^2{b}^2+t^2{a}^2-2stab[\cos (\alpha +\beta )\cos C-\sin (\alpha +\beta )\sin C]$$ $$D{F}^2=s^2{b}^2+t^2{a}^2-2stab[\cos (\alpha +\beta )\cos C-\sin (\alpha +\beta )\sin C].$$ $\frac{1}{2}ac\sin A=\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin E$ equals the area of $△ACE$, and $B{F}^2=B{D}^2=D{F}^2$. Moreover, we have $2ac\cos A=a^2+c^2-b^2$, $2ab\cos C=a^2+b^2-c^2$, $2bc\cos E=b^2+c^2-a^2$. Let $\cos (\alpha +\beta )=K$, then we have $$\begin{array}{rl}& s^2{a}^2+t^2{c}^2-stK(a^2+c^2-b^2)\\ & =s^2{b}^2+t^2{c}^2-stK(a^2+b^2-c^2)\\ & =s^2{c}^2+t^2{b}^2-stK(b^2+c^2-a^2)\end{array}$$ Now let $$s^2-t^2=x,2stK-s^2=y,t^2-2stK=z,$$ then we have $$x{a}^2+y{b}^2+z{c}^2=y{a}^2+z{b}^2+x{c}^2=z{a}^2+x{b}^2+y{c}^2(1)$$ From the equations (1), suppose first that $x=y=z$, then we have $s=t$ and $K=\frac{1}{2}$; hence $\alpha +\beta =60^{\circ }$. That is, three similar triangles $ABC$, $CDE$, $EFA$ are of a isosceles triangle whose two same angles are $30^{\circ }$. We exclude this case. Second, suppose $x,y,z$ are either all distinct or two of them are equal, then we get the equations $a=b=c$ by some calculations, which means that $△ACE$ is equilateral. We conclude that $△ACE$ is equilateral unless three similar triangles $ABC$, $CDE$, $EFA$ are of a isosceles triangle whose two same angles are $30^{\circ }$. $\square$