equilateral wanted if trisectors are concyclic - 2001 Estonia Open Junior 1.3

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equilateral wanted if trisectors are concyclic - 2001 Estonia Open Junior 1.3

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#1 h Sep 22, 2020, 3:53 PM

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Consider points $C_1, C_2$ on the side $AB$ of a triangle $ABC$ , points $A_1, A_2$ on the side $BC$ and points $B_1 , B_2$ on the side $CA$ such that these points divide the corresponding sides to three equal parts. It is known that all the points $A_1, A_2, B_1, B_2 , C_1$ and $C_2$ are concyclic. Prove that triangle $ABC$ is equilateral.

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#2 h Sep 27, 2020, 6:24 PM

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Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$ .
Then $A_{1}(\frac{c-2b}{3},0),A_{2}(\frac{2c-b}{3},0)$ and $B_{1}(\frac{2c}{3},\frac{a}{3}),B_{2}(\frac{c}{3},\frac{2a}{3})$ and $C_{1}(-\frac{b}{3},\frac{2a}{3}),C_{2}(-\frac{2b}{3},\frac{a}{3})$ .

The perpendicular bisector of $A_{1}A_{2}\ :\ x=\frac{c-b}{2}$ must be the same as the perpendicular bisector of $B_{2}C_{1}\ :\ x=\frac{c-b}{6}$ , so $x=0$ and $b=c$ .

The perpendicular bisector of $C_{1}C_{2}\ :\ y-\frac{a}{2}=-\frac{b}{a}(x+\frac{b}{2})$ cuts $x=0$ in the point $(0,\frac{a^{2}-b^{2}}{2a})$ .
The perpendicular bisector of $A_{1}C_{2}\ :\ y-\frac{a}{6}=\frac{c}{a}(x-\frac{c-4b}{6})$ cuts $x=0$ in the point $(0,\frac{a^{2}+3b^{2}}{6a})$ .
Both points on the y-axis must be the same, so $a=\sqrt{2}c$ .

Then $\triangle ABC$ is equilateral.

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