Problem 1

Six points are chosen on the sides of an equilateral triangle

ABC

A_1, A_2

BC

B_1, B_2

CA

and

C_1, C_2

AB

, such that they are the vertices of a convex hexagon

A_1A_2B_1B_2C_1C_2

with equal side lengths. Prove that the lines

A_1B_2

B_1C_2

and

C_1A_2

are concurrent.

Step 6 of 6: Three perpendicular bisectors meet at one center

Previous step Next step

In plain words

A line through two points that are each equidistant from two other points must be the perpendicular bisector of the segment joining those two other points.

C_1A_2\perp\text{-bisects }B_2C_2,\quad A_1B_2\perp\text{-bisects }C_2A_2,\quad B_1C_2\perp\text{-bisects }A_2B_2

Detailed analysis

Since the hexagon's sides at $C_1$ are equal, $C_1B_2=C_1C_2$ , so $C_1$ too is equidistant from $B_2$ and $C_2$ ; combined with $A_2B_2=A_2C_2$ from the previous step, both $A_2$ and $C_1$ lie on the perpendicular bisector of $B_2C_2$ , so line $C_1A_2$ is that perpendicular bisector. The same argument, cycled through the vertices, shows $A_1B_2$ and $B_1C_2$ are the perpendicular bisectors of $C_2A_2$ and $A_2B_2$ . These are exactly the three perpendicular bisectors of triangle $A_2B_2C_2$ , so $A_1B_2$ , $B_1C_2$ , $C_1A_2$ concur at its circumcenter.