2005 IMO Problems/Problem 1

Problem

Six points are chosen on the sides of an equilateral triangle $ABC$ : $A_1, A_2$ on $BC$ , $B_1$ , $B_2$ on $CA$ and $C_1$ , $C_2$ on $AB$ , 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.

Solution

Let $D = C_2A_1 \cap A_2B_1$ , and similarly define $E$ and $F$ . We claim that $DEF$ is equilateral.

Proof: Note that the vectors $\overrightarrow{A_1A_2}$ , $\overrightarrow{A_2B_1}$ , $\overrightarrow{B_1B_2}$ , $\overrightarrow{B_2C_1}$ , $\overrightarrow{C_1C_2}$ , $\overrightarrow{C_2A_1}$ add to 0. But we also have: $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0$ since these are at the sides of an equilateral triangle. This means that $\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0.$ These three unit vectors thus form an equilateral triangle.

Now $\angle A_1DB_1 = 60^{\circ} = \angle A_1CB_1$ , so quadrilateral $A_1DCB_1$ is cyclic. So $\angle CA_1D = \angle CB_1D$ and $\angle C_2A_1A_2 = \angle A_2B_1B_2$ . This means that triangles $C_2A_1A_2$ and $A_2B_1B_2$ are congruent, so $C_2A_2 = A_2B_2$ . But also, $C_2C_1 = C_1B_2$ , so $C_1A_2$ is the perpendicular bisector of $A_1B_1C_1$ . Similarly, $A_1B_2$ and $B_1C_2$ are also perpendicular bisectors. Therefore, they concur.

2005 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2005 IMO Problems/Problem 1

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