Track / Stage 7 / 67 of 300 #1467 of 2000

Problem 1467

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.1 Prove it

Let $ABC$ be an equilateral triangle with the side length equals $a+ b+ c$ . On the side $AB{}$ of the triangle $ABC$ points $C_1$ and $C_2$ are chosen, on the side $BC$ points $A_1$ and $A_2$ , arc chosen, and on the side $CA$ points $B_1$ and $B_2$ are chosen such that $A_1A_2 = CB_1 = BC_2 = a, B_1B_2 = AC_1 = CA_2 = b,
C_1C_2 = BA_1 = AB_2 = c $. Let the point$ A^{’} $be such that the triangle$ A^{'} B_2C_1 $is equilateral, and the points$ A $and$ A^{'} $lie on different sides of the line$ B_2C_1 $. Similarly, the points$ B^{’} $and$ C^{'} $are constructed (the triangle$ B^{'} C_2A_1 $is equilateral, and the points$ B $and$ B^{’} $lie on different sides of the line$ C_2A_1 $; the triangle$ C^{'} A_2B_1 $is equilateral, and the points$ C $and$ C^{'} $lie on different sides of the line$ A_2B_1 $). Prove that the triangle$ A^{'}B^{'}C^{'}$ is equilateral.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Identify the given conditions and setup:
- We have an equilateral triangle $\triangle ABC$ with side length $a + b + c$ .
- Points $C_1$ and $C_2$ are chosen on $AB$ , points $A_1$ and $A_2$ on $BC$ , and points $B_1$ and $B_2$ on $CA$ such that:
$A_1A_2 = CB_1 = BC_2 = a, \quad B_1B_2 = AC_1 = CA_2 = b, \quad C_1C_2 = BA_1 = AB_2 = c$
- Points $A'$ , $B'$ , and $C'$ are constructed such that $\triangle A'B_2C_1$ , $\triangle B'C_2A_1$ , and $\triangle C'A_2B_1$ are equilateral, with $A$ and $A'$ on different sides of $B_2C_1$ , $B$ and $B'$ on different sides of $C_2A_1$ , and $C$ and $C'$ on different sides of $A_2B_1$ .

2. Prove concurrency and angle properties:
- Note that lines $A_1B_2$ , $C_1A_2$ , and $B_1C_2$ concur at some point $P$ inside $\triangle ABC$ , and these lines meet each other at $60^\circ$ angles. This is due to the symmetry and equal partitioning of the sides of the equilateral triangle.

3. Cyclic quadrilaterals and angle chasing:
- Consider the quadrilateral $C_1B_2A'P$ . Since $\triangle A'B_2C_1$ is equilateral, $\angle PA'B_2 = 60^\circ$ and $\angle PA'C_1 = 60^\circ$ . Thus, $C_1B_2A'P$ is cyclic.
- Similarly, $\triangle B'C_2A_1$ and $\triangle C'A_2B_1$ being equilateral implies that $A_1C_2B'P$ and $B_1A_2C'P$ are cyclic.

4. Using Ptolemy's Theorem:
- By Ptolemy's Theorem on cyclic quadrilateral $PC_1B_2A'$ :
$PA' \cdot C_1B_2 + PB_2 \cdot C_1A' = PC_1 \cdot B_2A'$
- Since $\triangle PA_1A_2$ , $\triangle PB_1B_2$ , and $\triangle PC_1C_2$ are equilateral, we have:
$PA' = PB_2 - PC_1 \quad \text{(in directed lengths)}$

5. Summing directed lengths:
- By similar arguments for $PB'$ and $PC'$ :
$PB' = PC_2 - PA_1 \quad \text{and} \quad PC' = PA_2 - PB_1$
- Summing these directed lengths:
$PA' + PB' + PC' = (PB_2 - PC_1) + (PC_2 - PA_1) + (PA_2 - PB_1)$
- Since $\triangle PA_1A_2$ , $\triangle PB_1B_2$ , and $\triangle PC_1C_2$ are equilateral, the total sum $PA' + PB' + PC' = 0$ .

6. Conclusion:
- Since the sum of the directed lengths is zero, $\triangle A'B'C'$ must be equilateral.

$\blacksquare$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.