Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line segments $B_{1} C_{2}, C_{1} A_{2}$, and $A_{1} B_{2}$ are concurrent, and the perimeters of triangles $A B_{2} C_{1}, B C_{2} A_{1}$, and $C A_{2} B_{1}$ are all equal. Find all possible values of this common perimeter.
The only possible value of the common perimeter, denoted \( p \), is 1. 【 Synthetic approach (from author). We prove the converse of the problem first: Claim (\( p=1 \) implies concurrence) - Suppose the six points are chosen so that triangles \( A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1} \) all have perimeter 1. Then lines \( \overline{B_{1} C_{2}}, \overline{C_{1} A_{2}} \), and \( \overline{A_{1} B_{2}} \) are concurrent. 

![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-25.jpg?height=603&width=680&top_left_y=1389&top_left_x=688)

Hence the result follows by Brianchon's theorem. Now suppose \( p \neq 1 \). Let \( \overline{B_{2}^{\prime} C_{1}^{\prime}} \) be the dilation of \( \overline{B_{2} C_{1}} \) with ratio \( \frac{1}{p} \) at center \( A \), and define \( C_{2}^{\prime}, A_{1}^{\prime}, A_{2}^{\prime}, B_{1}^{\prime} \) similarly. The following diagram showcases the situation \( p < 1 \).

![](https://cdn.mathpix.com/cropped/2024_11_19_1cf0843c143ab4f13b3fg-26.jpg?height=774&width=861&top_left_y=244&top_left_x=603)

By the reasoning in the \( p=1 \) case, note that \( \overline{B_{1}^{\prime} C_{2}^{\prime}}, \overline{C_{1}^{\prime} A_{2}^{\prime}} \), and \( \overline{A_{1}^{\prime} B_{2}^{\prime}} \) are concurrent. However, \( \overline{B_{1} C_{2}}, \overline{C_{1} A_{2}}, \overline{A_{1} B_{2}} \) lie in the interior of quadrilaterals \( B C B_{1}^{\prime} C_{2}^{\prime}, C A C_{1}^{\prime} A_{2}^{\prime} \), and \( A B A_{1}^{\prime} B_{2}^{\prime} \), and these quadrilaterals do not share an interior point, a contradiction. 

Thus \( p \geq 1 \). Similarly, we can show \( p \leq 1 \), and so \( p=1 \) is forced (and achieved if, for example, the three triangles are equilateral with side length \( \frac{1}{3} \)).