Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

Example 1. (Gallacher, 1987, see Figure 4) In $\triangle P_{1} P_{2} P_{3}$ , let $p_{i}$ be the side opposite vertex $P_{i}$ , and $s_{i}$ be parallel to $p_{i}$ but not coincident with $p_{i}$ . Let $s_{i}$ intersect $p_{i-1}$ at $Q_{i}$ , and $Q_{i}$ divides $p_{i} p_{i+1}$ into two parts, with a ratio of 3. Prove: The lines $S_{1}, S_{2}$ , and $S_{3}$ are concurrent if and only if $\lambda_{1} \lambda_{2} \lambda_{3}-\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)=2$ .

Solution

Prove that if $P_{1}, P_{2}, P_{3}, Q_{1}, Q_{2}$ and $Q_{3}$ are replaced by $A_{1}, A_{2}, A_{3}, C_{3}, C_{1}$ and $C_{2}$ respectively, then $\lambda_{1}, \lambda_{2}, \lambda_{3}$ will be equal to $c_{3}$ , $c_{1}, c_{2}$ in equation (*). Taking $B_{1}, B_{2}, B_{3}$ as the ideal points of the lines $A_{2} A_{3}, A_{3} A_{1}$ , $A_{1} A_{2}$ respectively, then $b_{1}=b_{2}=b_{3}=-1$ . Thus, from equation (*), we can derive $c_{1} c_{2} c_{3}-\left(c_{1}+c_{2}+c_{3}\right)=2$ , which is equivalent to the equation to be proven.

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