Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 3.8 AMC 10/12 Prove it North Macedonia

A triangle $\triangle ABC$ is given together with a segment $PQ$ of length $t$ on the segment $BC$ , so that $P$ is between $B$ and $Q$ and $Q$ is between $P$ and $C$ . We draw parallel lines from the point $P$ to $AB$ and $AC$ which intersect $AC$ and $AB$ in $P_1$ and $P_2$ , respectively. We draw parallel lines from the point $Q$ to $AB$ and $AC$ which intersect $AC$ and $AB$ in $Q_1$ and $Q_2$ , respectively. Prove that the sum of the areas of $PQQ_1P_1$ and $PQQ_2P_2$ doesn't depend on the position of $PQ$ on $BC$ .

Solution

Solution. Let $D$ be the intersection of $PP_1$ and $QQ_2$ . Let us note that $P_1DQ_2P_2 = 2P_1\triangle ADP$ and $P_1Q_1DQ_1 = 2P_1\triangle ADQ$ . So now we have:
$\begin{aligned} P_1PQQ_2P_2 + P_1P_1Q_1Q &= P_1DQ_2P_2 + P_1Q_1DQ_1 + 2P_1\triangle PQD \\ &= 2P_1\triangle ADP + 2P_1\triangle ADQ + 2P_1\triangle PQD \\ &= 2P_1\triangle APQ = \overline{PQ} \cdot h_a \end{aligned}$

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