On a line $l$ there are three different points $A$, $B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $BQ$ intersects $BQ$ in $L$ and $BR$ in $T$. The line through $B$ perpendicular to $AR$ intersects $AR$ in $K$ and $AQ$ in $S$. (a) Prove that $P$, $T$, $S$ are collinear. (b) Prove that $P$, $K$, $L$ are collinear. (2nd Benelux Mathematical Olympiad 2010, Problem 3)
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Tags: geometry, geometry proposed
Luis González
03.05.2010 21:21
By Pascal theorem for the degenerate cyclic hexagon $AALKBB,$ the intersections $S \equiv a \cap BK,$ $T \equiv b \cap AL$ and $P' \equiv LK \cap AB$ are collinear. Let $C \equiv AL \cap BK$ and $D$ be the projection of $C$ onto $AB,$ then obviously $CD$ passes through the orthocenter $H$ of $\triangle ABC$ and $(A,B,D,P')=-1.$ But from $QA \parallel HD \parallel RB,$ it follows that $(A,B,D,P)=-1$ $\Longrightarrow$ $P \equiv P'$ $\Longrightarrow$ $P,T,S$ and $P,K,L$ are collinear.
Pikachu!!!
04.05.2010 20:41
Luis's solution is similar to mine. Here is a bit difference. By Pascal Theorem,(too) 1. for cyclic $AALKBB$, we have $ST,KL,AB$ are concurent, 2. for cyclic $AAKLBB$, we have $QR,KL,AB$ are concurent. Hence, we get that $ST,KL$ must pass through $P$
eze100
05.05.2010 06:07
What does it mean $(A,B,D,P')=-1$???
Dimitris X
05.05.2010 21:58
eze100 wrote: What does it mean $(A,B,D,P')=-1$??? It means that $(A,B,D,P')$ is a harmonic 4-tumble Dimitris
Lyub4o
18.08.2011 17:57
I don't really understand how did Pikachu!!! use ABLK to prove that QR;KL;AB at the same are concurent.