Problem 49: IZhO 2016, Day 2, Problem 2

by henderson, Sep 20, 2016, 2:14 PM

$\color{red}\bf{Problem \ 49}$ A convex hexagon $ABCDEF$ is given such that $AB\parallel DE,$ $BC\parallel EF$ and $CD\parallel FA.$ The point $M, N, K$ are common points of the lines $BD$ and $AE,$ $AC$ and $DF,$ $CE$ and $BF,$ respectively. Prove that perpendiculars drawn from $M, N, K$ to lines $AB, CD, EF,$ respectively, are concurrent. $($ Official solution

$\color{red}\bf{Official \ solution}$ $\color{green}\boxed{\color{blue}\bf{Lemma}}$ Let $T$ be the common point of extended legs $PS$ and $QR$ of trapezoid $PQRS.$ Then the radical axis of two circles with diameters $PR$ and $QS$ is the altitude of the triangle $TPQ$ drawn from $T.$

$\color{green}\boxed{\color{blue}\bf{Proof}}$ Consider the three circles: $\omega$ with diameter $PQ,$ $\omega _1$ with diameter $PR$ and
$\omega _2$ with diameter $QS.$ The common chord of $\omega$ and $\omega _1$ is the altitude drawn from $P$ to $QR$ and the common chord of $\omega$ and $\omega _2$ is the altitude drawn from $Q$ to $PS.$ The common point of these altitudes, that is, the orthocenter of $\triangle TPQ,$ has equal powers with respect to the circles $\omega _1$ and $\omega _2,$ and therefore belongs to their radical axis. Similarly, their radical axis contains the orthocentre of $\triangle TRS$ $($ to prove this, consider the circle with diameter $SR$ instead of $\omega$ $)$ $.$ Since the perpendicular drawn from $T$ to $PQ$ contains both of the orthocentres, it is the radical axis of $\omega _1$ and $\omega _2.$ $\quad$ $\square$

Applying the lemma to the trapezoid $ABDE,$ we obtain that the perpendicular drawn from $M$ to $AB$ is the radical axis of circles with diameters $AD$ and $BE.$ Considering the trapezoids $CDFA$ and $EFBC$ in the same way, we can conclude that the perpendicular drawn from $N$ to $CD$ is the radical axis of the circles with diameters $AD$ and $CF,$ and the perpendicular drawn from $K$ to $EF$ is the radical axis of the circles with diameters $CF$ and $BE.$ In conclusion, the perpendiculars drawn from $M, N, K$ to lines $AB, CD, EF,$ respectively, are the radical axis of three circles, so are concurrent. $\quad$ $\square$

$)$
(IZhO 2016)

This post has been edited 14 times. Last edited by henderson, Jan 5, 2017, 2:54 PM

geometry

izho

No Comments
(Post Your Comment)

Comment

0 Comments

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

henderson

Math Path

Problem 49: IZhO 2016, Day 2, Problem 2

by henderson, Sep 20, 2016, 2:14 PM

0 Comments