AoPS - Problem

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Problem

Source: Romania JBMO TST 2015 Day 1 Problem 5

Tags: geometry, quadrilateral, orthocenter, parallel

ComplexPhi

14.05.2015 10:01

Let $ABCD$ be a convex quadrilateral with non perpendicular diagonals and with the sides $AB$ and $CD$ non parallel . Denote by $O$ the intersection of the diagonals , $H_1$ the orthocenter of the triangle $AOB$ and $H_2$ the orthocenter of the triangle $COD$ . Also denote with $M$ the midpoint of the side $AB$ and with $N$ the midpoint of the side $CD$ . Prove that $H_1H_2$ and $MN$ are parallel if and only if $AC=BD$

TelvCohl

14.05.2015 15:41

My solution : Let $X, Y$ be the midpoint of $BC, DA$ , respectively . Since $H_1H_2$ is the Steiner line of complete quadrilateral $\{ AB, CD, AC, BD \}$ , so $H_1H_2$ is perpendicular to the Newton line $XY$ of the complete quadrilateral $\{ AB, CD, AC, BD \}$ . Since $MXNY$ is a parallelogram ( $MX=NY=\tfrac{1}{2} AC, MY=NX=\tfrac{1}{2} BD$ ) , so $H_1H_2 \parallel MN \Longleftrightarrow MN \perp XY \Longleftrightarrow MXNY$ is a rhombus $\Longleftrightarrow AC=BD$ . Q.E.D

kjytay

06.08.2015 04:36

Label the feet of the perpendiculars and extend them to meet at points

$X$ and

$Y$ as in the diagram: Define 4 circles:

$\Gamma_1 = \odot AA'GD, \Gamma_2 = \odot DCGF, \Gamma_3 = \odot EFCB, \Gamma_4 = \odot AEA'B$ (vaguely labelled in diagram as well). Consider the radical axis for each pair of circles. Note that for any 3 circles, the 3 radical axes they generate must meet at a point. By going through all the pairs of circles, we can conclude that

$XY$ is the radical axis of

$(\Gamma_2, \Gamma_4)$ . Since

$M$ and

$N$ are the centres of

$\Gamma_2$ and

$\Gamma_4$ respectively, we have

$XY \perp MN$ . Next, since

$AA'$ and

$CF$ are both perpendicular to

$BD$ , we have

$H_1X \parallel H_2Y$ . Similarly,

$H_1Y \parallel H_2X$ implying that

$H_1XH_2Y$ is a parallelogram. Finally, note that

$A'F = AC \cos AOB$ and

$EG = BD \cos AOB$ . Putting this all together:

$AC = BD \\ \Leftrightarrow A'F = EG \\ \Leftrightarrow Area(H_1XH_2Y) / H_1X = Area(H_1XH_2Y) / H_2X \\ \Leftrightarrow H_1X = H_2X \\ \Leftrightarrow H_1XH_2Y \text{is a rhombus} \\ \Leftrightarrow XY \perp H_1H_2 \\ \Leftrightarrow MN \parallel H_1H_2.$