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 , H1 the orthocenter of the triangle AOB and H2 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 H1H2 and MN are parallel if and only if AC=BD
Problem
Source: Romania JBMO TST 2015 Day 1 Problem 5
Tags: geometry, quadrilateral, orthocenter, parallel
TelvCohl
14.05.2015 15:41
My solution : Let X,Y be the midpoint of BC,DA, respectively . Since H1H2 is the Steiner line of complete quadrilateral {AB,CD,AC,BD} , so H1H2 is perpendicular to the Newton line XY of the complete quadrilateral {AB,CD,AC,BD} . Since MXNY is a parallelogram ( MX=NY=12AC,MY=NX=12BD ) , so H1H2∥MN⟺MN⊥XY⟺MXNY is a rhombus ⟺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: Γ1=⊙AA′GD,Γ2=⊙DCGF,Γ3=⊙EFCB,Γ4=⊙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 (Γ2,Γ4). Since M and N are the centres of Γ2 and Γ4 respectively, we have XY⊥MN.
Next, since AA′ and CF are both perpendicular to BD, we have H1X∥H2Y. Similarly, H1Y∥H2X implying that H1XH2Y is a parallelogram.
Finally, note that A′F=ACcosAOB and EG=BDcosAOB.
Putting this all together:
AC=BD⇔A′F=EG⇔Area(H1XH2Y)/H1X=Area(H1XH2Y)/H2X⇔H1X=H2X⇔H1XH2Yis a rhombus⇔XY⊥H1H2⇔MN∥H1H2.