assume ABCD a convex quadrilatral. P and Q are on BC and DC respectively such that angle BAP= angle DAQ .prove that [ADQ]=[ABP] ([ABC] means its area ) iff the line which crosses through the orthocenters of these traingles , is perpendicular to AC.
Problem
Source: Iran(2003)
Tags: geometry, geometry solved
grobber
21.03.2004 00:17
It's a bit boring to write it all. Let H<sub>b</sub> and H<sub>d</sub> be the orthocenters of ABP and ADQ respectively. Let B' and P' be the feet of the perpendiculars from B and P respectively in ABP and let D' and Q' be defined similarly for triangle ADQ. It's easy to show that if <BAP=<DAQ and their areas are equal then AP*AB'=AQ*AD'. Now consider the inversion of pole A and power AP*AB'=AQ*AD'=AH<sub>b</sub>*AP'=AH<sub>d</sub>*AQ'. This inversion turns H<sub>b</sub>H<sub>d</sub> into the circle which passes through A, P' and Q', and since CP' perpendicular to AP' and CQ' perpendicular to AQ', it means that the circle AP'Q' has AC as its diameter, and a line is always turned into a circle which passes through the pole s.t. the diameter of the circle which passes through the pole is perpendicular to the initial line, so, in our case, AC perpendicular to H<sub>b</sub>H<sub>d</sub>. This is only one of the implications, but the other one isn't hard to derive from this one: We know that the lines are perpendicular if the areas are equal, and if we move one of P and Q a bit, say P, the orthocenter H<sub>b</sub> moves along AP', which is fixed, H<sub>d</sub> is fixed, AC is fixed, so AC and H<sub>b</sub>H<sub>d</sub> can't be perpendicular anymore.
amfulger
21.03.2004 22:19
I have an inversion free solution. Let H1, H2 be the orthocenters of DAQ, BAP. Let R1, R2 be the circumradi of DAQ and BAP. Let DAQ=BAP=a. Let C1 be the angle ACD and C2 be the angle ACB. Let F be a point such as CF and BC are perpendicular and AF is paralel to DC. Let E be a point such as CE and CD are perpendicular and AE and BC are paralel. It is easy to see that C is the orthocenter of AEF and CFE=C2 and CEF=C1. We get CE/CF=sinC2/sinC1 It is well known that AH1=2R1cosa and AH2=2R2cosa. AH1H2 and CEF have AH1||CE and AH2||CF. C is the orthocenter of AEF, so AC and EF are perpendicular. It is clear that H1H2 and AC are perpendicular iff H1H2||EF iff AH1H2 and CEF are similar iff CE/CF=AH1/AH2 iff sinC2/sinC1=R1/R2 iff R1sinC1=R2sinC2. Now Area[DAQ]=Area[BAP] iff R1*AD*sina*sinD=R2*AB*sina*sinB iff R1*(AC*sinC1/sinD)*sina*sinD=R2*(AC*sinC2/sinB)*sina*sinB iff R1*sinC1=R2*sinC2. Problem solved. I'll do my best to offer you a picture.
amfulger
21.03.2004 23:24
If C=90<sup>0</sup> then the construction of E and F doesn't work. In this case, take M to be AH1 \cap CD and N=AH2 \cap BC. Then AMCN is a ractangle and H1AC=C2 and H2AC=C1. As before, we have [DAQ]=[BAP] iff R1sinC1=R2sinC2. H1H2 is perpendicular to AC iff AC is an altitude in AH1H2, iff AH1H2=C1 and AH2H1=C2, iff R1/R2=AH1/AH2=sinC2/sinC1 iff R1sinC1=R2sinC2.