Let l,m be two parallel lines in the plane. Let P be a fixed point between them. Let E,F be variable points on l,m such that the angle EPF is fixed to a number like α where 0<α<π2. (By angle EPF we mean the directed angle) Show that there is another point (not P) such that it sees the segment EF with a fixed angle too.
Problem
Source: Iran TST 2006
Tags: geometry, parallelogram, circumcircle, geometric transformation, reflection, conics, angle bisector
29.04.2006 16:57
Let n be the line through P which is parallel to ℓ,m, and let S,T be points on ℓ,m respectively such that ∠SPT=2∠EPF (1). Finally, let Q be the fourth vertex of the parallelogram SPTQ. We'll prove that the point we're looking for is Q, and that ∠EQF=π−∠EPF (∗). Since QF↦QE is a projective transformation, and so is the map sending a line through Q to the line which makes an angle of π−∠EPF with it (such that the two lines have the appropriate orientation, i.e. the same as the orientation of the angle ∠EPF), it suffices to check (∗) for three particular positions of E,F on ℓ,m. We'll take the first two to be E→∞,F→∞, where the conclusion is easy to draw, and we'll take the third one such that PF is the angle bisector of ∠QPT. Because of (1), PE is the angle bisector of ∠QPS. Since SE is the external angle bisector of ∠PSQ, it means that E is the P-excenter of PSQ, so QE must be the external angle bisector of ∠PQS, and thus orthogonal to the angle bisector of ∠PQS. This latter line is parallel to PF, the angle bisector of ∠QPT, so we have QE⊥PF. In the same way we show that QF⊥PE, so Q is the orthocenter of PEF, and (∗) folows.
29.04.2006 21:18
let circumcircle of PEF intersect ℓ at Q so we have:∠EQF=α therfore all QF lines are parallel to each other ,hence the locus of circumcenter will be a line,hence circumcircle pass thorugh the reflection of P wrt locus of circumcenter. So We are done
29.04.2006 22:44
amir2 wrote: let circumcircle of PEF intersect ℓ at Q so we have:∠EQF=α therfore all QF lines are parallel to each other ,hence the locus of circumcenter will be a line,hence circumcircle pass thorugh the reflection of P wrt locus of circumcenter. So We are done You are completely wrong! Just tell me why you conclude from QF moving parallel that the locus of circumcenter is a line? And of course your result is wrong for 0<α<π2 because just see the two cases where one of E,F goes to infinity, then the circumcenter will be the point of infinity that is perpendicular to PF or PE (The one that has no infinity point) But if these two points are not equal (which occurs when 0<α<π2) then the line you mentioned shall be the infinity line! But of course it is not!
08.05.2006 16:37
Take Q as in grobber's post. It is enough to show that the triangle △XQE is similar to △YFQ (play with angles around Q to show that ∠EQF is π−α). To prove the similarity notice that ∠EXQ=∠FYQ=α and that the triangles △PXE and △FPY are similar. From the latter we have XE/PY=XP/YF⇒XE/QX=QY/YF.
30.05.2006 11:52
I don't know if following is correct. Map E↦F is projective from line l to line m. So envelope EF is some conic. Since it is bounded betwen parallel lines m and l it is not parabola. It is quite obviously that m and l are tangents to this conic. Now, following is well known: Let l and m be tangents (they don't have to be parallel) on conic. If Fi is focus of this conic then directed angle ∠XFiY is constant, where X and Y are variable points on l and m such that XY is also tangent on this conic. Useing this one of focus F1 and F2 is for sure not equal P wich proves our statement. Note: From this we see that point P need not to be betwen lines l and m since in this case envelope EF is hyperbola. Please, give some comments. Thanks!
31.03.2011 14:22
initialy you can prove that the locus of H(Orthocenter of triangle EPF) is a line perpendicular to lines l and m named it n.now assume that the intersecion point of circumcircle of EHF and line n is T.now prove that the distance between T and line l is equal with distance between P and line m.consequently T is a point that sees EF with angle 180-EPF.