211. Pretty hard problem !
by Virgil Nicula, Jan 20, 2011, 1:53 PM
Proposed problem. A nonisosceles triangle 
is given and let 
be its altitudes with orthocenter 
.Denote the midpoints
of ![$ [BC],\ [CA],\ [AB]$](//latex.artofproblemsolving.com/e/d/d/eddb489088ad36feac80f18fa89c9127a586400d.png)
respectively. Denote 
. Prove that the points 
, 
, 
, 
are collinearly.
Lemma (memorize !). Let be a triangle. For 
denote 
. Then 
.
Generalization (Vittasko). A nonisosceles triangle
is given and let 
be the cevian triangle of the point w.r.t. 
, where 
, 
and
. Denote the midpoints of respectively. Denote . Prove that , , , are collinearly.
is given and let 
be its altitudes with orthocenter 
.Denote the midpoints
of ![$ [BC],\ [CA],\ [AB]$](http://latex.artofproblemsolving.com/e/d/d/eddb489088ad36feac80f18fa89c9127a586400d.png)
respectively. Denote 
. Prove that the points 
, 
, 
, 
are collinearly.Lemma (memorize !). Let
be a triangle. For 
denote 
. Then 
.
.
(in the Euler's circle of 
) : 
.
Denote 
. Apply Ceva's theorem to 
. Suppose 
.
. Using 
obtain 
.
in 
. Since 
obtain 
. Apply Menelaus' theorem to 
in 
. Since 
obtain 
, .e. 
.
, 
results 
, i.e. 
. In conclusion, the points Generalization (Vittasko). A nonisosceles triangle
is given and let 
be the cevian triangle of the point w.r.t. 
, where 
, 
and
. Denote the midpoints of respectively. Denote . Prove that , , , are collinearly.
and 
. I"ll prove that 
. Applying the Menelaus's theorem to transversal 
in 
, i.e. 
. Apply in the Ceva's theorem in 
, i.e. 
. From the relations 
obtain that 
. Applying the Menelaus' theorem to the transversal 
in 
, i.e. 
(because of 
). From the relations 
obtain that 
. From the relation 
obtain that 
.
, 
and 
(we say that two parallel lines, are concurrent at infinity ), where 
based on the Desarques' theorem, we conclude that 
and 
are perspective and so, we have that the line segments 
, 
and 
are concurrent at one point, so be it 
. Then, because of the collinearity of the points 
, 
and 
we conclude that the triangles 
and 
are also perspective 
, 
and 
. So, we conclude that the line segments 
, 
and 
are concurrent at one point. That is, the line segment 
passes through the point 
and 
and we will prove the collinearuty of 
and 
and since 
we have that 
. Since 
are in harmonic conjugation we obtain that 
. From the relations 
and 
obtain that 
. From the relation 
we conclude that 
and the proof is completed.This post has been edited 10 times. Last edited by Virgil Nicula, Nov 22, 2015, 4:11 PM
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