Given equilateral triangle, can we divide it by staight lines into 5 tiangles, equal to each others? If not, why? Best regards, Alexei [Non-text portions of...
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alex_geom@...
Sep 1, 2008 3:14 pm
16725
Dear Antreas, Suppose P is any point with barycentrics (p:q:r). MaMbMc is median triangle of ABC. HaHbHc is cevian triangle of P. Pa is concurrent point of...
Dear Andreas, In this case triangles (ABC, PaPbPc) are perspective with perspector: a^2*((a^6 - 3*a^2*b^2*c^2)*(b^2 + c^2) + 3*a^2*(b^6 + c^6) + (b^4 +...
Dear Tuan Probably is interesting the following generalization: Let MaMbMc, HaHbHc be the cevian triangles of two ISOGONAL conjugate points T, T*. Which is the...
Dear Andreas, Generally we can take MaMbMc as Cevian triangle of any point U=(u:v:w). With each U, the locus of P such that ABC and PaPbPc are perspective is...
Let L,Lb,Lc be three lines. To construct triangle ABC such that Lb,Lc be two of its sidelines and L its Euler line. (See my construction in the end of the...
Dear Cosmin! This is very nice problem. Can I use it for Sharygin's olympiad? Sincerely Alexey Dear friends, ...
Alexey.A.Zaslavsky
zasl@...
Sep 2, 2008 6:26 am
16731
How about a line L, and La, Lb, Lc its reflections in BC,CA,AB, resp.? Where is lying the Incenter of the triangle (La,Lb,Lc)? On the circumcircle of ABC? APH ...
Dear Alexey, You are free to use it in the Sharygin Olympiad, of course. Dear Antreas, Indeed, I used the fact you mentioned as a main lemma in the proof of my...
... Dear Martin, this appears as problem 311 in Geometriagon ( www.polarprof.org/geometriagon ) and it seems hard, as only one user solved it so far . And...
Dear Antreas ... Suppose that your line L intersects BC, CA, AB respectively at U, V, W and that A'B'C' is the triangle with sidelines La, Lb, Lc. It is known...
I get this conic as solution of a problem: x^2 (c^2 y^2 - a^2 y z + b^2 y z + c^2 y z + b^2 z^2) == a^2 y^2 z^2 I have seen that the conic is an hyperbola in...
Dear Francisco, ... This must be the isogonal transform of your conic... ... Consider the vertices of the tangential triangle... Best regards Bernard [Non-text...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200821index.html The editors Forum...
ForumGeom
ForumGeom@...
Sep 2, 2008 3:59 pm
16739
Dear Cosmin and Antreas! I found another proof. Let A', B', C' be the reflections of A, B, C in opposite sidelines. Then the lines YZ, ZX, XY pass through A',...
Alexey.A.Zaslavsky
zasl@...
Sep 3, 2008 6:00 am
16740
Dear Hyacinthists, Giovanni, Martin Lukarevski, Look at the sites http://blog.escolademestres.com/qedtexte/#comments ...
Let ABC be a triangle, L a line passing through P and Q, and La,Lb,Lc the parallels to L through A,B,C, resp. Let Ma,Mb,Mc be the Reflections of BC,CA,AB, in ...
Dear All My Friends, Given triangle ABC, one point P inside ABC with barycentrics (p : q : r), A'B'C' is Cevian triangle of P. A1 = intersection of two angle...
Let ABC be a triangle and A1B1C1, A2B2C2 the cevian triangles of two points P,Q, resp. Denote: A' := B1C1 /\ B2C2 B' := C1A1 /\ C2A2 C' := A1B1 /\ A2B2 A line...
I think that the theorem I posted is analogous to Poncelet Porism. PONCELET PORISM: Let (C1), (C2) be two circles. If there exists a triangle inscribed in (C1)...
I don't remember, but probably we have seen the following: Let ABC be a triangle, A'B'C' its orthic triangle and P = (x:y:z) a point. The circles PAA',...
Hi Antreas .. Orthic .. P' = (SB y^2 + SC z^2 - SA x (y + z) ) / SA :: Excentral .. P' = a (b c x^2 + 2 sa (b x z + c x y - a y z) - a (c y^2 + b z^2)) :: ...
Dear Antreas and Peter, If instead of orthic triangle we choose medial triangle A'B'C' and P = symmedian point then three circles PAA', PBB',PCC' concur at...
... Denote A = Lb /\ Lc If L = OI line, then the incenter I is the intersection point of L and the bisector of the angle of the lines (Lb,Lc). We construct the...
Let ABC be a triangle, Q, P two points, QaQbQc the pedal triangle of Q, and A',B',C' the orth. projections of P on QQa, QQb, QQc, resp. The triangles ABC,...
Dear Antreas and Peter, For generalization instead of orthic triangle we choose A'B'C' as Cevian triangle of U = (u : v : w) and P is any point with...
Old Theorem: The OH lines (Euler Lines) of the Triangles IBC,ICA,IAB are concurrent. Let's replace one of the points O,H with its reflections in the lines...
... It seems that it is an hyperbola ... In this case, repeating the same procedure, I get a locus that is not a conic. Am I correct ? Giovanni Artico...
... I think that the lines O'aHa, O'bHb, O'cHc are parallels to AI,BI,CI, resp. through the vertices A',B',C' of the cevian triangle of the isotomic conjugate...
