We give a 20th degree polynomial one of whose roots is one of the sidelengths of the triangle.
Happy New Year!
We give a 20th degree polynomial one of whose roots is one of the sidelengths of the triangle.
Happy New Year!
Antreas Hatzipolakis proposes the problem Euclid 3637 as follows:
Let ABC be a triangle, L the Euler line, P a point on L and A'B'C' the pedal triangle of P
Denote
A", B", C" = the orthogonal projections of A, B, C, on L, resp.
(Oa), (Ob), (Oc) = the circles with diameters A'A", B'B", C'C", resp.
The radical center of (Oa), (Ob), (Oc) lies on the NPC.
Which point is it in terms of P?
Here is the Mathematica code and its output to the solution of this problem and its generalization for any line $px + qy + rz =0$.
Solution to Euclid Problem 3637
We study by transformations some loci related to the centroid of degenerate triangles.
Some problems on degenerate triangles
This is my solution to Problem Euclid 2066 by Antreas Hatzipolakis. The solution contains a conic and a cubic and several new triangle centers.
The conic:
The cubic:
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