## Wednesday, January 2, 2013

### Problem 847: van Lamoen Circle, Triangle, Medians, Centroid, Six Circumcenters, Concyclic Points

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 847.

1. By Problem 846,
O1, O2, O3, O4 concyclic.

By symmetry,
O1, O2, O5, O6 concyclic
O3, O4, O5, O6 concyclic

Hence, O1, O2, O3, O4, O5, O6 concyclic.

2. To Jacob Ha:

There are 3 different circles, each pair with 2 common points. Based on your result these points are all part of the same circle. Why?

3. As in previous problem, let O1O2 and O3O4 meet at H.
Let HO5 meets circle O1O2O5O6 at P, and O3O4O5O6 at Q.

Then
HO1×HO2=HP×HO5
HO3×HO4=HQ×HO5

But we have HO1×HO2=HO3×HO4.
Therefore, HP=HQ.

Since H, P, Q, O5 are collinear,
thus P and Q coincide.

Hence, HO5 is the radical axis of circles O1O2O5O6 and O3O4O5O6.
Similarly, HO6 is the radical axis of circles O1O2O5O6 and O3O4O5O6.

But H, O5, O6 are not collinear, so the two circles coincide.
That is, O1, O2, O3, O4, O5, O6 concyclic.

4. we assign ω1=(O1,O2,O3,O4)
ω2=(O1,O2,O5,O6)
ω3=(O3,O4,O5,O6)
we suppose that ω1 is different from ω2

(O1O2)∩(O3O4)= H
(O1O2)∩(O5O6)= P that is different from H

The radical axis theorem states that the three radical axes (for each pair of circles) intersect in one point called the radical center, or are parallel.

The perbendicular lines over two intersecting lines, are intersecting themselves. From here we could say that:
ω1=ω2, and this way O1, O2, O3, O4, O5, O6 are concyclic.

Erina from NJ

5. Dear Friends, our work is on Six Concyclic Circumcenter with respect to incenter which we have published at MTRJ journal here : https://commons.hostos.cuny.edu/mtrj/wp-content/uploads/sites/30/2022/04/v14n1-Problem-Corner.pdf

If anyone know how to prove this conjecture then you can send your proof to the editor of this journal mentioned in above link