Question on circumcentre (medians related)?

Here are 2 pics necessary to follow the proof.

http://lm250.fr/Circumcenter.PDF

Let's call A' the middle of AG and similarly for B'...F'. We denote by O_1 the circumcenter of AEG, O_2 of ECG and O_3,O_4,O_5 O_6 for CDG, DBG, BFG

and AFG respectively.

Let us show that O_1O_2O_4O_5 is an isosceles trapezoïd.

The line (O_1O_2) is the mediatrix (or whatever you call it) of EG and (O_4O_5) is the mediatrix of GB. Therefore those 2 lines are parallel.

We shall denote vectors using {}

Since {GA} + {GB} + {GC} = 0 and

{B'E'} = {BE}/2 = (3/4)*{BG}, we have

{A'D'} + {B'E'} + {C'F'} = 0.

So we can construct (pic 2) a triangle IJK so that

{IJ} = {F'C'}, {JK} = {E'B'}, {KI} = {D'A'}

Let ☼ be the circumcircle of IJK. There is only one point M in the plane such that

MI ┴ IK and MJ ┴ JK.

This point M is on ☼ and is such that KM is a diameter of ☼.

Notice that there exist vectors U and V such that

U ┴ A'D' = IL and {O_1O_4} = {A'D'} + U = {IK} + U

V ┴ B'E' = KJ and {O_1O_4} = {E'B'} + V = {JK} + V.

It follows that U = {MI}, V = {MJ} and {O_1O_4} = {MK}.

(btw, I inverted purple and orange between the 2 pics... sorry)

Similarly if JL is another diameter of ☼, then {O_2O_5} = {JL}.

It follows that O_1O_2O_4O_5 is an isosceles trapezoïd and its vertices do lie on a circle.

Same for O_1O_6O_3O_4 and O_2O_3O_5O_6.

To conclude, it suffices to show that 2 of the 3 circumcircles of these trapezoids are equal.

Let's show for instance that that O_1O_2O_5O_6 lie on the same circle.

Using oriented line angles, we have in ☼,

(JL,JK) = (IL,IK) so that, since IL ┴ IJ

(O_2O_5,EB) = (IL,AD) = (IJ,AD) + pi/2 = (CF,AD) +pi/2.

Since EB ┴ O_1O_2, (O_2O_5,O_1O_2) = (CF,AD).

But since CF ┴ O_5O_6 and AD ┴ O_1O_6, we have

(O_2O_5,O_1O_2) = (O_5O_6,O_1O_6).

This in turn implies that O_1O_2O_5O_6 lie on the same circle, the same one that contains

O_1O_2O_4O_5 and O_2O_3O_5O_6. That's it.

In less than 24 hrs. I am due to leave USA for India. Yesterday, I thought as under. Assuming that the centroid is at the origin (0, 0), if A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices, the endpoints of medians through these vertices will have coordinates given by D(-x1/2, - y1/2), E(-x2/2, -y2/2), and F(-x3/2,-y3/2). Now, all the six circles will pass through origin (0, 0) and hence will have equations of the form x^2 + y^2 + 2gx + 2fy = 0. Plugging coordinates of A and F, I tried to find the center of the circle AOF. I did not proceed further as it was cumbersome. But since it is possible, by rotating, x1, x2, x3 and y1, y2, y3, the centers of the remaining five circles can be found easily. Then, it remains to prove that they lie on a common circle. Two pairs of three circles can easily be proved, and then proving that both pairs have the same radii shall solve the problem. I am posting this as an incomplete answer in case someone may complete it.

I received an email from Duke whom I had requested to attempt that he has just returned from vacation and he too has thought on the above lines as I understood from his letter. Hope he or someone completes the solution. I will not have internet connection in India immediately on return. I thought of posting my efforts in case it is useful for anyone trying to solve the problem.

Here is the animated visualization for your question,

http://www.gogeometry.com/geometry/triangle_median...

I attempted to do this by remembering that the centroid divides the medians in the ratio 2:1.. Also was thinking of going about by somehow showing that the hexahedron formed is cyclic.. Anyway, am in a time crunch. For your immediate need, here is a site that purports to have the proof - but I do not see anything here in the normal realm of an 8th standard problem/proof unless you are trying for iit/sat..

http://www.cut-the-knot.org/ctk/SixCircum.shtml

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Ok, found some more time. Refer to this link, maybe this can be used with an induction proof (2 concyclic triangles, 3, n C/Ts)

http://en.wikipedia.org/wiki/Japanese_theorem_for_...

The issue is that cyclic polygons have not been studied to the extent that the need to be.

Same as above. If you didn't want to use coordinate geometry you would need to use a lot of trig and you would end up with very very messy systems to solve. using the coordinate system would let you avoid most of the systems involving trig, but either way it'd be messy.

It could be lengthy, but solution lies in co-ordinate geometry. to make things simple, one apex of A,B.C can be origin and one side can be along x or y axis.

then you know co-ords of midpoints and pt. of intersection of medians, so you know co-ords of all points of all triangles, so you know the co-ords of their epicentres. then you can easily test whether they lie on circumference of the same circle.

it could be formidable and requires knowledge of some basic formulae.