(i) Let K the midpoint of the chord BC and $\theta = \angle OPA$. Using Pythagorean theorem for the right angle triangles $\triangle ABP, \triangle ACP$, we have $AB^2 + AC^2 + BC^2 = 2\ PA^2 + PB^2 + PC^2 + BC^2$ $PB^2 + PC^2 = \left(\frac{BC}{2} + PK\right)\left(\frac{BC}{2} - PK\right) = \frac{BC^2}{2} + 2\ PK^2$ $AB^2 + AC^2 + BC^2 = 2 (PA^2 + PK^2) + \frac 3 2\ BC^2$ The power of the point P to the larger circle (O, R) is $PB \cdot PC = (R + r)(R - r) = R^2 - r^2$ On the other hand, $PB \cdot PC = \left(\frac{BC}{2} + PK\right)\left(\frac{BC}{2} - PK\right) = \frac{BC^2}{4} - PK^2$ so that $\frac{BC^2}{4} - PK^2 = R^2 - r^2$ $BC^2 = 4(R^2 - r^2) + 4\ PK^2$ $AB^2 + AC^2 + BC^2 = 2 (PA^2 + PK^2) + 6(R^2 - r^2) + 6\ PK^2 =$ $= 2\ PA^2 + 8\ PK^2 + 6(R^2 - r^2)$ Obviously, $PA = 2r \cos \theta$ According to the solution of the problem PQR similar to DGH, the locus of midpoints K is a circle (Q, r/2) with the diameter OP = r. The argument used to solve the proposition (ii) later on could be used as well. Since $PK \perp PA$, $PK = r \cos (90^\circ - \theta) = r \sin \theta$ Substituting for PA, PK into the last formula for $AB^2 + AC^2 + BC^2$, $AB^2 + AC^2 + BC^2 = 8r^2 \cos^2 \theta + 8 r^2 \sin^2 \theta + 6(R^2 - r^2) = 6R^2 + 2r^2$ Thus the sum of squares $AB^2 + AC^2 + BC^2$ does not depend on the position of the point A on the smaller circle (O, r). (ii) The triangle $\triangle KLM$ is the medial triangle of the triangle $\triangle ABC$. Since the vertex A and the vertices B, C are on concentric circles (O, r), (O, R), respectively, the midpoint K and the midpoints L, M are also on concentric circles (Q, r/2), (Q, R/2), respectively. Comment: The point P just provides an anker (sic) of this fine configuration concentric circles .

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.................. M.T.

armpist wrote: It was a very interesting comment Yetti put at the end of his post, and here is the reason why I consider it interesting. As it is a case with most Olympiad problems, there usually exists a deeper current in them, and this one is a good example of such problems. Yetti delivered a reasonably good solution, taking the path that the problem composer wanted a problem-solver to take. But his comment at the end about anchor was highly unusual, because he was obviously making a joke that made him laugh in a horizontal position.? You are a magnificently split personality, which is why I put the comment there. armpist wrote: Somewhere else on this Geometry forum (Find Perimeter problem) Yetti asked me for explanation of some 3 points collinearity. ... I hoped that Yetti will realize the simple reasons of questioned collinearity and edit his question out. But no! He insists, claiming that without my personal proof the points are not collinear, and that's it. I trully belive that I can not upstage Darij in such explanation, so I decided ( que sera sera) that the points should stay triangular, I will not bring them in line. On top of that I disagree with the entire solution I gave. It is free for grabs, I'm not returning to that thread. The reasons for collinearity of the said points are indead simple, Ceva's theorem and the ratio of sides $1 : \sqrt 3$. I calculated the ratio and deduced the collinearity. You postulated the collinearity and used it to get the ratio. Thus you did not have even a sketch of a solution, as you claimed, not to mention the desired result in a closed form, regardless of the fact that the said points indeed are collinear. armpist wrote: Thank you. M.T. Does sfwc know that you are using his signature ? Yetti

yetti wrote: The reasons for collinearity of the said points are indead simple, Ceva's theorem and the ratio of sides $1 : \sqrt 3$. I calculated the ratio and deduced the collinearity. You postulated the collinearity and used it to get the ratio.