Consider 2 concentric circle radii $ R$ and  $ r$ ($ R > r$) with centre $ O.$ Fix $ P$ on the small circle and consider the variable chord $ PA$ of the small circle. Points $ B$ and $ C$ lie on the large circle; $ B,P,C$ are collinear and $ BC$ is perpendicular to $ AP.$

[b]i.)[/b] For which values of $ \angle OPA$ is the sum $ BC^2 \plus{} CA^2 \plus{} AB^2$ extremal?

[b]ii.)[/b] What are the possible positions of the midpoints $ U$ of $ BA$ and $ V$ of $ AC$ as $ \angle OPA$ varies?
1. **(i)** Let \( K \) be the midpoint of the chord \( BC \) and let \(\theta = \angle OPA\). Using the Pythagorean theorem for the right angle triangles \(\triangle ABP\) and \(\triangle ACP\), we have:

\[
AB^2 + AC^2 + BC^2 = 2 PA^2 + PB^2 + PC^2 + BC^2
\]

2. The power of the point \( P \) to the larger circle (O, R) is given by:

\[
PB \cdot PC = (R + r)(R - r) = R^2 - r^2
\]

3. On the other hand, using the midpoint \( K \):

\[
PB \cdot PC = \left(\frac{BC}{2} + PK\right)\left(\frac{BC}{2} - PK\right) = \frac{BC^2}{4} - PK^2
\]

4. Equating the two expressions for \( PB \cdot PC \):

\[
\frac{BC^2}{4} - PK^2 = R^2 - r^2
\]

5. Solving for \( BC^2 \):

\[
BC^2 = 4(R^2 - r^2) + 4 PK^2
\]

6. Substituting \( BC^2 \) back into the expression for \( AB^2 + AC^2 + BC^2 \):

\[
AB^2 + AC^2 + BC^2 = 2 (PA^2 + PK^2) + 6(R^2 - r^2) + 6 PK^2
\]

7. Simplifying further:

\[
AB^2 + AC^2 + BC^2 = 2 PA^2 + 8 PK^2 + 6(R^2 - r^2)
\]

8. Since \( PA = 2r \cos \theta \) and \( PK = r \sin \theta \):

\[
AB^2 + AC^2 + BC^2 = 8r^2 \cos^2 \theta + 8 r^2 \sin^2 \theta + 6(R^2 - r^2)
\]

9. Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[
AB^2 + AC^2 + BC^2 = 8r^2 + 6(R^2 - r^2) = 6R^2 + 2r^2
\]

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).

10. **(ii)** The triangle \(\triangle KLM\) is the medial triangle of the triangle \(\triangle ABC\). Since the vertex \( A \) and the vertices \( B \) and \( C \) are on concentric circles (O, r) and (O, R) respectively, the midpoint \( K \) and the midpoints \( L \) and \( M \) are also on concentric circles (Q, r/2) and (Q, R/2) respectively.

The final answer is \( \boxed{ 6R^2 + 2r^2 } \)