HMMT-Feb 2012 geo, Problem 10

Let $C$ denote the set of points $(x, y) \in \mathbb{R}^{2}$ such that $x^{2}+y^{2} \leq 1$. A sequence $A_{i}=\left(x_{i}, y_{i}\right) \mid i \geq 0$ of points in $\mathbb{R}^{2}$ is 'centric' if it satisfies the following properties: \begin{itemize} $A_{0}=\left(x_{0}, y_{0}\right)=(0,0), A_{1}=\left(x_{1}, y_{1}\right)=(1,0)$. For all $n \geq 0$, the circumcenter of triangle $A_{n} A_{n+1} A_{n+2}$ lies in $C$. \end{itemize} Let $K$ be the maximum value of $x_{2012}^{2}+y_{2012}^{2}$ over all centric sequences. Find all points $(x, y)$ such that $x^{2}+y^{2}=K$ and there exists a centric sequence such that $A_{2012}=(x, y)$.

Solution: Consider any triple of points $\triangle A_{n} A_{n+1} A_{n+2}$ with circumcenter $P_{n}$. By the Triangle Inequality we have $A_{n} P_{n} \leq A_{n} A_{0}+A_{0} P_{n} \leq A_{n} A_{0}+1$. Since $P_{n}$ is the circumcenter, we have $P_{n} A_{n}=P_{n} A_{n+2}$. Finally we have $A_{n+2} a_{0} \leq P_{n} A_{n+2}+1=A_{n} P_{n}+1 \leq A_{n} A_{0}+2$. Therefore $\sqrt{x_{n+2}^{2}+y_{n+2}^{2}} \leq \sqrt{x_{n}^{2}+y_{n}^{2}}+2$. It is also clear that equality occurs if and only if $A_{n}, A_{0}, P_{n}, A_{n+2}$ are all collinear and $P_{n}$ lies on the unit circle. From the above inequality it is clear that $\sqrt{x_{2012}^{2}+y_{2012}^{2}} \leq 2012$. So the maximum value of $K$ is $2012^{2}$. Now we must find all points $A_{2}$ that conforms to the conditions of the equality case. $P_{0}$ must lie on the unit circle, so it lies on the intersection of the unit circle with the perpendicular bisector of $A_{0} A_{1}$, which is the line $x=\frac{1}{2}$. Thus $P_{0}$ must be one of $\left(\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)$. From now on we assume that we take the positive root, as the negative root just reflects all successive points about the $x$-axis. If $P_{0}=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ then $A_{0}, P_{0}, A_{2}$ must be colinear, so $A_{2}=(1, \sqrt{3})$. Then since we must have $A_{0}, P_{2 n}, A_{2 n}, A_{2 n+2}$ colinear and $P_{2 n}$ on the unit circle, it follows that $P_{2 n}=(-1)^{n}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Then by induction we have $A_{2 n}=(-1)^{n+1}(n, n \sqrt{3})$. To fill out the rest of the sequence, we may take $A_{2 n+1}=(-1)^{n}(n+1,-n \sqrt{3})$ and $P_{2 n+1}=(-1)^{n+1}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ so that we get the needed properties. Therefore the answer is \[ A_{2012} \in\{(-1006,1006 \sqrt{3}),(-1006,-1006 \sqrt{3})\} \] $\fbox{(-1006,1006 \sqrt{3}),(-1006,-1006 \sqrt{3})}$.

Answer: (-1006,1006 \sqrt{3}),(-1006,-1006 \sqrt{3})