SUBIECTUL IV: On the right of $d$, points $A_{0}, A_{1}, \ldots, A_{50}$ are considered in this order, such that $A_{0} A_{1}=1 \mathrm{~cm}, A_{1} A_{2}=3 \mathrm{~cm}, A_{2} A_{3}=5 \mathrm{~cm}, \ldots, A_{49} A_{50}=99 \mathrm{~cm}$. Let $O$ be the midpoint of the segment $\left[A_{0} A_{50}\right]$.

a. Determine $p \in \mathbb{N}$ for which $O \in\left[A_{p} A_{p+1}\right]$

b. Do there exist two natural numbers $m$ and $n, 0 < m < n < 50$, such that $O \in\left[A_{m} A_{n}\right]$?

ETAPA LOCALĂ-21 FEBRUARIE 2016  Clasa a VI-a  BAREM DE EVALUARE ŞI NOTARE

- For any correct solution, even if it is different from the one in the marking scheme, the corresponding score is given.
- No fractional points are awarded, but intermediate scores can be given for partial solutions, within the limits of the score indicated in the marking scheme.
## SUBJECT IV

$$
\begin{array}{c|c}
\hline \text { a. } A_{0} A_{p}=p^{2} \Rightarrow A_{0} A_{50}=1+3+5+\cdots+99=50^{2}=2500 & 1 p \\
O-\text { midpoint of }\left[A_{0} A_{50}\right] \Rightarrow A_{0} O=1250 & 1 p \\
O \in\left[A_{p} A_{p+1}\right] \Leftrightarrow p^{2}<1250<(p+1)^{2} \Rightarrow p=35 \Rightarrow O \in\left[A_{35} A_{36}\right] & 1 p \\
\text { b. } O-\text { midpoint of }\left[A_{m} A_{n}\right] \text { if } A_{m} O=O A_{n} & \\
\left.\begin{array}{c}
A_{m} O=1250-m^{2} \\
O A_{n}=n^{2}-1250
\end{array}\right\} \Rightarrow 1250-m^{2}=n^{2}-1250 & \mathbf{1 p} \\
\Rightarrow n^{2}+m^{2}=\mathbf{2 5 0 0}, n^{2}+m^{2}=\mathbf{2 5} \cdot \mathbf{1 0 0}=\mathbf{1 6} \cdot \mathbf{1 0 0}+\mathbf{9} \cdot \mathbf{1 0 0}=\mathbf{4 0}^{2}+\mathbf{3 0}^{2} & \mathbf{1 p} \\
\Rightarrow O \text { is the midpoint of the segment }\left[A_{30} A_{40}\right] &
\end{array}
$$