PROBLEM 4. a) Consider the collinear points $A_{0}, A_{1}, A_{2}, \ldots, A_{2015}$ (in this order), such that $A_{0} A_{1}=1 \text{ cm}, A_{1} A_{2}=2 \text{ cm}, \ldots, A_{2014} A_{2015}=2015 \text{ cm}$. Calculate the distance between the points $A_{1000}$ and $A_{2015}$.

b) Let $A, B, C, D, E$ and $F$ be six points, such that any three of the given points are non-collinear. We color each of the segments determined by them either orange or violet, at random. Prove that there exists a triangle with vertices in three of the six points, which has all sides of the same color.[^0]

## NATIONAL MATHEMATICS OLYMPIAD  Local stage - 14.02. 2015  GRADING SCALE - Grade VI -
## PROBLEM 4.

\[
\begin{aligned}
& (1 p) \quad \text { a) } A_{1000} A_{2015}=A_{1000} A_{1001}+A_{1001} A_{1002}+\ldots+A_{2014} A_{2015} \\
& (1 p) \quad A_{1000} A_{2015}=1001 \text{ cm}+1002 \text{ cm}+\ldots+2015 \text{ cm}
\end{aligned}
\]

$(1 p)$

\[
(2 p) \quad A_{1000} A_{2015}=1530620 \text{ cm}
\]

(1p) b) Consider the 5 segments with one end at point $A$. According to the pigeonhole principle, at least three of these are of the same color.

(1p) Suppose, for example, that the segments $A B, A C$ and $A D$ are all of the same color, namely, violet.

If one side of the triangle $B C D$ is violet, it completes a violet triangle with two of the previous segments.

(1p) Otherwise, the triangle $B C D$ is orange, and the proof is complete.[^1]


[^0]:    ${ }^{1}$ The actual working time is 2 hours;

    ${ }^{2}$ All problems are mandatory;

    ${ }^{3}$ Each problem is graded from 0 to 7.

[^1]:    ${ }^{1}$ Each corrector awards an integer number of points;

    ${ }^{2}$ Any other correct solution is graded accordingly.