2. svibnja 2007.

## Solutions

Problem 1A-1. Find the real solutions of the system of equations:

$$
\begin{gathered}
x+y+z=2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)=1 \\
x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)=-6
\end{gathered}
$$
Solution. First, we will transform the equations to obtain an equivalent but simpler system.

By rearranging the second equation and then using the first equality, we get:

$$
\begin{gathered}
x^{2}+y^{2}+z^{2}+3 x y+3 y z+3 z x=1 \\
(x+y+z)^{2}+x y+y z+z x=1 \\
x y+y z+z x=-3
\end{gathered}
$$

Similarly, we handle the third equality:

$$
\begin{gathered}
x(x y+x z)+y(y z+x y)+z(x z+y z)=-6 \\
x(3+y z)+y(3+x z)+z(3+x y)=6 \\
x+y+z+x y z=2 \\
x y z=0
\end{gathered}
$$

Thus, the original system is equivalent to the system:

$$
\begin{gathered}
x+y+z=2 \\
x y+y z+z x=-3 \\
x y z=0
\end{gathered}
$$

From the last equation, it follows that $x=0$ or $y=0$ or $z=0$, reducing the system to three simpler systems:

$$
\begin{array}{lll}
x=0, & y z=-3, & y+z=2 \\
y=0, & x z=-3, & x+z=2 \\
z=0, & x y=-3, & x+y=2
\end{array}
$$

From which we obtain the solutions:

$$
(x, y, z) \in\{(0,3,-1),(0,-1,3),(3,0,-1),(-1,0,3),(3,-1,0),(-1,3,0)\}
$$

Problem 1A-2. On the rays $p$ and $q$ with a common origin $O$, points $A$ and $C$ (on $p$) and $B$ and $D$ (on $q$) are given. If the line $CD$ is parallel to the median of triangle $OAB$, prove that the line $AB$ is parallel to the median of triangle $OCD$.

Solution. Let $M$ be the intersection of the parallel line through point $D$ with ray $p$, and let $N$ be the intersection of the parallel line through point $A$ with ray $q$.

![](https://cdn.mathpix.com/cropped/2024_05_30_f50eee0e926d4c73fbe3g-06.jpg?height=363&width=583&top_left_y=612&top_left_x=742)

Due to the similarity of triangles $OAB$ and $OMD$, we have $\frac{|OM|}{|OA|}=\frac{|OD|}{|OB|}$, and due to the similarity of triangles $OCD$ and $OAN$, we have $\frac{|OA|}{|OC|}=\frac{|ON|}{|OD|}$.

Thus, $|OM| \cdot|OB|=|OA| \cdot|OD|$ and $|OA| \cdot|OD|=|OC| \cdot|ON|$, so $|OM| \cdot|OB|=|OC| \cdot|ON|$.

If the line $CD$ is parallel to the median of triangle $OAB$, then $\overline{AN}$ is the median of that triangle, so $N$ is the midpoint of segment $\overline{OB}$.

Now, because $|OB|=2|ON|$, from the previous equality it follows that $|OC|=2|OM|$, so $M$ is the midpoint of segment $\overline{OC}$, i.e., $\overline{DM}$ is the median of triangle $OCD$, which is what we needed to prove.

Problem 1A-3. a) Prove that a $4 \times 4$ tile can be colored in two colors such that for any choice of two rows and two columns, the four fields at the intersections of these rows and columns are not all colored the same.

b) Prove that the above property does not hold for a $5 \times 5$ tile.

## Solution.

a) One possible coloring is shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_f50eee0e926d4c73fbe3g-07.jpg?height=172&width=186&top_left_y=608&top_left_x=935)

b) Assume the opposite, i.e., that a $5 \times 5$ tile is colored in black and white such that the condition is satisfied.

In the first row, at least three fields are colored with one color (say black). Consider the columns where the first field is black:
![](https://cdn.mathpix.com/cropped/2024_05_30_f50eee0e926d4c73fbe3g-07.jpg?height=218&width=482&top_left_y=1088&top_left_x=794)

If all three fields in the second row were white, it would be impossible to fill the table further.

Actually, in each of the remaining four rows, these three columns must have exactly two white fields. Namely, if there were two (or three) black fields, the condition would not be satisfied.
![](https://cdn.mathpix.com/cropped/2024_05_30_f50eee0e926d4c73fbe3g-07.jpg?height=220&width=474&top_left_y=1609&top_left_x=790)

However, this is not possible because we only have three such possibilities (and no repetition is allowed).

Problem 1A-4. Determine the largest natural number $n$ such that $n^{2}+2007 n$ is a square of some natural number.

Solution. Let $n^{2}+2007 n=m^{2}$. Then $m=n+k$ for some integer $k$, and from $n^{2}+2007 n=(n+k)^{2}$ we get $n=\frac{k^{2}}{2007-2 k}$. Since $n$ is a natural number, it must be $2007-2 k>0$, with the condition that $k^{2}$ is divisible by $2007-2 k$. Thus, $k \leq 1003$. Since we are looking for the largest $n$, the numerator of the fraction should be as large as possible, and the denominator as small as possible, so the largest $n$ is obtained for $k=1003$. Then $n=\frac{1003^{2}}{1}=1006009$.