3-ча 1. Find a four-digit number that is a perfect square, and such that the first two digits are the same as each other and the last two digits are also the same.
Solution 1. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100a)$, so $b+100a=11x^2$ for some natural number $x$. Moreover, $100 \leqslant b+100a \leqslant 908$, which means $3 \leqslant x \leqslant 9$. By calculating the squares of the numbers $33, 44, \ldots, 99$, we find that exactly one of them has the required form: $88^2=7744$.

Part 2. Points $A, B, C$ are the vertices of an equilateral triangle inscribed in a circle. Point $D$ lies on the smaller arc $AB$. $\quad DC=AD+BD$. Prove.

Solution 2. First, note that $\angle ADC=\angle ABC=60^\circ$, since these angles subtend the same arc. Similarly, $\angle BDC=60^\circ$. Mark points $K$ and $L$ on segment $CD$ such that triangles $BDK$ and $ADL$ are equilateral. Triangles $ADB$ and $ALC$ are equal by two sides and the angle between them. Therefore, $CL=DB$, which means $DC=DL+LC=AD+BD$.