1.4. Let initially each island is inhabited by one colony, and let one of the islands have $d$ neighboring islands. What can the maximum possible number of colonies that can settle on this island be equal to?
Solution. Answer: $d+1$.

Example. We will prove that in a vertex of degree $d$, $d+1$ colonies can gather. Suspend the tree from this vertex as the root and prove that in each vertex from which $e$ edges go down, $e+1$ colonies can gather, conducting only migrations within its subtree. We will prove this by "induction from the lower vertices to the upper ones." For leaves, the statement is obvious (they already have one colony), and for any other vertex, it is sufficient to gather the required number of colonies in its immediate children, after which perform one migration for each of them.

Estimate. We will show that any available distribution of numbers on the tree can be obtained by organizing migrations so that each colony does not move further than to a neighboring island. From this, it will follow that the answer is no more than $d+1$.

We prove this by induction on the number of migrations. Base case - zero migrations, is obvious.

Inductive step. Suppose a migration is about to occur from island $v$. By the inductive hypothesis, all colonies on it are from it and neighboring islands. Since a migration is possible, by the inductive hypothesis, there are colonies on the island from at least $\operatorname{deg} v-1$ neighboring islands; we will send them back to their islands. If there is a colony from the remaining neighboring island, we will also send it back, and if there is a colony from the island itself, we will send it to any of the neighboring islands.

## Plot 2.

Two circles inscribed in an angle with vertex $R$ intersect at points $A$ and $B$. A line through $A$ intersects the smaller circle at point $C$ and the larger circle at point $D$. It turns out that $A B=A C=A D$.