Baltic Way 1991 · Problem 13
Combinatorics
An equilateral triangle is divided into 25 congruent triangles enumerated with numbers from 1 to 25 . Prove that one can find two triangles having a common side and with the difference of the numbers assigned to them greater than 3 .
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Topics
Invariants and monovariants · Pigeonhole and extremal arguments · Graph theory
Solutions
Solution
Solution:
Define the distance between two small triangles to be the minimal number of steps one needs to move from one of the triangles to the other (a step here means transition from one triangle to another having a common side with it). The maximum distance between two small triangles is and this maximum is achieved if and only if one of these lies at a corner of the big triangle and the other lies anywhere at the opposite side of it. Assume now that we have assigned the numbers to the small triangles so that the difference of the numbers assigned to any two adjacent triangles does not exceed . Then the distance between the triangles numbered and ; and ; and ; and must be equal to . However, this is not possible since it implies that either the numbers and or and are assigned to the same "corner" triangle.