Päevaülesanne

Juhuslik

Harjutuskomplekt

Balti Tee 1991 · Ülesanne 13

Kombinatoorika

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 .

Muuda valikut

Kui oled valmis

Ülevaatematerjal muutub kättesaadavaks järgmise päevaülesannete komplektiga.

Ülevaade

Teemad

Invariandid ja monovariandid · Dirichlet’ printsiip ja ekstremaalargumendid · Graafiteooria

Lahendused

Lahendus

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 88 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 1,…,251, \ldots, 25 to the small triangles so that the difference of the numbers assigned to any two adjacent triangles does not exceed 33. Then the distance between the triangles numbered 11 and 2525; 11 and 2424; 22 and 2525; 22 and 2424 must be equal to 88. However, this is not possible since it implies that either the numbers 11 and 22 or 2424 and 2525 are assigned to the same "corner" triangle.