Cesko-Slovacko-Poljsko 2013

Notice, that with a contamination of one cell, the perimeter of infected area decreases at least by 1. Let $k$ cells be infected at the beginning. Then the perimeter is at most $3k$. It takes $n^2-k$ contaminations to get the whole triangle infected. The perimeter of the (infected) area is then $3n$. Thus $3n\le 3k-(n^2-k)$, or $$k\ge \frac{n^2+3n}{4}.$$ For $n=12$ the estimate gives $k\ge 45$. The layout of 45 initially infected cells which cause the infection of the whole system could be as illustrated on Fig. 4.



Another option is to cover the triangle with three equilateral triangles of side 4 (covering one side of the big triangle) and the rest could be covered with three rhombs of side 4 (Fig. 5). Each of the smaller triangles could be infected with 7 cells, each of the rhombs with 8 cells. Possible initial state for infecting an equilateral triangle with side 4 and a rhomb with side 4 is on Fig. 6a and Fig. 6b, respectively. (Each triangle with side 4 can get infected itself independently of the rest area. To get an entire rhomb infected, we need the area to its left and upper side to be infected first.)