2 added 74 characters in body edited Feb 19 '18 at 15:26 Evargalo 3,57511 gold badge1010 silver badges2828 bronze badges As a starter: There are $$6*6=36$$ boxes' corners. Each diagonal connects two corners. Thus we have an upper bound of $$36/2=18$$ diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me. edit: as usual, my intuition proved wrong. See Jaap's nice solution. As a starter: There are $$6*6=36$$ boxes' corners. Each diagonal connects two corners. Thus we have an upper bound of $$36/2=18$$ diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me. As a starter: There are $$6*6=36$$ boxes' corners. Each diagonal connects two corners. Thus we have an upper bound of $$36/2=18$$ diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me. edit: as usual, my intuition proved wrong. See Jaap's nice solution. 1 answered Feb 19 '18 at 12:02 Evargalo 3,57511 gold badge1010 silver badges2828 bronze badges As a starter: There are $$6*6=36$$ boxes' corners. Each diagonal connects two corners. Thus we have an upper bound of $$36/2=18$$ diagonals. It is easy to achieve $$15$$ diagonals: just place NW-SE diagonals in each box in the first, third and fifth columns. Thus we have a lower bound of $$15$$ diagonals. My intuition is that it is impossible to stack more than 15 diagonals and that there should be an easy, visual, geometric proof for that. But that proof still escapes me.