Number of equal triangles in a chessboard

$$1\times 1$$ is cut and taken out from every corner of a $$8\times 8$$ chess board. At least, how many equal triangles (equal triangles means congruent triangles, and color is not important) can be drawn on the remaining figure?

What is the answer when we cut a $$1\times 1$$ from a single corner of chessboard?

For the first part of the question I cover chessboard by 20 equal triangles after removing corners like this:

Is less than this possible?

• I think the question is not quite complete. For any integer $n > 0$, a single (grey or white) square can be divided into an $n \times n$ grid, and each subsquare divided by a diagonal into a pair of triangles, yielding $2n^2$ triangles. Hence the whole board can be divided into $63 \cdot 2n^2$ equal triangles, for any $n$...but surely that's not what's intended. Can you clarify? – John Hughes Apr 9 '19 at 15:37
• Thanks for the point you brought up. Look at my edit please. – Ghartal Apr 9 '19 at 15:39
• The triangles I described were all congruent (45-45-90), so I don't believe your edit addresses my question. I'll be honest -- I have no idea what you're actually asking for here! – John Hughes Apr 9 '19 at 15:41
• @Ghartal: That doesn't really clarify; John Hughes's argument already produces arbitrarily many congruent triangles. – hmakholm left over Monica Apr 9 '19 at 15:41
• Thats exactly what I'm looking for. – Ghartal Apr 9 '19 at 15:52

1 Answer

Edit: This is now a full solution.

In the below picture, there must be a triangle $$R$$ which has an edge lying on the red segment, and a triangle $$B$$ lying on the blue segment. The two sides of $$R$$ and $$B$$ lying these colored segments segment cannot both be longer than $$1$$, else $$R$$ and $$B$$ would overlap. Without loss of generality, say that $$R$$'s side on the red segment is at most one.

Furthermore, considering the side of $$R$$ on the red edge as the base of $$R$$, the height of $$R$$ is at most $$7$$. Therefore, the area of each triangle is at most $$7/2$$. This implies that when the board has a single corner removed, the number of triangles is at least $$63/(7/2)=18$$. This is indeed achievable; tile this checkerboard with a corner removed with $$7\times 1$$ rectangles, then cut each rectangle in half diagonally.

Credits to antkam in the comments for coming up with this part of the solution.

For the first problem, first suppose that the height of the triangle is at most $$6$$. Then an area argument proves that at least $$20$$ triangles are necessary. If we did have height more then $$6$$, then the triangle would resemble the second picture. Both of the longer sides of the triangle would have length more than $$6$$. Now, considering the triangles covering the top edge of the chessboard, they would all have to have their small edge on the top (because the other edges are too long), and would be leaning to the left. It is easy to see that this is impossible, and the triangles would have to jut out the left edge of the board. Therefore, $$20$$ triangles is optimal.

• oh! here it is: the 3 edges are of lengths $\le 1, >6, >6$ respectively. so the top edge must be tiled by $\ge 6$ copies of the shortest edge, and in particular, the leftmost copy must be a 90 degree rotation of the red triangle (since that's the only way to fit the obtuse angle). ... maybe we can bring in algebra at this point and parametrize the red triangle tip and show the two triangles must intersect? – antkam Apr 12 '19 at 1:56
• ah, good, no need for algebra after all! if you consider the $6 \times 6$ square, the red triangle sticks to the right edge and out the left edge, while the other triangle sticks to the top edge and must stick out the bottom edge since they are 90 degrees rotations. therefore they intersect. whew! – antkam Apr 12 '19 at 2:03
• @antkam Brilliant! I will update my solution and credit your comments. – Mike Earnest Apr 12 '19 at 2:47