We can observe that if any of the points M, N, and P are inside the 1/2x1/2 square, you would get a bigger triangle by moving them to the side along the extensions of the existing edges. There are several ways to do this, but all follow the same logic. For example, in your sketch, extend MN until it crosses two edges and/or vertices of the square. Let's call the two cross-points M1 and N1. Triangle M1N1P has area bigger than or equal to that of triangle MNP, because M1N1>=MN (equal if M and N were already on the perimeter), and the height from P is unchanged. Then in triangle M1N1P extend M1P in the M1--->P direction, until it crosses an edge or vertex of the square. Let's call that point P1. Triangle M1N1P1 has area bigger than or equal to that of triangle M1N1P. because M1N1 is unchanged, but the height from P1 down to M1N1 is greater than or equal to that from P. So we have established that for any triangle with points strictly inside the square we can find one with a bigger area with points on the periphery. From here, we need to show that 3 points on the perimeter of the square cannot form a triangle with area bigger than 1/2 the area of the square. If you select any edge of the triangle, the third point which maximizes the area will be in a corner. Thus, at least one vertex of the triangle must lie on a vertex of the square. Let's place a square MNPQ in the cartesian plane with M(0,0), N(1,0), P(0,1), and Q(1,1) and fix M as the vertex of our triangle. Triangle MNP has an area of 1/2. We cannot improve on that: Moving N along (0,0) and (0,1) and P along (0,0) and (1,0) decreases the triangle area for obvious reasons - MN
We can observe that if any of the points M, N, and P are inside the 1/2x1/2 square, you would get a bigger triangle by moving them to the side along the extensions of the existing edges. There are several ways to do this, but all follow the same logic. For example, in your sketch, extend MN until it crosses two edges and/or vertices of the square. Let's call the two cross-points M1 and N1. Triangle M1N1P has area bigger than or equal to that of triangle MNP, because M1N1>=MN (equal if M and N were already on the perimeter), and the height from P is unchanged. Then in triangle M1N1P extend M1P in the M1--->P direction, until it crosses an edge or vertex of the square. Let's call that point P1. Triangle M1N1P1 has area bigger than or equal to that of triangle M1N1P. because M1N1 is unchanged, but the height from P1 down to M1N1 is greater than or equal to that from P.
So we have established that for any triangle with points strictly inside the square we can find one with a bigger area with points on the periphery. From here, we need to show that 3 points on the perimeter of the square cannot form a triangle with area bigger than 1/2 the area of the square. If you select any edge of the triangle, the third point which maximizes the area will be in a corner. Thus, at least one vertex of the triangle must lie on a vertex of the square. Let's place a square MNPQ in the cartesian plane with M(0,0), N(1,0), P(0,1), and Q(1,1) and fix M as the vertex of our triangle. Triangle MNP has an area of 1/2. We cannot improve on that:
Moving N along (0,0) and (0,1) and P along (0,0) and (1,0) decreases the triangle area for obvious reasons - MN