Maximum area of a triangle in a square

For a given square, consider 3 points on the perimeter to form a triangle. How to prove that:

  1. The maximum area of triangle is half the square’s.
  2. The maximum area of triangle occurs if and only if the chosen points are vertexes of the square.

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First, recall that the area of a triangle is $\frac{1}{2}bh$ where $b$ denotes the length of the base and $h$ the height. Keeping $h$ fixed, we may increase $b$ to increase the area of the triangle. Similarly, keeping $b$ fixed, we may increase $h$ to increase the area of the triangle. This shows that the vertices must lie along the boundary of the square.

Now, we claim that one vertex lies at the corner of the square. Suppose not. Then, pick any edge of the triangle. Consider the vertex not along this edge. By moving this vertex away from the edge, we may increase the height of the triangle, by the remarks above. Since a corner of the square will be the furthest distance from our selected edge, moving the opposite vertex to the corner of the square maximizes the area.

Now, let $S$ be a square in the plane. Let $s$ denote the length of the edges of $S$. Without loss of generality, we may suppose that $S$ lies in the first quadrant with one vertex at the origin. Consider a triangle in $S$ that maximizes area. By the above remarks, we may suppose that one vertex of the triangle lies at the origin.

Let $b$ be the base of the triangle and $h$ be the height of the triangle. By the above remarks on the construction of the triangle, $b = s$. Similarly, $h = s$. So the area of the triangle is $\frac{1}{2}bh = \frac{1}{2}s^2 = \frac{1}{2}Area(\Box)$. (You may want to argue these points more carefully, but they can be argued by using the ideas of the first two paragraphs.)

Note that the second statement is not true: Let the vertices of the sqaure be $(0,0), (1,0), (0,1)$, and $(1,1)$. Let the vertices of the triangle be $(0,0), (1,0)$ and $(\frac{1}{2},1)$. Then we still have $Area(\triangle) = \frac{1}{2}Area(\Box)$.

Here is an analytic solution.

Let $p_i$ be the vertices of the triangle. We can take the square to have side $1$, without loss of generality. So, we require $p_i \in S=[0,1]^2$. If $p_1$ lies in the interior of $S$, then $p_1$ can be moved in a direction perpendicular to, and away from, the line joining $p_2$ and $p_3$ until it touches the edge. This increases the area. Similarly for the other vertices. So we may assume that $p_i \in \partial S$, the boundary of the square.

If two vertices lie on the same edge, then they may be spread apart, increasing the area until they are at opposite ends of the edge. So, we may assume that each point lies on an edge that no other point lies on (the wording is awkward because a vertex lies on two edges).

So, by rotating if necessary, we may suppose that $p_1 = (x_1,0)$, $p_2 = (x_2,1)$, and $p_3 = (0,y)$, with $x_1,x_2,y \in [0,1]$.

The area is given by $A(x_1,x_2,y) = \frac{1}{2} | (p_1-p_3) \times (p_2-p_3)| = \frac{1}{2} |x_1(1-y)+yx_2|$. Since $p_i \in S$, we have $2 A(x_1,x_2,y) = x_1(1-y)+yx_2$.

It is easy to see that $A$ is a non-decreasing function of $x_1,x_2$, so it is maximized with $x_1 = x_2 =1$, which gives $2 A(1,1,y) = 1$, the desired result.

Since $A(1,1,y) = \frac{1}{2}$ for any $y \in [0,1]$, we see that the second statement cannot be true.