What are the possible values of its area?

Question

Each side of a square contains one of the points $(0,9), (8,5), (6,0),$ and $(-2,4)$. What are the possible values of its area? (The specified points are not vertices of the square.)

Attemp:

Answer 1

Let $A, B, C, D$ be the four points, as shown in your picture.

Firstly, we know that $A, C$ must lie on two opposite sides of the square, and $B, D$ on the other two opposite sides. This is because, if we form the two line segments by connecting the two pairs of points on opposite sides, then they must intersect in the interior of the square.

(This justifies that the position of the square should look like what you draw in the picture.)

For simplicity, I will use $C$-edge to indicate the edge of the square passing through the point $C$.

Now draw a line passing through $A$ and perpendicular to the line $BD$. Let $E$ be the intersection point of this line with the $C$-edge.

With simple geometry, we see that the length of $AE$ must be equal to the length of $BD$. This uniquely determines the point $E = (7, 10)$.

Since the $C$-edge passes through $C = (8, 5)$ and $E = (7, 10)$, its equation is $y = -5 x + 45$.

From this, it's easy to calculate the side length of the square as the distance from $A$ to the $C$-edge, which is $\frac{51}{\sqrt{26}}$.

So the area of the square is $\frac{2601}{26} \approx 100$.