A square is formed such that its opposite sides are tangential to the hyperbola $x^2-y^2=1$. Determine which of the following is/are true
(A) Side length of largest possible square is $2$
(B) Side length of largest possible square is $2\sqrt 2$.
(C) Side length of smallest possible square is $2^{\frac{3}{4}}$.
(D) Side length of smallest possible square is $2^{\frac{5}{4}}$.
Solution given
Length of Diagonal of square must be greater than or equal to distance between points of tangency.
Why above statement must be true? Can we prove this?
After drawing figure it seems correct but can we prove this mathematically.
The longest possible length between two points on a square is the length of the diagonal. Since the two points of tangency are, by hypothesis, on the square, the distance between them must certainly be less than or equal to the length of the diagonal.
To prove that the diagonal is the greatest length in the square, we can just consider an arbitrary pair of points in a square and consider the possible distances. If the two points are $(x_1,y_1)$ and $(x_2,y_2)$, then the squared distance between them is
$$(x_2-x_1)^2 + (y_2-y_1)^2 = a^2 + b^2$$
If $s$ is the side length of the square, then obviously the greatest purely vertical or purely horizontal distance in the square is $s$. This gives us an upper bound on the squared distance, when $a=b=s$, of
$$ s^2 + s^2 = 2s^2 $$
and, since the square root is increasing, the upper bound on the distance is $s\sqrt{2}$. To prove that this upper bound is a maximum, we just have to show that it can be achieved. Consider the two points to be on opposite corners of the square, and we calculate the distance between them (the diagonal of the square) to be $s\sqrt{2}$ as well.