### Instruction: Let $ABCD$ be a unit square. A hyperbola has its foci at the midpoints of $\overline{AB}$ and $\overline{CD},$ and passes through all the vertices of the square. Compute the distance between the two vertices of the hyperbola.### Response: This is an interesting problem.
I need to find the equation of the hyperbola and then the distance between its vertices.
I know that a hyperbola is defined by the property that the difference of the distances from any point on the curve to the foci is constant.
Let's call that constant $2a$, where $a$ is the semi-transverse axis of the hyperbola.
I also know that the midpoints of $\overline{AB}$ and $\overline{CD}$ are $(0, \frac{1}{2})$ and $(1, \frac{1}{2})$, respectively.
To find $a$, I can use the fact that the hyperbola passes through the vertex $A$.
The distance from $A$ to the focus $(0, \frac{1}{2})$ is $\frac{1}{2}$, and the distance from $A$ to the focus $(1, \frac{1}{2})$ is $\sqrt{\frac{5}{4}}$.
So the difference is $\sqrt{\frac{5}{4}} - \frac{1}{2} = 2a$.
Solving for $a$, I get $a = \frac{1}{4} \left(\sqrt{5} - 1\right)$.
Now I can write the equation of the hyperbola in standard form.
Since the foci are on the horizontal axis, the hyperbola opens left and right.
The equation is 
\[ \frac{(x - \frac{1}{2})^2}{a^2} - \frac{(y - \frac{1}{2})^2}{b^2} = 1, \]
where $b$ is the semi-conjugate axis of the hyperbola.
To find $b$, I can use another fact: the product of the slopes of the asymptotes of the hyperbola is $-\frac{b^2}{a^2}$.
The asymptotes are the extended diagonals of the square, so their slopes are $\pm 1$.
Therefore, 
\[ -\frac{b^2}{a^2} = -1 \implies b^2 = a^2 \implies b = a. \]
This simplifies the equation of the hyperbola to 
\[ \frac{(x - \frac{1}{2})^2}{a^2} - \frac{(y - \frac{1}{2})^2}{a^2} = 1 \implies (x - \frac{1}{2})^2 - (y - \frac{1}{2})^2 = a^2. \]
To find the vertices of the hyperbola, I need to find the points where the curve intersects the horizontal axis.
This means setting $y = \frac{1}{2}$ and solving for $x$.
Plugging in $y = \frac{1}{2}$, I get 
\[ (x - \frac{1}{2})^2 - (\frac{1}{2} - \frac{1}{2})^2 = a^2 \implies (x - \frac{1}{2})^2 = a^2 \implies x - \frac{1}{2} = \pm a. \]
Therefore, the vertices are 
\[ \left(\frac{1}{2} + a, \frac{1}{2}\right) \text{ and } \left(\frac{1}{2} - a, \frac{1}{2}\right). \]
The distance between them is $2a$, which is the same as the constant difference of the distances from the foci.
Substituting the value of $a$, I get the final answer.