Given a square with vertices: $\sqrt3+i,-1+i\sqrt3,-i-\sqrt3,1-i\sqrt3$
So that a circle that is centered in the origin is blocking the square.
My question is how can I find the vertices of a square (different one) that blocks this origin centerd circle such that his sides are parallel to the first square I mentioned.
Is there any easy way to solve it?
Thanks.
On
The given square is circumscribed by the circle $\;|z|=2\;$ (why?), and we want a new square circumscribing this circle, and from here that each of its sides will be of length $\;2\cdot2=4=\;$ the circle's diameter.
Both squares will be concentric, meaning the four diagonals of them will meet in one single point and with sides parallel.
An idea for finding the vertices of the new suqare (disclaimer: it is not claimed this is the only or even the fastest/simplest way to do it).
For example, take the middle point of the first square's "right" side $\;\ell:=[(1,-\sqrt3)\;,\;(\sqrt3,1)]\;$ is
$$M_r=\left(\frac{1+\sqrt3}2\;,\;\;\frac{1-\sqrt3}2\right)$$
the line through the above point and the origin is
$$y=\frac{1-\sqrt3}{1+\sqrt3}x=(\sqrt3-2)x$$
and the middle point of the new side parallel to $\;\ell\;$ will be on this line (why?) and at length $\;2\;$ from the origin, so we need a point
$$\left(x\;,\;(\sqrt3-2)x\right)\;\;s.t.\;\; x^2+(5-4\sqrt3)x^2=4$$
Solve this and you have your new square's right side's middle point, and then you can calculate the line on which this new side is on (you've the point and the line's slope. Why?), adn you need both vertices to be at a distance of $\;2\;$ from this new middle point...and etc.
Please note the above is basic (analytic and plain) geometry.
You started with a square $Q$, centered at the origin, whose diagonals have length $d$. The circle $C$ circumscribing this square has diameter $d$. You want to find the square $Q'$ circumscribing $C$ whose sides are parallel to the sides of $Q$.
Since $Q$ is centered at the origin, $Q'$ is just a rescaling of $Q$. In other words, there is some positive real number $\lambda$ with the property that $Q' = \{\lambda z : z \in Q\}$. For short, I'll just write $Q' = \lambda Q$.
To get $Q'$, all we have to do is find $\lambda$. To find $\lambda$, notice that the sides of $Q'$ must have length $d$. On the other hand, the sides of $Q$ have length $\frac{d}{\sqrt{2}}$, so the sides of $Q'$ will have length $\lambda \frac{d}{\sqrt{2}}$. That means you should pick $\lambda = \sqrt{2}$, just like you guessed.
@Semiclassical drew a nice picture of $Q$, $C$, and $Q'$. Looking at the right triangles formed by the extra lines they drew from the center, you can see quite directly that the diagonals of $Q$, the diameter of $C$, and the sides of $Q'$ are the same length.