Show that four vertices of a square cannot lie on four concentric circles, radii of which form an arithmetic sequence

Question

My teacher said it's solved using proof through contradiction. I've considered cases of the centre of the circle, but I lose geometry big time so not sure how to do this.

Answer 1

Without loss of generality, the common center of your four circles is $(0,0)$ and the radii are $r,r+1,r+2,r+3$. Furthermore, without loss of generality the point $A$ has coordinates $(r,0)$ and the point $C$ has coordinates $\frac{r+3}{t^2+1}(t^2-1,2t)$ (rational parametrization of a circle) which avoids some juggling with trigonometric functions. Then you can compute coordinates for $B$ and $D$ depending on $t$, and look at their norms. You won't be able to make them $r+1$ resp. $r+2$ for the same $t$ and $r$.

(Originally I had incorrectly assumed that $r$ would have to be $1$, in which case you won't be able to reach $\lVert B\rVert=2$ at all. The comment below by Ewan corrected my mistake.)

To be more specific, your other two corner points will have coordinates

\begin{align*} B&= \frac1{2(t^2+1)}\begin{pmatrix} 2 r t^{2} + 2 r t + 3 t^{2} + 6 t - 3 \\ 2 r t - 3 t^{2} + 2 r + 6 t + 3 \end{pmatrix} \\ D&= \frac1{2(t^2+1)}\begin{pmatrix} 2 r t^{2} - 2 r t + 3 t^{2} - 6 t - 3 \\ 2 r t + 3 t^{2} - 2 r + 6 t - 3 \end{pmatrix} \end{align*}

Now if you want to achieve $\lVert B\rVert=r+1$ and $\lVert D\rVert=r+2$ then you eventually obtain the following system of equations:

\begin{align*} 4 r^{2} t + 2 r t^{2} + 12 r t + 7 t^{2} + 2 r + 7 &= 0 \\ 4 r^{2} t + 2 r t^{2} + 12 r t - t^{2} + 2 r - 1 &= 0 \end{align*}

This system of equations results has no real solutions. (Its four complex solutions can be summarized as $r\in\{0,3\}, t=\pm i$.)

You might wonder whether it's possible to cover the smallest and the largest circle with two adjacent points, instead of two opposite ones. I.e. try for $\lVert A\rVert=r,\lVert B\rVert=r+3$. There might be more obvious reasons against this, but if in doubt you can still do a computation as above.

Answer 2

Assume the contrary that there are indeed a square whose vertices are lying on 4 concentric circles whose radii forming an arithmetic progression.

Scale everything and rotate the coordinate axis such the common center of the circles is the origin $(0,0)$ and the center of the square is $(1,0)$.

It is clear, we can choose a point $(u,v)$ in the first quadrant such that the 4 vertices of the square are

$$(1 + u, v),\; (1 - v, u ),\;(1 - u, -v), \;(1 + v, -u )$$

Their distances to the center of circle will be given by $\sqrt{ \Delta \pm 2u }$ and $\sqrt{ \Delta \pm 2v }$ where $\Delta = 1 + u^2 + v^2$.

First consider the case $u \ge v \ge 0$. Since the four radii are distinct, we can get rid of the equality cases and find $u > v > 0$. This leads to

$$\sqrt{\Delta - 2u} < \sqrt{\Delta - 2v} < \sqrt{\Delta + 2v} < \sqrt{\Delta + 2u}$$

If these distances form an arithmetic progression $r < r+\alpha < r+2\alpha < r+3\alpha$, we will have

$$ \begin{align} 2u - 2v &= (\Delta - 2v) - (\Delta - 2u) = (r+\alpha)^2 - r^2 = 2\alpha r + \alpha^2\\ \text{ AND }\quad 2u - 2v &= (\Delta + 2u) - (\Delta + 2v) = (r+3\alpha)^2 - (r+2\alpha)^2 = 2\alpha r + 5\alpha^2 \end{align} $$

These two equalities together leads to $\alpha^2 = 0 \iff \alpha = 0$. ie. the contradiction that the four radii are not distinct.

When $v \ge u \ge 0$, the situation is similar. This means it is impossible for the distances of the 4 vertices of the square forming an arithmetic progression.