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Professor Halfbrain has recently made a fascinating discovery on quadrilaterals in the plane.

Halfbrain's quadrilateral theorem: Let $ABCD$ be a plane quadrilateral that possesses an incircle and a circumcircle. If the center of the incircle coincides with the center of the circumcircle, then $ABCD$ is a square.

Question: Is this theorem indeed true, or has the professor once again made one of his mathematical blunders?

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It is

true

Proof:

Let's first draw the concentric incircle and circumcircle. We draw the quadrilateral starting with a vertex on the outer circle and proceeding clockwise. The edge leaving that vertex must be tangent to the inner circle and so is forced.
The next vertex is where this line intersects the outer circle. Now, we're in the same situation as before, but rotated. So, the resulting shape must be rotationally-symmetric (edge-transitive).
If it returns to the initial vertex without crossing itself, it must be a regular polygon. If it has four edges, it's a square.

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The theorem is...

... not true. There are an infinite number of concave quadrangles which also bicentric with a incentre-circumcentre distance of zero, each one associated with a specific out-radius:in-radius ratio $R/r$, where $R/r < \sqrt{2}$. According to Fuss' Theorem, only in the special case where $R/r = \sqrt{2}$ does a square solution appear.

Proof:

To construct a concave solution, within a given circle, draw two arbitrary chords $AB$ and $AD$ from $A$, such that $\angle BAD < 60^{\circ}$. Then, draw the incircle such that the chords $AB$ and $AD$ are both tangent to the incircle. Finally, choose a point $C$ on the incircle to complete the quadrangle $ABCD$.

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    $\begingroup$ Does a concave quadrilateral have a circumcircle? I understand that to mean that all four points lie on that circle, i.e. it is a cyclic quadrilateral. $\endgroup$
    – xnor
    Sep 21, 2015 at 9:37
  • $\begingroup$ By the way, your construction didn't work for me. Also, Fuss theorem applies to any bicentric quad. If you plus $d=0$ in it, You get $R/r= \sqrt{2}$ which only holds for a square. $\endgroup$
    – Rohcana
    Sep 21, 2015 at 9:53
  • $\begingroup$ Yes, Fuss' Theorem only applies to simple convex quadrilaterals. For the concave case, as xnor has noitced, I took the liberty to generalize the incircle and circumcircle to the maximum inscriptable circle and the minimum bounded circle, respectively (otherwise both are hilariously undefined for the concave polygon). $\endgroup$
    – Hackiisan
    Sep 21, 2015 at 10:00

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