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December 30, 2011

The results of Fuss and Carlitz for bicentric quadrilaterals

Filed under: mathematics — ckrao @ 4:47 am

Any triangle has both an incircle (tangent to its sides) and circumcircle (through its vertices).

The distance d between the centres O, I of these circles is related to their radii r (for the incircle) and R (for the circumcircle) via Euler’s theorem in geometry:

\displaystyle d^2 = R^2 - 2Rr

or the equivalent formula

\displaystyle \frac{1}{R-d} + \frac{1}{R+d} = \frac{1}{r}.

A convex quadrilateral however need not have an incircle or circumcircle in general. For it to have an incircle (i.e. a tangential quadrilateral), its pairs of opposite sides must have the same sum. For it to have a circumcircle (i.e. a cyclic quadrilateral), its opposite angles must be supplementary. One that satisfies both of these properties is called a bicentric quadrilateral.

Remarkably, if we define d, r and R for a bicentric quadrilateral analogous to the triangular case, we have formulas similar to Euler’s due to Nicolaus Fuss and Leonard Carlitz:

Carlitz: \displaystyle d^2 = R^2 - 2Rr.\mu,

where \displaystyle \mu = \sqrt{\frac{(ab+cd)(ad + bc)}{(a+c)^2(ac+bd)}} and a,b,c,d are the quadrilateral’s side lengths.

Fuss: \displaystyle \frac{1}{(R-d)^2} + \frac{1}{(R+d)^2} = \frac{1}{r^2}

Isn’t it cool how similar these formulas look for the triangular and quadrilateral cases?! Both of these results (and indeed Euler’s theorem) follow from the intersecting chords theorem. Fuss’s theorem can be shown by proving that both sides of the equation are equal to \displaystyle \frac{1}{AI^2} + \frac{1}{CI^2}. Carlitz’s identity can be shown via the preliminary result that \mu = IE/BE, where E is the intersection of line AI with the circumcircle. A beautiful proof of Fuss’s theorem due to Salazar (using little more than a short angle chase and the theorems of Pythagoras and Apollonius) is found in [1] while that of Carlitz can be seen from p154 of [2].

By Poncelet’s Porism, it is true that if we start with two circles of radius r and R with distance d between their centres such that the Fuss equation \frac{1}{(R-d)^2} + \frac{1}{(R+d)^2} = \frac{1}{r^2} is satisfied, then a bicentric quadrilateral with those parameters may be constructed. Simply start at any point on the circumcircle and successively draw tangents to the incircle to generate the other three vertices of the quadrilateral. This was how I drew the above bicentric quadrilateral.


[1] A. Bogomolny, Fuss’ Theorem from Interactive Mathematics Miscellany and Puzzles, Accessed 30 December 2011.

[2] O. Calin, Euclidean and Non-Euclidean Geometry: a metric approach, available at


1 Comment »

  1. Science is tremendous, Mathematics is magic !

    Comment by Ali Toumi — May 25, 2013 @ 5:41 pm | Reply

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