Can a quadrilateral whose sides (in some order) are in arithmetic progression have an inscribed circle?

Question

Can the sides of a quadrilateral (not necessarily taken in order) be in arithmetic progression in which a circle can be inscribed?

I tried to think of it, but I can only think of a condition in which all the sides are equal. Are there other conditions possible?

Answer 1

For a quadrilateral with sides $a,b,c,d$ (in that order) to have an inscribed circle, a necessary${}^{\color{blue}{[1]}}$ condition is $a+c=b+d$. If $a,b,c,d$ form a non-trivial A.P, then it is impossible for that quadrilateral to have an inscribed circle.

It turns out the condition is also sufficient${}^{\color{blue}{[2]}}$. Given four numbers $a, b, c, d$. If $a+c = b +d$, there are quadrilaterals with sides in that order and having an inscribed circle. In particular, this means given any AP $a, a+x, a+2x, a+3x$; there is a quadrilateral with sides $a, a+x, a+3x, a+2x$ (in that order)${}^{\color{blue}{[3]}}$.

As an example, consider the right trapezoid $ABCD$ with vertices at $$A (1,2), B : (-2,2), C : ( -2, -2 ), D : (4,-2)$$ Its sides are $AB = 3, BC = 4, CD = 6, DA = 5$. They form an AP $3,4,5,6$ but not in order. One can verify that $x^2+y^2 = 2^2$ is an inscribed circle of this right trapezoid.

In general, a quadrilateral which can be inscribed by a circle is known as tangential quadrilateral. Its wiki entry has other necessary and sufficient conditions for a quadrilateral to be tangential.

Notes/References

• $\color{blue}{[1]}$ - The necessary condition is known as Pitot theom, first proved by Henri Pitot in 1725.
• $\color{blue}{[2]}$ - Pointed out by Jean Marie. The converse of Pitot's theorem was proved by Jakob Steiner in $1846$.
• $\color{blue}{[3]}$ - Pointed out by Aretino in comment.