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Let (ABCD) be a (convex) quadrilateral that is partitioned into finitely many quadrilaterals. Assume:

  1. Every quadrilateral in the partition is cyclic.
  2. No vertex of any quadrilateral of the partition lies on a side of (ABCD).
  3. No vertex of any quadrilateral of the partition lies on a side of another quadrilateral of the partition (so the pieces meet only along entire sides, with no T-junctions).

Prove that the outer quadrilateral (ABCD) is also cyclic.

A figure for the case of five and eigth pieces is shown below: enter image description here enter image description here

I tried to use induction on the number of quadrilaterals in the partition, but this did not lead anywhere.

Lee Mosher
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    See Miquel's six-circles theorem. It's a classical angle chase. – darij grinberg Aug 29 '25 at 19:32
  • I need to prove this statement in the general case — for an arbitrary number of quadrilaterals in the partition. –  Aug 29 '25 at 19:42
  • Ah! I had misread your post based on your picture. This looks like it is what Sanjay Ramassamy, Miquel dynamics for circle patterns, arXiv:1709.05509v2 is about, at least if the decomposition of the quadrilateral is "combinatorially grid-like". But I have never thought about formalizing this in full generality. – darij grinberg Aug 29 '25 at 20:00
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    I think the geometric part of the proof will still boil down to the simple fact (a consequence of the inscribed angle theorem) that if $o,a,b,c,d$ are five lines, then the quadrilateral formed by the lines $a,b,c,d$ (in this order) is cyclic if and only if $\measuredangle\left(o,a\right) + \measuredangle\left(o,c\right) = \measuredangle\left(o,b\right) + \measuredangle\left(o,d\right)$ (where angles are directed modulo $180^\circ$). This makes it very easy to combine cyclicities using the telescope principle. But this needs to be combined with some graph theory (cycle spaces perhaps). – darij grinberg Aug 29 '25 at 20:04
  • Ah, I see. In graph-theoretical terms, the whole picture is a graph (vertices = vertices; edges = sides), and the little quadrilaterals and the big quadrilateral are a set of chosen $4$-cycles in this graph. Fix an arbitrary line $o$, and assign to each edge $e$ the weight $\measuredangle\left(o,e\right)$. Then, each edge belongs to exactly two chosen $4$-cycles, and all but one of the chosen $4$-cycles has the property that the alternating sum of the edge-weights along this cycle is $0$. The claim is that the remaining chosen $4$-cycle (namely, the big quadrilateral) also has this property. – darij grinberg Aug 29 '25 at 20:08
  • And now it is clear how the claim can be proved: You just need to show that the sum of the alternating sums of the edge-weights along all chosen $4$-cycles (each traversed in counterclockwise order) is $0$. But this should be easy from the fact that each edge belongs to exactly two chosen $4$-cycles. The only missing part is showing that its two contributions to the sum appear with opposite signs. This requires some topological carefulness I am not currently able to provide. – darij grinberg Aug 29 '25 at 20:11
  • The figure with the number of partitions equal to $8$ is shown above. –  Aug 29 '25 at 20:24
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    Looks like $\angle B,\angle D<\frac{\pi}{2}$ hence $B+D<\pi$ hence neither of those big quadrilaterals are cyclic. Are all of the smaller ones cyclic, or are those just arbitrary partitions of quadrilaterals into smaller ones? In fact, I think the latter graph is impossible if the big and all the smaller quadrilaterals are supposed to be cyclic. – coiso Sep 01 '25 at 05:29
  • @coiso At the latter graph, the middle small quadrilateral at the vertex $C$ is not cyclic. – Alex Ravsky Sep 01 '25 at 05:55
  • @coiso It is just about arbitrary partitions of quadrilaterals into smaller ones, under the second and third conditions. These pictures are only for illustration. –  Sep 01 '25 at 07:11

2 Answers2

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We answer your question affirmatively as follows. Let $G$ be the plane graph naturally formed by the partition. Since each face of $G$ has an even number of edges, $G$ is bipartite, see, for instance, here. Fix any bipartition $V=V_1\cup V_2$ of the set $V$ of the vertices of $G$. For each vertex $v$ of $G$ let $s(v)$ be the sum of the angles at $v$ of all inner faces of $G$ incident to $v$ (so $s(v)=2\pi$ for each inner vertex $v$ of $G$). For $i=1,2$ let $S_i=\sum_{v\in V_i} s(v)$. Since each partition quadrilateral is cyclic, the sum of the opposite angles at each inner face $F$ of $G$ is equal to $\pi$. Since both pairs of opposite vertices of $F$ belong to one part of the bipartition, the face $F$ contributes $\pi$ to each of $S_i$. So $S_1=S_2$. Since $s(v)=2\pi$ for each inner vertex $v$ of $G$, $A,C$ belong to one part of the bipartition, and $B,D$ belong to its other part, $s(A)+s(C)\equiv s(B)+s(D)\pmod {2\pi}$. But $s(A)+s(B)+s(C)+s(D)=2\pi,$ so $2\pi>|s(A)+s(C)-(s(B)+s(D))|$ and thus the latter value is $0$, therefore the quadrilateral $ABCD$ is cyclic.

Alex Ravsky
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    Sir, didn’t you close your account? – Andre Lin Sep 01 '25 at 02:19
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    The one thing I wasn't able to figure out is if $q$ can be even at all (given the smaller quads are cyclic). I wonder if excluding T-junctions prevents $q$ from being even somehow. – coiso Sep 01 '25 at 07:19
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    The theorem looks a bit surprising, since the original source of the problem states that the claim should hold both for even and odd numbers of quadrilaterals in the partition. However, the part of the proof that I have read so far seems clear and raises no doubts. –  Sep 01 '25 at 07:22
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    @AndreLin Right, so please do not consider me as an SE member. I can stop to write here at any moment, but I still do this to help people. For instance, my MSE questions/answers ratio is more than $100$, but I do not write at Meta or do not elect moderators. – Alex Ravsky Sep 01 '25 at 11:32
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    @AlexRavsky Thanks for your reply and the help to us – Andre Lin Sep 01 '25 at 11:47
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Draw your quadrilaterals on the Argand diagram of the complex plane, so that each point corresponds to a complex number.

Then any polygon $WXYZ$ is cyclic iff $\frac{(w-x)(y-z)}{(x-y)(z-w)} \in \mathbb R$. This follows from the usual angle-chasing, and the fact that the (directed) angle between two vectors is the complex argument of their quotient.

In your first diagram, naming the other points $E, F, G,$ and $H$, we're given that $ABFE, BCGF, CDHG, DAEH,$ and $EFGH$ are cyclic. Then

$$\frac{(a-b)(f-e)}{(b-f)(e-a)}\cdot\frac{(c-g)(f-b)}{(b-c)(g-f)}\cdot\frac{(c-d)(h-g)}{(d-h)(g-c)}\cdot\frac{(a-e)(h-d)}{(d-a)(e-h)}\cdot\frac{(f-g)(h-e)}{(e-f)(g-h)} \in \mathbb R$$

which upon simplifying and adding some factors of $-1$ as appropriate,

$$\frac{(a-b)(c-d)}{(b-c)(d-a)} \in \mathbb R$$

This is probably easier to visualise with a drawing. Labeling an edge in blue when it appears on top of the fraction, and in red on the bottom:

the diagram from the question with the edges marked in alternating red and blue colours

... every quadrilateral that we know to be cyclic has two blue edges and two red edges in alternating colours, every interior edge appears exactly once in each colour (so in the product, they cancel out), and the edges that don't cancel are exactly the sides of the exterior quadrilateral, showing that it is cyclic.

Can you label any partition of the quadrilateral in this way? Yes! As noted in the other answer, the graph of this partition is bipartite. You can partition it into two sets $V_1$ and $V_2$ such that every edge goes from one to the other. Then, for each quadrilateral, traverse it in a clockwise order, labelling an edge in blue when it goes from $V_1$ to $V_2$ and in red when it goes from $V_2$ to $V_1$. If it's an interior edge, it'll be cancelled out by the quadrilateral on the other side, which we traverse in the opposite direction. If it's an exterior edge, it'll be left with one blue label when it goes clockwise from $V_2$ to $V_1$ and one red otherwise (the opposite of the usual order, as if the large quadrilateral were inside-out).

There's a related result you can show if the polygons are not just cyclic, but pairs of harmonic conjugates. That is when $\frac{(w-x)(y-z)}{(x-y)(z-w)} = -1$. If you can partition a quadrilateral $ABCD$ into pairs of harmonic conjugates in this way, then $AC$ and $BD$ are also harmonic conjugates.

Toph
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  • Hmm, at first glance, this seems to work for both even and odd $n$. –  Sep 01 '25 at 08:28
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    @Mathph I think so too. Are there any partitions of a quadrilateral into an even number of cyclic quadrilaterals? (Without T-junctions.) – Toph Sep 01 '25 at 08:35
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    Yes, there is a schematic diagram in the post for $n = 8$ –  Sep 01 '25 at 08:40
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    I saw, but the comments say that not all the little quadrilaterals are cyclic. Is there a diagram like it where all the quadrilaterals in the partition are cyclic? Whether or not the big one is. – Toph Sep 01 '25 at 08:46
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    I agree this is weird -- this should be equivalent to Alex Ravsky's proof, just with different notation. So either I've made a mistake or he has... or there's no solution for even $n$ and we're both right. – Toph Sep 01 '25 at 08:54
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    I checked your proof. It is OK (but, as I understand and see at the picture, in the last sentence we have the same order of the colored edges both for the large and for the small quadrilaterals). A good job, +1. I also fixed the stupid error in my proof, so now it answers OP's question affirmatively too. I am sorry for the error. – Alex Ravsky Sep 01 '25 at 12:13