I suppose this should be in C ShortList rather than G tastymath75025 wrote: Call the ordered pair of distinct circles $(\omega, \gamma)$ scribable if there exists a triangle with circumcircle $\omega$ and incircle $\gamma$. Prove that among $n$ distinct circles there are at most $(n/2)^2$ scribable pairs. Proposed by Daniel Liu Consider the graph $\mathcal{G}$ with vertices as the circles and an edge between them if and only if they form a scribable pair. Claim: $\mathcal{G}$ is triangle-free. (Proof) Let $A, B, C$ be three circles in the plane. Note that if each pair of these circles is scribable, then we can suppose that $A$ is contained in $B$ which is contained in $C$. For any point $P$ on $C$, draw tangents $\overline{PX}, \overline{PY}$ to $B$ with $X, Y$ on $C$. Poncelet's Porism ensures that $\overline{XY}$ is tangent to $B$. Also draw tangents $\overline{PZ}, \overline{PW}$ to $A$ with $Z, W$ on $C$ and argue likewise to see that $\overline{WZ}$ is tangent to $A$. Finally, choose $P$ to be on the line joining the centres of $A, B$. Then, we see that $P, X, Z, W, Y$ are concyclic in that order; so $\overline{XY}$ separates $\overline{WZ}$ and the region contained by $B$; contradicting that $\overline{WZ}$ is tangent to $A$ which lies inside $B$. $\blacksquare$ Apply Turan's theorem to conclude that $\mathcal{G}$ has no more than $\tfrac{n^2}{4}$ edges. $\blacksquare$

Congrats to Daniel Liu for writing such a neat hybrid problem!

Interesting follow-up question: is the bound sharp, i.e., does equality ever hold?

anantmudgal09 wrote: Interesting follow-up question: is the bound sharp, i.e., does equality ever hold? When $n\leq 5$, we can get $\left\lfloor\frac{n^2}{4} \right\rfloor$, but for $n\geq 6$ it seems like making a $K_{3,3}$ is impossible which means the equality cannot hold.