## Problem B3

What is the largest number of the quadrilaterals formed by four adjacent vertices of an convex n-gon that can have an inscribed circle?


## Solution

Answer: $[\mathrm{n} / 2]$.

Take a regular n-gon and slice off alternate corners (until [n/2] corners have been cut). Specifically, if the vertices are $A_{1}, A_{2}, \ldots, A_{n}$, then we slice off the corners at $A_{1}, A_{3}, A_{5}, \ldots, A_{m}$, where $m=n-1$ if $n$ is even, or $n-$ 2 if $n$ is odd. At each corner $A_{i}$ which we slice, we take the inscribed circle to the triangle $A_{i-1} A_{i} A_{i+1}$, draw a tangent to it parallel to $\mathrm{A}_{\mathrm{i}-1} \mathrm{~A}_{\mathrm{i}+1}$ and cut along the tangent. This procedure shows that $[\mathrm{n} / 2]$ can be achieved.

To show that it is optimal it is sufficient to show that if A, B, C, D, E are adjacent vertices of any polygon and $\mathrm{ABCD}$ has an inscribed polygon, then $\mathrm{BCDE}$ does not. Since $\mathrm{ABCD}$ has an inscribed polygon, $\mathrm{AD}+\mathrm{BC}$ $=A B+C D$ (if the inscribed circle touches at $X$ on $A B$ and $Y$ on $A D$, then $A X=A Y$. That and the three similar equations give the result). So if $\mathrm{BCDE}$ also has an inscribed polygon, then $\mathrm{BE}+\mathrm{CD}=\mathrm{BC}+\mathrm{DE}$. Hence (adding) $\mathrm{AD}+\mathrm{BE}=\mathrm{AB}+\mathrm{DE}$. But the diagonals $\mathrm{AD}$ and $\mathrm{BE}$ must meet at some point $\mathrm{X}$. Then $\mathrm{AD}+$ $\mathrm{BE}=\mathrm{AX}+\mathrm{XD}+\mathrm{BX}+\mathrm{XE}=(\mathrm{AX}+\mathrm{XB})+(\mathrm{DX}+\mathrm{XE})>\mathrm{AB}+\mathrm{DE}$. Contradiction.

## 28th USAMO 1999

