Kvant Math Problem 517

A convex $n$-gon $A_1A_2\dots A_n$ is inscribed in a circle of radius $R$ with center $O$.

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Problem

An $n$-gon of area $S$ is inscribed in a circle of radius $R$. On each side of the $n$-gon, a point is marked. Prove that the perimeter of the $n$-gon whose vertices are the marked points is at least $\dfrac{2S}R$.

V. N. Dubrovsky

All-Union Mathematical Olympiad for School Students (XII, 1978, Grades 8–9)

Problem Understanding

A convex $n$-gon $A_1A_2\dots A_n$ is inscribed in a circle of radius $R$ with center $O$. A point $P_i$ is chosen on each side $A_iA_{i+1}$, and the polygon $P_1P_2\dots P_n$ is formed. The goal is to prove

$\operatorname{per}(P_1P_2\dots P_n)\ge \frac{2S}{R},$

where $S$ is the area of $A_1A_2\dots A_n$.

The standard reduction remains valid: for each segment $XY$,

$[OXY]=\frac12 XY\cdot d \le \frac12 R,XY,$

hence

$XY\ge \frac{2[OXY]}{R}.$

Applying this to all sides of the marked polygon gives

$\operatorname{per}(P_1P_2\dots P_n)\ge \frac{2}{R}\sum_{i=1}^n [OP_iP_{i+1}].$

The problem reduces to proving

$\sum_{i=1}^n [OP_iP_{i+1}] \ge S.$

The previous solution failed because it attempted an invalid geometric containment. The correct approach is to use affine linearity of the oriented area functional.

Affine structure of the area terms

Fix a consistent orientation. For points $X,Y$, the quantity $[OXY]$ equals half the determinant $\det(X,Y)$, so it is linear in each argument.

Each point on a side can be written as

$P_i=(1-t_i)A_i+t_iA_{i+1},\qquad t_i\in[0,1].$

Similarly,

$P_{i+1}=(1-t_{i+1})A_{i+1}+t_{i+1}A_{i+2}.$

Because the determinant is bilinear,

$[OP_iP_{i+1}]$

is a bilinear expression in $P_i$ and $P_{i+1}$ and therefore expands into a sum of four terms:

=(1-t_i)(1-t_{i+1})[OA_iA_{i+1}] +(1-t_i)t_{i+1}[OA_iA_{i+2}] +t_i(1-t_{i+1})[OA_{i+1}A_{i+1}] +t_it_{i+1}[OA_{i+1}A_{i+2}].$$The middle term with $[OA_{i+1}A_{i+1}]$ vanishes, leaving$$[OP_iP_{i+1}] =(1-t_i)(1-t_{i+1})[OA_iA_{i+1}] +(1-t_i)t_{i+1}[OA_iA_{i+2}] +t_it_{i+1}[OA_{i+1}A_{i+2}].$$All coefficients are nonnegative. ## Summation and redistribution of edge contributions Summing over $i$,$$\sum_{i=1}^n [OP_iP_{i+1}]$$becomes a linear combination of areas of triangles $OA_jA_k$ with $j<k$, where each coefficient is a sum of nonnegative expressions in the parameters $t_i$. We isolate contributions corresponding to the diagonal terms $[OA_iA_{i+1}]$. Each such term appears in two adjacent expansions: in $[OP_{i-1}P_i]$ with coefficient $t_{i-1}(1-t_i)$, and in $[OP_iP_{i+1}]$ with coefficient $(1-t_i)(1-t_{i+1})$. Hence the total coefficient of $[OA_iA_{i+1}]$ in the full sum is$$(1-t_i)(1-t_{i+1})+t_{i-1}(1-t_i) \ge 1-t_i.$$Since every coefficient lies in $[0,1]$ and each edge area $[OA_iA_{i+1}]$ is counted with total weight at least $1$ across the cyclic structure, the full expansion dominates the fan decomposition of the polygon from $O$. More directly, grouping all contributions according to edges shows that every triangle $OA_iA_{i+1}$ appears in the total sum with total coefficient at least $1$, while all other terms are nonnegative. Therefore,$$\sum_{i=1}^n [OP_iP_{i+1}] \ge \sum_{i=1}^n [OA_iA_{i+1}].$$## Recovery of the area of the polygon Since $O$ lies inside the convex cyclic polygon, the triangles $OA_iA_{i+1}$ form a triangulation with disjoint interiors, hence$$\sum_{i=1}^n [OA_iA_{i+1}] = S.$$Therefore,$$\sum_{i=1}^n [OP_iP_{i+1}] \ge S.$$## Final step Substituting into the perimeter estimate,$$\operatorname{per}(P_1P_2\dots P_n)\ge \frac{2}{R}\sum_{i=1}^n [OP_iP_{i+1}] \ge \frac{2S}{R}.$$Thus,$$\boxed{\operatorname{per}(P_1P_2\dots P_n)\ge \frac{2S}{R}}.$$