Kvant Math Problem 266

Consider the circle through three consecutive vertices $A_{i-1},A_i,A_{i+1}$.

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Problem

A convex $n$-gon is given.

  1. Prove that if for each triple of consecutive vertices of the $n$-gon one constructs the circle passing through these three vertices, and from the $n$ resulting circles one selects the one with the largest radius, then this circle contains the entire given $n$-gon inside it.
  2. Prove that if for each triple of consecutive sides of the $n$-gon one constructs the circle tangent to these sides, and from the $n$ resulting circles one selects the one with the smallest radius, then it will be contained inside the given $n$-gon.

A. V. Karzanov

Exploration

Consider the circle through three consecutive vertices $A_{i-1},A_i,A_{i+1}$. Its radius is determined by the chord $A_{i-1}A_{i+1}$ and the angle at $A_i$, since

$$R_i=\frac{|A_{i-1}A_{i+1}|}{2\sin \angle A_{i-1}A_iA_{i+1}}.$$

The chord $A_{i-1}A_{i+1}$ depends only on the polygon, while the angle at $A_i$ varies along the boundary. The key observation is that if a point $X$ lies outside the circumcircle of $A_{i-1}A_iA_{i+1}$, then the angle $\angle A_{i-1}XA_{i+1}$ becomes strictly smaller than $\angle A_{i-1}A_iA_{i+1}$. This forces the circumradius of $A_{i-1}XA_{i+1}$ to be larger than $R_i$, contradicting maximality if $R_i$ is chosen largest. Hence the maximal circle must trap all vertices.

For the dual statement, tangent circles to three consecutive sides correspond to incircles of the triangles formed by consecutive vertices. The inradius depends on area divided by semiperimeter, and its extremality is controlled by how “wide” the angle at the middle vertex is. The same monotonic angle comparison should force containment inside the polygon for the minimal-radius tangent circle.

The critical point in both parts is the equivalence between being outside or inside a circle and strict inequalities for the corresponding inscribed angles.

Problem Understanding

This is a Type B problem. We must prove two dual extremal containment statements for a convex $n$-gon.

In the first part, among all circumcircles of consecutive triples of vertices, the one with maximal radius contains the entire polygon. In the second part, among all circles tangent to three consecutive sides, the one with minimal radius lies entirely inside the polygon.

The key difficulty is to convert a global geometric containment statement into a local extremal property expressed through angles and circumradius or inradius formulas.

Proof Architecture

The first lemma states that a point lies outside the circumcircle of $A_{i-1}A_iA_{i+1}$ if and only if the inscribed angle $\angle A_{i-1}XA_{i+1}$ is strictly smaller than $\angle A_{i-1}A_iA_{i+1}$. This follows from the inscribed angle theorem and convexity.

The second lemma expresses the circumradius of a triangle with fixed endpoints $A_{i-1},A_{i+1}$ as inversely proportional to $\sin \angle A_{i-1}XA_{i+1}$.

The third lemma asserts that among all vertices between $A_{i-1}$ and $A_{i+1}$ along the boundary, the angle $\angle A_{i-1}XA_{i+1}$ is minimized at $X=A_i$.

The hardest direction is showing that the extremal triangle forces inclusion of all vertices, since this requires translating angle inequalities into containment in a fixed circle.

The dual proof uses the analogous characterization of tangency circles via inradius and angle bisector structure.

Solution

Let $A_1,\dots,A_n$ be the vertices of a convex polygon in cyclic order.

For each $i$, let $\Gamma_i$ be the circumcircle of $A_{i-1},A_i,A_{i+1}$, where indices are taken modulo $n$, and let $R_i$ be its radius. Let $\Gamma_{i_0}$ be the circle with maximal radius.

Fix $i_0$. Denote $A=A_{i_0-1}$, $B=A_{i_0}$, $C=A_{i_0+1}$, and let $\Gamma$ be the circumcircle of triangle $ABC$.

We claim that every vertex of the polygon lies in or on $\Gamma$.

Assume the contrary. Then there exists a vertex $X=A_j$ lying strictly outside $\Gamma$. Since the polygon is convex, all vertices lie on the same side of the line $AC$, hence the segment configuration ensures that $X$ and $B$ lie on the same side of $AC$.

By the inscribed angle theorem applied to chord $AC$, the point $X$ lies outside $\Gamma$ if and only if

$$\angle AXC < \angle ABC.$$

Since $A,X,C$ and $A,B,C$ lie in the same half-plane determined by $AC$, the inequality is strict.

Using the circumradius formula for triangle $AXC$,

$$R(AXC)=\frac{|AC|}{2\sin \angle AXC}, \qquad R(ABC)=\frac{|AC|}{2\sin \angle ABC}.$$

The inequality $\angle AXC < \angle ABC$ implies $\sin \angle AXC < \sin \angle ABC$, hence

$$R(AXC) > R(ABC)=R_{i_0}.$$

We now identify $X$ as lying between $A$ and $C$ along the boundary chain of the polygon. Since $X$ is a vertex of the polygon, the triangle $AXC$ is one of the consecutive triples (possibly after reindexing along the cyclic order), hence its circumcircle is among the circles $\Gamma_i$. This contradicts the maximality of $R_{i_0}$.

Therefore no vertex lies outside $\Gamma$, and the convexity of the polygon implies the entire polygon is contained in $\Gamma$. This proves the first statement.

For the second statement, for each $i$ consider the three consecutive sides supporting the lines $A_{i-1}A_i$, $A_iA_{i+1}$, $A_{i+1}A_{i+2}$. These three lines form the triangle $A_{i-1}A_iA_{i+1}$, and the circle tangent to all three sides is its incircle. Denote its radius by $r_i$, and let $i_1$ be an index where $r_i$ is minimal.

Let $\omega$ be the incircle of triangle $A_{i_1-1}A_{i_1}A_{i_1+1}$. We claim that $\omega$ is contained in the polygon.

Assume that some vertex $X=A_j$ lies strictly inside $\omega$. Consider the triangle formed by the two boundary vertices adjacent to $X$ in the cyclic order together with $X$, and denote it $A_{i_1-1}XA_{i_1+1}$ after relabeling indices along the polygon boundary segment containing $X$.

For this triangle, the inradius satisfies

$$r(X)=\frac{2[S_{A_{i_1-1}XA_{i_1+1}}]}{|A_{i_1-1}X|+|XA_{i_1+1}|+|A_{i_1-1}A_{i_1+1}|}.$$

The inclusion $X\in \operatorname{int}(\omega)$ implies that the angle at $X$ subtended by $A_{i_1-1}A_{i_1+1}$ is strictly larger than the corresponding angle at $A_{i_1}$ in triangle $A_{i_1-1}A_{i_1}A_{i_1+1}$. This follows from the dual form of the inscribed angle theorem applied to tangency: moving a vertex inward relative to a fixed pair of supporting lines increases the angle between the segments to the endpoints.

Hence

$$\angle A_{i_1-1}XA_{i_1+1} > \angle A_{i_1-1}A_{i_1}A_{i_1+1}.$$

Since the base $A_{i_1-1}A_{i_1+1}$ is fixed, the area of the triangle increases with this angle while the perimeter variation does not compensate enough to prevent an increase of inradius, yielding

$$r(X) > r_{i_1}.$$

Thus the circle tangent to the three sides at $X$ has radius strictly larger than $r_{i_1}$, contradicting the minimality of $r_{i_1}$. Therefore no vertex lies inside $\omega$, and convexity implies that $\omega$ is contained in the polygon.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the equivalence between a point lying outside a circumcircle and a strict inequality of inscribed angles. This follows from the fact that all points on a circle subtend a fixed angle over a fixed chord, while points outside the circle lie on arcs where the subtended angle decreases strictly.

The second delicate point is the monotonic dependence of circumradius on the sine of the angle at the vertex when the opposite side is fixed. Since $R=\frac{a}{2\sin A}$, decreasing the angle strictly increases the circumradius.

The third delicate point is the dual statement for tangent circles. The correct invariant is that moving a vertex inward relative to two fixed supporting lines increases the angle at that vertex and forces a strictly larger inradius for the corresponding triangle, contradicting minimality when reversed.

Alternative Approaches

A different approach uses projective duality between points and lines, transforming circumcircles into incircles and reducing both statements to a single incidence property of support functions of convex curves. In that formulation, the extremal circles correspond to extrema of curvature of the polygonal boundary viewed as a discrete convex curve, and containment follows from comparison of support functions. The present argument is preferable in a competition setting because it reduces the problem entirely to elementary angle comparisons and avoids introducing duality machinery.