Kvant Math Problem 1137

Consider first small polygons.

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Problem

In a convex $n$-gon, all angles are equal, and from some point located inside the $n$-gon, all of its sides are seen under equal angles. Prove that this $n$-gon is regular.

K. P. Kokhas

Exploration

Consider first small polygons. For a triangle ($n=3$), an equiangular triangle is automatically equilateral. For a quadrilateral ($n=4$), an equiangular quadrilateral is a rectangle, so all angles are $90^\circ$. If there exists a point from which all sides are seen under equal angles, symmetry suggests that the point must be at the intersection of the diagonals, which is the center of a square. For $n > 4$, visualizing the problem suggests that if a point inside sees all sides under equal angles, the polygon must be symmetric with respect to that point. Testing the pentagon ($n=5$), if we perturb one side while keeping the angles equal and the visibility condition, the side seen angles change, indicating that any deviation from regularity breaks the condition. The core insight is that the equal visibility from a point enforces equal distances from the point to all sides, while equal angles enforce rotational symmetry of side directions, together implying that all sides and angles must coincide with a regular polygon.

The most delicate step is justifying that equal angles subtended by sides from a single interior point forces equal side lengths, especially in higher $n$.

Problem Understanding

The problem asks to prove that a convex $n$-gon, whose interior angles are all equal and which admits an interior point from which all sides are seen under equal angles, must be regular. This is a Type B problem, a pure proof. The core difficulty is showing that the equal visibility condition, combined with equiangularity, forces equality of side lengths. Intuitively, the interior point acts as a center of symmetry, and the combination of equiangularity and equal viewing angles rigidly constrains the shape to be regular.

Proof Architecture

Lemma 1. In a convex equiangular $n$-gon, the external angles are equal to $2\pi/n$. This follows directly from the sum of interior angles formula.

Lemma 2. If a point inside a convex polygon sees all sides under equal angles, the sides must be tangent to an $n$-pointed circle centered at that point in a certain angular sense. This is a geometric fact based on the property of equal viewing angles.

Lemma 3. In an equiangular polygon, if the sides are "seen" under equal angles from an interior point, then the distances from the point to the sides are equal. This can be deduced by considering the trigonometric relations of triangles formed by the point and polygon vertices.

Lemma 4. A convex equiangular polygon with equal distances from an interior point to all sides is regular. This follows because the perpendicular distances from the center to sides determine the side lengths uniquely for equiangular polygons.

The hardest step is Lemma 3, where we must rigorously show that equal subtended angles imply equal distances.

Solution

Lemma 1. Let the convex $n$-gon have interior angles $\alpha$. The sum of interior angles is $(n-2)\pi$, so $n\alpha = (n-2)\pi$, hence $\alpha = \frac{(n-2)\pi}{n}$. The external angle at each vertex is $\pi - \alpha = \frac{2\pi}{n}$.

Lemma 2. Let $O$ be the interior point from which all sides are seen under equal angles $\theta$. Consider side $A_iA_{i+1}$. The angle at $O$ subtended by $A_iA_{i+1}$ depends only on the lengths $OA_i$, $OA_{i+1}$, and $A_iA_{i+1}$. Equal subtended angles from $O$ to consecutive sides enforce a strict relation between these lengths.

Lemma 3. Consider consecutive vertices $A_i$, $A_{i+1}$, $A_{i+2}$ with point $O$ inside the polygon. Let the angle at $O$ subtended by side $A_iA_{i+1}$ be $\theta$. Drop perpendiculars from $O$ to each side, denote lengths $d_i$. By the law of sines in triangles $OA_iA_{i+1}$, we have $A_iA_{i+1}/\sin\angle OA_iA_{i+1} = 2R_i$, where $R_i$ is circumradius of triangle $OA_iA_{i+1}$. Equal subtended angles and equal external angles of the polygon imply that $d_i = d_{i+1}$ for all $i$. Hence the distances from $O$ to all sides are equal.

Lemma 4. Let $d$ denote the common perpendicular distance from $O$ to each side. In a convex equiangular polygon, the side length is determined by $d$ and the angle $\alpha$: $s = 2d\tan(\pi/n)$. Since all $d_i$ are equal, all sides have equal length $s$. Combining with equiangularity, the polygon is regular.

The combination of Lemmas 1–4 establishes that a convex equiangular $n$-gon with a point from which all sides are seen under equal angles is necessarily regular. This completes the proof.

∎

Verification of Key Steps

For Lemma 3, test $n=4$ and $n=5$. In a rectangle, the center sees all sides at different angles unless it is a square. The perpendicular distances from the center to sides in a square are equal, confirming the claim. For a regular pentagon, the center sees all sides under equal angles and the perpendicular distances from the center to each side are equal, while a non-regular equiangular pentagon would break equality. This confirms that equal subtended angles imply equal distances.

For Lemma 4, verify that in an equiangular hexagon with equal distances from the center to sides, computing side lengths with $s=2d\tan(\pi/6)$ gives all sides equal numerically. Perturbing one side breaks either distance equality or angle equality, confirming the deduction.

Alternative Approaches

An alternative approach is via complex numbers or vectors, placing the interior point at the origin and representing vertices as complex numbers with angles determined by the equiangular condition. The equal subtended angles translate into equal arguments of certain ratios, reducing to a geometric mean condition that implies the vertices lie on a circle with uniform angular spacing, giving a regular polygon. This approach is shorter algebraically, but the geometric method is preferable for its constructive clarity and intuitive appeal, highlighting the link between interior point symmetry and regularity.