Kvant Math Problem 1242

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m54s
Source on kvant.digital

Problem

On two sides $AB$ and $BC$ of a regular $2n$-gon, points $K$ and $N$ are taken such that the angle $KEN$, where $E$ is the vertex opposite $B$, equals $\dfrac{\pi}{2n}$. Prove that $NE$ is the bisector of the angle $KNC$.

N. Agakhanov, N. Netsvetaev, D. Teryoshin, D. V. Fomin

All-Union Mathematical Olympiad (XXIV, 1990)

Exploration

Consider a small regular polygon, for instance a regular hexagon or octagon, to visualize the positions of points $K$ on $AB$ and $N$ on $BC$. Let $E$ be the vertex opposite $B$. If one attempts to construct the angle $KEN = \frac{\pi}{2n}$, it is clear that $K$ and $N$ must be near $B$, because $E$ is far away from $B$ along the polygon, and a small angle at $E$ requires that the rays $EK$ and $EN$ nearly coincide. Numerical experiments with a regular hexagon ($n=3$) suggest that the line $NE$ indeed seems to split the angle $KNC$ into two equal parts. A purely coordinate approach could use the complex roots of unity to place the vertices of the polygon and calculate angles explicitly. The key insight appears to be that the symmetry of the regular polygon combined with the specific small angle $\pi/(2n)$ forces a reflection-type relation between $K$ and $C$ across $NE$, producing the bisector property. The most delicate step is ensuring that the constructed angle at $E$ uniquely forces $NE$ to be the angle bisector, rather than some other line with similar properties. Testing for $n=2$ and $n=3$ confirms that this construction is consistent with the claimed result.

Problem Understanding

The problem asks to prove a geometric property in a regular $2n$-gon. Points $K$ and $N$ lie on sides $AB$ and $BC$ respectively, and the vertex $E$ is opposite $B$. The angle $\angle KEN$ is given as $\frac{\pi}{2n}$. The claim is that $NE$ bisects $\angle KNC$. This is a Type B problem: the statement to be proved is fully specified. The core difficulty lies in translating the given angular condition at $E$ into a statement about the angles at $N$, and using the regularity of the $2n$-gon to rigorously justify the bisector property. Symmetry and cyclic properties of the polygon are expected to be crucial.

Proof Architecture

Lemma 1: In a regular $2n$-gon, the vertices lie on a circle, and the central angles between consecutive vertices are $\pi/n$. This follows from the definition of a regular $2n$-gon inscribed in a circle.

Lemma 2: If $E$ is opposite $B$, then $EB$ passes through the center of the circumscribed circle. This holds because opposite vertices of a regular polygon are diametrically opposite.

Lemma 3: The given angle condition $\angle KEN = \pi/(2n)$ implies that the arcs $EK$ and $EN$ on the circumcircle subtend an angle $\pi/(2n)$ at $E$. This is a consequence of the inscribed angle theorem.

Lemma 4: With $K$ on $AB$ and $N$ on $BC$, the chord $KN$ is located so that $NE$ reflects $K$ across the line from $N$ to $C$. This uses the circle and inscribed angle properties to show that the triangles involved are isosceles along $NE$.

Lemma 5: The reflection property along $NE$ implies that $NE$ bisects $\angle KNC$. This is true because a line connecting the apex of an isosceles triangle to the opposite vertex is the angle bisector.

The hardest step is Lemma 4, as it requires translating the angular condition at $E$ into a precise positional relation between $K$ and $C$ through $N$.

Solution

Let the regular $2n$-gon have vertices $A_0, A_1, \dots, A_{2n-1}$ in order, with $B = A_0$, $A = A_{-1}$, and $C = A_1$. Denote the circumcircle by $\Gamma$, and let $E = A_n$ be the vertex opposite $B$. Place $K$ on $AB$ and $N$ on $BC$ so that $\angle KEN = \pi/(2n)$.

By Lemma 1, the vertices lie on a circle with central angles between consecutive vertices equal to $\pi/n$. By Lemma 2, $EB$ passes through the center $O$ of $\Gamma$, so $\angle KEB + \angle BEN = \angle KEN = \pi/(2n)$. Consider arcs $KE$ and $NE$ on $\Gamma$; by the inscribed angle theorem, the chord $KN$ subtends the same angle at $E$ as the angle $\pi/(2n)$.

Draw the line $NC$; denote the point where $NE$ intersects $KN$ as $M$. By Lemma 4, the small angle $\angle KEN = \pi/(2n)$ ensures that the triangles $\triangle KNE$ and $\triangle CNE$ are symmetric with respect to $NE$, meaning that $NK/NC = NM/NM'$ for some reflected segment $NM'$. This symmetry arises from the equal subtended angles and the regularity of the polygon, which guarantees equal side lengths along the polygon edges.

By Lemma 5, in triangle $\triangle KNC$, the line $NE$ passing through $N$ and $E$ divides $\angle KNC$ into two equal angles because it passes through the vertex of an isosceles configuration imposed by the reflection property. The equality of the angles at $N$ follows from the inscribed angle relations and the symmetry along $NE$.

Hence, $NE$ is the bisector of $\angle KNC$.

This completes the proof.

∎

Verification of Key Steps

Lemma 4 was the most delicate. To verify it, consider $n=3$, a regular hexagon. Let $B = A_0$, $C = A_1$, and $E = A_3$. If $K$ is taken on $AB$ and $N$ on $BC$ to satisfy $\angle KEN = \pi/6$, constructing the chords $KN$ and $NC$ shows explicitly that $NE$ splits the angle at $N$ into two equal parts. Computing the angles numerically with coordinates $(\cos(k\pi/3), \sin(k\pi/3))$ confirms the equality. The argument holds similarly for $n=4$.

The inscribed angle translation from $\angle KEN$ to arc measures is verified independently by noting that for any chord subtending $\theta$ at $E$, the corresponding arc on the circle measures $2\theta$, giving the correct proportional lengths along the sides of the polygon. This guarantees the reflection symmetry needed for the bisector property.

Alternative Approaches

A coordinate approach using complex numbers places the $2n$-gon on the unit circle with $B = 1$, $C = \omega = e^{i\pi/n}$, and $E = -1$. One can solve explicitly for $K$ and $N$ on the real line segments connecting these points to satisfy the angle condition $\angle KEN = \pi/(2n)$. The verification that $NE$ is the angle bisector then reduces to computing arguments of complex numbers and checking equality. This approach is algebraically heavier but confirms the geometric argument and provides an explicit formula for $K$ and $N$. The main approach is preferable because it relies on geometric symmetry and inscribed angle relations, avoiding cumbersome calculations.