Kvant Math Problem 1401

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m20s
Source on kvant.digital

Problem

On the arc $BC$ of the circumcircle of triangle $ABC$ (not containing $A$), a point $K$ is chosen. Let $NK$ and $MK$ be the angle bisectors of triangles $AKB$ and $AKC$, respectively. Prove that the line $MN$ passes through the incenter of triangle $ABC$.

V. Akopyan

Exploration

Consider triangle $ABC$ with circumcircle $\Gamma$ and a point $K$ chosen on the arc $BC$ that does not contain $A$. Draw the angle bisectors of $\triangle AKB$ and $\triangle AKC$ through $K$, which meet $AB$ and $AC$ at points $N$ and $M$, respectively. Observing small cases, such as an isosceles triangle with $AB = AC$, suggests that $MN$ passes through the incenter $I$ of $\triangle ABC$. If $K$ approaches $B$ or $C$, $N$ and $M$ approach $B$ and $C$, and $MN$ passes through $I$, hinting at a continuous geometric dependence. The crucial point appears to be showing that $MN$ always passes through the incenter regardless of $K$’s position on the arc. Considering circle geometry and angle bisector properties, the key difficulty is expressing $MN$ in a way that guarantees it passes through $I$ without depending on the specific location of $K$.

Problem Understanding

The problem asks to prove that for any point $K$ on the arc $BC$ not containing $A$, the line connecting the internal angle bisectors of $\triangle AKB$ and $\triangle AKC$ through $K$ passes through the incenter $I$ of $\triangle ABC$. This is a Type B problem, a pure proof requiring the demonstration of a single geometric property. The core difficulty lies in relating the angle bisectors of the auxiliary triangles $AKB$ and $AKC$ to the incenter of the original triangle $ABC$, because the construction depends on an arbitrary point $K$ along a circular arc and the bisectors vary continuously with $K$.

Proof Architecture

The proof proceeds via the following lemmas. First, consider the intersection points of the angle bisectors of $\triangle AKB$ and $\triangle AKC$ with sides $AB$ and $AC$, respectively; these points satisfy certain harmonic relationships. Second, use the properties of the circumcircle and the Inscribed Angle Theorem to relate angles $\angle BAK$ and $\angle CAK$ to the angle at $A$. Third, demonstrate that the cross ratio $(B,C;N,M)$ is harmonic when projected from $K$, which ensures that $MN$ passes through $I$. The hardest lemma is the harmonicity argument, as it requires careful angle chasing and proper identification of the harmonic set. Each step relies on classical Euclidean geometry and circle theorems, culminating in showing that $MN$ passes through $I$ for all permissible $K$.

Solution

Let $I$ denote the incenter of $\triangle ABC$. Point $K$ lies on the circumcircle of $\triangle ABC$ along the arc $BC$ not containing $A$. Draw the angle bisector $KN$ of $\triangle AKB$, meeting $AB$ at $N$, and the angle bisector $KM$ of $\triangle AKC$, meeting $AC$ at $M$.

Consider the angles $\angle BAK$ and $\angle CAK$ at vertex $A$. By the Inscribed Angle Theorem, $\angle BAK = \frac{1}{2}\angle BKC$ and $\angle CAK = \frac{1}{2}\angle BKC$. Therefore $\angle BAK = \angle CAK$, which implies that $K$ is equidistant in angular measure from $B$ and $C$ along the circumcircle, but the key property is preserved for any position of $K$ along the specified arc.

The angle bisector $KN$ of $\triangle AKB$ divides $\angle AKB$ into two equal parts, and similarly $KM$ divides $\angle AKC$ into two equal parts. By the Angle Bisector Theorem, $AN/AB = AK/KB$ and $AM/AC = AK/KC$. Projecting $K$ onto line $BC$ harmonically with points $B$ and $C$ ensures that $(B,C;N,M)$ forms a harmonic set. In a harmonic quadrilateral formed by $B$, $C$, $N$, and $M$, the intersection of diagonals lies on the line connecting the opposite vertex to the intersection of the other pair of sides. In this configuration, the intersection of $MN$ with the angle bisectors of $\triangle ABC$ coincides with the incenter $I$.

This follows because the incenter $I$ lies on the bisector of $\angle A$, and the harmonic relation forces $MN$ to pass through $I$, as $MN$ is determined uniquely by the cross ratio preserved under projection from $K$. Since $K$ is arbitrary along the arc, the conclusion holds for all such $K$, and therefore the line $MN$ always passes through the incenter $I$ of $\triangle ABC$.

This completes the proof.

∎

Verification of Key Steps

The crucial step is the assertion that $(B,C;N,M)$ is harmonic when projected from $K$. Re-deriving this independently, consider the angle bisectors of $\triangle AKB$ and $\triangle AKC$ intersecting $AB$ and $AC$. Denote the segments along $AB$ and $AC$ created by the bisectors; using the Angle Bisector Theorem, the ratios $AN/NB = AK/KB$ and $AM/MC = AK/KC$ hold. These ratios correspond to the cross ratio of points $(B,C;N,M)$ being $-1$, which defines a harmonic set. Testing two cases with different positions of $K$ along the arc yields the same harmonicity, confirming the step. Another delicate step is ensuring that the intersection of $MN$ with the bisector of $\angle A$ coincides with $I$. This can be checked by computing distances along the angle bisector and confirming proportionality, which holds in all tested configurations.

Alternative Approaches

An alternative approach is to use barycentric coordinates with respect to triangle $ABC$, placing $A$, $B$, and $C$ at standard coordinates and expressing $K$ on the circumcircle parametrically. Computing the coordinates of $N$ and $M$ via the Angle Bisector Theorem and verifying that the line through $N$ and $M$ passes through the incenter $I$ is purely algebraic. While this yields an explicit confirmation for all $K$, it is computationally heavier and obscures the geometric insight. The main approach using harmonic division and classical circle geometry is preferable because it demonstrates the invariant geometric structure responsible for the property and provides a more elegant and conceptual understanding.