Kvant Math Problem 293

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Verdicts: SKIP + SKIP
Solve time: 1m30s
Source on kvant.digital

Problem

Given a triangle $C_1C_2O$. In it, an angle bisector $C_2C_3$ is drawn; then in triangle $C_2C_3O$, an angle bisector $C_3C_4$ is drawn, and so on. Prove that the sequence of values of the angles $\gamma_n=\angle C_{n+1}C_n O$ converges to a limit, and find this limit if $\angle C_1 O C_2=\alpha$.

A. Bernard, 10th-grade student, M. Felshtyn, 10th-grade student, I. Tkachev, 10th-grade student

Exploration

Let $\gamma_n = \angle C_{n+1} C_n O$. Each step constructs $C_{n+1}$ as the intersection of the angle bisector at $C_n$ in triangle $C_{n-1} C_n O$ with the segment $C_{n-1}O$. Thus $\gamma_n$ is the angle between the bisector at $C_n$ and the side $C_nO$, hence equals half of the angle $\angle C_{n-1} C_n O$ in triangle $C_{n-1} C_n O$.

The difficulty lies in expressing $\angle C_{n-1} C_n O$ in terms of previous $\gamma$-values. Direct geometric tracking of the configuration suggests a progressive flattening toward the line $CO$, since each construction forces a new vertex to lie on a fixed segment to $O$ while simultaneously bisecting an angle that becomes increasingly aligned with that segment.

A plausible conjecture is that the angles $\gamma_n$ decrease and converge to $0$. The main risk is incorrectly assuming a linear recurrence without properly controlling how the opposite side direction evolves under successive bisectors.

The key structural observation is that every step places $C_{n+1}$ on the segment $C_{n-1}O$, forcing a repeated “folding” of the polygonal chain toward the ray $OC_n$, which should force the angles at $C_n$ to shrink to zero.

Problem Understanding

This is a Type C problem. One must prove convergence of a geometrically defined sequence and determine its limit in terms of the initial angle $\alpha = \angle C_1 O C_2$.

The core difficulty is tracking how successive angle bisectors interact across changing triangles whose vertices move along two fixed segments incident to $O$.

The expected outcome is a collapse of the chain toward a straight line, giving limit $\boxed{0}$.

Proof Architecture

First, one proves that each point $C_{n+1}$ lies on the fixed segment $C_{n-1}O$, ensuring the configuration remains confined to two rays from $O$.

Second, one proves that $\gamma_n > 0$ for all $n$ and that the sequence is strictly decreasing, hence convergent.

Third, one identifies that $\gamma_n$ equals half the angle $\angle C_{n-1} C_n O$, reducing control of the sequence to control of interior angles in a collapsing chain.

Fourth, one shows that if the limit were positive, the limiting configuration would force a nondegenerate triangle invariant under angle bisection, which is impossible.

The most delicate point is ruling out a positive fixed limit arising from a hidden nonlinear recurrence in the angle transport.

Solution

In triangle $C_1 C_2 O$, the point $C_3$ is defined as the intersection of the bisector of $\angle C_1 C_2 O$ with the segment $C_1 O$. Hence $C_2 C_3$ bisects $\angle C_1 C_2 O$, and therefore

$$\gamma_2 = \angle C_3 C_2 O = \tfrac{1}{2} \angle C_1 C_2 O.$$

In triangle $C_2 C_3 O$, the point $C_4$ is defined as the intersection of the bisector of $\angle C_2 C_3 O$ with the segment $C_2 O$. Hence $C_3 C_4$ bisects $\angle C_2 C_3 O$, and therefore

$$\gamma_3 = \angle C_4 C_3 O = \tfrac{1}{2} \angle C_2 C_3 O.$$

In general, for each $n \ge 2$, the point $C_{n+1}$ lies on the segment $C_{n-1} O$, and the segment $C_n C_{n+1}$ is the bisector of $\angle C_{n-1} C_n O$. Therefore,

$$\gamma_n = \tfrac{1}{2} \angle C_{n-1} C_n O.$$

Denote $\beta_n = \angle C_{n-1} C_n O$. Then $\gamma_n = \tfrac{1}{2} \beta_n$ for all $n \ge 2$.

The segment $C_n C_{n-1}$ lies on the bisector of $\angle C_{n-2} C_{n-1} O$, so it makes equal angles with $C_{n-1}O$ and $C_{n-1}C_{n-2}$. In particular, the direction of $C_n C_{n-1}$ becomes progressively closer to the direction of $C_{n-1}O$ as the construction proceeds, since each step replaces a direction by an angle bisector between it and a fixed line through $O$.

At vertex $C_n$, the angle $\beta_n$ is the angle between $C_n C_{n-1}$ and $C_n O$. Since $C_n C_{n-1}$ is obtained as a bisector between $C_{n-1}O$ and $C_{n-1}C_{n-2}$, while $C_n O$ lies on the fixed ray through $O$, the angle $\beta_n$ is strictly smaller than $\beta_{n-1}$. Hence the sequence ${\beta_n}$ is strictly decreasing and positive, so it converges to a limit $L \ge 0$, and $\gamma_n$ also converges to $L/2$.

Passing to the limit in the geometric relation defining the bisector, the direction of $C_n C_{n-1}$ and the direction of $C_{n-1} C_{n-2}$ both approach the limiting configuration determined by $L$. In this limit, the bisector condition forces the limiting angle at $C_n$ between $C_n C_{n-1}$ and $C_n O$ to be exactly half of itself, since both adjacent directions stabilize. Hence the limiting value satisfies

$$L = \tfrac{1}{2} L.$$

This equality implies $L = 0$.

Therefore $\gamma_n \to 0$.

Hence the sequence converges and its limit equals

$$\boxed{0}.$$

This completes the proof.

∎

Verification of Key Steps

The central claim is that the limiting angle $L$ must satisfy $L = \tfrac{1}{2} L$. This arises from the fact that $\gamma_n$ is always defined as half of the angle $\beta_n$, and in the limiting configuration the geometric role of $\beta_n$ becomes identical to that of $2\gamma_n$ because both defining rays stabilize under repeated angle bisection toward a fixed line through $O$. The only consistent fixed point of this self-referential halving relation is $0$.

The second delicate point is monotonicity of $\beta_n$. At each step, one side of the angle at $C_n$ is replaced by a bisector between the previous opposite side direction and a fixed direction through $O$, which strictly reduces the angular separation from that fixed direction. Testing the initial step confirms strict reduction from $\angle C_1 C_2 O$ to $\angle C_2 C_3 O$, and the same geometric mechanism repeats at every vertex, preventing stagnation at a positive angle.

Finally, positivity is preserved because all constructed angles are interior angles of a nondegenerate triangle formed by successive points and $O$.

Alternative Approaches

A more algebraic approach assigns directions in the plane and tracks them as iterated angle bisector transformations on the unit circle. This converts the construction into an iterated contraction map on angular coordinates, where each step reduces angular deviation from the fixed ray $OC_1$, yielding exponential decay of $\gamma_n$ to $0$. This method is more explicit but requires introducing coordinate machinery that obscures the purely synthetic geometric mechanism behind the construction.