Kvant Math Problem 239

Let points $A$ and $B$ be fixed on the plane, and let $C$ lie on the perpendicular bisector of segment $AB$, since it must satisfy $|AC| = |BC|$.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m12s
Source on kvant.digital

Problem

On the plane, two points $A$ and $B$ are given. Let $C$ be a point equidistant from $A$ and $B$. Construct a sequence of points $C_1=C$, $C_2$, $C_3$, $\ldots$, $C_n$, $C_{n+1}$, $\ldots$ where $C_{n+1}$ is the circumcenter of triangle $AC_nB$. For which position of the point $C$

  1. does the point $C_n$ fall at the midpoint of segment $AB$ (in this case, $C_{n+1}$ and subsequent members of the sequence are not defined)?
  2. does the point $C_n$ coincide with $C$?

N. Chernov, 10th grade student (Kryvyi Rih)

Exploration

Let points $A$ and $B$ be fixed on the plane, and let $C$ lie on the perpendicular bisector of segment $AB$, since it must satisfy $|AC| = |BC|$. The sequence is defined recursively: $C_1 = C$ and $C_{n+1}$ is the circumcenter of triangle $AC_nB$. The first question asks for the position of $C$ so that at some step $n$, $C_n$ lands exactly at the midpoint $M$ of $AB$. The second question asks for positions where the sequence becomes stationary, $C_n = C$.

Consider simple geometric configurations. Placing $C$ far from $AB$ along the perpendicular bisector seems to "pull" the circumcenter closer to $AB$, suggesting a contraction toward $M$. If $C$ lies on the line perpendicular to $AB$ at $M$, then $C_2$ lies somewhere between $C$ and $M$. This hints that the sequence is effectively a one-dimensional iteration along this perpendicular line. The key difficulty is describing the iteration in a tractable form and determining its fixed points and finite-time hits at $M$.

Exploring a coordinate model may help. Place $A=(-1,0)$, $B=(1,0)$, then $M=(0,0)$. Any $C$ equidistant from $A$ and $B$ has coordinates $(0,h)$. Compute $C_2$, the circumcenter of triangle $A C_1 B$. The circumcenter lies at the intersection of perpendicular bisectors. The perpendicular bisector of $AB$ is $x=0$, so $C_2$ must have $x=0$. Similarly, the perpendicular bisector of $A C_1$ passes through $((0 + -1)/2, (h + 0)/2) = (-0.5, h/2)$ with slope $-1/m$, giving $C_2 = (0, h/2)$. This suggests that each step halves the distance to $M$, so $C_n = (0, h/2^{,n-1})$. This immediately gives a clear pattern for both questions: the sequence hits $M$ in finite steps if $h = 0$ initially (then $C_1 = M$), and otherwise it approaches $M$ asymptotically.

Problem Understanding

The problem requires classifying the positions of $C$ along the perpendicular bisector of $AB$ with respect to the recursive circumcenter sequence. This is Type A: we must find all $C$ that satisfy the finite-time hit or stationarity. The core difficulty is formalizing the geometric recursion and translating it into a computable sequence. Intuitively, the perpendicular bisector reduces the problem to a one-dimensional contraction, making the midpoint the only candidate for finite-time termination and the only fixed point for stationarity.

The conjectured answers are: for the first question, $C = M$; for the second, the same $C = M$, because any other point is strictly drawn toward $M$ but never reaches it in a finite number of steps.

Proof Architecture

Lemma 1. If $A=(-1,0)$, $B=(1,0)$, any point $C$ equidistant from $A$ and $B$ has coordinates $(0,h)$. Sketch: this follows from the equation $|AC| = |BC|$.

Lemma 2. If $C_n = (0,h_n)$, then $C_{n+1} = (0,h_{n+1})$ with $h_{n+1} = h_n/2$. Sketch: the circumcenter of triangle $A C_n B$ lies on $x=0$ and is the midpoint of $C_n M$ along the perpendicular.

Lemma 3. The sequence $h_n$ converges to $0$ and hits $0$ in finite steps only if $h_1 = 0$. Sketch: geometric series $h_n = h/2^{,n-1}$ implies $h_n \to 0$ asymptotically unless $h=0$.

Hardest step: justifying that the iteration along the perpendicular bisector indeed halves the distance each time. Verification must carefully compute the perpendicular bisectors and their intersection.

Solution

Place $A=(-1,0)$ and $B=(1,0)$; then the midpoint is $M=(0,0)$. Any point $C$ equidistant from $A$ and $B$ lies on the perpendicular bisector of $AB$, which is the line $x=0$, so $C=(0,h)$ for some real number $h$. Consider the sequence $C_1 = C$, $C_{n+1}$ as the circumcenter of triangle $A C_n B$.

The circumcenter of a triangle lies at the intersection of the perpendicular bisectors of its sides. The perpendicular bisector of $AB$ is $x=0$. The perpendicular bisector of $A C_n$ passes through the midpoint of $A C_n$, which is $(-1/2, h_n/2)$, and has slope $1$, since $A C_n$ is vertical distance $h_n$ and horizontal distance $1$, giving slope of the perpendicular $1$. The equation of this perpendicular bisector is $y - h_n/2 = 1(x + 1/2)$, so at $x=0$, it gives $y = h_n/2 + 1/2$. Correction: carefully recompute slope.

$A C_n$ has coordinates $A=(-1,0)$, $C_n=(0,h_n)$. Its slope is $m = (h_n - 0)/(0 - (-1)) = h_n / 1 = h_n$. The perpendicular slope is $-1/m = -1/h_n$. The midpoint is $(-1/2, h_n/2)$. The perpendicular bisector equation is $y - h_n/2 = -1/h_n (x + 1/2)$. At $x=0$, $y = h_n/2 - 1/(2h_n)$. Wait, check units.

Better: treat algebraically. Circumcenter of triangle $(-1,0)$, $(1,0)$, $(0,h)$ lies at $(0, h/2)$: indeed, the perpendicular bisector of $AB$ is $x=0$. The perpendicular bisector of $A C$ passes through midpoint $(-0.5,h/2)$, slope $-1/h$, line: $y - h/2 = -1/h (x + 0.5)$. At $x=0$, $y - h/2 = -1/(2h)$? That is not consistent dimensionally. Use vector method.

Better approach: triangle with base $AB$ horizontal and apex $C=(0,h)$. The circumcenter $O$ has $x=0$ by symmetry. Let $y = k$. Distance from $O=(0,k)$ to $A=(-1,0)$ equals distance to $C=(0,h)$. Compute: $(0+1)^2 + (k-0)^2 = (0-0)^2 + (k - h)^2$, so $1 + k^2 = (k - h)^2 = k^2 - 2 h k + h^2$, thus $1 + k^2 = k^2 - 2 h k + h^2$, so $1 = -2 h k + h^2$, so $2 h k = h^2 -1$, so $k = (h^2 -1)/(2h)$.

Check: $k = (h^2 -1)/(2h)$. This gives the exact iteration $h_{n+1} = (h_n^2 -1)/(2 h_n)$.

Finite-time hit at midpoint $M=(0,0)$ requires $h_n = 0$ for some $n$. Then the recursion formula gives $0 = (h_{n-1}^2 -1)/(2 h_{n-1})$, so $h_{n-1}^2 -1 =0$, so $h_{n-1} = \pm 1$. Hence $C = (0,1)$ or $(0,-1)$ leads to $C_2 = M$.

Stationarity $C_{n+1} = C_n$ requires $h = (h^2 -1)/(2h)$, solving $2h^2 = h^2 -1$, so $h^2 = -1$, impossible over real numbers. Hence no nontrivial fixed point; the sequence never becomes stationary except asymptotically at $M$.

Therefore, for the first part, $C = (0,1)$ or $C=(0,-1)$ produces $C_2 = M$ and terminates there. For the second part, no real point produces $C_n = C$ for all $n$.

Verification of Key Steps

The critical step is computing the circumcenter along $x=0$. Using coordinates $A=(-1,0)$, $B=(1,0)$, $C=(0,h)$, we verify $|O A| = |O C|$ with $O=(0,k)$: $|O A|^2 = (0+1)^2 + (k-0)^2 = 1 + k^2$, $|O C|^2 = (0-0)^2 + (k - h)^2 = (k-h)^