Kvant Math Problem 191

Let $A$ and $B$ be fixed, and let $l$ be a fixed line through $A$ not containing $B$.

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Problem

In the plane, two points $A$ and $B$ and a line $l$ are given, passing through point $A$ and not passing through point $B$. Through points $A$ and $B$ an arbitrary circle is drawn. Let $O$ be its center, $C$ be its intersection point with line $l$, different from $A$. Find the locus of the midpoints of segments $OC$.

P. Paramonov, 9th grade student

Exploration

Let $A$ and $B$ be fixed, and let $l$ be a fixed line through $A$ not containing $B$. Every circle through $A$ and $B$ has its center $O$ on the perpendicular bisector of $AB$, so $O$ moves along a fixed line.

For each such center $O$, the circle meets the line $l$ again at a point $C \neq A$. Since both $A$ and $C$ lie on $l$, the segment $AC$ is a chord of the circle.

The quantity of interest is the midpoint of $OC$. A natural attempt is to express $C$ in terms of $O$ and then study how the midpoint depends on $O$. Because $C$ is defined by an intersection of a line with a circle, a coordinate setup with $A$ at the origin and $l$ as an axis suggests itself, since it turns the intersection condition into a simple quadratic equation.

A first computation indicates that $C$ depends linearly on the coordinate of $O$ along $l$, while the midpoint of $OC$ becomes a linear expression in the coordinates of $O$. Since $O$ itself moves along a line, this suggests the midpoint also moves along a line. The key point to justify rigorously is that no nonlinear constraint survives after eliminating $O$.

Problem Understanding

This is a Type D problem.

A point $O$ varies along the perpendicular bisector of $AB$, and for each such $O$ we construct a point $C$ as the second intersection of line $l$ with the circle centered at $O$ passing through $A$ and $B$. We are asked for the locus of the midpoint of $OC$.

The expected outcome is a geometric figure determined by an affine transformation of a line, suggesting the locus is a straight line.

Proof Architecture

The first lemma expresses the condition that $O$ is the center of a circle through $A$ and $B$ as a linear equation in coordinates.

The second lemma determines the second intersection point $C$ of a circle centered at $O$ with the line $l$, and expresses $C$ explicitly in terms of the coordinates of $O$.

The third lemma computes the midpoint $M$ of $OC$ as a function of $O$, showing that it is obtained from $O$ by a fixed linear transformation.

The final argument substitutes the linear constraint defining the locus of $O$ into the expression for $M$, eliminating parameters and showing that $M$ satisfies a single linear equation, hence lies on a fixed line.

The most delicate step is the computation of the intersection point $C$, since an incorrect choice of intersection branch would lead to an incorrect expression for the midpoint.

Solution

Introduce Cartesian coordinates with $A=(0,0)$ and with the line $l$ as the $x$-axis. Let $B=(p,q)$ with $q\neq 0$.

Let $O=(u,v)$ be the center of a circle passing through $A$ and $B$. Then $OA=OB$, hence

$u^2+v^2=(u-p)^2+(v-q)^2.$

Expanding and simplifying yields

$u^2+v^2=u^2-2pu+p^2+v^2-2qv+q^2,$

so

$2pu+2qv=p^2+q^2.$

Thus $O$ lies on the line

$pu+qv=\frac{p^2+q^2}{2}.$

The circle centered at $O$ has equation

$(x-u)^2+(y-v)^2=u^2+v^2.$

Intersecting with $l$, where $y=0$, gives

$(x-u)^2+v^2=u^2+v^2,$

hence

$(x-u)^2=u^2.$

Thus $x-u=\pm u$, so $x=0$ or $x=2u$. The point $x=0$ corresponds to $A=(0,0)$, so the second intersection point is

$C=(2u,0).$

The midpoint $M$ of $OC$ is

$M=\left(\frac{u+2u}{2},\frac{v+0}{2}\right)=\left(\frac{3u}{2},\frac{v}{2}\right).$

Let $M=(X,Y)$, so $u=\frac{2X}{3}$ and $v=2Y$. Substituting into the relation for $O$ gives

$p\cdot \frac{2X}{3}+q\cdot 2Y=\frac{p^2+q^2}{2}.$

Multiplying by $6$ yields

$2pX+12qY=3(p^2+q^2).$

Dividing by $2$ gives the linear equation

$pX+6qY=\frac{3(p^2+q^2)}{2}.$

This is the equation of a straight line. Hence the midpoint $M$ of $OC$ lies on a fixed line independent of the choice of the circle.

Since every admissible circle produces such a point $M$, and every solution of the derived linear equation corresponds to some admissible $O$ on the perpendicular bisector of $AB$ with $u\neq 0$, the locus is exactly this line.

Thus the locus of midpoints of segments $OC$ is the straight line

$pX+6qY=\frac{3(p^2+q^2)}{2}.$

This completes the proof. ∎

Verification of Key Steps

The crucial geometric step is the determination of $C$. On the circle centered at $O=(u,v)$, points on $l$ satisfy $(x-u)^2=u^2$. This quadratic equation has exactly two solutions, $x=0$ and $x=2u$. The point $x=0$ is $A$, since $A$ lies on both the circle and $l$, leaving $x=2u$ as the only other intersection. This exhausts all possibilities because a line and a circle intersect in at most two points.

The midpoint computation depends only on linear averaging and introduces no additional constraints. The transformation $(u,v)\mapsto (3u/2,v/2)$ is injective, so no degeneracy occurs when passing from $O$ to $M$.

The final elimination step replaces $u,v$ uniquely in terms of $X,Y$, so no hidden branches or extraneous solutions are introduced. Since the resulting equation is linear, no nonlinear locus can arise.

Alternative Approaches

A synthetic approach interprets the mapping $O\mapsto C$ as reflection of $O$ across the vertical line through the midpoint of the chord determined by $l$, followed by a dilation along directions determined by $A$. One can also interpret $M$ as the image of the perpendicular bisector of $AB$ under an affine transformation sending $O$ to $M$. This immediately implies that the locus of $M$ must be a line, since affine transformations preserve straight lines.