Kvant Math Problem 1557

Let $O$ be the center of the given circle.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m57s
Source on kvant.digital

Problem

$A$ and $B$ are two given points on a circle. Find the locus of the midpoints of chords of this circle that intersect the segment $AB$.

I. F. Sharygin

Moscow LIX Mathematical Olympiad 1996, Tournament of Towns (Spring, 1996)

Exploration

Let $O$ be the center of the given circle. Any chord of a circle has the property that the segment from $O$ to its midpoint is perpendicular to the chord, so each midpoint $M$ determines a unique chord, namely the chord lying on the line through $M$ perpendicular to $OM$.

Thus the problem can be rephrased in terms of points $M$ inside the circle: such a point belongs to the locus if and only if the line through $M$ perpendicular to $OM$ intersects the segment $AB$.

A line intersects a segment if and only if the endpoints of the segment lie on opposite sides of the line or one endpoint lies on the line. Therefore the condition becomes a sign condition comparing the positions of $A$ and $B$ relative to the line through $M$ perpendicular to $OM$.

The boundary cases occur when this line passes through $A$ or through $B$, which translates into right angles involving $OA$, $OB$, and $OM$. This strongly suggests that circles with diameters $OA$ and $OB$ play a role, since right-angle conditions in a circle are characterized by Thales’ theorem.

The key is to convert the “intersection with $AB$” condition into a clean geometric inequality describing a region bounded by those two Thales circles.

Problem Understanding

This is a Type A problem.

We are given a fixed circle with center $O$ and two fixed points $A$ and $B$ on it. We consider all chords of the circle whose supporting lines intersect the segment $AB$, and we take the midpoints of all such chords. We are asked to describe the locus of these midpoints.

The midpoint of any chord uniquely determines that chord, so the problem reduces to describing which points $M$ inside the circle correspond to chords whose line intersects $AB$. The answer will be a region inside the circle bounded by arcs of circles determined by right-angle conditions involving $OA$ and $OB$.

Proof Architecture

The proof uses the following steps.

First, for any point $M$ inside the circle, the midpoint of the corresponding chord is characterized by the chord being perpendicular to $OM$.

Second, the chord through $M$ intersects the segment $AB$ if and only if $A$ and $B$ lie on opposite sides of the line through $M$ perpendicular to $OM$, which is equivalent to a sign condition involving oriented distances.

Third, this sign condition can be rewritten using scalar products, reducing the condition to inequalities involving angles $\angle OMA$ and $\angle OMB$.

Fourth, the boundary cases correspond to $\angle OMA = 90^\circ$ or $\angle OMB = 90^\circ$, which characterize circles with diameters $OA$ and $OB$.

The most delicate point is the equivalence between the intersection condition and the sign condition, since it requires careful handling of degenerate cases when the line passes through $A$ or $B$.

Solution

Let $O$ be the center of the given circle. For a point $M$ inside the circle, there exists exactly one chord having midpoint $M$, namely the chord lying on the line through $M$ perpendicular to $OM$, since in a circle the radius to the midpoint of a chord is perpendicular to the chord.

Denote this line by $\ell(M)$. Then $M$ belongs to the required locus if and only if the line $\ell(M)$ intersects the segment $AB$.

A line intersects a segment if and only if the endpoints of the segment lie on opposite sides of the line or at least one endpoint lies on the line. Hence the condition is that the points $A$ and $B$ do not lie strictly on the same side of $\ell(M)$.

Introduce a signed distance to $\ell(M)$. Since $\ell(M)\perp OM$, a normal vector to $\ell(M)$ is directed along $OM$. Therefore the signed distance from a point $X$ to $\ell(M)$ is proportional to the scalar product $(X-M)\cdot OM$. Consequently, $A$ and $B$ lie on opposite sides of $\ell(M)$ or one lies on it if and only if

$$((A-M)\cdot OM),((B-M)\cdot OM)\le 0.$$

We now interpret the factors geometrically. The expression $(A-M)\cdot OM$ is nonpositive precisely when the angle between $MA$ and $MO$ is at least $90^\circ$, since

$$(A-M)\cdot OM \le 0 \quad \Longleftrightarrow \quad \angle AMO \ge 90^\circ.$$

Similarly,

$$(B-M)\cdot OM \le 0 \quad \Longleftrightarrow \quad \angle BMO \ge 90^\circ.$$

Thus the condition becomes that $\angle AMO \ge 90^\circ$ or $\angle BMO \ge 90^\circ$ (allowing equality). Therefore the locus consists of all points $M$ inside the circle for which at least one of the angles $\angle AMO$ or $\angle BMO$ is right or obtuse.

By Thales’ theorem, the condition $\angle AMO = 90^\circ$ is equivalent to $M$ lying on the circle with diameter $AO$, and similarly $\angle BMO = 90^\circ$ is equivalent to $M$ lying on the circle with diameter $BO$. Moreover, the inequality $\angle AMO \ge 90^\circ$ describes the closed region inside the original circle bounded by the arc of the circle with diameter $AO$ connecting $A$ to the circle boundary, and similarly for $B$.

Hence the locus is the union of the two closed regions inside the given circle determined by the inequalities

$$\angle AMO \ge 90^\circ \quad \text{or} \quad \angle BMO \ge 90^\circ.$$

Equivalently, it is the set of all points $M$ in the given circle lying on or inside at least one of the circles with diameters $OA$ and $OB$.

Therefore the required locus is the union of the two circular regions inside the given circle bounded by the arcs of the circles with diameters $OA$ and $OB$.

Verification of Key Steps

The first delicate point is the uniqueness of the chord corresponding to a midpoint $M$. This follows because in a circle the perpendicular from the center to a chord bisects it, so any chord with midpoint $M$ must be perpendicular to $OM$, and conversely the line through $M$ perpendicular to $OM$ intersects the circle in a symmetric pair of points.

The second delicate point is the equivalence between intersection of $\ell(M)$ with $AB$ and the sign condition. If $A$ and $B$ lie strictly on the same side of $\ell(M)$, then the entire segment $AB$ lies in that half-plane and does not intersect the line. If they lie on opposite sides or one lies on the line, continuity of the segment ensures at least one intersection point, establishing equivalence.

The third delicate point is the conversion of the scalar product condition into the angular condition. Since $(A-M)\cdot OM=0$ characterizes perpendicularity of $MA$ and $MO$, the sign of the scalar product distinguishes acute and obtuse angles, yielding the stated equivalence without ambiguity.

Alternative Approaches

One can also describe the locus purely synthetically by fixing $M$ and studying the pencil of lines through $M$ perpendicular to $OM$, then characterizing those lines intersecting $AB$ via projective arguments on the circle. Another approach replaces scalar products with inversion centered at $M$, turning the perpendicularity condition into a simpler collinearity statement, though the resulting region description ultimately reduces to the same pair of Thales circles.