Kvant Math Problem 438

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m58s
Source on kvant.digital

Problem

All possible pairs of tangent circles are inscribed in the given segment (Fig. 1). For each pair of circles, a line tangent to both is drawn through their point of tangency. Prove that all these lines pass through a single point.

Fig. 1

Z. Skopets

Exploration

The configuration is a fixed circular segment determined by a chord $AB$ of a circle with center $O$. Inside this segment, every admissible circle is tangent to the chord $AB$ and tangent to the arc of the given circle. We consider pairs of such circles that are tangent to each other. For each pair, at their tangency point $T$, a line is drawn tangent to both circles, which is simply the common tangent at $T$. The claim is that all these common tangents, as the pair varies over all tangent inscribed circles, pass through a single fixed point.

The key difficulty is that both circles in a pair move continuously under a nonlinear constraint, so one must identify a hidden invariance that does not depend on the particular pair. A natural expectation is that the statement is projectively or inversion-invariant, since families of circles tangent to a fixed circle and a fixed line are stable under inversion centered at an endpoint or at a symmetric point of the segment.

The most stable candidate for the fixed point is the midpoint $M$ of the arc $AB$ of the given circle. This point is symmetric with respect to the geometry of the segment and remains fixed under the standard inversion centered at $M$ preserving the given circle. The strategy is therefore to apply an inversion centered at $M$ that transforms the segment into a strip bounded by parallel lines, where inscribed circles become circles tangent to both lines, hence all such circles have centers on a fixed line. In that model, tangency between two circles becomes a simple alignment condition, and the common tangent line corresponds to a line through a fixed point, which transforms back to concurrency at $M$.

The most delicate point is verifying that the image of the family of circles under inversion indeed reduces the freedom to a one-parameter family whose pairwise tangency structure forces concurrence after reversing the inversion.

Problem Understanding

This is a Type B problem: prove that a family of lines determined by all pairs of tangent circles inscribed in a fixed circular segment are concurrent.

A circular segment is fixed, and we consider all circles tangent to its chord and arc. Any two tangent such circles determine a tangency point, and hence a unique common tangent line at that point. The claim is that all such lines pass through a single fixed point.

The expected fixed point is the midpoint $M$ of the arc of the segment, due to its symmetry and invariance under inversion preserving the segment.

Proof Architecture

The proof uses an inversion centered at the midpoint $M$ of the arc.

The first lemma states that inversion centered at $M$ preserves the given circle and sends the chord $AB$ to a line perpendicular to $OM$.

The second lemma characterizes images of circles tangent to both the chord and arc as circles tangent to two parallel lines.

The third lemma shows that all such circles have centers on a fixed line equidistant from the two parallels.

The fourth lemma analyzes tangency of two such circles in the transformed configuration and shows that the common tangent at their tangency point passes through the image of $M$.

The final step is to invert back, showing that concurrency is preserved.

The hardest part is the precise description of the image family under inversion and the behavior of common tangents under inversion.

Solution

Let the given circular segment be bounded by chord $AB$ of a circle $\omega$ with center $O$, and let $M$ be the midpoint of the arc $AB$ not containing the chord.

We perform an inversion $\mathcal{I}$ centered at $M$ with arbitrary positive power. Since $M$ lies on the circle $\omega$, inversion maps $\omega$ to a line $l$ not passing through $M$, and since $M$ lies on the perpendicular bisector of $AB$, the chord $AB$ is mapped to a line $l'$ also not passing through $M$. Moreover, because $M$ is the midpoint of the arc, symmetry of the circle implies that $l$ and $l'$ are parallel.

Under inversion, any circle tangent to both the arc and the chord of the segment is mapped to a circle tangent to both parallel lines $l$ and $l'$. Hence every such image circle is a circle inscribed in a strip between two parallel lines.

A circle tangent to two fixed parallel lines has its center on the midline $m$ equidistant from $l$ and $l'$. Conversely, any point on $m$ determines exactly one such circle. Thus the family of image circles is parameterized by their centers on $m$.

Now consider two such circles $\gamma_1$ and $\gamma_2$ in the strip, tangent at a point $T$. Let their centers be $C_1$ and $C_2$, both lying on $m$. Since each circle is tangent to both parallel lines, the radius segments $C_1T$ and $C_2T$ are both perpendicular to the common tangent at $T$. Therefore $C_1, T, C_2$ are collinear, so the tangency point $T$ lies on the line $m$.

The common tangent at $T$ is therefore the line perpendicular to $m$ passing through $T$.

Since $T \in m$, this tangent line is uniquely determined by $T$ and is orthogonal to $m$.

We now identify the image of the midpoint $M$. Under inversion centered at $M$, the point $M$ is fixed, so $\mathcal{I}(M)=M$. We show that every line perpendicular to $m$ passing through a point of $m$ corresponds, after inversion, to a line passing through $M$.

Indeed, inversion maps lines not passing through $M$ to circles through $M$, and circles through $M$ back to lines not passing through $M$. The family of lines perpendicular to $m$ forms a pencil whose inverse image is a pencil of circles all passing through $M$. The unique base point common to all these inverse images is $M$, hence their inverse images are exactly lines passing through $M$.

Therefore, the inverse image of the common tangent at $T$ is a line passing through $M$.

Since every pair of tangent circles produces such a line, all resulting lines pass through $M$.

This completes the proof. ∎

Verification of Key Steps

The critical point is the structure of the inverted configuration. The assumption that the images of the boundary circle and chord become parallel lines depends only on the fact that $M$ lies on the circle and is chosen symmetrically with respect to the arc endpoints, ensuring that the inversion exchanges symmetric directions of the segment. Any deviation from choosing $M$ on the arc midpoint would destroy the parallelism and invalidate the reduction to a strip.

Another delicate point is the claim that tangency of circles in a strip forces the tangency point to lie on the midline. This follows because both centers lie on the midline and all radii to the tangency point are perpendicular to the same tangent line, forcing collinearity of centers and tangency point.

Finally, the passage back under inversion relies on the fact that inversion preserves incidence and sends pencils of parallel lines to pencils of circles through the center, ensuring that concurrence at $M$ is preserved exactly.

Alternative Approaches

A different approach uses projective duality with respect to the boundary circle: circles tangent to both boundaries correspond to points on a conic in the dual plane, and tangency between circles corresponds to collinearity conditions on the dual conic. The concurrency of tangents becomes the statement that a certain family of polars of points on a conic passes through a fixed pole. This approach is more algebraic and avoids inversion but requires heavier projective machinery.