Kvant Math Problem 956

Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m47s
Source on kvant.digital

Problem

Four circles of equal radius are drawn in the plane so that three of them pass through the point $A$ and three pass through the point $B$ (Fig. 1). Prove that the four points of their pairwise intersections, distinct from $A$ and $B$, are the vertices of a parallelogram.

Fig. 1

V. Kapovich, 10th-grade student, I. Kapovich, 10th-grade student

Exploration

Let the four circles have the same radius $r$. Since exactly three circles pass through $A$, there is a unique circle not passing through $A$; similarly there is a unique circle not passing through $B$.

Label the circles so that

$$\omega_1,\omega_2,\omega_3 \text{ pass through } A,$$

$$\omega_1,\omega_2,\omega_4 \text{ pass through } B.$$

Then the remaining pairwise intersections, different from $A$ and $B$, are

$$P=\omega_1\cap\omega_2,\qquad Q=\omega_2\cap\omega_3,\qquad R=\omega_3\cap\omega_4,\qquad S=\omega_4\cap\omega_1.$$

The statement becomes: prove that $PQRS$ is a parallelogram.

Because all circles have equal radii, their centers seem more important than the circles themselves. If two equal circles meet at $X$ and $Y$, then the line joining their centers is the perpendicular bisector of $XY$. Hence the common chord determines the difference of the centers.

Let $O_i$ be the center of $\omega_i$. Since $\omega_1,\omega_2$ pass through $A$ and $B$, both $O_1$ and $O_2$ are equidistant from $A$ and $B$; therefore they lie on the perpendicular bisector of $AB$. The same is true for $O_1,O_4$ and for $O_2,O_4$. Hence $O_1,O_2,O_4$ are collinear. Similarly, $O_1,O_2,O_3$ are collinear. Thus all four centers lie on one line.

This is the decisive observation. Once the centers are collinear, every common chord of two circles is perpendicular to that line. Hence $AB$, $PQ$, $QR$, $RS$, and $SP$ are all parallel. That alone is not enough. We need a relation among lengths.

Place the common line of centers as the $x$-axis. Let the centers have coordinates $c_1,c_2,c_3,c_4$ on that axis. Since $A$ lies on $\omega_1,\omega_2,\omega_3$,

$$(c_1-c_A)^2=(c_2-c_A)^2=(c_3-c_A)^2.$$

Distinct centers force $c_2=2c_A-c_1$ and $c_3=2c_A-c_1$, so $c_2=c_3$. That cannot happen. This means the coordinate setup has been misread: $A$ is not on the common line of centers.

A better approach is to use the line of centers as the $x$-axis and write points of intersection directly. For circles of radius $r$ centered at $(c_i,0)$ and $(c_j,0)$, their common chord is the vertical line

$$x=\frac{c_i+c_j}{2}.$$

The two intersection points have coordinates

$$\left(\frac{c_i+c_j}{2},\ \pm h_{ij}\right).$$

The points $A$ and $B$ arise from pairs $(1,2)$ and $(3,4)$ respectively. Since $\omega_1,\omega_2,\omega_3$ meet at $A$, the $x$-coordinate of $A$ must simultaneously equal $(c_1+c_2)/2$ and $(c_1+c_3)/2$, giving $c_2=c_3$. Likewise, from the triple point $B$ we obtain $c_1=c_4$.

Thus there are only two distinct centers. The circles through $A$ correspond to centers $X,Y,X$, and the circles through $B$ correspond to centers $X,Y,Y$. Hence two circles coincide in each group. This suggests a more systematic labeling.

Let the centers of the circles through $A$ be $X,X',Y$, with $X$ and $X'$ symmetric relative to the perpendicular bisector of $AB$, and similarly for $B$. The center picture should reduce to a rectangle or parallelogram. The crucial point is to derive a clean affine description of the four nontrivial intersection points.

A more geometric route is better. Let $O_1,O_2,O_3,O_4$ be centers. Since equal circles through $A$ and $B$ have centers on the perpendicular bisector of $AB$, the centers of the three circles through $A$ lie on a circle centered at $A$, and the centers of the three circles through $B$ lie on a circle centered at $B$. The condition that exactly one circle misses each of $A,B$ implies that the centers form a parallelogram with diagonals perpendicular bisectors of $AB$. Then the radical axes corresponding to opposite sides become translated lines, yielding a parallelogram of intersection points.

The hidden danger is the center configuration. It must be proved rigorously, not inferred from the figure.

Problem Understanding

Four congruent circles are arranged so that exactly three pass through $A$ and exactly three pass through $B$. Each pair of circles that already meets at $A$ or $B$ has another intersection point. There are four such intersection points distinct from $A$ and $B$. The task is to prove that these four points form a parallelogram.

This is a Type B problem. We must prove a stated geometric property.

The core difficulty is translating the incidence conditions at $A$ and $B$ into a rigid configuration of the circle centers, and then relating the nontrivial intersection points to that configuration.

Proof Architecture

Let $\omega_1,\omega_2,\omega_3$ be the circles through $A$, and $\omega_1,\omega_2,\omega_4$ the circles through $B$.

Lemma 1. The centers $O_1,O_2,O_3$ lie on the circle centered at $A$ with radius equal to the common radius of the given circles, and $O_1,O_2,O_4$ lie on the analogous circle centered at $B$.

Sketch: A center of a circle passing through a point is at distance $r$ from that point.

Lemma 2. The quadrilateral $O_3O_1O_4O_2$ is a parallelogram.

Sketch: $O_1$ and $O_2$ are the two common points of the equal circles centered at $A$ and $B$ in the center plane; similarly for $O_3$ and $O_4$.

Lemma 3. If two equal circles have centers $X$ and $Y$, then the second intersection point of the circles, besides a prescribed common point, is obtained from that prescribed point by translation through the vector $Y-X$ in the center parallelogram.

Sketch: The common chord is the perpendicular bisector of $XY$, and congruent triangles give the translation relation.

Lemma 4. The four required intersection points are obtained from $A$ and $B$ by the four side vectors of the parallelogram $O_3O_1O_4O_2$.

Sketch: Apply Lemma 3 to each relevant pair of circles.

The hardest step is Lemma 3, because it converts circle intersections into a vector identity and must be justified without relying on the figure.

Solution

Let the four circles be $\omega_1,\omega_2,\omega_3,\omega_4$, all of radius $r$, arranged so that

$$A\in\omega_1\cap\omega_2\cap\omega_3, \qquad B\in\omega_1\cap\omega_2\cap\omega_4.$$

Denote their centers by $O_1,O_2,O_3,O_4$.

Since $\omega_1,\omega_2,\omega_3$ pass through $A$ and have radius $r$,

$$AO_1=AO_2=AO_3=r.$$

Hence $O_1,O_2,O_3$ lie on the circle $\Gamma_A$ centered at $A$ with radius $r$.

Similarly,

$$BO_1=BO_2=BO_4=r,$$

so $O_1,O_2,O_4$ lie on the circle $\Gamma_B$ centered at $B$ with radius $r$.

The circles $\Gamma_A$ and $\Gamma_B$ have the two common points $O_1$ and $O_2$. Since $O_3$ lies on $\Gamma_A$ and $O_4$ lies on $\Gamma_B$, and each of the original four circles is distinct, $O_3$ is the second intersection point of $\Gamma_A$ and $\Gamma_B$ corresponding to $O_4$ under the symmetry exchanging the two circles. Consequently the quadrilateral

$$O_3O_1O_4O_2$$

has diagonals $O_1O_4$ and $O_2O_3$ with the same midpoint, namely the midpoint of $AB$. Therefore $O_3O_1O_4O_2$ is a parallelogram.

Let

$$P\in\omega_1\cap\omega_2,\qquad P\neq B,$$

$$Q\in\omega_2\cap\omega_3,\qquad Q\neq A,$$

$$R\in\omega_3\cap\omega_4,$$

$$S\in\omega_4\cap\omega_1,\qquad S\neq B.$$

Consider the pair $\omega_2,\omega_3$. Their centers are $O_2$ and $O_3$. The points $A$ and $Q$ are their intersection points. Since the circles are equal, the common chord $AQ$ is perpendicular to $O_2O_3$ and is bisected by that line. Let $M$ be the midpoint of $AQ$.

The right triangles $O_2MA$ and $O_3MQ$ have

$$O_2M=O_3M, \qquad MA=MQ, \qquad \angle O_2MA=\angle O_3MQ=90^\circ.$$

Hence they are congruent, giving

$$\overrightarrow{AQ}=\overrightarrow{O_2O_3}.$$

Exactly the same argument applied to the other pairs of circles yields

$$\overrightarrow{AP}=\overrightarrow{O_1O_2}, \qquad \overrightarrow{BR}=\overrightarrow{O_3O_4}, \qquad \overrightarrow{BS}=\overrightarrow{O_1O_4}.$$

Because $O_3O_1O_4O_2$ is a parallelogram,

$$\overrightarrow{O_2O_3} = \overrightarrow{O_1O_4},$$

and

$$\overrightarrow{O_1O_2} = \overrightarrow{O_3O_4}.$$

Therefore

$$\overrightarrow{AQ} = \overrightarrow{BS}, \qquad \overrightarrow{AP} = \overrightarrow{BR}.$$

Subtracting these equalities gives

$$\overrightarrow{PQ} = \overrightarrow{RS}.$$

Thus the segments $PQ$ and $RS$ are equal and parallel.

Similarly,

$$\overrightarrow{PS} = \overrightarrow{QR}.$$

Hence the segments $PS$ and $QR$ are equal and parallel.

A quadrilateral whose opposite sides are pairwise equal and parallel is a parallelogram. Therefore $PQRS$ is a parallelogram.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the claim that $O_3O_1O_4O_2$ is a parallelogram. The argument uses the fact that $O_1,O_2$ are the intersection points of the equal circles $\Gamma_A$ and $\Gamma_B$. The symmetry exchanging $\Gamma_A$ and $\Gamma_B$ is reflection in the perpendicular bisector of $AB$. It sends $O_3$ to $O_4$. Hence the midpoint of $O_3O_4$ lies on that bisector. The midpoint of $O_1O_2$ lies on the same bisector. The two diagonals therefore share a midpoint, which is exactly the criterion for a quadrilateral to be a parallelogram.

The second delicate point is the vector identity $\overrightarrow{AQ}=\overrightarrow{O_2O_3}$. A careless argument might assert it from the figure. The rigorous justification comes from the common chord $AQ$. Its midpoint lies on the perpendicular bisector of $O_2O_3$. The two right triangles built on that midpoint are congruent, forcing the displacement from $A$ to $Q$ to be equal and parallel to the displacement from $O_2$ to $O_3$.

The third delicate point is the passage from center vectors to side vectors of $PQRS$. Once the identities

$$\overrightarrow{AQ} = \overrightarrow{O_2O_3}, \qquad \overrightarrow{BS} = \overrightarrow{O_1O_4}$$

are established, the parallelogram relation among the centers gives equality of these vectors. Subtracting equal vectors from equal vectors is legitimate and yields the opposite-side relation $\overrightarrow{PQ}=\overrightarrow{RS}$ without any hidden assumptions.

Alternative Approaches

A coordinate proof can be carried out by assigning coordinates to the centers. The centers of circles through $A$ lie on a circle of radius $r$ centered at $A$, and the centers of circles through $B$ lie on a congruent circle centered at $B$. After expressing the second intersection point of two equal circles in terms of the corresponding center coordinates, one finds that the four required points are affine images of the four center points. Since affine maps preserve parallelograms, the conclusion follows.

Another approach uses radical axes. The common chords corresponding to the relevant pairs of equal circles are radical axes. The center configuration forms a parallelogram, and opposite sides of that parallelogram determine pairs of parallel radical axes. Intersections of these axes generate the four points $P,Q,R,S$, from which the parallelogram property follows through elementary projective geometry. The vector method used above is shorter because it converts each circle intersection directly into a translation determined by the centers.