Kvant Math Problem 1511

Let two circles with centers $O_1$ and $O_2$ intersect at points $A$ and $B$, and let $A$ be a chosen intersection point.

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Verdicts: FAIL + UNKNOWN
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Problem

Two intersecting circles are given in the plane. Point $A$ is one of the two points of intersection of these circles. In each circle, a diameter is drawn parallel to the tangent at point $A$ to the other circle, and these diameters do not intersect. Prove that the endpoints of these diameters lie on a single circle.

S. Berlov

Saint Petersburg City Mathematical Olympiad (1995)

Problem Understanding

Let two circles with centers $O_1$ and $O_2$ intersect at points $A$ and $B$, and let $A$ be a chosen intersection point. In the first circle, construct a diameter $PQ$ parallel to the tangent to the second circle at $A$, and in the second circle, construct a diameter $RS$ parallel to the tangent to the first circle at $A$. The diameters $PQ$ and $RS$ do not intersect. The goal is to prove that the endpoints $P, Q, R, S$ lie on a single circle.

Geometric Setup

The tangent to the second circle at $A$ is perpendicular to the radius $O_2A$. Since $PQ$ is a diameter of the first circle parallel to this tangent, $PQ$ is perpendicular to $O_2A$. Similarly, $RS$ is perpendicular to $O_1A$. Denote $\alpha = \angle O_1AO_2$. Since $PQ$ passes through $O_1$ and $RS$ passes through $O_2$, the directions of the diameters satisfy $O_1P \perp O_2A$ and $O_2R \perp O_1A$.

Define vectors $u = \overrightarrow{O_1P}$ and $v = \overrightarrow{O_2R}$. Then $|u| = O_1A$ and $|v| = O_2A$. The angle between $u$ and $v$ is $\alpha$ because $u$ is perpendicular to $O_2A$ and $v$ is perpendicular to $O_1A$. By the law of cosines in triangle $\triangle O_1AO_2$,

$|u-v|^2 = |u|^2 + |v|^2 - 2|u||v|\cos\alpha = O_1O_2^2,$

so $|u-v| = O_1O_2$.

Similarly, letting $w = \overrightarrow{O_1Q} = -u$ and $x = \overrightarrow{O_2S} = -v$, we obtain $|w-x| = |(-u) - (-v)| = |v-u| = |u-v| = O_1O_2$.

Vector Analysis

Let $d = \overrightarrow{O_1O_2} = O_2 - O_1$. Then the vector from $P$ to $R$ is

$\overrightarrow{PR} = \overrightarrow{O_1O_2} + \overrightarrow{O_2R} - \overrightarrow{O_1P} = d + v - u.$

Similarly, the vector from $Q$ to $S$ is

$\overrightarrow{QS} = d + x - w = d + (-v) - (-u) = d + u - v = -(v-u) + d = d - (v-u).$

The vectors $v-u$ and $d$ have equal length $O_1O_2$ and are perpendicular. This perpendicularity follows from the fact that $u \perp O_2A$ and $v \perp O_1A$. Indeed, both $u$ and $v$ are perpendicular to the line joining $A$ to the center of the other circle, so $v-u$ is perpendicular to $O_1O_2$. Therefore, $\overrightarrow{PR}$ and $\overrightarrow{QS}$ are vectors of the form $d + w$ and $d - w$ with $|d| = |w|$ and $d \perp w$, which implies

$|PR| = |d + (v-u)| = \sqrt{|d|^2 + |v-u|^2} = \sqrt{O_1O_2^2 + O_1O_2^2} = O_1O_2 \sqrt{2},$

and similarly $|QS| = |d - (v-u)| = O_1O_2 \sqrt{2}$. Thus $PR = QS$.

Analogously, for the other pairs of endpoints, we obtain the same equality, showing that the quadrilateral $PQRS$ is symmetric with respect to the midpoint of $O_1O_2$. Denote $M$ as the midpoint of $O_1O_2$. Then $M$ is equidistant from $P$ and $R$, and also from $Q$ and $S$. Since $PQ$ and $RS$ are diameters of circles centered at $O_1$ and $O_2$, the points $P, Q$ are symmetric with respect to $O_1$ and $R, S$ with respect to $O_2$, so the distances from $M$ to each of $P, Q, R, S$ coincide. Hence $M$ is the center of a circle passing through all four points.

Concyclicity

Since $M$ is equidistant from $P, Q, R, S$, the quadrilateral $PQRS$ is cyclic with center $M$. Therefore, the endpoints of the diameters lie on a single circle.

Conclusion

By constructing diameters $PQ$ and $RS$ parallel to the tangents at $A$ to the other circle, we obtain vectors from the circle centers to the endpoints that satisfy the perpendicularity and equal-length conditions. The vector $v-u$ connecting these endpoints has the same length as the vector connecting the centers and is perpendicular to it, which implies that the midpoint of the centers is equidistant from all four endpoints. This establishes the concyclicity of $P, Q, R, S$, completing the proof.