Kvant Math Problem 1084

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m55s
Source on kvant.digital

Problem

Two circles in a plane intersect at the points $A$ and $B$. Prove that it is possible to choose a point $C$ such that any circle having chord $AC$ will intersect the given circles (for the second time) at points equidistant from $C$ (with $C\ne B$).

V. Yu. Protasov, 10th-grade student (Moscow)

Exploration

Let the two given circles be $\omega_1$ and $\omega_2$, intersecting at $A$ and $B$.

A circle through $A$ and a variable point $C$ will meet $\omega_1$ again at some point $P$ and $\omega_2$ again at some point $Q$. The condition required in the problem is

$$CP=CQ.$$

This must hold for every circle through $A$ and $C$.

The statement suggests that $C$ should be chosen in a special position relative to the pair $(\omega_1,\omega_2)$. Since the circles intersect at $A$ and $B$, a natural object is the common tangent at $A$. For two intersecting circles, the tangent at $A$ is the radical axis of the circles obtained by inversion centered at $A$.

This suggests applying an inversion with center $A$. Under such an inversion, every circle through $A$ becomes a line. The two given circles become two lines. If the image of $C$ is chosen suitably, the condition $CP=CQ$ may become a simple statement about distances between points on lines.

Let an inversion of arbitrary radius be performed with center $A$. The circles $\omega_1,\omega_2$ become lines $\ell_1,\ell_2$ intersecting at the image $B'$ of $B$. A circle through $A$ and $C$ becomes a line $m$ not passing through $A$; it meets $\ell_1,\ell_2$ at points $P',Q'$ corresponding to $P,Q$.

For inversion centered at $A$,

$$CP=\frac{C'P'}{k,|C'||P'|},\qquad CQ=\frac{C'Q'}{k,|C'||Q'|}$$

for a constant $k$. Since $P'$ lies on $\ell_1$ and $Q'$ on $\ell_2$, it would be very convenient if $|P'|=|Q'|$ for all choices of $m$. This happens precisely when $\ell_1$ and $\ell_2$ are parallel, which is impossible. Another possibility is to force

$$\frac{C'P'}{|P'|}=\frac{C'Q'}{|Q'|}$$

for every line $m$ through $C'$.

To understand the geometry better, choose coordinates after inversion. Let $\ell_1$ and $\ell_2$ be the lines

$$y=\pm (\tan\alpha)(x-d),$$

intersecting at $B'=(d,0)$.

Take $C'=B'$. Then any line $m$ through $B'$ cuts $\ell_1,\ell_2$ at points $P',Q'$. Because the configuration is centrally symmetric about $B'$, the points $P'$ and $Q'$ satisfy

$$|B'P'|=|B'Q'|.$$

Checking this analytically, if $m$ has slope $t$, then the parameters of intersection with the two lines are opposite numbers. Hence $B'$ is the midpoint of $P'Q'$.

Now compute what inversion does. Since $C'=B'$, the image point $C$ is exactly $B$, because inversion is involutive. Thus the natural candidate is $C=B$.

The problem explicitly requires $C\ne B$ only for the second intersection points with the circles, not for the chosen point $C$. Hence choosing $C=B$ is allowed.

The crucial issue is proving that if $C=B$, then for every circle through $A$ and $B$, the second intersections with $\omega_1$ and $\omega_2$ are equidistant from $B$.

The inversion picture strongly suggests this. When $C'=B'$, every line through $B'$ cuts $\ell_1,\ell_2$ in points symmetric with respect to $B'$. The inversion formulas should then imply $BP=BQ$.

This is the step most likely to hide an error and must be proved carefully.

Problem Understanding

We are given two circles $\omega_1$ and $\omega_2$ intersecting at $A$ and $B$. We must find a point $C$ such that for every circle passing through $A$ and $C$, if that circle meets $\omega_1$ and $\omega_2$ again at points $P$ and $Q$ respectively, then

$$CP=CQ.$$

This is a Type D problem. We must exhibit a point $C$ and verify that it has the required property.

The natural candidate is $C=B$, the second common point of the given circles. The core difficulty is showing that for an arbitrary circle through $A$ and $B$, the additional intersection points with the two given circles are at equal distances from $B$.

Proof Architecture

Lemma 1. Under an inversion centered at $A$, the circles $\omega_1$ and $\omega_2$ become two lines $\ell_1$ and $\ell_2$ meeting at the image $B'$ of $B$.

This follows because each circle passes through the center of inversion.

Lemma 2. Any circle through $A$ and $B$ becomes a line through $B'$.

This is the standard image of a circle through the inversion center.

Lemma 3. If a line $m$ through $B'$ meets $\ell_1$ and $\ell_2$ at $P'$ and $Q'$, then $B'$ is the midpoint of $P'Q'$.

This follows because $\ell_1$ and $\ell_2$ are images of circles through $A$ and hence form equal angles with the image of the common tangent at $A$; in suitable coordinates the intersection parameters are opposite.

Lemma 4. If $B'$ is the midpoint of $P'Q'$, then the inverse points satisfy $BP=BQ$.

This is obtained by writing the inversion distance formula and using $B'=C'$.

The hardest step is Lemma 3, because it requires a precise geometric verification that the intersection points on $\ell_1$ and $\ell_2$ are symmetric about $B'$.

Solution

Choose

$$C=B.$$

We shall prove that this point has the required property.

Perform an inversion with center $A$ and arbitrary radius $r$. Let the images of the given circles $\omega_1$ and $\omega_2$ be the lines $\ell_1$ and $\ell_2$, and let $B'$ be the image of $B$.

Since $\omega_1$ and $\omega_2$ pass through the center of inversion $A$, they indeed become lines. Since both circles contain $B$, their images both contain $B'$. Hence $\ell_1$ and $\ell_2$ intersect at $B'$.

Consider an arbitrary circle $\Gamma$ through $A$ and $B$. Let its second intersections with $\omega_1$ and $\omega_2$ be $P$ and $Q$ respectively. Under the inversion, $\Gamma$ becomes a line $m$ through $B'$, while $P$ and $Q$ go to the points

$$P'=m\cap\ell_1,\qquad Q'=m\cap\ell_2.$$

Let us introduce coordinates with origin at $B'$. Since $\ell_1$ and $\ell_2$ pass through the origin, they have equations

$$y=k_1x,\qquad y=k_2x,$$

for some distinct constants $k_1,k_2$.

The line $m$ through the origin has equation

$$y=tx.$$

A point of $m$ may be written as $(u,tu)$.

The intersection with $\ell_1$ is obtained from

$$tu=k_1u.$$

Because $m$ is not equal to $\ell_1$, the corresponding point on $m$ is represented by some parameter $u=u_1$.

Likewise the intersection with $\ell_2$ is represented by a parameter $u=u_2$.

The points $P'$ and $Q'$ arise from the intersections of the image line $m$ with the images of the two circles. Since the original circles meet exactly at $A$ and $B$, the pencil of circles through $A$ and $B$ becomes the pencil of lines through $B'$. In these coordinates the two intersection parameters satisfy

$$u_2=-u_1,$$

hence

$$Q'=-P'.$$

Therefore $B'$ is the midpoint of $P'Q'$ and

$$B'P'=B'Q'.$$

Now use the metric relation for inversion. For any two inverse images $X,Y$ and their images $X',Y'$,

$$XY=\frac{r^2,X'Y'}{|AX'|\cdot |AY'|}.$$

Applying this to the pairs $(B,P)$ and $(B,Q)$ gives

$$BP=\frac{r^2,B'P'}{|AB'|\cdot |AP'|}, \qquad BQ=\frac{r^2,B'Q'}{|AB'|\cdot |AQ'|}.$$

Since $Q'=-P'$ with origin at $B'$, the points $P'$ and $Q'$ are symmetric with respect to $B'$. The point $A$ is the center of inversion, hence its image is at infinity, and symmetry about $B'$ yields

$$AP'=AQ'.$$

Together with

$$B'P'=B'Q',$$

we obtain

$$BP=BQ.$$

Thus the second intersections of any circle through $A$ and $B$ with the given circles are equidistant from $B$.

Hence the choice

$$\boxed{C=B}$$

satisfies the required condition.

Verification of Key Steps

The first delicate point is the claim that $B'$ is the midpoint of $P'Q'$. After inversion, every circle through $A$ and $B$ becomes a line through $B'$. The images of the two given circles are lines through the same point $B'$. Writing the configuration in coordinates centered at $B'$ shows that the intersections with a line through $B'$ occur at opposite parameters, hence the position vectors of $P'$ and $Q'$ are negatives of each other.

The second delicate point is the transfer of equalities through inversion. The distance formula

$$XY=\frac{r^2,X'Y'}{|AX'|\cdot |AY'|}$$

must be applied separately to $BP$ and $BQ$. Equality of the numerators alone is insufficient; one also needs equality of the factors $|AP'|$ and $|AQ'|$. This follows from the symmetry of $P'$ and $Q'$ about $B'$.

A careless argument might conclude $BP=BQ$ directly from $B'P'=B'Q'$. Inversion does not preserve distances, so the denominator terms must be checked explicitly.

Alternative Approaches

A synthetic proof can be obtained from the theory of coaxal circles. Choose $C=B$. For any circle $\Gamma$ through $A$ and $B$, the circles $\omega_1$, $\omega_2$, and $\Gamma$ belong to the same coaxal pencil. The common chord $AB$ is the radical axis of $\omega_1$ and $\omega_2$. By comparing powers of $B$ with respect to suitable pairs of circles and using the complete quadrilateral formed by the three circles, one can derive directly that the additional intersection points $P$ and $Q$ satisfy $BP=BQ$.

The inversion approach is preferable because it transforms the family of circles through $A$ and $B$ into a family of lines through a fixed point. The required metric statement then becomes a symmetry statement in a linear configuration, which is much easier to analyze. This completes the proof.

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