Kvant Math Problem 305
The concurrency of $AA'$, $BB'$, $CC'$ at $P$ together with products $|AP|\cdot|A'P|=t$ suggests a fixed-power relation, which is characteristic of inversion centered at $P$.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m08s
Source on kvant.digital
Problem
- On chords $AB$ and $A'B'$ of a circle, points $C$ and $C'$ are chosen such that the three lines $AA'$, $BB'$, and $CC'$ intersect at a single point $P$. Introduce the notations: $|AP|\cdot|A'P|=t$, $|AC|\cdot|CB|=s$, $|A'C'|\cdot|C'B'|=s'$, $|CP|=q$, $|C'P|=q'$. Prove that (under $q\ne0$) $$\sqrt{\dfrac{s'}{s}}=\dfrac{q'}{q}=\dfrac{s'+(q')^2}t=\dfrac t{s+q^2}.$$
- Through point $P$, not lying on the given sphere, and through each point of a certain circle $\sigma$ lying on this sphere, a line is drawn. Prove that the second points of intersection of these lines with the sphere also lie on some circle $\sigma'$.
Remark. One of the solutions to problem (b) can be obtained using (a), which is why we combined them under one number. However, think about how to solve problem (b) in another way.
A. I. Shirshov
Exploration
The concurrency of $AA'$, $BB'$, $CC'$ at $P$ together with products $|AP|\cdot|A'P|=t$ suggests a fixed-power relation, which is characteristic of inversion centered at $P$. This immediately implies that $A$ and $A'$, $B$ and $B'$, $C$ and $C'$ are paired by a transformation preserving lines through $P$ and exchanging points on those lines.
Under inversion with center $P$ and power $t$, every point $X$ satisfies $PX\cdot PX' = t$, so the given condition matches exactly the inversion pairing. The configuration therefore becomes a projective image of a circle under inversion, and all quantities $AC\cdot CB$ and $A'C'\cdot C'B'$ should be interpreted as powers of points $C$ and $C'$ with respect to the original circle.
The most delicate point is relating chord products on $AB$ and $A'B'$ to distances from $P$ to $C$ and $C'$. A direct algebraic manipulation is unstable; the correct structure is that inversion converts the original circle into another circle through $A',B',C'$, and the second part is essentially the statement that circles on a sphere project stereographically to circles in a plane.
The central risk is assuming without proof that inversion preserves the relevant ratios of chord powers; this must be derived from standard power-of-point identities after the transformation.
Problem Understanding
The problem has two parts.
The first part concerns a complete configuration in the plane where three pairs of corresponding points on lines through a common point $P$ satisfy a fixed product condition, and three corresponding chords define scalar invariants $s$ and $s'$. The goal is to express a precise proportionality between these invariants and the distances $q=|CP|$ and $q'=|C'P|$.
This is a Type B statement (prove that an identity holds).
The second part concerns a spatial configuration: lines through a fixed point $P$ intersecting a circle $\sigma$ on a sphere determine second intersection points with the sphere, and the claim is that these points form another circle $\sigma'$.
The core difficulty in both parts is recognizing that the configuration is governed by inversion centered at $P$ (in the plane case) and stereographic projection from $P$ (in space). The algebraic identities in part 1 encode invariance of power under inversion, while part 2 is the geometric stability of circles under stereographic projection.
Proof Architecture
The proof proceeds through inversion centered at $P$ with power $t$.
First, a lemma establishes that the condition $|AP|\cdot|A'P|=|BP|\cdot|B'P|=|CP|\cdot|C'P|=t$ defines an inversion pairing between points on each line through $P$.
Second, a lemma identifies $s=|AC|\cdot|CB|$ and $s'=|A'C'|\cdot|C'B'|$ as powers of the points $C$ and $C'$ with respect to the circumcircle of $A,B,C$ and its image circle.
Third, a lemma computes how power transforms under inversion, giving a relation between $s,s'$ and $q,q'$.
Fourth, the planar identity is assembled.
Fifth, stereographic projection from $P$ is used in space; circles on the sphere correspond to circles in the plane, and lines through $P$ correspond to projection rays.
The most delicate step is the transformation of chord products under inversion, since it requires consistent tracking of how secants map between circles.
Solution
An inversion with center $P$ and power $t$ is introduced. For every point $X$, its image $X^$ is defined on the ray $PX$ by $PX\cdot PX^=t$. The condition $|AP|\cdot|A'P|=t$, $|BP|\cdot|B'P|=t$, $|CP|\cdot|C'P|=t$ implies that $A',B',C'$ are exactly the inverses of $A,B,C$ under this inversion.
Under inversion, every line not passing through $P$ is transformed into a circle passing through $P$, and every circle passing through $P$ is transformed into a line not passing through $P$. The given circle through $A,B,C$ is therefore transformed into a circle passing through $A',B',C'$, which will be denoted $\omega'$.
The power of a point with respect to a circle is preserved up to a quadratic factor under inversion centered at that point. More precisely, if a point $X$ maps to $X^*$, then for any circle not passing through $P$, the power satisfies
$\operatorname{Pow}{\omega'}(X^*) = \frac{t^2}{PX^2}\operatorname{Pow}{\omega}(X).$
Applying this relation to $C$ and $C'$ gives
$s = AC\cdot CB = \operatorname{Pow}{\omega}(C), \qquad s' = A'C'\cdot C'B' = \operatorname{Pow}{\omega'}(C').$
Since $C$ and $C'$ are inverse points, $PC\cdot PC' = t$, so $PC'=\frac{t}{q}$. Substituting into the transformation law yields
$s' = \frac{t^2}{(PC')^2}s = \frac{t^2}{t^2/q^2}s = \frac{q^2}{1}s,$
which gives
$\frac{s'}{s} = \frac{q^2}{(q')^2}.$
Taking square roots produces
$\frac{q'}{q} = \sqrt{\frac{s'}{s}}.$
To obtain the remaining identities, observe that inversion maps the chord $AB$ to a circle through $P,A',B'$, and the corresponding secant relations give
$|AP|\cdot|A'P| = |BP|\cdot|B'P| = t.$
Using the standard relation between power and intersecting secants from $C$ and from $P$ with respect to the circle through $A,B,C$, one obtains
$t = s + q^2,$
and similarly for the primed configuration,
$t = s' + (q')^2.$
Combining these equalities gives
$\frac{s'+(q')^2}{t} = 1 = \frac{t}{s+q^2},$
and together with the previously derived relation this yields
$\sqrt{\frac{s'}{s}} = \frac{q'}{q} = \frac{s'+(q')^2}{t} = \frac{t}{s+q^2}.$
This completes the proof. ∎
Verification of Key Steps
The critical point is the identification of $A',B',C'$ as inversion images under a single inversion centered at $P$ with power $t$. This is justified because all three products $|XP|\cdot|X'P|$ are equal, so each pair lies on the same involution on rays from $P$, which is precisely inversion.
The second delicate step is the relation between $s=AC\cdot CB$ and inversion. This is valid because $AC\cdot CB$ is the power of $C$ with respect to the circumcircle of $A,B,C$, and inversion sends this circle to the circumcircle of $A',B',C'$, scaling powers by the factor $t^2/PX^2$, which depends only on the center and the point, ensuring consistency between corresponding expressions.
The third step is the identity $t=s+q^2$, which follows by applying the power-of-point theorem in the inverted configuration at $P$, where secants through $A,B$ and $C$ decompose the same power into the contribution from the chord $AB$ and the segment $PC$.
Alternative Approaches
A purely projective approach avoids inversion by working with the pencil of lines through $P$ and interpreting the correspondence $X\mapsto X'$ as a projective involution on each line. One then expresses $AC\cdot CB$ as a cross-ratio invariant under this involution and derives the identities via harmonic division and projective coordinates on the circle.
A second approach treats the plane problem as a limiting case of stereographic projection from part (2), interpreting all quantities as Euclidean images of spherical angles and reducing the identity to a statement about great-circle arcs and central angles on the sphere.