IMO 2010 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m58s

Problem

Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Author: Marcin E. Kuczma, Poland

Exploration

Let $\Gamma$ be the circumcircle of $ABC$, and let $K,L,M$ be the second intersections of $AP,BP,CP$ with $\Gamma$. The condition $SC=SP$ suggests a circle centered at $S$ passing through $C$ and $P$, which makes $S$ a candidate for a radical-axis type configuration involving $\Gamma$ and this new circle.

Since $S$ lies on $AB$ and is defined via the tangent at $C$, the classical power-of-a-point relation gives $SA \cdot SB = SC^2$. Combining this with $SC=SP$ yields $SA \cdot SB = SP^2$, which suggests that $S$ has equal power with respect to the circumcircle $\Gamma$ and the circle centered at $P$ through $S$, or equivalently that $S$ lies on a radical axis of two circles naturally connected to $P$ and $\Gamma$.

The desired conclusion $MK=ML$ is symmetric in $K$ and $L$ and refers to equal chords in $\Gamma$, so it is equivalent to showing that $M$ lies on the perpendicular bisector of $KL$, or that $M$ is equidistant from $K$ and $L$. Since $K,L,M$ are intersections of lines through $P$ with $\Gamma$, a natural approach is to convert the statement into an equality of angles subtending equal arcs, or into a projective involution on $\Gamma$ induced by the pencil at $P$.

The most promising route is to interpret $S$ as a power center forcing a harmonic relation between $A,B,C,P$ on $\Gamma$ after inversion or via spiral similarity at $M$. The central difficulty is linking the metric condition $SC=SP$ to a projective symmetry on $\Gamma$ that forces $MK=ML$.

Problem Understanding

This is a Type B problem: a conditional geometric implication must be proven. We are given a triangle $ABC$ with circumcircle $\Gamma$, a point $P$ inside the triangle, and intersections $K,L,M$ of lines $AP,BP,CP$ with $\Gamma$ again. The tangent at $C$ meets $AB$ at $S$, and we assume the metric condition $SC=SP$. The goal is to prove that $MK=ML$.

Geometrically, $K,L,M$ are the antipodal-type projections of $P$ onto the circumcircle along the cevians $AP,BP,CP$. The condition $SC=SP$ ties $P$ to the tangent geometry at $C$, strongly suggesting a hidden symmetry centered at $S$ relating $P$ and $C$ relative to $\Gamma$.

The difficulty is that $K$ and $L$ come from different cevians, so any equality involving $MK$ and $ML$ requires a global symmetry of the circle induced indirectly by $S$ and $P$, rather than a direct angle chase.

Proof Architecture

First, we will establish the classical tangent-secant power relation $SA \cdot SB = SC^2$, and combine it with the hypothesis $SC=SP$ to deduce $SA \cdot SB = SP^2$.

Second, we introduce the circle $\omega$ with center $P$ passing through $S$, so $PS=PC$ will not be assumed but instead $PS=SC$ gives $S \in \omega$. We will show that $A$ and $B$ have equal power with respect to $\omega$ in a way that forces a projective involution on line $AB$ with fixed structure determined by $C$.

Third, we prove that the pencil of lines through $P$ induces a projective involution on $\Gamma$ pairing $K$ and $L$ symmetrically with respect to a diameter through $M$.

Fourth, we show that this involution fixes $M$ in the sense that $M$ lies on the symmetry axis of the chord $KL$, implying $MK=ML$.

The most delicate step is the transition from the metric condition at $S$ to a projective involution on $\Gamma$ involving $P$, since it requires coordinating power-of-point relations between two circles.

Solution

Let $\Gamma$ be the circumcircle of $ABC$. Let the tangent at $C$ meet $AB$ at $S$. By the classical tangent-secant theorem applied to point $S$ with respect to $\Gamma$, one has

$SA \cdot SB = SC^2.$

The hypothesis $SC = SP$ transforms this identity into

$SA \cdot SB = SP^2.$

Define $\omega$ as the circle centered at $P$ with radius $PS$. Since $SP = SC$, the point $C$ lies on $\omega$.

Lemma 1

For the fixed point $S$ on line $AB$, the equality $SA \cdot SB = SP^2$ implies that $A$ and $B$ are inverse points with respect to the circle $\omega$.

Proof. The condition $SA \cdot SB = SP^2$ is equivalent to the statement that the power of $S$ with respect to $\omega$ equals $SA \cdot SB$. Since inversion in $\omega$ sends a point $X$ on line $AB$ to the unique point $X'$ on $AB$ satisfying $SX \cdot SX' = SP^2$, it follows that $A$ and $B$ are exchanged by inversion in $\omega$. ∎

This establishes that $\omega$ induces a harmonic symmetry on line $AB$, and any geometric construction involving $P$ and $\Gamma$ must respect this inversion symmetry centered at $S$.

Lemma 2

The inversion with respect to $\omega$ maps the circumcircle $\Gamma$ to itself.

Proof. The circle $\Gamma$ passes through $A$ and $B$, which are exchanged by inversion in $\omega$. The image of a circle under inversion is a circle or line, and since $\Gamma$ does not pass through the center $P$ of inversion, its image is a circle passing through the images of $A$ and $B$, namely $B$ and $A$. Hence the image coincides with $\Gamma$. ∎

This establishes that inversion in $\omega$ preserves $\Gamma$ as a set.

Lemma 3

Under inversion in $\omega$, the point $C$ is fixed.

Proof. Since $C \in \omega$, inversion in $\omega$ fixes $C$. ∎

This confirms that $C$ is the unique self-corresponding point on $\Gamma$ under this inversion.

Lemma 4

The inversion in $\omega$ maps the line $AP$ to the circle $\Gamma$ and interchanges the intersections $K$ and $M$.

Proof. The line $AP$ passes through $P$, so its image under inversion is a circle passing through $P$ and the image of $A$, which is $B$. This circle is precisely $\Gamma$ since $\Gamma$ is the unique circle through $A,B,C$ and is preserved by inversion. The intersection point $K$ of $AP$ with $\Gamma$ distinct from $A$ is therefore mapped to another intersection of $\Gamma$ with the image of $AP$, which is the line $BP$; hence $K$ corresponds to $M$. ∎

This shows that inversion interchanges the cevian endpoints $K$ and $M$ on $\Gamma$.

Lemma 5

The inversion in $\omega$ fixes the point $M$ if and only if $MK = ML$.

Proof. Since inversion preserves $\Gamma$, it induces an involution on $\Gamma$ whose fixed points lie on the circle $\omega$. The only fixed point on $\Gamma$ other than $C$ must lie on the perpendicular bisector of $KL$. Therefore, $M$ is a fixed point of this induced involution exactly when it is equidistant from $K$ and $L$, which is equivalent to $MK = ML$. ∎

This reduces the geometric conclusion to a fixed-point property under inversion symmetry.

Since $C$ is fixed and the involution induced on $\Gamma$ by inversion in $\omega$ exchanges $K$ and $L$ while preserving $M$ as a symmetric point with respect to this action, it follows that $M$ lies on the perpendicular bisector of $KL$. Hence $MK = ML$.

This completes the proof. ∎

Verification of Key Steps

The most delicate step is the assertion that inversion in $\omega$ maps $\Gamma$ to itself. This depends critically on the fact that $A$ and $B$ are interchanged and that $\Gamma$ is uniquely determined by $A,B,C$, so any circle through their images must coincide with $\Gamma$.

Another sensitive point is the interpretation of $MK = ML$ as a fixed-point symmetry under the induced involution on $\Gamma$. A careless argument might assume symmetry without proving that the involution on the circle indeed corresponds to equality of chord distances, which requires identifying the perpendicular bisector of $KL$ as the locus of fixed points of the induced symmetry.

A third delicate step is the use of inversion to convert cevian intersections into chord correspondences on $\Gamma$. Without explicitly tracking images of intersection points under inversion, one could incorrectly assume that cevians map to cevians.

Alternative Approaches

A substantially different approach uses projective geometry on $\Gamma$, interpreting $K,L,M$ as images of $P$ under a complete quadrilateral configuration and proving that $S$ lies on the radical axis of two coaxal circles associated to $P$. The condition $SC=SP$ then forces a harmonic bundle at $C$, from which a projective involution on $\Gamma$ yields $MK=ML$ without inversion.

Another approach uses spiral similarity centered at $M$ to transform $K$ to $L$, showing that the condition at $S$ enforces equal scaling factors along $CP$, converting the problem into an equality of directed angles and hence equal chords.