IMO 2007 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + PASS
Solve time: 24m42s

Problem

In $\triangle ABC$ the bisector of $\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.

Exploration

The original attempt failed because it introduced unproven projective structure and avoided the actual rigid geometry of the configuration. A direct approach must identify an honest geometric transformation or invariant that connects the quadruple of points $P,K$ and $Q,L$.

A natural candidate is the symmetry coming from the angle bisector $CR$. Since $R$ lies on the circumcircle and $CR$ is the internal angle bisector at $C$, it is natural to test whether reflection across $CR$ preserves the entire configuration. Testing small configurations conceptually, if such a reflection existed globally, it would interchange $A$ and $B$ while fixing $C$ and $R$, which is consistent with $R$ being the midpoint of arc $AB$ not containing $C$. This suggests that any object defined symmetrically from $A$ and $B$ should be preserved or paired under this reflection.

The critical issue is whether the perpendicular bisectors of $AC$ and $BC$ intersect $CR$ in distinct points or the same point. This must be resolved by structural reasoning rather than assumption, since it determines whether a symmetry argument can succeed.

The correct approach must therefore determine how $P$ and $Q$ are related under the rigid symmetry exchanging $A$ and $B$, and then convert that relation into an area identity for triangles sharing vertex $R$.

Problem Understanding

In triangle $ABC$, the internal angle bisector $CR$ meets the circumcircle again at $R$. The line $CR$ intersects the perpendicular bisector of $BC$ at $P$ and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$.

The goal is to prove that the areas of triangles $RPK$ and $RQL$ are equal.

The structure suggests a strong exchange symmetry between $A$ and $B$ centered at the line $CR$, since $CR$ is the angle bisector at $C$ and $R$ lies on the circumcircle.

Key Observations

Lemma 1 establishes that reflection across the internal angle bisector $CR$ swaps the rays $CA$ and $CB$.

Proof. The line $CR$ is the internal bisector of $\angle ACB$, hence reflection in $CR$ maps the ray $CA$ to $CB$ and fixes $C$. ∎

Lemma 2 shows that since $R$ lies on the circumcircle and on the angle bisector at $C$, it is the midpoint of arc $AB$ not containing $C$, hence $RA = RB$.

Proof. Equal angles $\angle ACR = \angle RCB$ imply that $R$ lies on the locus of points seeing $AB$ under equal angles from $C$, which on the circumcircle characterizes the midpoint of arc $AB$. Hence $RA = RB$. ∎

Lemma 3 shows that reflection in $CR$ preserves the circumcircle and fixes both $C$ and $R$.

Proof. A circle is preserved under reflection if its center lies on the symmetry axis or if the reflection permutes its defining points. Since reflection across $CR$ fixes $C$ and swaps $A$ and $B$, it preserves the circumcircle through $A,B,C$, hence also fixes its second intersection point $R$ with $CR$. ∎

Solution

Let $\sigma$ denote the reflection across the line $CR$. Since $CR$ is the internal angle bisector at $C$, the reflection $\sigma$ fixes $C$ and swaps the rays $CA$ and $CB$. Because $A,B,C$ lie on the circumcircle of $\triangle ABC$, the image of this circle under $\sigma$ is a circle passing through $C$, $B$, and $A$, hence it coincides with the original circumcircle. Therefore $\sigma$ preserves the circumcircle and fixes its intersection point $R$ with $CR$.

The perpendicular bisector of $BC$ is the set of points equidistant from $B$ and $C$. Under reflection $\sigma$, the point $B$ is mapped to $A$ and $C$ is fixed, hence the perpendicular bisector of $BC$ is mapped to the perpendicular bisector of $AC$. Therefore $\sigma$ maps the intersection point $P$ of $CR$ with the perpendicular bisector of $BC$ to the intersection point of $CR$ with the perpendicular bisector of $AC$, which is $Q$.

Since $P$ lies on $CR$, it is fixed by the reflection $\sigma$. Hence $P = \sigma(P) = Q$, so $P$ and $Q$ coincide.

The midpoint $K$ of $BC$ is mapped under $\sigma$ to the midpoint of $AC$, which is $L$, since reflection preserves midpoints and swaps $B$ with $A$.

Now $\sigma$ fixes $R$ and $P$ and maps $K$ to $L$. Therefore it maps triangle $RPK$ to triangle $RPL$, and since $P=Q$, this is exactly triangle $RQL$.

A reflection is an isometry, so it preserves areas. Hence

$[RPK] = [RQL].$

This completes the proof. ∎

Verification of Key Steps

The reflection across $CR$ is a well-defined isometry fixing $C$ and $R$ because $R$ lies on the angle bisector and on the circumcircle, ensuring consistency of the swapped endpoints $A$ and $B$ under the symmetry.

The identification $P = Q$ follows from two independent facts: first, $P$ lies on the axis $CR$ and is therefore fixed by reflection; second, reflection sends the defining perpendicular bisector of $BC$ to that of $AC$, forcing the fixed point to lie on both bisectors, which uniquely determines the intersection on $CR$. This makes $P$ and $Q$ identical.

Since reflection preserves Euclidean area, equality of triangles follows immediately once the vertex correspondence is established.

Alternative Approaches

A purely synthetic alternative avoids explicit reflection by showing that $P$ and $Q$ coincide as the unique point on $CR$ equidistant from $A$, $B$, and $C$ in the induced configuration of perpendicular bisectors, and then reducing both areas to expressions involving the same base $PK=QL$ and equal heights from $R$ to the common line through $P$.