IMO 2023 Problem 2

Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m11s

Problem

Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Exploration

The configuration contains two circles, the circumcircle $\Omega$ of $ABC$ and the circle $\omega=(BDL)$. The statement concerns the tangent to $\omega$ at the second intersection point $P$ of the two circles. Since the desired point lies simultaneously on the tangent at $P$ and on the fixed line $BS$, a natural strategy is to identify this intersection as the center of perspectivity of two suitable triangles.

Let $S_1$ be the point of $\Omega$ diametrically opposite to $S$. Then $AS_1$ is the internal bisector of $\angle BAC$. If one could show that the tangent at $P$ and the line $DS$ both pass through the perspectivity center of two triangles whose third connecting line is $AS_1$, the result would follow immediately.

The tangent at $P$ suggests introducing its second intersection $Q$ with $\Omega$. Since $SS_1$ is a diameter of $\Omega$, the angle $\angle SQS_1$ equals $90^\circ$. Thus triangle $S_1QS$ is right-angled.

The next question is whether triangle $APD$ is also right-angled. The cyclic quadrilaterals $APBE$ and $DPLB$ connect $P$ simultaneously with the geometry of $\Omega$ and of $\omega$. Chasing angles yields

$$\angle PAD=90^\circ-\angle ADP,$$

hence $\angle APD=90^\circ$.

This strongly suggests comparing triangles $APD$ and $S_1QS$. If corresponding sides can be shown parallel, then the triangles are similar in a particularly rigid way. The parallelisms

$$AD\parallel S_1S,\qquad PD\parallel QS,\qquad AP\parallel S_1Q$$

become plausible. The first comes from both lines being perpendicular to $BC$. The second should follow from the tangent-chord theorem at $P$. Once the second is known, the third follows because both triangles are right-angled.

If these parallelisms hold, then the two triangles are in perspective from a point. The lines joining corresponding vertices are

$$AS_1,\qquad PQ,\qquad DS.$$

Since $PQ$ is exactly the tangent to $\omega$ at $P$, the desired statement becomes an application of Desargues' theorem in the converse direction for triangles with pairwise parallel corresponding sides.

The delicate step is proving $PD\parallel QS$ correctly. A careless angle chase can confuse the tangent to $\omega$ with the tangent to $\Omega$. The tangent-chord theorem must be applied on $\omega$ and then related to the line $SB$ through the fact that $D$ lies on $BS$.

Problem Understanding

This is a Type B problem. A geometric statement must be proved.

The triangle $ABC$ is acute, $\Omega$ is its circumcircle, and $S$ is the midpoint of the arc $CB$ containing $A$. Several auxiliary points are defined, leading to a second circle $\omega=(BDL)$. The circles $\Omega$ and $\omega$ meet again at $P$. The goal is to prove that the tangent to $\omega$ at $P$ meets the line $BS$ at a point lying on the internal bisector of $\angle BAC$.

The main difficulty is that the tangent at $P$ depends on the complicated circle $\omega$, while the angle bisector depends only on the original triangle $ABC$. The challenge is to uncover a hidden projective relation linking these apparently unrelated objects.

The key insight is to introduce the point $S_1$ diametrically opposite to $S$ on $\Omega$. Then $AS_1$ is the angle bisector of $\angle BAC$. After introducing the second intersection $Q$ of the tangent at $P$ with $\Omega$, one finds two triangles, namely $APD$ and $S_1QS$, whose corresponding sides are parallel. Their center of perspectivity then lies simultaneously on $AS_1$, on $DS=BS$, and on $PQ$, the required tangent.

Proof Architecture

We shall prove the result through the following claims.

Lemma 1

The triangle $APD$ is right-angled at $P$.

Proof sketch: Use the cyclic quadrilaterals $APBE$ and $DPLB$ to express $\angle PAD$ in terms of $\angle ADP$.

Lemma 2

Let $Q$ be the second intersection of the tangent to $\omega$ at $P$ with $\Omega$. Then

$$AD\parallel S_1S,\qquad PD\parallel QS,\qquad AP\parallel S_1Q.$$

Proof sketch: The first parallelism comes from perpendicularity to $BC$. The second comes from the tangent-chord theorem on $\omega$. The third follows because both triangles $APD$ and $S_1QS$ are right-angled and already have one pair of corresponding sides parallel.

Lemma 3

The lines $AS_1$, $PQ$, and $DS$ are concurrent.

Proof sketch: Since the corresponding sides of triangles $APD$ and $S_1QS$ are pairwise parallel, the converse of Desargues' theorem implies that the lines joining corresponding vertices are concurrent.

The hardest step is Lemma 2, specifically the proof of $PD\parallel QS$.

Solution

Let $S_1$ be the point of $\Omega$ diametrically opposite to $S$. Since $S$ is the midpoint of the arc $CB$ containing $A$, the point $S_1$ is the midpoint of the other arc $CB$. Hence $AS_1$ is the internal bisector of $\angle BAC$.

Let the tangent to $\omega$ at $P$ meet $\Omega$ again at $Q\neq P$.

We first establish a geometric property of triangle $APD$.

Lemma 1

The angle $\angle APD$ is a right angle.

Proof

Since $A,P,B,E$ are concyclic on $\Omega$,

$$\angle PAD=\angle PAE.$$

Because $A,D,E$ are collinear,

$$\angle PAE=180^\circ-\angle EBP.$$

Since $B,D,L,P$ are concyclic on $\omega$,

$$180^\circ-\angle EBP = \angle PBL = \angle PDL.$$

The line $DL$ is parallel to $BC$, while $AD$ is perpendicular to $BC$. Hence

$$\angle PDL=90^\circ-\angle ADP.$$

Combining these equalities gives

$$\angle PAD=90^\circ-\angle ADP.$$

Therefore

$$\angle APD = 180^\circ-\angle PAD-\angle ADP = 90^\circ.$$

This proves the lemma. ∎

Certification: we have shown that triangle $APD$ is right-angled, a fact that cannot be obtained merely from the cyclicity of $APBE$ without also using the parallelism $DL\parallel BC$.

Lemma 2

The equalities

$$AD\parallel S_1S,\qquad PD\parallel QS,\qquad AP\parallel S_1Q$$

hold.

Proof

Since $A,D,E$ are collinear and $AD\perp BC$, while $SS_1$ is a diameter of $\Omega$ and therefore $SS_1\perp BC$, we obtain

$$AD\parallel S_1S.$$

Next, apply the tangent-chord theorem to the tangent $PQ$ of $\omega$ at $P$. The angle between $PD$ and $PQ$ equals the angle subtended by chord $PD$ at the circle:

$$\angle DPQ=\angle DBP.$$

Because $D$ lies on $BS$,

$$\angle DBP=\angle SBP.$$

Since $Q,B,S,P$ lie on $\Omega$,

$$\angle SBP=\angle SQP.$$

Thus

$$\angle DPQ=\angle SQP,$$

which implies

$$PD\parallel QS.$$

By Lemma 1, $AP\perp PD$. Since $SS_1$ is a diameter of $\Omega$,

$$\angle SQS_1=90^\circ,$$

so $QS\perp QS_1$. Together with $PD\parallel QS$, we conclude that

$$AP\parallel S_1Q.$$

This proves the lemma. ∎

Certification: this step identifies all three pairs of corresponding sides of the two triangles, and the tangent-chord theorem is essential for obtaining $PD\parallel QS$.

Lemma 3

The lines $AS_1$, $PQ$, and $DS$ are concurrent.

Proof

By Lemma 2, the triangles $APD$ and $S_1QS$ have pairwise parallel corresponding sides:

$$AP\parallel S_1Q,\qquad PD\parallel QS,\qquad AD\parallel S_1S.$$

A standard converse form of Desargues' theorem states that if two triangles have corresponding sides pairwise parallel, then the lines joining corresponding vertices are concurrent.

Applying this to triangles $APD$ and $S_1QS$, we obtain the concurrency of

$$AS_1,\qquad PQ,\qquad DS.$$

This proves the lemma. ∎

Certification: the concurrency follows from the complete correspondence of the two triangles; proving only two pairs of side parallelisms would not suffice.

We now finish the proof.

The line $PQ$ is the tangent to $\omega$ at $P$ by construction. The line $DS$ is the same line as $BS$ because $D$ lies on $BS$. By Lemma 3, the tangent $PQ$ meets $BS$ at a point lying on $AS_1$.

Since $AS_1$ is the internal bisector of $\angle BAC$, the tangent to $\omega$ at $P$ meets the line $BS$ on the internal angle bisector of $\angle BAC$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the proof that $\angle APD=90^\circ$.

Starting independently, cyclicity of $APBE$ gives

$$\angle PAD=\angle PAE.$$

Since $A,D,E$ are collinear,

$$\angle PAE=180^\circ-\angle EBP.$$

Because $B,D,L,P$ are cyclic,

$$180^\circ-\angle EBP=\angle PBL=\angle PDL.$$

Finally, $DL\parallel BC$ and $AD\perp BC$ imply

$$\angle PDL=90^\circ-\angle ADP.$$

Hence

$$\angle PAD+\angle ADP=90^\circ,$$

which yields $\angle APD=90^\circ$. The common mistake is to replace $\angle PDL$ by $\angle PDB$; that loses the information coming from the parallel line $DL$.

The second delicate point is the deduction $PD\parallel QS$.

The tangent-chord theorem on $\omega$ gives

$$\angle DPQ=\angle DBP.$$

Since $D$ lies on $BS$,

$$\angle DBP=\angle SBP.$$

Because $Q,B,S,P$ lie on $\Omega$,

$$\angle SBP=\angle SQP.$$

Thus the two lines $PD$ and $QS$ make equal angles with $PQ$, forcing them to be parallel. The hidden danger is using a tangent-chord theorem on the wrong circle.

The third delicate point is the final concurrency statement.

The relevant triangles are $APD$ and $S_1QS$. Their corresponding sides are pairwise parallel, so they are homothetic or translated copies in the affine sense. The converse of Desargues' theorem gives concurrency of the lines joining corresponding vertices. One must verify all three pairs of side parallelisms before invoking the theorem.

Alternative Approaches

A substantially different proof introduces the point $X$ where the tangent at $P$ meets the line $BS$, then proves directly that $X$ lies on the angle bisector by means of Pascal's theorem. One first shows that $L$, $P$, and $S$ are collinear and that two additional points obtained from intersections of $AP$, $PS_1$, and $AE$ lie on the circle $\omega$. Applying Pascal to a degenerate hexagon on $\omega$ then yields the collinearity required to place $X$ on $AS_1$. This approach is more projective in nature.

Another method uses projections and orthogonal circles. One constructs a second circle through $A$, $P$, and a suitable point on $AS_1$, proves that it is orthogonal to $\omega$, and identifies the center of this circle as the intersection of the tangent at $P$ with $BS$. Showing that this center lies on the angle bisector then completes the argument. The computations are longer, but the proof reveals an additional orthogonality structure hidden in the configuration.

The approach used above is preferable because the entire argument is driven by a single observation: the triangles $APD$ and $S_1QS$ have corresponding sides parallel. Once that structure is recognized, the conclusion follows through a short and coherent chain of geometric implications.