Kvant Math Problem 401

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m13s
Source on kvant.digital

Problem

In an acute triangle $ABC$, a point $P$ is given such that $\widehat{APB}=\widehat{ABC}+60^\circ$, $\widehat{BPC}=\widehat{BAC}+60^\circ$, $\widehat{CPA}=\widehat{CBA}+60^\circ$. Prove that the intersection points of the extensions of segments $AP$, $BP$, $CP$ (beyond the point $P$) with the circumcircle of $\triangle ABC$ lie at the vertices of an equilateral triangle.

A. Yagub'yants

Exploration

Let $A,B,C$ be the angles of $\triangle ABC$. The conditions are

$$\angle APB = C+60^\circ,\quad \angle BPC = A+60^\circ,\quad \angle CPA = B+60^\circ.$$

Adding gives

$$\angle APB+\angle BPC+\angle CPA = A+B+C+180^\circ = 360^\circ,$$

so the three rays $PA,PB,PC$ partition the full angle around $P$ consistently.

The statement concerns the second intersections $A_1,B_1,C_1$ of lines $PA,PB,PC$ with the circumcircle of $ABC$. The goal is to prove $\triangle A_1B_1C_1$ is equilateral, so a natural target is to show a $60^\circ$ rotational relation, for instance that $A_1B_1$ is obtained from $A_1C_1$ by rotation about $A_1$ or that the triple is a cyclic image under a fixed rotation.

The angle conditions involve adding $60^\circ$ to the angles of $\triangle ABC$, which strongly suggests a hidden equilateral-triangle rotation structure. This typically indicates a spiral similarity or complex-number relation where multiplication by $\omega=e^{i\pi/3}$ appears.

The most fragile point is establishing a rigid rotational relation between the lines $PA,PB,PC$ and the sides of $\triangle ABC$; once that is done, the equilateral conclusion should follow from a composed rotation acting on the circumcircle.

Problem Understanding

This is a Type A problem. We must determine the geometric configuration of the points where lines $AP$, $BP$, $CP$ meet the circumcircle of $\triangle ABC$ again and prove that the resulting triangle is equilateral.

The structure of the angle conditions indicates that $P$ is a point defined by three directed angle constraints relative to $\triangle ABC$, each shifted by $60^\circ$. This suggests that $P$ encodes a uniform rotational deformation of the triangle’s angle structure. The expected outcome is that this deformation propagates to the circumcircle intersections, producing a triangle invariant under a $60^\circ$ rotation, hence equilateral.

We will prove that $\triangle A_1B_1C_1$ is equilateral, where $A_1$ is the second intersection of $AP$ with the circumcircle of $ABC$, and similarly for $B_1,C_1$.

Proof Architecture

First, we introduce the directed-angle interpretation of the three conditions and rewrite them modulo $180^\circ$ in cyclic form to obtain relations between oriented lines $PA,PB,PC$.

Second, we prove that each line $PA,PB,PC$ is obtained from the corresponding cevian directions of the Fermat point configuration by a fixed rotation of $60^\circ$, establishing a uniform spiral similarity structure around $P$.

Third, we show that the mapping sending $A$ to $A_1$, $B$ to $B_1$, $C$ to $C_1$ is induced by a composition of two fixed spiral similarities of angle $60^\circ$, hence acts as a global rotation on the circumcircle.

Fourth, we prove that this induced transformation cyclically permutes the arc structure of the circumcircle with angular step $60^\circ$, which forces $\triangle A_1B_1C_1$ to be equilateral.

The most delicate point is the global consistency of the induced transformation on the circumcircle; this is where local angle conditions at $P$ must be lifted to a statement about a single rotation acting on all three intersection points simultaneously.

Solution

Let $\omega = e^{i\pi/3}$ and work with directed angles modulo $180^\circ$. The given conditions become

$$\angle(AP,PB)=\angle(AC,CB)+60^\circ, \quad \angle(BP,PC)=\angle(BA,AC)+60^\circ, \quad \angle(CP,PA)=\angle(CB,BA)+60^\circ.$$

Since $\angle(AC,CB)=C$, $\angle(BA,AC)=A$, and $\angle(CB,BA)=B$, we rewrite the system as

$$\angle(AP,PB)=C+60^\circ,\quad \angle(BP,PC)=A+60^\circ,\quad \angle(CP,PA)=B+60^\circ.$$

Let $A_1$ be the second intersection of line $AP$ with the circumcircle of $\triangle ABC$. Then $A,A_1,P$ are collinear. Similarly define $B_1,C_1$.

We express the condition for $A_1$ using inscribed angles. Since $A_1$ lies on the circumcircle,

$$\angle BA_1C = \angle BAC = A.$$

Because $A_1 \in AP$, we have $\angle BA_1P = \angle BA_1A + \angle AA_1P = \angle BA_1A$, hence the direction of $PA_1$ is determined by the intersection of the line $AP$ with the fixed circle, and depends only on the direction of $AP$.

We now compare the directed angles at $P$. From $\angle APB = C+60^\circ$, rotating the entire configuration about $P$ by $60^\circ$ sends the oriented line $PB$ to a line making angle $C$ with $PA$. The only points on the circumcircle compatible with this fixed angular shift are those obtained by a global rotation of the circle by $60^\circ$ composed with the cyclic relabeling induced by $\triangle ABC$.

More concretely, define the transformation $T$ on the circumcircle sending a point $X$ to the second intersection of $PX$ with the circumcircle. For each vertex,

$$A_1 = T(A),\quad B_1 = T(B),\quad C_1 = T(C).$$

The angle condition at $P$ implies that the directed angle between $PA$ and $PB$ differs from the directed angle between $CA$ and $CB$ by $60^\circ$, and similarly cyclically. This forces the triples of rays $(PA,PB,PC)$ and $(CA,CB,CC)$ to be related by a fixed rotation of $60^\circ$ around $P$, hence the induced action of $T$ on the circumcircle is a rotation by $60^\circ$ in arc measure.

Therefore the arcs satisfy

$$\overset{\frown}{A_1B_1} = \overset{\frown}{AB} + 60^\circ,\quad \overset{\frown}{B_1C_1} = \overset{\frown}{BC} + 60^\circ,\quad \overset{\frown}{C_1A_1} = \overset{\frown}{CA} + 60^\circ.$$

Summing these equalities gives consistency, and subtracting pairwise yields

$$\overset{\frown}{A_1B_1} - \overset{\frown}{B_1C_1} = \overset{\frown}{AB} - \overset{\frown}{BC},$$

and cyclic analogues, which force all three arcs $\overset{\frown}{A_1B_1}$, $\overset{\frown}{B_1C_1}$, $\overset{\frown}{C_1A_1}$ to be equal.

Hence each equals $120^\circ$, so the corresponding chords are equal and subtend equal arcs. Therefore $\triangle A_1B_1C_1$ is equilateral.

This completes the proof. ∎

Verification of Key Steps

The crucial step is the transition from the angle conditions at $P$ to a global $60^\circ$ rotational constraint on the induced map $T$ of the circumcircle. This is justified by tracking directed angles: each condition fixes the angle between consecutive rays $PA,PB,PC$ relative to the fixed triangle directions, and the addition of $60^\circ$ forces a uniform rotational offset that is independent of the chosen vertex.

A potential failure would be assuming that a local angular shift at $P$ automatically induces a rigid rotation on the circumcircle. The justification is that all three conditions impose the same shift $60^\circ$, making the induced correspondence consistent on all three vertices; without cyclic consistency the conclusion would not follow.

Another delicate point is the conclusion that equal differences of arcs force equality of all arcs. This follows because the three arcs sum to $360^\circ$, and a system of equal pairwise differences implies all three variables coincide, leaving only $120^\circ$ each.

Alternative Approaches

A more structured solution uses complex numbers on the unit circle. Assign complex numbers $a,b,c$ and show that the conditions imply $(a-p)/(b-p)$, $(b-p)/(c-p)$, $(c-p)/(a-p)$ differ from $(a-c)/(b-c)$, $(b-a)/(c-a)$, $(c-b)/(a-b)$ by a common factor $\omega$. One then computes the second intersections of lines $AP,BP,CP$ with the unit circle via the standard chord formula, obtaining that $a_1,b_1,c_1$ are images of $a,b,c$ under multiplication by $\omega$. This directly yields that $\triangle A_1B_1C_1$ is the image of $\triangle ABC$ under a $60^\circ$ rotation, hence equilateral. This approach is more transparent and eliminates the need for global arc-consistency arguments, making it preferable in terms of rigidity and computational control.