IMO 2023 · Problem 6

Let \(ABC\) be an equilateral triangle. Let \(A_1,B_1,C_1\) be interior points of \(ABC\) such that \(BA_1=A_1C\), \(CB_1=B_1A\), \(AC_1=C_1B\), and \[ \angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ. \] Let \(BC_1\) and \(CB_1\) meet at \(A_2\), let \(CA_1\) and \(AC_1\) meet at \(B_2\), and let \(AB_1\) and \(BA_1\) meet at \(C_2\). Prove that if triangle \(A_1B_1C_1\) is scalene, then the three circumcircles of triangles \(AA_1A_2\), \(BB_1B_2\) and \(CC_1C_2\) all pass through two common points.

(Note: a scalene triangle is one where no two sides have equal length.)

Official text of the problem. Official problem paper (PDF)

Solution

Let \(O\) be the centre of \(ABC\). Choose coordinates, with circumradius \(1\), such that \[ A=(1,0),\qquad B=\left(-\frac12,\frac{\sqrt3}{2}\right),\qquad C=\left(-\frac12,-\frac{\sqrt3}{2}\right). \]

Set \[ \alpha=\angle A_1BC,\qquad \beta=\angle B_1CA,\qquad \gamma=\angle C_1AB. \] The three given isosceles triangles yield \[ \angle BA_1C=180^\circ-2\alpha, \] and the analogous identities cyclically. Hence \[ \alpha+\beta+\gamma=30^\circ. \] All three angles are positive, so each is less than \(30^\circ\).

Because \(BA_1=A_1C\), the point \(A_1\) lies on the median from \(A\). The inequality \(\alpha<30^\circ\) places it between \(O\) and the midpoint of \(BC\). The same holds cyclically. Thus, identifying points with their position vectors, we can write \[ A_1=-xA,\qquad B_1=-yB,\qquad C_1=-zC, \qquad 0<x,y,z<\frac12. \] Moreover, \[ \tan\alpha=\frac{1-2x}{\sqrt3},\qquad \tan\beta=\frac{1-2y}{\sqrt3},\qquad \tan\gamma=\frac{1-2z}{\sqrt3}. \] Substituting these into the tangent addition formula for \(\alpha+\beta+\gamma=30^\circ\) and simplifying gives \[ \boxed{x+y+z-xyz=1.} \tag{1} \]

Since \(A\cdot B=B\cdot C=C\cdot A=-\tfrac12\), we have \[ A_1B_1^2=x^2+xy+y^2. \] In particular, \[ A_1B_1^2-A_1C_1^2=(y-z)(x+y+z), \] with analogous identities cyclically. The scalene hypothesis therefore implies that \(x,y,z\) are pairwise distinct.

Solving the equations of \(BC_1\) and \(CB_1\) gives \[ A_2= \left( \frac{y+z+2yz}{2(1-yz)}, \frac{\sqrt3(z-y)}{2(1-yz)} \right). \tag{2} \] For completeness, those two line equations, in coordinates \((u,v)\), are \[ \sqrt3(z-1)u-(z+1)v+\sqrt3z=0, \] \[ \sqrt3(1-y)u-(1+y)v-\sqrt3y=0. \] Since \(1-yz>0\) and \(y\ne z\), the point \(A_2\) exists and is not on \(AA_1\). Thus \(AA_1A_2\) is nondegenerate; the same reasoning applies cyclically.

We now obtain equations of the three circumcircles. Define \[ \begin{aligned} d_x&=x(y+z)+4yz+2x-y-z,\\ d_y&=y(z+x)+4zx+2y-z-x,\\ d_z&=z(x+y)+4xy+2z-x-y. \end{aligned} \] A useful identity is \[ d_y-d_z=3(1-x)(y-z), \tag{3} \] together with its cyclic counterparts.

For a variable point \(P\), put \[ R=|P|^2,\qquad U=P\cdot A,\qquad V=P\cdot B,\qquad W=P\cdot C, \] and define \[ \begin{aligned} F_A(P)&=R+(x-1)U+\frac{d_x}{3(z-y)}(V-W)-x,\\ F_B(P)&=R+(y-1)V+\frac{d_y}{3(x-z)}(W-U)-y,\\ F_C(P)&=R+(z-1)W+\frac{d_z}{3(y-x)}(U-V)-z. \end{aligned} \tag{4} \]

We claim that \(F_A(P)=0\) is precisely the circumcircle of \(AA_1A_2\). Indeed, if \(P=(u,v)\), then \(U=u\) and \(V-W=\sqrt3v\), so this equation is \[ u^2+v^2+(x-1)u+\frac{d_x}{\sqrt3(z-y)}v-x=0. \] It is a circle equation, and substitution gives \[ F_A(A)=F_A(A_1)=0. \] Using (2), direct substitution also gives \[ F_A(A_2) = \frac{(y+z+2yz)(x+y+z-xyz-1)}{(1-yz)^2} =0 \] by (1). Since these three points are noncollinear, the claim follows. Cyclically, \(F_B=0\) and \(F_C=0\) are the other two required circumcircles. Denote them by \(\omega_A,\omega_B,\omega_C\).

The crucial relation is \[ \boxed{(y-z)F_A+(z-x)F_B+(x-y)F_C=0} \tag{5} \] as an identity in \(P\). To verify it, the coefficient of \(R\) and the constant term cancel immediately. The coefficient of \(U\) is \[ (y-z)(x-1)+\frac{d_y-d_z}{3}=0 \] by (3), and those of \(V,W\) vanish cyclically.

Because \(x,y,z\) are pairwise distinct, all three coefficients in (5) are nonzero. Consequently, every common point of any two of the circles lies on the third.

It remains to prove that some pair has two distinct real common points. This does not follow from (5) alone, so we supply a geometric argument.

The three circles are distinct: their equations in (4) all have quadratic part \(|P|^2\), but their constant terms \(-x,-y,-z\) are distinct. Furthermore, \[ F_A(O)=-x<0,\qquad F_B(O)=-y<0,\qquad F_C(O)=-z<0, \] so \(O\) lies strictly inside all three disks bounded by the circles.

Suppose that no pair of circles meets in two distinct points. Any two of their disks must then be nested. Indeed, two disks with intersecting interiors, neither containing the other, have centre distance \(d\) and radii \(r_1,r_2\) satisfying \[ |r_1-r_2|<d<r_1+r_2, \] which makes their boundary circles intersect in two distinct points.

Thus the three disks are pairwise nested, and one contains the other two. That largest disk contains all three vertices \(A,B,C\), hence, by convexity, the whole triangle \(ABC\). But its boundary also contains its corresponding point among \(A_1,B_1,C_1\). This is impossible: each of those points is strictly inside \(ABC\), and therefore strictly inside any disk containing \(ABC\).

Hence two of the circles meet in two distinct points. By (5), both points lie on the third circle, proving the assertion.

The solution was checked against the official IMO shortlist solutions.