IMO 2021 · Problem 3

Let \(D\) be an interior point of the acute triangle \(ABC\) with \(AB > AC\) so that \(\angle DAB = \angle CAD\). The point \(E\) on the segment \(AC\) satisfies \(\angle ADE = \angle BCD\), the point \(F\) on the segment \(AB\) satisfies \(\angle FDA = \angle DBC\), and the point \(X\) on the line \(AC\) satisfies \(CX = BX\). Let \(O_1\) and \(O_2\) be the circumcentres of the triangles \(ADC\) and \(EXD\), respectively. Prove that the lines \(BC\), \(EF\), and \(O_1O_2\) are concurrent.

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

Solution

Apply a similarity so that \[ A=(0,0),\qquad D=(1,0). \] Put \(u=\cos\frac{\angle BAC}{2}\) and \(v=\sin\frac{\angle BAC}{2}\). Choosing the orientation appropriately, write \[ B=(bu,bv),\qquad C=(cu,-cv), \] where \(b=AB>c=AC>0\), \(u,v>0\), and \(u^2+v^2=1\). Also write \[ E=(eu,-ev),\qquad F=(fu,fv). \] The prescribed angles are positive, so \(e,f>0\).

The intersection of \(AD\) with \(BC\) has first coordinate \(2bcu/(b+c)\). Since \(D\) is interior to the triangle, \[ \Delta:=2bcu-b-c>0. \]

Both \(\angle BCD\) and \(\angle DBC\) are acute. Thus the angle conditions imply \(1-eu>0\) and \(1-fu>0\). Computing the tangents of those equal angles and cancelling \(v\) gives \[ \frac{e}{1-eu} = \frac{\Delta}{u(b-c)+c^2-bc(u^2-v^2)}, \] \[ \frac{f}{1-fu} = \frac{\Delta}{u(c-b)+b^2-bc(u^2-v^2)}. \] For example, the numerator and denominator on the first right-hand side, before cancelling \(v\), are respectively the determinant and dot product of \(B-C\) and \(D-C\).

Cross-multiplication, using \(u^2+v^2=1\), yields \[ ce(b+c-2u)=\Delta,\qquad bf(b+c-2u)=\Delta. \] Since \(ce>0\) and \(\Delta>0\), the common factor \(b+c-2u\) is positive. Consequently, we may put \[ k=ce=bf>0. \] The preceding equations also give \[ k(b+c-2u)=2bcu-b-c, \] or equivalently, \[ (1+k)(b+c)=2u(bc+k). \tag{1} \]

For variable coordinates \((r,s)\), the equations of \(BC\) and \(EF\) are \[ BC:\quad \frac{(b+c)r}{u}-\frac{(b-c)s}{v}=2bc, \] \[ EF:\quad \frac{(b+c)r}{u}+\frac{(b-c)s}{v}=2k. \] The second equation follows by substituting \(E\) and \(F\), using \(ce=bf=k\). Since \(b>c\), these lines have a unique intersection \(T\). Adding their equations and using (1), we obtain \[ T_x=\frac{u(bc+k)}{b+c}=\frac{1+k}{2}. \tag{2} \] Moreover, because \(T\in BC\), comparison of first coordinates gives \[ T=C+\frac{k-c^2}{b^2-c^2}(B-C). \tag{3} \]

We now regard points as their position vectors and use the usual dot product. Set \[ \mathbf w=(u,-v), \] so that \(C=c\mathbf w\) and \(E=e\mathbf w\). Write \(X=z\mathbf w\), where \(z\) is a signed real number. Expanding \(XB^2=XC^2\) gives \[ 2z\,\mathbf w\cdot(B-C)=b^2-c^2. \tag{4} \] Since the angle at \(C\) is acute, \[ (A-C)\cdot(B-C)>0. \] As \(A-C=-c\mathbf w\), this means \(\mathbf w\cdot(B-C)<0\). Hence (4) implies \(z<0\). In particular, \(z\ne0\) and \(X\ne E\); together with \(D\notin AC\), this ensures that both circumcircles are nondegenerate.

Equations (3) and (4) imply \[ 2T\cdot\mathbf w =2c+\frac{k-c^2}{z}. \tag{5} \]

The circumcentre \(O_1\), being equidistant from \(A,D,C\), satisfies \[ 2(O_1)_x=1,\qquad 2O_1\cdot\mathbf w=c. \tag{6} \] Similarly, equality of the distances from \(O_2\) to \(E\) and \(X\) gives \[ 2O_2\cdot\mathbf w=e+z. \] Using equality of its distances to \(D\) and \(E\) then gives \[ 2(O_2)_x=1+ez. \tag{7} \]

Define \[ Q=O_1+\frac{z}{c}(T-O_1). \] By (2), (6), and \(k=ce\), \[ 2Q_x =1+\frac{z}{c}\bigl(2T_x-1\bigr) =1+\frac{zk}{c} =1+ez. \] Likewise, by (5) and (6), \[ \begin{aligned} 2Q\cdot\mathbf w &=c+\frac{z}{c}\bigl(2T\cdot\mathbf w-c\bigr)\\ &=c+\frac{z}{c}\left(c+\frac{k-c^2}{z}\right)\\ &=e+z. \end{aligned} \] Because \(v>0\), a point is uniquely determined by its first coordinate and its dot product with \(\mathbf w=(u,-v)\). Comparing with (7), we conclude that \(Q=O_2\). Thus \[ O_2=O_1+\frac{z}{c}(T-O_1). \]

Finally, \(ez\ne0\), so (6)–(7) show that \(O_1\ne O_2\). Since \(z\ne0\), the displayed relation proves that \(T\in O_1O_2\). By construction, \(T\in BC\cap EF\), establishing the required concurrency. \(\square\)

The solution was checked against the official IMO shortlist solutions.