IMO 2011 · Problem 6

Let \(ABC\) be an acute triangle with circumcircle \(\Gamma\). Let \(\ell\) be a tangent line to \(\Gamma\), and let \(\ell_a,\ell_b\) and \(\ell_c\) be the lines obtained by reflecting \(\ell\) in the lines \(BC,CA\) and \(AB\), respectively. Show that the circumcircle of the triangle determined by the lines \(\ell_a,\ell_b\) and \(\ell_c\) is tangent to the circle \(\Gamma\).

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

Solution

Apply a similarity so that \(\Gamma\) is the unit circle in the complex plane and \(\ell\) has equation \(\operatorname{Re}z=1\). Let \(a,b,c\) be the complex coordinates of \(A,B,C\), and put \[ s=a+b+c,\qquad p=abc,\qquad d=1-\operatorname{Re}s. \]

The orthocenter of \(ABC\) has coordinate \(s\). Indeed, \[ \operatorname{Re}\bigl((s-a)\overline{(c-b)}\bigr) =\operatorname{Re}(b\bar c-c\bar b)=0, \] so \(s\) lies on the altitude from \(A\), and similarly on the other altitudes. Because \(ABC\) is acute, its orthocenter lies inside the triangle, hence inside \(\Gamma\). Thus \[ |s|<1,\qquad d>0. \] Also, none of \(a+b,b+c,c+a\) is zero: otherwise a side of \(ABC\) would be a diameter, giving a right angle.

Reflection in \(BC\) is the map \[ z\longmapsto b+c-bc\bar z, \] since this is an orientation-reversing isometry fixing \(b\) and \(c\). Consequently, the equation of \(\ell_a\) is \[ \operatorname{Re}(b+c-bc\bar z)=1, \] or, equivalently, \[ \operatorname{Re}\left(a\left(\frac z p+1\right)\right)=-d. \tag{1} \] The equations of \(\ell_b,\ell_c\) follow cyclically. In particular, none of these lines contains \(-p\).

Consider the bijective transformation \[ w=F(z)=-\frac{dp}{z+p}, \qquad z=-p\left(1+\frac d w\right), \] between \(\mathbb C\setminus\{-p\}\) and \(\mathbb C\setminus\{0\}\). By (1), the image of \(\ell_a\) satisfies \[ \operatorname{Re}\frac a w=1 \quad\Longleftrightarrow\quad \left|w-\frac a2\right|=\frac12. \] Thus \(F(\ell_a)\) is the circle \(\omega_a\) with diameter joining \(0\) to \(a\), with \(0\) omitted. Define \(\omega_b,\omega_c\) similarly.

The distinct circles \(\omega_b,\omega_c\) intersect at \[ 0\quad\text{and}\quad m_a=\frac{b+c}{2}, \] because \[ \left|m_a-\frac b2\right| =\left|m_a-\frac c2\right|=\frac12. \] Since \(m_a\ne0\), the lines \(\ell_b,\ell_c\) have the unique intersection \[ X=F^{-1}(m_a). \] Cyclically, their other intersections are \[ Y=F^{-1}\left(\frac{c+a}{2}\right), \qquad Z=F^{-1}\left(\frac{a+b}{2}\right). \] The three midpoints are distinct, so \(X,Y,Z\) are distinct. Hence the three reflected lines form a nondegenerate triangle.

The three midpoints lie on the circle \[ \mathcal N:\quad \left|w-\frac s2\right|=\frac12, \] since, for example, \(m_a-s/2=-a/2\). Moreover, \(0\notin\mathcal N\), because \(|s|<1\).

Its inverse image under \(F\) is a genuine circle. To verify this directly, put \(u=(z+p)/p\), so \(w=-d/u\). The equation of \(\mathcal N\) becomes \[ |su+2d|=|u|, \] which, on completing the square, is \[ \left|u-\frac{2d\bar s}{1-|s|^2}\right| =\frac{2d}{1-|s|^2}. \tag{2} \] The radius is positive, and \(u=0\) does not satisfy (2). Since \(u\) differs from \(z\) by a rotation and translation, this defines a circle \(\mathcal K\) avoiding \(-p\). It passes through \(X,Y,Z\), so it is their circumcircle.

Finally, using \(|p|=1\), we obtain \[ |z|=1 \quad\Longleftrightarrow\quad |w+d|=|w| \quad\Longleftrightarrow\quad \operatorname{Re}w=-\frac d2. \] Thus \(F\) maps \(\Gamma\setminus\{-p\}\) onto the line \[ L:\quad \operatorname{Re}w=-\frac d2. \] The distance from the center \(s/2\) of \(\mathcal N\) to \(L\) is \[ \left|\frac{\operatorname{Re}s+d}{2}\right|=\frac12, \] which equals the radius of \(\mathcal N\). Therefore \(L\) and \(\mathcal N\) have exactly one common point. This point is nonzero, since \(d>0\).

Taking inverse images, and recalling that \(-p\notin\mathcal K\), shows that \(\Gamma\) and \(\mathcal K\) have exactly one common point. They are distinct circles, because \(-p\in\Gamma\) but \(-p\notin\mathcal K\). Hence they are tangent. \(\square\)

The solution was checked against the official IMO shortlist solutions.