IMO 2000 · Problem 6

\(A_1A_2A_3\) is an acute-angled triangle. The foot of the altitude from \(A_i\) is \(K_i\) and the incircle touches the side opposite \(A_i\) at \(L_i\). The line \(K_1K_2\) is reflected in the line \(L_1L_2\). Similarly, the line \(K_2K_3\) is reflected in \(L_2L_3\) and \(K_3K_1\) is reflected in \(L_3L_1\). Show that the three new lines form a triangle with vertices on the incircle.

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

Solution

Let \(I,O\) be the incenter and circumcenter, and let \(r,R\) be the inradius and circumradius. Define points \(P_1,P_2,P_3\) by \[ \overrightarrow{IP_i}=-\frac rR\,\overrightarrow{OA_i}. \] Then \(IP_i=r\), so all three points lie on the incircle. Moreover, the map \[ X\longmapsto I-\frac rR(X-O) \] is a similarity of nonzero ratio. Thus \(P_1,P_2,P_3\) are distinct and noncollinear.

We shall prove that reflection in \(L_1L_2\) takes the line \(K_1K_2\) to \(P_1P_2\). The other two assertions will follow cyclically.

Write \[ A=A_1,\qquad B=A_2,\qquad C=A_3,\qquad \gamma=\angle ACB,\qquad q=\cos\gamma>0. \] Let \(\sigma\) be reflection in the internal angle bisector \(CI\), and let \(\tau\) be reflection in \(L_1L_2\).

First we determine the reflected line. For this calculation, take \(C\) as the origin and regard points as position vectors. Since \(\sigma\) interchanges the rays \(CA,CB\), the altitude projections give \[ K_1=q\,\sigma(A),\qquad K_2=q\,\sigma(B). \] Indeed, \(CK_1=CA\cos\gamma\) and \(CK_2=CB\cos\gamma\).

Let \(M\) be the midpoint of \(L_1L_2\). Symmetry about \(CI\) shows that \(M\in CI\) and \(L_1L_2\perp CI\). Put \(t=\gamma/2\). The right triangles \(CIL_1\) and \(CML_1\) yield \[ CL_1=CI\cos t,\qquad CM=CL_1\cos t=CI\cos^2t. \] Consequently, in our vector coordinates, \[ 2M=(1+q)I. \] Because the two reflection axes are perpendicular and meet at \(M\), \[ \tau(X)=2M-\sigma(X). \] For example, with \(CI\) as the horizontal axis, this identity is simply \((x,y)\mapsto(2CM-x,y)\). Therefore \[ \begin{aligned} \tau(K_1) &=2M-\sigma\bigl(q\,\sigma(A)\bigr)\\ &=(1+q)I-qA =I-q(A-I), \end{aligned} \] and similarly \[ \tau(K_2)=I-q(B-I). \] Thus the reflected line is the image of \(AB\) under the homothety with center \(I\) and ratio \(-q\).

To identify this line precisely, let \(\mathbf n\) be the unit normal to \(AB\) pointing into the triangle. Tangency of the incircle gives \[ \mathbf n\cdot(A-I)=\mathbf n\cdot(B-I)=-r. \] Hence the reflected line has equation \[ \mathbf n\cdot(X-I)=rq. \tag{1} \]

Since the triangle is acute, \(O\) lies inside it. Its distance from \(AB\) is \(R\cos\gamma=Rq\): the chord \(AB\) subtends the central angle \(2\gamma\), so bisecting the isosceles triangle \(OAB\) gives this distance. Thus \[ \mathbf n\cdot(O-A)=\mathbf n\cdot(O-B)=Rq. \] By the definitions of \(P_1,P_2\), \[ \mathbf n\cdot(P_1-I) =\frac rR\,\mathbf n\cdot(O-A) =rq, \] and likewise \(\mathbf n\cdot(P_2-I)=rq\). Therefore both points satisfy (1), and the reflected line is exactly \(P_1P_2\).

Cyclically, the other reflected lines are \(P_2P_3\) and \(P_3P_1\). These form the nondegenerate triangle \(P_1P_2P_3\), whose vertices all lie on the incircle, as required.

The solution was checked against Evan Chen's published IMO solution notes.