IMO 2024 · Problem 4

Let \(ABC\) be a triangle with \(AB<AC<BC\). Let the incentre and incircle of triangle \(ABC\) be \(I\) and \(\omega\), respectively. Let \(X\) be the point on line \(BC\) different from \(C\) such that the line through \(X\) parallel to \(AC\) is tangent to \(\omega\). Similarly, let \(Y\) be the point on line \(BC\) different from \(B\) such that the line through \(Y\) parallel to \(AB\) is tangent to \(\omega\). Let \(AI\) intersect the circumcircle of triangle \(ABC\) again at \(P\ne A\). Let \(K\) and \(L\) be the midpoints of \(AC\) and \(AB\), respectively.

Prove that \(\angle KIL+\angle YPX=180^\circ\).

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

Solution

Let \(D=AI\cap BC\), and let \(Q\) be the reflection of \(A\) across \(I\).

First we establish the positions of these points. Applying the angle bisector theorem in triangles \(ABC\) and \(ABD\), respectively, gives \[ BD=\frac{AB\cdot BC}{AB+AC}, \qquad \frac{AI}{ID}=\frac{AB}{BD} =\frac{AB+AC}{BC}>1. \] Consequently, \(A,I,D,Q\) lie in that order on their common line. Also, \(P\) lies beyond \(D\) on the ray \(AD\), since \(D\) is inside the circumcircle. Thus \(P\) and \(Q\) are on the opposite side of \(BC\) from \(A\).

The half-turn about \(I\) preserves \(\omega\) and takes each tangent to its distinct parallel tangent. It therefore takes the lines \(AC\) and \(AB\) to \(QX\) and \(QY\), respectively. In particular, \[ QX\parallel AC,\qquad QY\parallel AB. \] The negative homothety centred at \(D\) taking \(A\) to \(Q\) consequently takes \(C\) to \(X\) and \(B\) to \(Y\). Hence \(X,D,Y\) occur in that order along the direction from \(B\) to \(C\). Moreover, \(X\ne B\) and \(Y\ne C\): either equality would give a third tangent from \(B\) or \(C\) to the incircle, in addition to its two adjacent side lines.

The homothety centred at \(A\) with ratio \(2\) takes \(I,K,L\) to \(Q,C,B\), respectively. Thus \[ \angle KIL=\angle CQB. \tag{1} \]

If \(Q=P\), the negative homothety already described gives \[ \angle YPX=\angle YQX=\angle BAC. \] Since \(A,B,P,C\) are concyclic and \(A,P\) lie on opposite sides of \(BC\), \[ \angle CQB=\angle CPB=180^\circ-\angle BAC. \] Together with (1), this proves the result in this case.

Now suppose \(Q\ne P\). All angle congruences in the following calculation are directed modulo \(180^\circ\). Since \(QX\parallel AC\), the points \(A,P,Q\) are collinear, and \(A,B,C,P\) are concyclic, \[ \angle BXQ \equiv \angle BCA \equiv \angle BPA \equiv \angle BPQ. \] Therefore \(B,X,Q,P\) are concyclic. Similarly, \[ \angle CYQ \equiv \angle CBA \equiv \angle CPA \equiv \angle CPQ, \] so \(C,Y,Q,P\) are concyclic.

Using these two cyclic quadrilaterals and the collinearity of \(B,C,X,Y\), we obtain \[ \begin{aligned} \angle CQB &\equiv \angle CQP+\angle PQB\\ &\equiv \angle CYP+\angle PXB\\ &\equiv -\angle YPX \pmod{180^\circ}. \end{aligned} \]

To interpret this as an equality of ordinary angles, orient the figure with \(B\) to the left of \(C\) and \(A\) above \(BC\). Both \(P,Q\) are below \(BC\), and \(X\) is to the left of \(Y\). Thus the counterclockwise directed angles \(\angle CQB\) and \(\angle YPX\) are precisely their ordinary angles, each strictly between \(0^\circ\) and \(180^\circ\). Their sum, being a multiple of \(180^\circ\), must therefore equal \(180^\circ\).

Finally, (1) gives \[ \boxed{\angle KIL+\angle YPX=180^\circ}. \]

The solution was checked against the official IMO shortlist solutions.