IMO 2026 · Problem 2

Let \(ABC\) be a triangle and let points \(M\) and \(N\) be the midpoints of sides \(AB\) and \(AC\), respectively. Let points \(K\) and \(L\) be chosen strictly inside triangles \(BMC\) and \(BNC\), respectively, such that \(K\) lies strictly inside triangle \(ABL\) and \(L\) lies strictly inside triangle \(AKC\). Suppose that \[ \angle KBA=\angle ACL,\qquad \angle LBK=\angle LNC,\qquad\text{and}\qquad \angle LCK=\angle BMK. \] Let \(O\) be the circumcentre of triangle \(AKL\). Prove that \(OM=ON\).

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

Solution

Let \(D=BK\cap AC\) and \(E=CL\cap AB\). Since \(K,L\) lie inside \(ABC\), these intersections lie in the open sides \(AC,AB\), respectively.

We use directed angles modulo \(180^\circ\). To justify their use, orient \(ABC\) counterclockwise. The interior conditions give the counterclockwise ray orders \[ BC,BL,BK,BA \qquad\text{and}\qquad CA,CL,CK,CB. \] Also, \(MK\) lies inside the counterclockwise sector from \(MB\) to \(MC\), and \(NL\) inside that from \(NB\) to \(NC\). Thus all the given angle equalities hold as directed-angle equalities.

The first equality shows that triangles \(ABD\) and \(ACE\) are similar: they have the same angle at \(A\), and \(\angle ABD=\angle ACE\). Consequently, \[ AB\cdot AE=AC\cdot AD=:p>0. \] By the converse of the secant power theorem, \(B,C,D,E\) lie on a circle \(\omega_0\).

Define two further circles \[ \omega_1=(BLN),\qquad \omega_2=(CKM). \] They are well-defined because \(L\) is strictly inside \(BNC\), and \(K\) is strictly inside \(BMC\).

If \(D\ne N\), the second angle equality gives \[ \angle LBD=\angle LNC=\angle LND, \] so \(D\in\omega_1\). If \(D=N\), that equality becomes \(\angle LBN=\angle LNC\); by the converse tangent–chord theorem, \(AC\) is tangent to \(\omega_1\) at \(N\). Thus, in either case, the intersections of \(\omega_1\) with \(AC\), counted with multiplicity, are \(D,N\).

Similarly, if \(E\ne M\), the third equality gives \[ \angle ECK=\angle BMK=\angle EMK, \] so \(E\in\omega_2\). If \(E=M\), the same equality says that \(AB\) is tangent to \(\omega_2\) at \(M\). Hence its intersections with \(AB\), counted with multiplicity, are \(E,M\).

Write \(\pi_i(X)\) for the power of \(X\) with respect to \(\omega_i\). The secant and tangent power formulas now give \[ \pi_0(A)=p,\qquad \pi_1(A)=AD\cdot AN=\frac p2,\qquad \pi_2(A)=AE\cdot AM=\frac p2. \tag{1} \]

Let \(\omega=(AKL)\), and write \(\pi(X)\) for its power function. This circle is well-defined since \(K\) lies strictly inside \(ABL\), so \(A,K,L\) are noncollinear. We claim that, for every point \(X\), \[ \pi(X)=\pi_1(X)+\pi_2(X)-\pi_0(X). \tag{2} \]

Indeed, a circle with centre \(T\) and radius \(r\) has power function \[ |X-T|^2-r^2. \] Therefore \[ H(X)=\pi_1(X)+\pi_2(X)-\pi_0(X)-\pi(X) \] is affine in the coordinates of \(X\), since its quadratic terms cancel. Equation (1) gives \(H(A)=0\). The circles \(\omega_0,\omega_1\) share the distinct points \(B,D\), so their powers agree along \(BD\), in particular at \(K\). As \(K\in\omega_2\cap\omega\), we obtain \(H(K)=0\). Likewise, \(\omega_0,\omega_2\) share \(C,E\), and \(L\in CE\cap\omega_1\cap\omega\), giving \(H(L)=0\). An affine function vanishing at three noncollinear points vanishes identically, proving (2).

Finally, the difference of two power functions is affine, so its value at a midpoint is the average of its endpoint values. Using (2), \(M\in\omega_2\), and \(N\in\omega_1\), we get \[ \begin{aligned} \pi(M) &=(\pi_1-\pi_0)(M)\\ &=\frac{(\pi_1-\pi_0)(A)+(\pi_1-\pi_0)(B)}2 =-\frac p4,\\[2mm] \pi(N) &=(\pi_2-\pi_0)(N)\\ &=\frac{(\pi_2-\pi_0)(A)+(\pi_2-\pi_0)(C)}2 =-\frac p4. \end{aligned} \] If \(R\) is the radius of \(\omega\), this says \[ OM^2-R^2=ON^2-R^2. \] Hence \(\boxed{OM=ON}\).

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