2026 IMO Problem 2 — Natural-Language Solution

Problem

Let $ABC$ be a triangle, and let $M$ and $N$ be the midpoints of $AB$ and $AC$ , respectively. Let $K$ 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 \angle LCK=\angle BMK.

Let $O$ be the circumcenter of triangle $AKL$ . Prove that

OM=ON.

Solution

Let $A'$ be the reflection of $A$ in $O$ , so $O$ is the midpoint of $AA'$ . Since $M$ and $N$ are the midpoints of $AB$ and $AC$ , respectively, a midpoint calculation gives

OM=\frac12 A'B, \qquad ON=\frac12 A'C.

It is therefore enough to prove that $A'$ is equidistant from $B$ and $C$ .

Coordinates forced by the interior conditions

Use the oriented basis

b=B-A,\qquad c=C-A.

The strict-interior hypotheses imply that $A,B,C$ are not collinear, so the oriented area $[b,c]$ is nonzero. Positive barycentric coordinates obtained from $K\in\mathrm{int}(BMC)$ and $L\in\mathrm{int}(AKC)$ give numbers $k_b,k_c,l_b,l_d$ such that

K-A=k_b b+k_c c, \qquad L-A=l_b b+l_d c.

To make the inequalities transparent, write $L-A=\lambda(K-A)+\mu c$ , where $\lambda,\mu>0$ and $\lambda+\mu<1$ . Then $l_b=\lambda k_b$ and $l_d=\lambda k_c+\mu$ . The coordinates of $K$ in $BMC$ similarly give $k_b,k_c>0$ and $k_b+k_c<1$ . Consequently

0<k_b<1,\qquad 0<l_d<1,\qquad k_b l_d-k_c l_b>0.

Indeed, the last determinant equals $k_b\mu>0$ . Put $\nu=1-\lambda-\mu>0$ and $\Delta=[b,c]\ne0$ . The condition $K\in\mathrm{int}(ABL)$ also gives

K-B=\alpha(A-B)+\beta(L-B)

for some $\alpha,\beta>0$ . Directly expanding the six oriented areas in the basis $(b,c)$ now gives

\begin{aligned} [K-B,A-B]&=k_c\Delta, & [A-C,L-C]&=l_b\Delta,\\ [L-B,K-B]&=\alpha l_d\Delta, & [L-N,C-N]&=\frac{l_b}{2}\Delta,\\ [L-C,K-C]&=\nu k_b\Delta, & [B-M,K-M]&=\frac{k_c}{2}\Delta. \end{aligned}

Every displayed coefficient is positive. Therefore the two angles in each of the three given equalities have the same oriented sign. Equality of their ordinary angles can consequently be upgraded, without a sign ambiguity, to

\measuredangle KBA=\measuredangle ACL, \quad \measuredangle LBK=\measuredangle LNC, \quad \measuredangle LCK=\measuredangle BMK.

Translating the angle and circle conditions

For two ordered pairs of nonzero vectors, equality of their oriented angles gives the division-free identity

[r,s]\langle u,v\rangle=[u,v]\langle r,s\rangle,

where $[u,v]$ denotes the oriented area of $u,v$ , and $\langle u,v\rangle$ denotes their Euclidean inner product. Expanding every ray in the basis $(b,c)$ , the three oriented-angle equalities therefore produce three polynomial identities in $k_b,k_c,l_b,l_d$ and the three Gram-matrix entries $\langle b,b\rangle,\langle b,c\rangle,\langle c,c\rangle$ .

More explicitly, identify a coefficient pair $(\alpha,\beta)$ with the vector $\alpha b+\beta c$ . In each of the following rows, the four entries are the coefficient pairs of $(u,v,r,s)$ in that order:

\begin{aligned} &((k_b-1,k_c),(-1,0),(0,-1),(l_b,l_d-1)),\\ &((l_b-1,l_d),(k_b-1,k_c),(l_b,l_d-\tfrac12),(0,\tfrac12)),\\ &((l_b,l_d-1),(k_b,k_c-1),(\tfrac12,0),(k_b-\tfrac12,k_c)). \end{aligned}

These are respectively the rays in the three oriented equalities above, expressed in the common basis. For $i=1,2,3$ , let $E_i$ be the left-hand side minus the right-hand side of the determinant-inner-product identity after substituting the $i$ th four-tuple. Thus $E_1=E_2=E_3=0$ , and all three equations follow by direct bilinear expansion, without dividing by an inner product or tangent that might vanish.

Put

z=A'-A.

Because $O$ is equidistant from $A,K,L$ , expanding $OA=OK$ and $OA=OL$ gives

\langle z,K-A\rangle=\lVert K-A\rVert^2, \qquad \langle z,L-A\rangle=\lVert L-A\rVert^2.

Define

\begin{aligned} C_K&=\langle z,K-A\rangle-\lVert K-A\rVert^2,\\ C_L&=\langle z,L-A\rangle-\lVert L-A\rVert^2. \end{aligned}

Then $C_K=C_L=0$ . Direct expansion and collection of terms gives the polynomial identity

\begin{aligned} &(2k_b l_b+3k_b l_d-k_b+k_c l_b+2k_c l_d-k_c-l_b-l_d)E_1\\ &\quad+4(k_b+k_c)(1-l_d)E_2-4(l_b+l_d)(1-k_b)E_3\\ &\quad+2(1-k_b)(1-l_d)(l_b+l_d)C_K\\ &\quad-2(1-k_b)(1-l_d)(k_b+k_c)C_L\\ &=(1-k_b)(1-l_d)(k_b l_d-k_c l_b)\\ &\qquad\cdot \left( 2(\langle z,b\rangle-\langle z,c\rangle) -(\langle b,b\rangle-\langle c,c\rangle) \right). \end{aligned}

The left-hand side is zero because all five defining equations vanish. Therefore

(1-k_b)(1-l_d)(k_b l_d-k_c l_b) \Bigl( 2(\langle z,b\rangle-\langle z,c\rangle) -(\langle b,b\rangle-\langle c,c\rangle) \Bigr)=0.

The first three factors are strictly positive by the interior-coordinate inequalities. Hence

2(\langle z,b\rangle-\langle z,c\rangle) =\lVert b\rVert^2-\lVert c\rVert^2.

Expanding squared norms now yields

\lVert z-b\rVert^2=\lVert z-c\rVert^2.

Since $z-b=A'-B$ and $z-c=A'-C$ , this is exactly $A'B=A'C$ . The midpoint observation at the start then gives

\boxed{OM=ON}.