Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

All the angles in the solution are directed modulo $\pi$. The following notation is used throughout the proof:

$2\alpha$, $2\beta$, $2\gamma$ measures of the angles $BAC$, $CBA$, $ACB$, respectively;

$2a$, $2b$, $2c$ lengths of the sides $BC$, $CA$, $AB$, respectively;

$I$ incentre of the triangle $ABC$;

$M_A$, $M_B$, $M_C$ midpoints of the sides $BC$, $CA$, $AB$, respectively;

$X_Y$ orthogonal projection of $X$ on the line $YI$, for all $X, Y \in \{A, B, C\}$;

$\omega_A$, $\omega_B$, $\omega_C$ circumcircles of the triangles $M_A B_C C_B$, $M_B C_A A_C$, $M_C A_B B_A$, respectively, centred at $A'$, $B'$, $C'$, respectively.

Since the angle $AA_B B$ is right, the segments $M_C A = M_C B = M_C A_B = c$ (see Fig. 3), so $\angle M_C A_B B = \angle A_B B M_C C = \angle C B A_B = \beta$; this means that the lines $M_C A_B$ and $BC$ are parallel, so $A_B$ lies on the line $M_B M_C$. So, $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, $(M_C, M_A, B_C, B_A)$ are quartets of collinear points.

Fig. 3

Let $A''$ be the incentre of the triangle $AM_C M_B$ (in other words, $A''$ is the midpoint of $AI$). We show that $A''$ lies on both $\omega_B$ and $\omega_C$. For that, notice first that the points $A, B, A_B$, and $B_A$ lie on the circle on diameter $AB$; hence $\angle A_B B_A A = \angle A_B B A = \beta$. Next, the points $A_B, A_C, M_B$, and $M_C$ lie on a line parallel to $BC$, so $\angle A_B M_C A'' = \angle M_B M_C A'' = \beta = \angle A_B B_A A$. This means that $A''$ lies on $\omega_C$. Similarly, $A''$ lies on $\omega_B$.

Fig. 4

Let $X$ be the second point of intersection of $\omega_B$ and $\omega_C$. By the preceding, $\angle M_B X A'' = \angle M_B C_A A'' = \alpha$ and similarly $\angle A'' X M_C = \alpha$. This yields $\angle M_B X M_C = \angle M_B X A'' + \angle A'' X M_C = 2\alpha = \angle M_B M_A M_C$, which shows that $X$ lies also on the circumcircle $\omega$ of the triangle $M_A M_B M_C$, which is the nine-point circle of the triangle $ABC$. Denote the center of $\omega$ by $O'$.

Now the lines $A''X, M_BX$, and $M_CX$ are the radical axes of the circles $\omega_B, \omega_C$, and $\omega$. Since $XA''$ forms equal angles with $M_BX$ and $M_CX$, the triangle $O'B'C'$ formed by the centres of these circles has equal angles at $B'$ and $C'$; therefore, $O'B' = O'C'$. A similar argument shows that $O'B' = O'A'$, and $O'$ is consequently the circumcentre of the triangle $A'B'C'$.

With reference to the notation in Solution 1, let $O$ and $O'$ be the circumcentre and the centre of the nine-point circle of the triangle $ABC$, respectively. As in the previous solution, usage is made of the fact that $(M_A, M_B, C_A, C_B)$, $(M_B, M_C, A_B, A_C)$, and $(M_C, M_A, B_C, B_A)$ are quartets of collinear points, and $M_A B_C = M_A C_B = a$, $M_B A_C = M_B C_A = b$, and $M_C A_B = M_C B_B = c$.

Fig. 5

We claim that $A'O' = IO/2$; similarly, $B'O' = IO/2 = C'O'$, whence the required result. To prove the claim, notice that each vector $v$ is uniquely determined by its projections on the lines $AB$ and $AC$. The signed lengths of these projections will be denoted $\mathrm{pr}_c v$ and $\mathrm{pr}_b v$, respectively, the rays $AB$ and $AC$ emanating from $A$ being considered positive.

The points $O'$ and $A'$ are the circumcentres of the triangles $M_A M_B M_C$ and $M_A B_C C_B$, respectively. Project onto $M_A M_C$, to get $\mathrm{pr}_b A'O' = \mathrm{pr}_b (M_A O' - M_A A') = (-b/2 + a/2) = (a-b)/2$. Similarly, $\mathrm{pr}_c A'O' = (a-c)/2$. On the other hand, $\mathrm{pr}_b IO = \mathrm{pr}_b (AO - AI) = b - (b+c-a) = a-c$; similarly, $\mathrm{pr}_c IO = a-b$. This means that the vector $A'O'$ reflected in the bisectrix $AI$ of the angle $BAC$ is equal to the vector $IO/2$, hence the claim.