NMO Selection Tests for JBMO

We shall make use of the following:

Lemma. Let $ABC$ be a triangle and $MN$ be a chord of its circumcircle which intersects side $[AB]$ in $D$ and side $[AC]$ in $E$. Then $\frac{DM}{DN} : \frac{EM}{EN} = \frac{BM}{BN} : \frac{CM}{CN}$.

Proof of the lemma. Using the law of sines in triangles $BDM$ and $BDN$, we have
$$\frac{DM}{\sin(\angle ABM)} = \frac{BM}{\sin(\angle BDM)} \quad \text{and} \quad \frac{DN}{\sin(\angle ABN)} = \frac{BN}{\sin(\angle BDN)}.$$
Since $\angle BDM + \angle BDN = 180^\circ$, we have $\sin(\angle BDM) = \sin(\angle BDN)$, so
$$\frac{DM}{DN} = \frac{BM}{BN} \cdot \frac{\sin(\angle ABM)}{\sin(\angle ABN)}.$$
Analogously, we get $\frac{EM}{EN} = \frac{CM}{CN} \cdot \frac{\sin(\angle ACM)}{\sin(\angle ACN)}$. But $\angle ABM = \angle ACM$ and $\angle ABN = \angle ACN$, so $\frac{DM}{DN} : \frac{EM}{EN} = \frac{BM}{BN} : \frac{CM}{CN}$. $\square$

Returning to our problem, let $D'$ and $E'$ be the reflections of $D$ and $E$ about $O$ and $A'$ be the second intersection point of $BE'$ with the circumcircle of $ABC$. Also, let $D''$ be the intersection of lines $A'C$ and $MN$.

From the lemma, we have
$$\frac{DM}{DN} : \frac{EM}{EN} = \frac{BM}{BN} : \frac{CM}{CN} = \frac{D''M}{D''N} : \frac{E'M}{E'N}.$$
Since $DM = D'N$, $DN = D'M$, $EM = E'N$ and $EN = E'M$, we conclude that $\frac{D'M}{D'N} = \frac{D''M}{D''N}$, hence $D' = D''$.

Now, $OF$ is midsegment in $\triangle BB'E$, hence $\angle BOF = \angle BB'X$; $OG$ is midsegment in $\triangle CC'D$, hence $\angle COG = \angle CC'X$. But clearly $\angle BB'X + \angle CC'X = \angle BAC$ and $\angle BOC = 2\angle BAC$, therefore
$$\angle FOG = \angle BOC - (\angle BOF + \angle COG) = 2\angle BAC - \angle BAC = \angle BAC,$$

Since $[OF]$ is a midsegment of the triangle $BEE'$ and $[OG]$ is a midsegment of the triangle $CDD'$, we get that $OF \parallel BA'$ and $OG \parallel CA'$, so $\angle FOG = \angle BA'C = \angle BAC$.

Let $B'$ be the point diametrically opposed to $B$ and $C'$ the point diametrically opposed to $C$; the triangle $ABC$ being acute, $B'$ is on the minor arc $AC$, and $C'$ is on the minor arc $AB$. Let $X$ be an arbitrary point on the minor arc $BC$. Applying Pascal's theorem to the hexagram $ABB'XC'C$ shows that points $O = BB' \cap CC'$, $D = AB \cap C'X$, $E = AC \cap B'X$ are collinear – on Pascal's line, which is the support line of a diameter.

Conversely, if a diameter intersects the sides $[AB]$ and $[AC]$ at $D$ and $E$, respectively, then the lines $B'E$ and $C'D$ will meet at a point $X$ situated on the minor arc $BC$ (consider $X$ only as the intersection point of the line $B'E$ with the circle, and apply Pascal's theorem; $D' = AB \cap C'X$ will be collinear with $O, E$, hence, it will be the intersection of the diameter with the line $AB$, which means that $D'$ is in fact $D$).