75th NMO Selection Tests

Let $H$ be the orthocentre of triangle $ABC$, let $N$ be the midpoint of the segment $AH$ and let $D$ the foot of the perpendicular from $B$ to $AC$. We first prove that $\angle BNF=90^{\circ }$. Let $F^{\prime }$ be the intersection of line $AC$ and the perpendicular at $N$ to $BN$. The points $B,N,D$, and $F^{\prime }$ lie on the circle on diameter $BF$, so $$\angle NB{F}^{\prime }=\angle ADN=\angle DAN=90^{\circ }-\angle BCA=\angle ABO,$$ where the second equality holds on account of $DN$ being median in the right triangle $ADH$. Since $\angle NB{F}^{\prime }=\angle ABO$, lines $BN$ and $BO$ are isogonal with respect to angle $\angle ABF$. Similarly, lines $AO$ and $AN$ are isogonal with respect to angle $\angle BAC$,

$$\angle A{F}^{\prime }O=\angle B{F}^{\prime }N=\angle BDN=\angle NHD=\angle ACB.$$ Hence $O{F}^{\prime }\parallel BC$, so $F\equiv F^{\prime }$, showing that $\angle BNF=90^{\circ }$.

Next, we prove that $N,F$ and $M$ are collinear. Since $F\equiv F^{\prime }$, points $O$ and $N$ are isogonal conjugates in triangle $ABF$, so $$\angle NFB=\angle AFO=\angle ACB.$$ This implies $NF$ is tangent to circle $BFC$ and hence $N,F,M$ are collinear, as desired. Let $AB$ and $A^{\prime }M$ cross at $E$. Note that $E$ is the reflection of $A$ across $B$. Since $AK=4AB$, point $E$ is the midpoint of the segment $AK$. In triangle $AEH$, line $BN$ is a midline, so $BN\parallel EH$. As $BN$ and $NM$ are perpendicular, so are $EH$ and $NM$. Moreover, $NH\perp EM$, hence $H$ is the orthocentre of the triangle $ENM$, implying $MH\perp EN$. Finally, in triangle $AKH$, line $EN$ is a midline, so $EN\parallel KH$ and hence $MH\perp KH$; that is, $H$ lies on the circle on diameter $KM$, as required.

Alternative solution for $\angle BNF=90^{\circ }$. We use the same notations as in the previous solution. Consider the foot $P$ of the perpendicular from $F$ to $BC$. Then $$2AN=2NH=AH=2\mathrm{dist}(O,BC)=2\mathrm{dist}(F,BC)=2FP,$$ so the segments $NH$ and $FP$ are parallel and have equal lengths, implying that $NFPH$ is a parallelogram. Similarly, $ANPF$ is also a parallelogram. Therefore, $HB\perp NP$, and since $NH\perp BC$, we deduce that $H$ is the orthocentre of the triangle $BNP$. Consequently, $HP\perp BN$, and since $HP\parallel NF$, it follows that $\angle BNF=90^{\circ }$.