75th NMO Selection Tests

Let ${\ell }_D,{\ell }_E,{\ell }_F$ be the Simson lines of $D,E,F$ with respect to triangle $ABC$ and let $X={\ell }_E\cap {\ell }_F$, $Y={\ell }_F\cap {\ell }_D$ and $Z={\ell }_D\cap {\ell }_E$. Let $D^{\prime },E^{\prime },F^{\prime }$ be the midpoints of $HD,HE,HF$, respectively, and let $H^{\prime }$ be the orthocentre of $△$.

Note that ${\ell }_D$ passes through $D^{\prime }$. This is a well-known fact about Simson lines. For the sake of completeness we provide a proof: Let $D_B$ and $D_C$ be the reflections of $D$ in $AB$ and $AC$, respectively. As the quadrangles $AHB{D}_B$ and $AHC{D}_C$ are both cyclic, $\angle AH{D}_B+\angle AH{D}_C=\angle AB{D}_B+\angle AC{D}_C=\angle ABD+\angle ACD=180^{\circ }$, so $D_B,H,D_C$ are collinear. A factor $\frac{1}{2}$ homothety from $D$ settles the case.

Claim 1. The quadrangles $H{E}^{\prime }X{F}^{\prime },H{F}^{\prime }Y{D}^{\prime },H{D}^{\prime }Z{E}^{\prime }$ are cyclic. Proof. Let $r_E$ and $r_F$ be the rays from $A$ isogonal in $\angle BAC$ to rays $AE$ and $AF$, respectively. Note that $r_E\perp {\ell }_E$ and $r_F\perp {\ell }_F$ to write $\angle E^{\prime }X{F}^{\prime }=\angle (r_E,r_F)=180^{\circ }-\angle EAF=\angle EDF=180^{\circ }-\angle E^{\prime }H{F}^{\prime }$. Hence $H{E}^{\prime }X{F}^{\prime }$ is cyclic. Similarly, $H{F}^{\prime }Y{D}^{\prime }$ and $H{D}^{\prime }Z{E}^{\prime }$ are both cyclic.

Claim 2. $H$ is the circumcentre of $△$. Proof. As $H$ is also the orthocentre of triangle $D^{\prime }{E}^{\prime }{F}^{\prime }$, it follows that $\angle HYZ=\angle H{F}^{\prime }{D}^{\prime }=\angle H{E}^{\prime }{D}^{\prime }=\angle HZY$, so $HY=HZ$. Similarly, $HX=HY$ and the claim follows.

As $D^{\prime },E^{\prime },F^{\prime }$ lie on the nine-point circle of triangle $DEF$, the midpoint $O^{\prime }$ of $OH$ is the circumcentre of triangle $D^{\prime }{E}^{\prime }{F}^{\prime }$. Proving that $H^{\prime }$ lies on the circle on diameter $OH$ is equivalent to proving $O^{\prime }{H}^{\prime }=O^{\prime }H$. The conclusion is then a consequence of Claim 3 below applied to $△$ and points $D^{\prime },E^{\prime },F^{\prime }$.

Claim 3. Let $ABC$ be a triangle with orthocentre $H$ and circumcentre $O\ne H$. Let $X,Y,Z$ be points on the sides $BC,CA,AB$, respectively, such that circles $AYZ,BZX,CXY$ are concurrent at $O$. Then the circumcentre of triangle $XYZ$ lies on the perpendicular bisectrix of $OH$.

Proof. Note that triangles $YOZ$ and $BHC$ are similar and the like. Then so are $XYZO$ and $ABCH$. Letting $O^{\prime }$ be the circumcentre of triangle $XYZ$, it follows that $XYZO{O}^{\prime }$ and $ABCHO$ are also similar. In particular, triangles $XO{O}^{\prime }$ and $AHO$ are similar. Varying $X$ along the line $BC$ shows that $O^{\prime }$ is the image of $X$ under a spiral-similarity from $O$. Consequently, $O^{\prime }$ lies on some fixed line $\ell$.

As $AH$ and $BC$ are perpendicular, the angle formed by $\ell$ and $OH$ is equal to the angle formed by $OH$ and the image of $BC$ under a rotation through $\angle AHO=90^{\circ }$. Letting $X,Y,Z$ be the midpoints of $BC,CA,AB$, respectively, it follows that $\ell$ passes through the midpoint of $OH$. Combining these two observations, it follows that $\ell$ is perpendicular bisectrix of $OH$, as desired. This ends the proof and completes the solution.

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Alternative solution.

As usual, the complex coordinate of a point in the plane is denoted by the corresponding lower case letter. Use the notation in Solution 1 and let the circumcircle of triangles $ABC$ and $DEF$ be centred at the origin and have unit radius. Thus, $a,b,c,d,e,f$ all have a unit absolute value, i.e., $a\bar{a}=b\bar{b}=\dots =f\bar{f}=1$ and $$h=a+b+c=d+e+f=\frac{1}{2}(a+b+c+d+e+f).$$