Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Vietnam

Let $ABC$ be an acute and non-isosceles triangle with $\angle A = 45^\circ$ . The altitudes $AD, BE, CF$ of triangle $ABC$ are concurrent at orthocenter $H$ . The line $EF$ meets the line $BC$ at $P$ . Point $I$ is the midpoint of segment $BC$ and $IF$ cuts $PH$ at $Q$ .
1. Prove that $\angle IQH = \angle AIE$ .
2. Let $K$ be the orthocenter of triangle $AEF$ and $(J)$ is the circumcircle of the triangle $KPD$ . The line $CK$ intersects circle $(J)$ at $G$ (different to $K$ ) and the line $IG$ intersect circle $(J)$ at $M$ (different to $G$ ), the line $JC$ intersect the circle with the diameter $BC$ at $N$ (different to $C$ ). Prove that $G, M, N, C$ are cyclic.

Solution

1. Without loss of generality, we can assume that $AB < AC$ , then $B$ will lie between points $P, C$ . The other case can be proved similarly.
First, we will prove that the line $PH$ is perpendicular to $AI$ .
Let $U, V$ be the midpoints of two segments $AH, IH$ , respectively, then $UV \parallel AI$ .
It is easy to check $(P, D, B, C) = -1$ , so based on the properties of harmonic division, we have $PB \cdot PC = PD \cdot PI$ .
We also have $PE \cdot PF = PB \cdot PC$ then $PE \cdot PF = PD \cdot PI$ or point $P$ lies on the radical axis of the circle with diameter $AH$ (center $U$ ) and the circle with diameter $IH$ (center $V$ ).
Furthermore, point $H$ also lies on the radical axis of these circles so $PH \perp UV$ .
Hence, we have $PH \perp AI$ .
Because $\angle BAC = 45^\circ$ then $\angle EIF = 90^\circ$ , this implies the result
$\angle IQH = 90^\circ - \angle AIF = \angle EIF - \angle AIF = \angle AIE.$
Therefore, we have $\angle IQH = \angle AIE$ (Q.E.D).

2. We can see that
$\angle EKF + \angle ECF = \angle EKF + \angle EAF = 180^\circ$
so $K$ belongs to the circle with diameter $BC$ . Because $D, P, B, C$ are harmonic so $ID \cdot IP = IC^2$ , but $IM \cdot IG = ID \cdot IP$ (equal to the power of point $I$ to the circle $(J)$ ) then $IM \cdot IG = IC^2$ , so we have
$\triangle IMC \sim \triangle ICG\ (c.g.c) \ \text{so}\ \angle IMC = \angle ICG = \angle ICK = 45^\circ. \quad (1)$

Let $T$ be the midpoint of the segment $PD$ then $CB \cdot CT = CD \cdot CP = CK \cdot CG$ or the quadrilateral $GTBK$ is cyclic and because $\angle BKG = 90^\circ$ , then we also have $\angle GTD = 90^\circ$ or $GT \perp PD$ .

Triangle $JPD$ is isosceles with $JP = JD$ and point $T$ is the midpoint of segment $PD$ then $JT$ is perpendicular to $PD$ .
Hence, $G, J, T$ are collinear and $\angle KGJ = 45^\circ$ .
On the other hand, because $CN \cdot CJ = CB \cdot CT = CK \cdot CG$ then the quadrilateral $KNJG$ is cyclic and this implies that
$\angle JNG = \angle JKG = \angle JGK = 45^\circ. \quad (2)$
From the equation (1) and (2), we have $\angle GMC = \angle GNC = 135^\circ$ or four points $G, M, N, C$ are cyclic (Q.E.D)

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