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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.4 National Olympiad Prove it Saudi Arabia

Let $ABC$ be an acute, non-isosceles triangle and $(O)$ be its circumcircle (with center $O$ ). Denote by $G$ the centroid of the triangle $ABC$ , by $H$ the foot of the altitude from $A$ onto the side $BC$ and by $I$ the midpoint of $AH$ . The line $IG$ intersects $BC$ at $K$ .
1. Prove that $CK = BH$ .
2. The ray $GH$ intersects $(O)$ at $L$ . Denote by $T$ the circumcenter of the circle $(BHL)$ . Prove that $AO$ and $BT$ intersect on the circle $(O)$ .

Solution

1)
Let $M$ be the midpoint of $BC$ then $G \in AM$ and $\frac{AG}{AM} = \frac{2}{3}$ . Take the point $K'$ on $BC$ such that $M$ is the midpoint of $HK'$ , then $AM$ is the median of triangle $AHK'$ and $G$ is its centroid.
Then $K'G$ is the median of triangle $AHK'$ or $K'G$ passes through the midpoint of $AH$ . This implies that $K \equiv K'$ and we have $BH = CK$ .

2)
The line passes through $A$ and parallel to $BC$ intersects $(O)$ at $E$ different from $A$ . Then by the symmetry through the perpendicular bisector of $BC$ , it is easy to check that $AHKE$ is a rectangle. Since $\frac{GA}{GM} = \frac{AE}{HM} = 2$ , we have $H, G, E, L$ are collinear.
Then $\angle BLH = \angle BLE = \angle BCE = \angle ABC$ which implies that
$\angle ABT = \angle ABC + \angle CBT = \angle BLH + \angle CBT = 90^\circ.$
Thus if we denote $D = BT \cap (O)$ then $AD$ is the diameter of $(O)$ , then $O \in AD$ . Therefore, $BT$ and $AO$ intersect at a point that belongs to $(O)$ . $\square$

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