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Thai Mathematical Olympiad

Geometry Difficulty 5.8 AIME, harder Prove it Thailand

Let $A$ , $B$ , $C$ be three distinct points on a unit circle. Let $G$ and $H$ be the centroid and the orthocenter of the triangle $ABC$ , respectively. Let $F$ be the midpoint of the segment $GH$ . Evaluate $|AF|^2 + |BF|^2 + |CF|^2$ .

Solution

Define a coordinate system with the origin at the center of the circle. We can see that $\vec{H} = \vec{A} + \vec{B} + \vec{C}$ and $\vec{G} = \frac{1}{3}(\vec{A} + \vec{B} + \vec{C})$ .
Thus, $\vec{F} = \frac{\vec{G} + \vec{H}}{2} = \frac{2}{3}(\vec{A} + \vec{B} + \vec{C})$ . We now have
$\begin{aligned} |AF|^2 + |BF|^2 + |CF|^2 &= (\vec{A} - \vec{F}) \cdot (\vec{A} - \vec{F}) + (\vec{B} - \vec{F}) \cdot (\vec{B} - \vec{F}) + (\vec{C} - \vec{F}) \cdot (\vec{C} - \vec{F}) \\ &= |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 - 2(\vec{A} + \vec{B} + \vec{C}) \cdot \vec{F} + 3\vec{F} \cdot \vec{F} \\ &= |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 - (2(\vec{A} + \vec{B} + \vec{C}) - 3\vec{F}) \cdot \vec{F} \\ &= |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 = 3. \end{aligned}$

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