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Geometry Difficulty 5.8 AIME, harder Prove it Thailand

Let AA, BB, CC be three distinct points on a unit circle. Let GG and HH be the centroid and the orthocenter of the triangle ABCABC, respectively. Let FF be the midpoint of the segment GHGH. Evaluate ∣AF∣2+∣BF∣2+∣CF∣2|AF|^2 + |BF|^2 + |CF|^2.

Solution

Define a coordinate system with the origin at the center of the circle. We can see that H⃗=A⃗+B⃗+C⃗\vec{H} = \vec{A} + \vec{B} + \vec{C} and G⃗=13(A⃗+B⃗+C⃗)\vec{G} = \frac{1}{3}(\vec{A} + \vec{B} + \vec{C}).
Thus, F⃗=G⃗+H⃗2=23(A⃗+B⃗+C⃗)\vec{F} = \frac{\vec{G} + \vec{H}}{2} = \frac{2}{3}(\vec{A} + \vec{B} + \vec{C}). We now have
∣AF∣2+∣BF∣2+∣CF∣2=(A⃗−F⃗)⋅(A⃗−F⃗)+(B⃗−F⃗)⋅(B⃗−F⃗)+(C⃗−F⃗)⋅(C⃗−F⃗)=∣A⃗∣2+∣B⃗∣2+∣C⃗∣2−2(A⃗+B⃗+C⃗)⋅F⃗+3F⃗⋅F⃗=∣A⃗∣2+∣B⃗∣2+∣C⃗∣2−(2(A⃗+B⃗+C⃗)−3F⃗)⋅F⃗=∣A⃗∣2+∣B⃗∣2+∣C⃗∣2=3. \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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