January 2, 2025
On a circle with center O and radius \(15\sqrt3\), three points are chosen randomly, call them \(A\), \(B\), \(C\). Suppose triangle \(ABC\) has orthocenter \(H\), compute the expected value of \(|OH|^2\).
\(\boxed{2025}\)
Let \(G\) be the centroid. By Sylvester’s theorem, \[\overrightarrow{OH} = 3 \cdot \overrightarrow{OG}, \] Hence \(|OH|^2 = 9 |OG|^2\).
It can be shown that \(E(|OG|^2) = \dfrac{R^2}{3}\) where \(R\) is the radius of circumcircle.
Consider the case when \(R=1\), i.e. \(ABC\) is circumscribed in a unit circle. Let \(A\), \(B\), \(C\) have coordinates \((1, 0),(\cos \alpha, \sin \alpha), (\cos \beta, \sin \beta)\). Then \[|OG|^2 = \dfrac{1}{9} ( (1 + \cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2) = \dfrac{1}{9} (3 + 2\cos(\alpha - \beta) + 2\cos \alpha + 2\cos\beta ) \]
Therefore \(E(|OH|^2) = 9E(|OG|^2) = 3R^2 = \boxed{2025}\).