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Choose \(P, Q,\) and \(R\) uniformly at random within any convex region \(C\). By considering the event that four randomly chosen points form a triangle, show that the mean area of the shaded region is three times the mean area of \(\triangle PQR\).
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By considering the event that a fourth randomly chosen point \(S\) results in the four points \(P, Q, R, S\) having a convex hull that is a triangle, note that this event occurs precisely when \(S\) lies in one of four disjoint regions: the interior of \(\triangle PQR\) or one of the three wedges extending from each vertex of \(\triangle PQR\) away from the opposite side (with these wedges intersected with the convex region \(C\)).
The probability of this event is \(4 \times \mathbb{E}[\text{area}(\triangle PQR)] / \text{area}(C)\), by symmetry over which point is interior to the triangle formed by the other three.
This probability also equals \(\mathbb{E}[(\text{area}(\triangle PQR) + \text{area of the three wedges}) / \text{area}(C)]\).
Equating these expressions yields \(\mathbb{E}[\text{area}(\triangle PQR) + \text{area of the three wedges}] = 4 \mathbb{E}[\text{area}(\triangle PQR)]\), so \(\mathbb{E}[\text{area of the three wedges}] = 3 \mathbb{E}[\text{area}(\triangle PQR)]\). |
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