IMC 2019 · Day 2 · P10

We will show that the quadrilateral has larger probability.

Let $D = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \le 1 \}$. Denote the random points by $X_1, \dots, X_{2019}$ and let $$\begin{align*} p &= \mathsf{P} \bigl( C \text{ is a triangle with vertices } X_1, X_2, X_3 \bigr), \\ q &= \mathsf{P} \bigl( C \text{ is a convex quadrilateral with vertices } X_1, X_2, X_3, X_4 \bigr). \end{align*}$$ By symmetry we have $\mathsf{P} \bigl( C \text{ is a triangle} \bigr) = \binom{2019}{3} p$, $\mathsf{P} \bigl( C \text{ is a quadrilateral} \bigr) = \binom{2019}{4} q$ and we need to prove that $\binom{2019}{4} q > \binom{2019}{3} p$, or equivalently $p < \frac{2016}{4} q = 504 q$.

Note that $p$ is the average over $X_1, X_2, X_3$ of the following expression: $$u(X_1, X_2, X_3) = \mathsf{P} \bigl( X_4 \in \triangle X_1 X_2 X_3 \bigr) \cdot \mathsf{P} \bigl( X_5, X_6, \dots, X_{2019} \in \triangle X_1 X_2 X_3 \bigr),$$ and $q$ is not less than the average over $X_1, X_2, X_3$ of $$v(X_1, X_2, X_3) = \mathsf{P} \bigl( X_1, X_2, X_3, X_4 \text{ form a convex quad.} \bigr) \cdot \mathsf{P} \bigl( X_5, X_6, \dots, X_{2019} \in \triangle X_1 X_2 X_3 \bigr).$$ Thus it suffices to prove that $u(X_1, X_2, X_3) \le 500\,v(X_1, X_2, X_3)$ for all $X_1, X_2, X_3$. It reads as $\operatorname{area}(\triangle X_1 X_2 X_3) \le 500 \operatorname{area}(\Omega)$, where $\Omega = \{ Y : X_1, X_2, X_3, Y \text{ form a convex quadrilateral} \}$. Assume the contrary, i.e., $\operatorname{area}(\triangle X_1 X_2 X_3) > 500 \operatorname{area}(\Omega)$.

Let the lines $X_1 X_2$, $X_1 X_3$, $X_2 X_3$ meet the boundary of $D$ at $A_1, A_2, A_3, B_1, B_2, B_3$; these lines divide $D$ into 7 regions as shown in the picture; $\Omega = D_4 \cup D_5 \cup D_6$.

By our indirect assumption, $$\operatorname{area}(D_4) + \operatorname{area}(D_5) + \operatorname{area}(D_6) = \operatorname{area}(\Omega) < \frac{1}{500} \operatorname{area}(D_0) < \frac{1}{500} \operatorname{area}(D) = \frac{\pi}{500}.$$ From $\triangle X_1 X_3 B_3 \subset \Omega$ we get $X_3 B_3 / X_3 X_2 = \operatorname{area}(\triangle X_1 X_3 B_3) / \operatorname{area}(\triangle X_1 X_2 X_3) < 1/500$, so $X_3 B_3 < \frac{1}{500} X_2 X_3 < \frac{1}{250}$. Similarly, the lengths segments $A_1 X_1, B_1 X_1, A_2 X_2, B_2 X_2, A_3 X_2$ are less than $\frac{1}{250}$.

The regions $D_1, D_2, D_3$ can be covered by disks with radius $\frac{1}{250}$, so $$\operatorname{area}(D_1) + \operatorname{area}(D_2) + \operatorname{area}(D_3) < 3 \cdot \frac{\pi}{250^2}.$$ Finally, it is well-known that the area of any triangle inside the unit disk is at most $\frac{3\sqrt3}{4}$, so $\operatorname{area}(D_0) \le \frac{3\sqrt3}{4}$.

But then $$\sum_{i=0}^{6} \operatorname{area}(D_i) < \frac{3\sqrt3}{4} + 3 \cdot \frac{\pi}{250^2} + \frac{\pi}{500} < \operatorname{area}(D),$$ contradiction.