HMMT-Feb 2024 guts, Problem 35

Barry picks infinitely many points inside a unit circle, each independently and uniformly at random, $P_{1}, P_{2}, \ldots$ Compute the expected value of $N$, where $N$ is the smallest integer such that $P_{N+1}$ is inside the convex hull formed by the points $P_{1}, P_{2}, \ldots, P_{N}$. Submit a positive real number $E$. If the correct answer is $A$, you will receive $\lfloor 100 \cdot \max (0.2099-|E-A|, 0)\rfloor$ points.

Solution: Solution: Clearly, $N \geq 3$, and let's scale the circle to have area 1 . We can see that the probability to not reach $N=4$ is equal to the probability that the fourth point is inside the convex hull of the past three points. That is, the probability is just one minus the expected area of those $N$ points. The area of this turns out to be really small, and is around 0.074 , and so $(1-0.074)$ of all sequences of points make it to $N=4$. The probability to reach to the fifth point from there should be around $(1-0.074)(1-0.074 \cdot 2)$, as any four points in convex configuration can be covered with 2 triangles. Similarly, the chance of reaching $N=6$ should be around $(1-0.074)(1-0.074 \cdot 2)(1-0.074 \cdot 3)$, and so on. Noting that our terms eventually decay to zero around term $1 / 0.074=13$, our answer should be an underestimate. In particular, we get \[ 3+(1-0.074)(1+(1-0.074 \cdot 2)(1+(1-0.074 \cdot 3)(1+\cdots))) \approx 6.3 \] Guessing anything slightly above this lower bound should give a positive score. Here is a Python code that simulates the result. $\fbox{6.54}$.

Answer: 6.54