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IMC / 2024 / Problems / Day 1, P5

IMC 2024 · Day 1 · P5

killer

Let n>dn > d be positive integers. Choose nn independent, uniformly distributed random points x1,,xnx_1, \dots, x_n in the unit ball BRdB \subset \mathbb{R}^d centered at the origin. For a point pBp \in B denote by f(p)f(p) the probability that the convex hull of x1,,xnx_1, \dots, x_n contains pp. Prove that if p,qBp, q \in B and the distance of pp from the origin is smaller than the distance of qq from the origin, then f(p)f(q)f(p) \ge f(q).

(proposed by Fedor Petrov, St Petersburg State University)

Solution (official)

By radial symmetry of the distribution, f(p)f(p) depends only on op|op| (the distance between oo and pp), so, we may assume that pp lies on the segment between oo and qq. For points x1,,xnx_1, \dots, x_n and xBx \in B denote by fx(x1,,xn)f_x(x_1, \dots, x_n) the indicator function of the event “xx is in the convex hull of x1,,xnx_1, \dots, x_n”. The claim follows from the following deterministic inequality fp(±x1,,±xn)fq(±x1,,±xn),(1)\tag{1} \sum f_p(\pm x_1, \dots, \pm x_n) \geqslant \sum f_q(\pm x_1, \dots, \pm x_n), where x1,,xnBx_1, \dots, x_n \in B are arbitrary points in general position and the summations are over all 2n2^n choices of signs (here oo is identified with the origin, that is, xx and x-x are symmetric with respect to oo). Indeed, taking the expectation in (1) over independent random uniform x1,,xnx_1, \dots, x_n, we get 2nf(p)2nf(q)2^n f(p) \geqslant 2^n f(q). (To be specific, here “general position” means that for any point set A{±x1,,±xn,p,q}A \subset \{\pm x_1, \dots, \pm x_n, p, q\}, which does not contain simultaneosuly xix_i and xi-x_i, is not contained in an (affine) (A2)(|A| - 2)-dimensional plane. This holds with probability 1.)

To prove (1), we use the following formula for the characteristic function χP\chi_P of the convex polyhedron PRdP \subset \mathbb{R}^d: if P1,,PkP_1, \dots, P_k are all facets of PP, and QiQ_i is the convex hull of oo and PiP_i, then χP=±χQi\chi_P = \sum \pm \chi_{Q_i}, where the sign is plus if oo and PP are on the same side of PiP_i, and minus otherwise. Indeed, for every point pp in general position look how the ray opop intersects the boundary of PP and realize that for at most two summands the contribution of the RHS at point pp is non-zero, and the total contribution equals 1 when pp is inside PP and 0 (possibly as 0=110 = 1 - 1) otherwise. Use this formula for every polyhedron PP with nn vertices y1,,yny_1, \dots, y_n, where each yiy_i is ±xi\pm x_i. These polyhedrons are simplicial (all facets are simplices) because of the general position condition. Sum up over all 2n2^n such PP, we get the expression of PχP\sum_P \chi_P as a linear combination of χS\chi_S, where SS are simplices formed by oo and some dd points in {±x1,,±xn}\{\pm x_1, \dots, \pm x_n\} (not containing xix_i and xi-x_i simultaneously).

For proving (1), it suffices to verify that all coefficients of χS\chi_S in this linear combination are positive (since two sides of (1) are the values of the sum PχP\sum_P \chi_P at pp and qq). Let's find a coefficient of χS\chi_S, where, say, SS is a simplex with vertices o,x1,,xdo, x_1, \dots, x_d. The plane α\alpha through x1,,xdx_1, \dots, x_d partitions Rd\mathbb{R}^d onto two parts H+H^+ (containing oo) and HH^- (not containing oo). For every pair {xi,xi}\{x_i, -x_i\} with i>di > d, either both points belong to H+H^+, or one belongs to HH^- and another to H+H^+. χS\chi_S goes with the plus sign for PP with vertices x1,,xdx_1, \dots, x_d and other vertices from H+H^+, and with the minus sign for PP with vertices x1,,xdx_1, \dots, x_d and other vertices from HH^-. It is immediate that there are at least as many pluses as minuses.

How the field did

contestants scored
397
average (of 10)
0.12
solved (≥ 80%)
0.8%
near-0 (≤ 10%)
97.5%
discrimination
0.25

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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