IMC / 2004 / Problems / Day 2, P9

IMC 2004 · Day 2 · P9

easy

Let $D$ be the closed unit disk in the plane, and let $p_1, p_2, \dots, p_n$ be fixed points in $D$ . Show that there exists a point $p$ in $D$ such that the sum of the distances of $p$ to each of $p_1, p_2, \dots, p_n$ is greater than or equal to 1.

Solution (official)

considering as vectors, thoose $p$ to be the unit vector which points into the opposite direction as $\sum\limits_{i=1}^{n} p_i$ . Then, by the triangle inequality, $\sum_{i=1}^{n} |p - p_i| \ge \left| np - \sum_{i=1}^{n} p_i \right| = n + \left| \sum_{i=1}^{n} p_i \right| \ge n..$

How the field did

contestants scored

176

average (of 20)

10.89

solved (≥ 80%)

51.1%

near-0 (≤ 10%)

40.9%

discrimination

0.52

Score distribution (field cohort)

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

IMC 2004 · Day 2 · P9

How the field did

Score distribution (field cohort)

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