Let n be an odd positive integer. On some field, n gunmen are placed such that all pairwise distances between them are different. At a signal, every gunman takes out his gun and shoots the closest gunman. Prove that: (a) at least one gunman survives; (b) no gunman is shot more than 5 times; (c) the trajectories of the bullets do not intersect.


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Question Number 13893 by Tinkutara last updated on 24/May/17

$Let n be an odd positive integer . On$ $some field, n gunmen are placed such$ $that all pairwise distances between$ $them are different . At a signal, every$ $gunman takes out his gun and shoots$ $the closest gunman . Prove that :$ $(a) at least one gunman survives;$ $(b) no gunman is shot more than 5$ $times;$ $(c) the trajectories of the bullets do$ $not intersect .$
Commented by prakash jain last updated on 25/May/17

$Part (a)$ $Let gunmen be a_{1}, a_{2}, . . . ., a_{n}$ $and let us distances are a_{i j} .$ $Since all distances are different$ $there exist minimum a_{i j} for some$ $value of i and j . Since a_{i} will$ $shoot a_{j} and a_{j} will shoot a_{i} .$ $Case A : a_{k} (k \neq i, k \neq j) fires at$ $a_{i} or a_{j} .$ $In this case a_{i} or a_{j} get shot twice .$ $Since there are only n bullets and$ $at least person got shot twice then$ $there will be one person who does$ $get shot at all . (p i g e o n h o l e p r i n c i p l e)$ $Case B : No one else fires at a_{i} or a_{j}$ $In this case we take a_{i} and a_{j} out of field$ $and consider remaining (n - 2) gunmen .$ $Continue this process finally there will$ $be 3 gunmen left where only case A$ $is possible .$
Commented by prakash jain last updated on 25/May/17
$Part (b) shot<5 times. Consider a_i is shot 6 times by a_j_1 ,...,a_j_6 . consider hexagon formed by a_j_1 ...a_j_6 . A: point a_i is inside the hexagom Make a traingle with a_i and one side of hexagon say a_j_1 a_(j_2 .) a_j_1 a_i <a_j_1 a_j_2 a_j_2 a_i <a_j_1 a_j_2 a_j_1 a_j_2 is largest side so ∠a_j_1 aa_j_2 >60° So six sides of hexagon spawn an angle >60×6=360° not possible. case B: a_i is outside the hexagon. easy to prove a_i a_j_k >a_j_k a_j_l for some k,l ∈{1,2,..6} than a_i is not shot 6 times.$
$Part (b) shot < 5 times .$ $Consider a_{i} is shot 6 times$ $by a_{j_{1}}, . . ., a_{j_{6}} .$ $consider hexagon formed by$ $a_{j_{1}} . . . a_{j_{6}} .$ $A : point a_{i} is inside the hexagom$ $Make a traingle with a_{i} and one$ $side of hexagon say a_{j_{1}} a_{j_{2} .}$ $a_{j_{1}} a_{i} < a_{j_{1}} a_{j_{2}}$ $a_{j_{2}} a_{i} < a_{j_{1}} a_{j_{2}}$ $a_{j_{1}} a_{j_{2}} is largest side so ∠ a_{j_{1}} {a a}_{j_{2}} > 60 °$ $So six sides of hexagon spawn an$ $angle > 60 \times 6 = 360 ° not possible .$ $case B : a_{i} is outside the hexagon .$ $easy to prove a_{i} a_{j_{k}} > a_{j_{k}} a_{j_{l}} for$ $some k, l \in {1, 2, . . 6} than a_{i} is not$ $shot 6 times .$
Commented by prakash jain last updated on 25/May/17

$Part (c)$ $Let a_{i} s h o o t a_{j} a n d a_{k} s h o o t s a_{l} .$ $Let us say a_{i} a_{j} intersects with a_{k} a_{l}$ $Let a_{i} s h o o t a_{j} a n d a_{k} s h o o t s a_{l} .$ $Consider quadilateral a_{i} a_{k} a_{j} a_{l}$ $so a_{i} a_{j} and a_{k} a_{l} are diagonals .$ $a_{i} a_{k} > a_{i} a_{j}$ $a_{i} a_{l} > a_{i} a_{j}$ $a_{k} a_{i} > a_{k} a_{l}$ $a_{k} a_{j} > a_{k} a_{l}$ $Will continue to prove contradiction .$
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