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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-gun




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.
$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}.\:\mathrm{On} \\ $$$$\mathrm{some}\:\mathrm{field},\:{n}\:\mathrm{gunmen}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{all}\:\mathrm{pairwise}\:\mathrm{distances}\:\mathrm{between} \\ $$$$\mathrm{them}\:\mathrm{are}\:\mathrm{different}.\:\mathrm{At}\:\mathrm{a}\:\mathrm{signal},\:\mathrm{every} \\ $$$$\mathrm{gunman}\:\mathrm{takes}\:\mathrm{out}\:\mathrm{his}\:\mathrm{gun}\:\mathrm{and}\:\mathrm{shoots} \\ $$$$\mathrm{the}\:\mathrm{closest}\:\mathrm{gunman}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{gunman}\:\mathrm{survives}; \\ $$$$\left(\mathrm{b}\right)\:\mathrm{no}\:\mathrm{gunman}\:\mathrm{is}\:\mathrm{shot}\:\mathrm{more}\:\mathrm{than}\:\mathrm{5} \\ $$$$\mathrm{times}; \\ $$$$\left(\mathrm{c}\right)\:\mathrm{the}\:\mathrm{trajectories}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bullets}\:\mathrm{do} \\ $$$$\mathrm{not}\:\mathrm{intersect}. \\ $$
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_(ij) .  Since all distances are different  there exist minimum a_(ij)  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≠i,k≠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. (pigeonhole principle)  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.
$$\mathrm{Part}\:\left({a}\right) \\ $$$$\mathrm{Let}\:\mathrm{gunmen}\:\mathrm{be}\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,….,{a}_{{n}} \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{us}\:\mathrm{distances}\:\mathrm{are}\:{a}_{{ij}} . \\ $$$$\mathrm{Since}\:\mathrm{all}\:\mathrm{distances}\:\mathrm{are}\:\mathrm{different} \\ $$$$\mathrm{there}\:\mathrm{exist}\:\mathrm{minimum}\:{a}_{{ij}} \:\mathrm{for}\:\mathrm{some} \\ $$$$\mathrm{value}\:\mathrm{of}\:{i}\:\mathrm{and}\:{j}.\:\mathrm{Since}\:{a}_{{i}} \:\mathrm{will} \\ $$$$\mathrm{shoot}\:{a}_{{j}} \:\mathrm{and}\:{a}_{{j}} \:\mathrm{will}\:\mathrm{shoot}\:{a}_{{i}} . \\ $$$$\mathrm{Case}\:\mathrm{A}:\:{a}_{{k}} \left({k}\neq{i},{k}\neq{j}\right)\:\mathrm{fires}\:\mathrm{at} \\ $$$$\:\:\:\:\:\:\:\:{a}_{{i}} \:\mathrm{or}\:{a}_{{j}} . \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:{a}_{{i}} \:\mathrm{or}\:{a}_{{j}} \:\mathrm{get}\:\mathrm{shot}\:\mathrm{twice}. \\ $$$$\mathrm{Since}\:\mathrm{there}\:\mathrm{are}\:\mathrm{only}\:{n}\:\mathrm{bullets}\:\mathrm{and} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{person}\:\mathrm{got}\:\mathrm{shot}\:\mathrm{twice}\:\mathrm{then} \\ $$$$\mathrm{there}\:\mathrm{will}\:\mathrm{be}\:\mathrm{one}\:\mathrm{person}\:\mathrm{who}\:\mathrm{does} \\ $$$$\mathrm{get}\:\mathrm{shot}\:\mathrm{at}\:\mathrm{all}.\:\left({pigeonhole}\:{principle}\right) \\ $$$$\mathrm{Case}\:\mathrm{B}:\:\mathrm{No}\:\mathrm{one}\:\mathrm{else}\:\mathrm{fires}\:\mathrm{at}\:{a}_{{i}} \:\mathrm{or}\:{a}_{{j}} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{take}\:{a}_{{i}} \:\mathrm{and}\:{a}_{{j}} \:\mathrm{out}\:\mathrm{of}\:\mathrm{field} \\ $$$$\mathrm{and}\:\mathrm{consider}\:\mathrm{remaining}\:\left({n}−\mathrm{2}\right)\:\mathrm{gunmen}. \\ $$$$ \\ $$$$\mathrm{Continue}\:\mathrm{this}\:\mathrm{process}\:\mathrm{finally}\:\mathrm{there}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{3}\:\mathrm{gunmen}\:\mathrm{left}\:\mathrm{where}\:\mathrm{only}\:\mathrm{case}\:\mathrm{A} \\ $$$$\mathrm{is}\:\mathrm{possible}. \\ $$$$ \\ $$
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.
$$\mathrm{Part}\:\left({b}\right)\:\mathrm{shot}<\mathrm{5}\:\mathrm{times}. \\ $$$$\mathrm{Consider}\:{a}_{{i}} \:\mathrm{is}\:\mathrm{shot}\:\mathrm{6}\:\mathrm{times} \\ $$$$\mathrm{by}\:{a}_{{j}_{\mathrm{1}} } ,…,{a}_{{j}_{\mathrm{6}} } . \\ $$$$\mathrm{consider}\:\mathrm{hexagon}\:\mathrm{formed}\:\mathrm{by} \\ $$$${a}_{{j}_{\mathrm{1}} } …{a}_{{j}_{\mathrm{6}} } .\: \\ $$$$\mathrm{A}:\:\mathrm{point}\:{a}_{{i}} \:\mathrm{is}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{hexagom} \\ $$$$\mathrm{Make}\:\mathrm{a}\:\mathrm{traingle}\:\mathrm{with}\:{a}_{{i}} \:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{hexagon}\:\mathrm{say}\:{a}_{{j}_{\mathrm{1}} } {a}_{{j}_{\mathrm{2}} .} \\ $$$${a}_{{j}_{\mathrm{1}} } {a}_{{i}} <{a}_{{j}_{\mathrm{1}} } {a}_{{j}_{\mathrm{2}} } \\ $$$${a}_{{j}_{\mathrm{2}} } {a}_{{i}} <{a}_{{j}_{\mathrm{1}} } {a}_{{j}_{\mathrm{2}} } \\ $$$${a}_{{j}_{\mathrm{1}} } {a}_{{j}_{\mathrm{2}} } \:\mathrm{is}\:\mathrm{largest}\:\mathrm{side}\:\mathrm{so}\:\angle{a}_{{j}_{\mathrm{1}} } {aa}_{{j}_{\mathrm{2}} } >\mathrm{60}° \\ $$$$\mathrm{So}\:\mathrm{six}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{hexagon}\:\mathrm{spawn}\:\mathrm{an} \\ $$$$\mathrm{angle}\:>\mathrm{60}×\mathrm{6}=\mathrm{360}°\:\mathrm{not}\:\mathrm{possible}. \\ $$$$\mathrm{case}\:\mathrm{B}:\:{a}_{{i}} \:\mathrm{is}\:\mathrm{outside}\:\mathrm{the}\:\mathrm{hexagon}. \\ $$$$\mathrm{easy}\:\mathrm{to}\:\mathrm{prove}\:{a}_{{i}} {a}_{{j}_{{k}} } >{a}_{{j}_{{k}} } {a}_{{j}_{{l}} } \:\mathrm{for} \\ $$$$\mathrm{some}\:{k},{l}\:\in\left\{\mathrm{1},\mathrm{2},..\mathrm{6}\right\}\:\mathrm{than}\:{a}_{{i}} \:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{shot}\:\mathrm{6}\:\mathrm{times}. \\ $$
Commented by prakash jain last updated on 25/May/17
Part (c)  Let a_i  shoot a_j  and a_k  shoots a_l .  Let us say a_i a_j  intersects with a_k a_l   Let a_i  shoot a_j  and a_k  shoots 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.
$$\mathrm{Part}\:\left({c}\right) \\ $$$$\mathrm{Let}\:{a}_{{i}} \:{shoot}\:{a}_{{j}} \:{and}\:{a}_{{k}} \:{shoots}\:{a}_{{l}} . \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:{a}_{{i}} {a}_{{j}} \:\mathrm{intersects}\:\mathrm{with}\:{a}_{{k}} {a}_{{l}} \\ $$$$\mathrm{Let}\:{a}_{{i}} \:{shoot}\:{a}_{{j}} \:{and}\:{a}_{{k}} \:{shoots}\:{a}_{{l}} . \\ $$$$\mathrm{Consider}\:\mathrm{quadilateral}\:{a}_{{i}} {a}_{{k}} {a}_{{j}} {a}_{{l}} \\ $$$$\mathrm{so}\:{a}_{{i}} {a}_{{j}} \:\mathrm{and}\:{a}_{{k}} {a}_{{l}} \:\mathrm{are}\:\mathrm{diagonals}. \\ $$$${a}_{{i}} {a}_{{k}} >{a}_{{i}} {a}_{{j}} \\ $$$${a}_{{i}} {a}_{{l}} >{a}_{{i}} {a}_{{j}} \\ $$$${a}_{{k}} {a}_{{i}} >{a}_{{k}} {a}_{{l}} \\ $$$${a}_{{k}} {a}_{{j}} >{a}_{{k}} {a}_{{l}} \\ $$$$\mathrm{Will}\:\mathrm{continue}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{contradiction}. \\ $$

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