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Re: [EMHL] a quadrilateral problem

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  • Nikolaos Dergiades
    Dear Mustafa your problem ... is with other words the 1st problem of the 18 IMO (1976). If a = AB, b = BD, c = DC, a+b+c=12 x = angABD, y = angBDC then
    Message 1 of 1 , Jul 31, 2003
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      Dear Mustafa

      your problem
      >ABCD is a convex quadrilateral with area 18.
      >If |AB|+|BD|+|DC|=12 then |AC|=?
      >This problem was asked in a math olympiad elimination in
      > Turkey.I couldn't solve it. Could you also try?
      >Are these given informations enough to solve?


      is with other words the 1st problem of the 18 IMO (1976).
      If a = AB, b = BD, c = DC, a+b+c=12
      x = angABD, y = angBDC then
      36 = 2(ABCD)
      = ab*sin(x)+bc*sin(y)
      <= b(a+c) <= ((a+b+c)/2)² = 36
      So we must have only equality i.e.
      sin(x) = sin(y) = 1, b = a+c = 6.
      Hence x = y = 90 and if the parallel from A to BD
      meets CD at E the right angled triangle AEC
      has AE = b = 6, EC = a+c = 6 and AC = 6sqrt(2).

      Best regards
      Nikolaos Dergiades
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