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  • .../math> (a radius of the circle) is 5, we can conclude that <math>\triangle BMO</math> is a 3-4-5 right triangle. Since <math>D</math> and line <math>AN</ Since <math>BO = 5</math>, <math>BM = 3</math>, and <math>\angle BMO</math> is right, <math>MO=4</math>. From quadrilateral <math>MNPO</math>, w
    20 KB (3,497 words) - 14:37, 27 May 2024
  • ...irc - (180^\circ - \angle{ACB}) = \angle{ACB}</math> and from <math>\angle{BMO} = \angle{MBO} =</math> <math>\angle{ABC}</math>. {{IMO box|year=2004|before=First Problem|num-a=2}}
    5 KB (941 words) - 23:51, 18 November 2023
  • '''2017 BMO''' problems and solutions. The 34th BMO was held in Ohrid, Former Yugoslav Republic of Macedonia on 2-7 May 2017. * [[2017 BMO Problems]]
    569 bytes (62 words) - 15:25, 17 September 2017
  • '''2007 BMO''' problems and solutions. The 24th BMO was held in Rhodes, Greece in 2007. * [[2007 BMO Problems]]
    531 bytes (56 words) - 15:26, 17 September 2017