• 锐角△ABC中,O,G,H分别是外心、重心、垂心.设外心到三边距离和为d外,重心到三边距离和为d重,垂心到三边距离和为d垂.求证:1?d垂+2?d外=3?d重.试题及答案-解答题-云返教育

    • 试题详情

      锐角△ABC中,O,G,H分别是外心、重心、垂心.设外心到三边距离和为d,重心到三边距离和为d,垂心到三边距离和为d
      求证:1?d
      +2?d=3?d

      试题解答


      见解析
      证明:设△ABC外接圆半径为1,三个内角记为A,B,C.
      易知d
      =OO1+OO2+OO3=cosA+cosB+cosC,
      ∴2d
      =2(cosA+cosB+cosC).①
      ∵AH
      1=sinB?AB=sinB?(2sinC)=2sinB?sinC,
      同样可得BH
      2=2sinC?sinA,CH3=2sinA?sinB.
      ∴3d
      =△ABC三条高的和=2?(sinB?sinC+sinC?sinA+sinA?sinB) ②,
      BH
      sin∠BCH
      =2,
      ∴HH
      1=cosC?BH=2?cosB?cosC.
      同样可得HH
      2,HH3
      ∴d
      =HH1+HH2+HH3=2(cosB?cosC+cosC?cosA+cosA?cosB) ③,
      ∴①+③,得1?d
      +2?d=2(cosA+cosB+cosC)+2(cosB?cosC+cosC?cosA+cosA?cosB),
      =2(cosA+cosB+cosC+cosB?cosC+cosC?cosA+cosA?cosB),
      观察①、②、③,可得(cosB?cosC+cosC?cosA+cosA?cosB)+(cosA+cosB+cosC)=sinB?sinC+sinC?sinA+sinA?sinB.
      则1?d
      +2?d=3?d
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