Saturday, November 29, 2014

Geometry Problem 1064: Triangle, Orthocenter, Altitudes, Circumradius, Sum of the Squares of Sides

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1064: Triangle, Orthocenter, Altitudes, Circumradius, Sum of the Squares of Sides.

3 comments:

  1. Let BB’ is the diameter of circumcircle of ABC
    We have AH//B’C ( both lines ⊥ to BC)
    And CH//AB’ ( both line ⊥ to AB)
    So AHCB’ is a parallelogram => B’C=AH=a1 and AB’=CH=c1
    In right triangle BCB’ we have BC^2+B’C^2= a^2+a1^2= BB’^2= 4R^2
    Similarly we also have b^2+B1^2=4R^2 and c^2+c1^2= 4R^2
    So a^2+b^2+c^2+a1^2+b1^2+c1^2= 12 R^2

    ReplyDelete
  2. Fie A2 punctul diametral punctului A, B2 punctul diametral punctului B si C2 punctul diametral punctului C=>
    HAB2C ,HBC2A, HBA2 paralelograme=>a.a+a1.a1=4R.R,b.b+b1.b1=R.R ,c.c+c1.c1=R.R si in final relatia didedemonstrat,am aplicat teorema lui pitagora in triunghiurile dreptunghice BCB2,ACC2,ABA2

    ReplyDelete
  3. Let O be the center of the circle.
    In ΔBOC, a² = 2R²(1−cos2A)
    In ΔAOC, b² = 2R²(1−cos2B)
    In ΔAOB, c² = 2R²(1−cos2C)

    Reflect H in the lines BC, CA and AB to get H₁, H₂ and H₃ resp.
    In ΔAOH₂, a₁² = 2R²(1+cos2A)
    In ΔBOH₃, b₁² = 2R²(1+cos2B)
    In ΔCOH₁, c₁² = 2R²(1+cos2C)

    Summing up the six equations,
    a² + b² + c² + a₁² + b₁² + c₁² = 12R²

    ReplyDelete