Thursday, December 10, 2015

Geometry Problem 1171: Area of a Triangle in terms of the three altitudes or heights

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

Click the figure below to view more details of problem 1171.

Online Math: Geometry Problem 1171: Area of a Triangle in terms of the three altitudes or heights.

3 comments:

  1. I use subscript 1 2 3 instead.

    Since K = 1/2 ah₁ = 1/2 bh₂ = 1/2 ch₃
    a = 2K/h₁
    b = 2K/h₂
    c = 2K/h₃

    Substitude into the result of the last problem, done.

    ReplyDelete
  2. Let p, q, r be the reciprocals of the altitudes drawn from A, B, C respectively.

    So a = 2Kp, b = 2Kq and c = 2Kr

    Per the result of Problem 1170 and my proof therein

    16K^2,= 2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4- b^4 - c^4 = (a+b+c)(a+b-c)(a-b+c)(-a+b+c) where I have used the difference between 2 squares to simplify the algebra

    This is = to 16K^4 (p+q+r)(p+q-r)(p-q+r)((-p+q+r) from which the result follows by dividing both sides of the equation by 16K^4 and taking the reciprocal and then the square root of both sides

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. calculate K.(1/ha+1/hb+1/hc)= 1/2(a+b+c)= p
    similarly K(-1/ha+1/hb+1/hc)= 1/2(-a+b+c)=p-a
    ....replace these value to the right hand side of the expression, we will get the result

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.