Sunday, September 17, 2017

Geometry Problem 1346: Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, Distance

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

Details: Click on the figure below.

Geometry Problem 1346: Equilateral Triangle, Point on the Incircle, Altitude, Perpendicular, Sum of Squares, Distance.

5 comments:

  1. An attempt at the solution, but feel like not a complete one

    Join BD, AD and CD and consider the individual triangles, BDA, ADC, CDB and let the side of the eqilateral triangle be 'a'
    S(BDA) = 0.5*a*d
    S(ADC) = 0.5*a*e
    S(CDB) = 0.5*a*f

    As we know S(ABC) =S(BDA)+S(ADC)+S(CDB)
    => 0.5*a*h = 0.5*a*(d+e+f)
    => h = d+e+f

    From the above, we can conclude that D is independent and can be any point on the incircle.
    Assuming D on the altitude and at the tip of the incircle, we have
    r = h/3 ( r-Radius of incircle)
    f = 2r = 2h/3
    d = e = h/6 ( Since BD = h/3 )

    Therefore d^2+e^2+f^2 = h^2/36+h^2/36+4h^2/9 = 18h^2/36 = h^2/2

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    Replies
    1. h = d+e+f holds true for any point inside the triangle

      Delete
  2. I too deduced that h = d+e+f using the same reasoning but could not proceed further

    So we need to prove that de+ef+fd = h^2/4 to arrive at the result.

    Only issue is the above proof does not use the fact that D can be any point on the incircle

    However I'm sure Antonio has a different straightforward proof. Antonio will u confirm do that we may continue to try?

    ReplyDelete
  3. In other words h = d+e+f is true for any point inside the equilateral triangle not just for a point on the in circle

    ReplyDelete
  4. Let O be the centre of the incircle. Let BO meet AC at X and let Y be the foot of the perpendicular from O to DG

    Since = S(ABC) = S(ABD) + S(ACD) + S(BCD), we infer easily that
    h = d+e+f …(1)

    Now O is also the centroid of ∆ABC,
    so OX = h/3 = OD.
    Also DY = f – h/3
    It is also not difficult to see that,
    OY= f/√3 + 2d/√3 - h/√3

    Hence applying Pythagoras to ∆ODY,
    h2/9 = (f-h/3)2 + (f+2d-h)2/3 which simplifies to

    4(f2 + d2) + h2 = 4(dh + fh – df)…..(2)

    Similarly we can show that
    4(f2 + e2) + h2 = 4(eh + fh – ef)…..(3)

    (2) + (3) gives us,

    4(d2 + e2 + f2) + 4f2 + 2h2
    = 4h(d+e+f) + 4f(h-d-e)
    = 4h2 + 4f2 from (1)

    Upon simplification,

    d2 + e2 + f2 = h2/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete