Monday, August 19, 2024

Geometry Problem 1579: Involving perpendicular diameters, tangents, and triangle areas.

Challenging Geometry Puzzle: Problem 1579. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1579

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2 comments:

  1. Let GH = h, GF = t, GE = q BE = p and radius = r

    Apply Pythagoras many times

    h^2 + 4 = p^2 + q^2 ...... (1) (Triangles GHB & BEG)
    h^2 + 64 = t^2 ................(2) (Triangle FGH)
    t^2 + r^2 = (r+p)^2 + q^2..(3) (Triangles OFG & OEG)

    (2) - (1) t^2 - p^2 - q^2 = 60 ....(4)
    (3) - (4) r^2 + p^2 + q^2 = (r+p)^2 + q^ - 60

    From which upon simplifying
    2rp = 60

    So Area S(BCE) = 1/2 rp = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Alternate solution using similar triangles

    Let FB and GE extended meet at P and let BE = p and radius = r
    Let < PBE = < ABF = < OFB = @

    Since GHBE is concyclic, < EGH = @
    Since FOEG is concylic, < AOF = 2@ = < FGE
    So < HGF = HGP = @

    Hence H is the midpoint of FP and BP = 6

    Now Triangles ABF & BEP are similar

    So p/6 = 10/2r which yields
    pr = 30 and therefore

    Area S(BCE) = 1/2 rp = 15

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete