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Geometry

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A circle \Gamma cuts the sides of a equilateral triangle ABC at 6 distinct points. Specifically, \Gamma intersects AB at points D and E such that A, D, E, B lie in order. \Gamma intersects BC at points F and G such that B, F, G, C lie in order. \Gamma intersects CA at points H and I such that C, H, I, A lie in order. If |AD| =3, |DE| =39, |EB| = 6 and |FG| = 21 , what is the value of |HI|^2 ?

  • Geometry - ,

    From the secant-secant rule,

    FB*21 = 6*39, so FB=78/7
    AD+DE+EB = 3+39+6 = 48
    CG+21+FB = 48, so CG = 111/7

    CH*HI = CG*CF = 111/7 * 258/7 = 28638/49
    AI*HI = 3*39 = 117
    AI+IH+HC = 48

    117/HI + HI + 28638/49HI = 48

    now "just" solve for HI

    The fractions look nasty, so you better check my arithmetic.

  • Geometry - ,

    Oops. I was multiplying the wrong segments:

    FB(FB+21) = 6*45
    FB = 9
    FB+FG+CG=48 so CG = 18

    CH*(CH+HI) = 18*39 = 702
    AI*(AI+HI) = 3*42 = 126
    AI+HI+CH = 48

    Things look better now

  • Geometry - ,

    Yes, solving gives HI^2= 792.

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