Asked by John

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 ?
Answered by Steve
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.


Answered by Steve
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
Answered by Abcd
Yes, solving gives HI^2= 792.
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