Posted by John on Wednesday, March 6, 2013 at 11:59pm.
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  Steve, Thursday, March 7, 2013 at 10:51am
From the secantsecant 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  Steve, Thursday, March 7, 2013 at 11:01am
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  Abcd, Thursday, March 7, 2013 at 12:05pm
Yes, solving gives HI^2= 792.
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