Proposed Problem
Click the figure below to see the complete problem 418 about Triangle, Incircle, Inradius, Equal Tangent circles, Radius.
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Complete Problem 418
Level: High School, SAT Prep, College geometry
Friday, January 15, 2010
Problem 418: Triangle, Incircle, Inradius, Equal Tangent circles, Radius
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join A to D & to O, C to E & to O ( D & O on bisector )
ReplyDeletejoin D to E, meet r at P
r meet AC at T
mark FP = z, AT = y
▲DPO ~ ▲ATO, => (x+z)/y = (r-x)/r (1)
▲PEO ~ ▲TCO, => (x-z)/(b-y) = (r-x)/r (2)
from (1) & (2)=>
(x+z)/y = (x-z)/(b-y)
xb-xy+zb-zy = xy-zy
xb+zb = 2xy
x+z = 2xy/b (3)
substitute (3) at (1)
(2xy/b)/y = (r-x)/r
2x/b = (r-x)/r
2xr = br-bx
(2r+b)x = br
x = br/(2r+b)
I the incenter
ReplyDeletethe two circles touch AC at G and H
AG+HC=b-2x
[AIC]=[ADG]+[CEH]+[GDEH]+[DIE]
br=x(b-2x)+4x²+2x(r-x)
br=x(b+2r)
x=br/(b+2r)
.-.
Let L, M, N be the feet of the Perpendiculars from D, E, I upon AC respectively.
ReplyDeleteDenote AL = y, CM = z.
Note AN = s - a, CN = s - c
DL // IN implies x:r = y:(s-a)
EM // IN implies x:r = z:(s-c)
From b = y + 2x + z, we have
rb = ry + rz + 2rx = x(s-a)+ x(s-c) + 2rx
=> rb = x(b + 2r), x = br/(b +2r)