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)