Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 703.

## Monday, December 12, 2011

### Problem 703: Triangle, Incircle, Tangency point, Diameter, Congruence

Labels:
circle,
congruence,
diameter,
incenter,
incircle,
tangency point,
triangle

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http://img851.imageshack.us/img851/4052/problem703.png

ReplyDeleteFrom E ,draw B’C’//to BC ( see picture)

Draw incircle of triangle AB’C’ with center O’ and tangent point D’

Let OD=r, O’D’=r’ , 2p’=perimeter of triangle AB’C’

1. Note that circle O is excircle of tri. AB’C’

So AO’/AO=r’/r=O’D’/OD , m(AO’D’)=m(AOD) (O’D’//OD)

Triangle AO’D’ similar to tri. AOD ( case SAS)

So m(O’AD’)=m(OAD) and 3 points A,D’, D are collinear.

2. We have B’E=p’-AB’ ( property of excircle O)

And C’O’=p’-AB’ ( property of incircle O’)

So B’E=C’D’ and BF=CD

Peter Tran

a trigonometric solution

ReplyDeletediameter DE // to altitude AH

in triangle DEF: tan E=DF/2r

in triangle AHF: tan E=HF/AH

HF=HD+DF

DF=2r.HD/(AH-2r)

after the substitutions:

HD=(s-c)-bcosC

AH=bsinC

2r=2S/s=absinC/s

DF=b-c

BF=CD=s-c

To Anonymous

ReplyDeleteRefer to above solution. It is not clear to me how we get DF=b-c from "HD=(s-c)-bcosC

AH=bsinC

2r=2S/s=absinC/s ". Please explain .

A synthetic proof:-

ReplyDeleteNote: Please draw a figure for clearly following the content, below:

Let the outward drawn perpendicular to BC at F meet AO extended at O’.

Let AL be the altitude from A upon BC. Let M be the midpoint of AL. Let AL = h

We claim that O’ is indeed the ex-centre opposite to A.

DE ∥ AL (each being ⊥ to BC).

∴ In ⊿ALF, the median FM which bisects AL also bisects ED (hence passes through O).

Now ∆AO’F ~ ∆AOE, and, ∆FEO ~ ∆FAM.

So O’F:r = O’F:OE = AF:AE = AF:(AF – EF)

= 1 : [1 – (EF:AF)] = 1: [1 – (OE:AM)]

= 1 : [1 – {r:[(½)h]}]

= 1 : [1 – (2r: h)]

∴ O’F = r : [1 – (2r: h)]

= 1 : [(1:r) – (2:h)] = 1 : [(s:∆) – (a:∆)] = ∆ : (s – a) = r'(the radius of the excircle opposite to A)

So F is precisely the point where the ex-circle opposite to A touches BC.

Hence BF = s – c = CD

to peter tran

ReplyDeleteit is in the equation

DF=2r.HD/(AH-2r)

that i substitut 2r,HD,AH and cosC=(a²+b²-c²)/2ab

to peter tran 2

ReplyDeletesorry! sign error

DF= -(b-c)=c-b

http://img29.imageshack.us/img29/5830/k0we.png

ReplyDelete