## Monday, December 12, 2011

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

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

Click the figure below to see the complete problem 703. 1. http://img851.imageshack.us/img851/4052/problem703.png
From 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)
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

2. a trigonometric solution
diameter 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

3. To Anonymous

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

4. A synthetic proof:-
Note: 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

5. to peter tran
it is in the equation
DF=2r.HD/(AH-2r)
that i substitut 2r,HD,AH and cosC=(a²+b²-c²)/2ab

6. to peter tran 2
sorry! sign error
DF= -(b-c)=c-b

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