## Thursday, April 16, 2009

### Problem 30. Right Triangle, Incenter, Inradius

Proposed Problem

Level: High School, SAT Prep, College geometry

1. From F draw FD perpendicular to AB meeting the latter in D. Now ED = r, angle EAB = angle EAF = A/2, r = AE sin(A/2) and AE = AFcos(A/2) which gives us: r/sin(A/2) = AFcos(A/2) or AF = r/sin(A/2)cos(A/2) = 2r/sin(A). Finally, BG = FD = AFsin(A) = 2r/sin(A) * sin(A) = 2r.
Ajit: ajitathle@gmail.com

2. Let G be an intersecting point of the segment AB and the extension of FE and let
P,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.
Then we see that BQ=EP=EQ=r and
two triangles EPG and ERF are congruence.
Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r.

3. Let G be an intersecting point of the segment AB and the extension of FE and let
P,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.
Then we see that BQ=EP=EQ=r and
two triangles EPG and ERF are congruence.
Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r

4. Through E draw MN ∥ AB with M on AF and N on BG.
∠AEM = ∠EAB = A/2 -> ME = MA
∠MEF = 90° - A/2 = ∠MFE -> ME = MF
AB ∥ MEN ∥ FG
So BN = NG
But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°
-> EN = BN
So BN = NG = EN = r
Hence BG = 2r
(-> denotes "implies")

5. Slight Correction:
in the line
"But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°"
" ∠BEN = ∠EBN = 45° "

6. Extend EF to cut AB at M; since M is the symmetrical of F about E and FG||AB, FG is tangent to the incircle and thus we are done.
Best regards,
Stan Fulger

7. Since FG is parallel to AB we can see that FE bisects < AFG. So E is the ex centre of Tr. FGC. Hence EG bisects right < BGF and so < EGB = 45. But < EBG = 45 and so BG= 2r where r is the in radius = altitude of isoceles Tr. BEG

Sumith Peiris
Moratuwa
Sri Lanka