## Thursday, February 28, 2013

### Problem 861: Triangle, Incircle, Parallel, Tangent, Metric Relations

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

Click the figure below to see the complete problem 861.

1. http://img16.imageshack.us/img16/3367/problem861.png
Let r= radius of incircle and L is tangent point of GH to incircle
Connect OG,OA,OE,OH,OF,OC
Note that triangles AOG and HOG are right triangles and OE and OF are altitudes.
Relation in right triangle AOG give OE^2=r^2= EG.AE= 7
Relation in right triangle HOC give OF^2=r^2=GH.HC=2.x=7
So x=7/2

1. Why triangles AOG and HOG are right triangles ?

2. see link below for the sketch
https://photos.app.goo.gl/UAI7OSNynfSzZVoa2

since HL, HC and CD tangent to circle O
so ∠(FHO) =1/2 ∠(LHF) and ∠(OCF)= 1/2∠(FCD)
note that ∠(LHF) supplement to ∠(FCD)
so ∠(FHO)+ ∠(OCF)= 90 degrees

2. Let is OE=OF=r
angle DEM= angle DFM=90 degrees , why MD is diameter of circle
ED is perpendicular of AO and EM is perpendicular of GO. So, angle GOA=90 degrees.
Similarly , angle HOC=90 degrees.
From the right triangle GOA we have, OE^2=AE.EG=7 or, r^2=7 (1)
From the right triangle COH we have, OF^2=CF.FH=7 o r, r^2=2x (2)
From (1),(2) ,3x=7 , x=7/2
http://img268.imageshack.us/img268/5052/74920469.png

3. Let BG=y, BH=z.

Since BG/BA = BH/BC = GH/AC and BE = BF
y/(y+8) = z/(z+2+x) = 3/(7+x) and y+1 = z+2

xy+yz+2y = yz+8z
3x+3z+6 = xz+7z
xy+7y = 3y+24
y+1 = z+2

Solving those equations, we have
x=7/2, y=16/5, z=11/5

4. Proiectam G si H pe AC si daca folosim Teorema lui Pitagora obtinem (x+2)(x+2)-(x-2)(x-2)=8.8-6.6,de unde x=7/2

5. Let BG = p so that BH = p-1 since BE = BF

From similar triangles
p/(p+8) = (p-1)/(p+1+x) = 3/((7+x)

Each ratio = 1/(7-x) where I have used the algebraic identity that if a/b = c/d then each ratio = (a-c) /(b-d)

So 3(7-x) = 7+x and so x = 7/2

Sumith Peiris
Moratuwa
Sri Lanka