## Tuesday, February 5, 2013

### Problem 853: Square, Arc, Quadrant, Triangle, Perpendicular, Metric Relations

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

Click the figure below to see the complete problem 853.

1. Let FG=x, and FG meet AD at H.

Then
GC=2x, AH=BG=5−2x, FH=5−x, AF=5.

By Pythagoras theorem,
AH^2 + FH^2 = AF^2
(5-2x)^2 + (5−x)^2 = 5^2
x^2 − 6x + 5 = 0
(x − 1)(x − 5) = 0

Since x<5, thus x=1.
i.e. FG=1.

2. Let CF=x, FG=y and CE meet AD at H.Then ,2EC=CH and AH=5.
By Pythagoras theorem of the triangle HCD , 2EC=5sqrt(5)
CB^2=CF.CH , 25=x. 2EC
Triangles FGC,HCD are similar.So, x/CH=y/5 ,x=CH.y/5
Therefore 25=(2EC) . CH.y/5. =y.(2EC)^2/5 .So ,y=1
Image: http://img802.imageshack.us/img802/7513/geogebra7.png

3. Reflect D along AB to have D', such that D'AD is the diameter of the circle.
By pyth. thm. on CDD', we have CD' = 5*sqrt(5)
By power of circle, CB*CB = CF*CP => CF = sqrt5.
Since triangle(CGF)~triangle(CBE), CG:GF = CB:CE = 2:1,
so again by pyth. thm. on triangle(CGF), FG = 1

4. FC=sqrt(5)*FG by Pythagoras. EC meets AD at P such that PD=5*2 and is a diameter of circle A because triangle PDC is similar to CBE which has a leg to leg ratio of 2. Because PD is diameter, <PFD=90. Combined with <GCF=<CDF, we know that triangle GCF similar to CDF so GF/(sqrt(5)*GF)=(sqrt(5)*GF)/5, making GF=1.