# Gogeometry Problem 285

From Gogeometry

Using :

• R radius of CircleO, and r of CircleF
• Define
• ro radius of CircleO , rd of CircleD, rc of CircleC and rf of CircleF
• Define
• G intersection of OA and CircleD
• H intersection of CircleD and CircleF
• I intersection of CircleC and CircleF
• J intersection of CircleO and CircleF
• CircleD if tangent to CircleF => D, H and F are aligned and DF=rd+r
• CircleC if tangent to CircleF => C, I and F are aligned and CF=rc+r
• CircleO if tangent to CircleF => O, F and J are aligned and OF=ro-r
• Let’s say R=6 regardless of the units
• OB=6
• OC=3
• From Pb 284:
• Therefore
• OD=4
• OF=6-r
• DF=2+r
• CF=3+r

• Is ODFC a rectangle?
• If it is true, then, obviously (DF=OC) and therefore r=1

• Let’s suppose ODFC is not a rectangle…
• Define D’ and C’ respectively orthogonal projection of D and C on OA and OB
• Define m=DD’ and n=CC’
• We have 4 different cases

CASE A : D’ and C’ are respectively in OD and OC

Proposition 13 Euclide Book II => OF^2=OD^2+DF^2-2 OD DD’

• (6-r)^2=(4)^2+(2+r)^2-2(4)m
• 36-12r+r^2=16+4+4r+r^2-8m
• 16+8m=16r
• 2+m=2r
• m=2r-2

Proposition 13 Euclide Book II => OF^2=OC^2+CF^2-2 OC CC’

• (6-r)^2=(3)^2+(3+r)^2-2(3)n
• 36-12r+r^2=9+9+6r+r^2-6n
• 18+6n=18r
• 3+n=3r
• n=3r-3
• OD’FC’ is a rectangle=>
• OC’^2+OD’^2=OF^2
• (3-n)^2+(4-m)^2=(6-r)^2
• (3- 3r+3)^2+(4- 2r+2)^2=(6-r)^2
• (6-3r)^2+(6-2r)^2=(6-r)^2
• 36-36r+9r^2+36-24r+4r^2=36-12r+r^2
• 12r^2-48r+36=0
• r^2-4r+3=0
• Delta=16-4×3=4
• r=(4+-2)/2
• r=3 or r=1
• If r=3 => m=4 flat triangle : impossible
• Therefore r=1 => m=0 and n=0
• Therefore ODFC is a rectangle

CASE B : D’ in OD and C in OC’

• OD’FC’ is a rectangle=>
• OC’^2+OD’^2=OF^2
• (3-n)^2+(4-m)^2=(6-r)^2
• (3- 3r+3)^2+(4- 2r+2)^2=(6-r)^2
• (6-3r)^2+(6-2r)^2=(6-r)^2
• 36-36r+9r^2+36-24r+4r^2=36-12r+r^2
• 12r^2-48r+36=0
• r^2-4r+3=0
• Delta=16-4×3=4
• r=(4+-2)/2
• r=3 or r=1
• If r=3 => m=4 flat triangle
• Therefore r=1 => m=0 and n=0
• Therefore ODFC is a rectangle

CASE C : D in OD’ and C’ in OC

• OD’FC’ is a rectangle=>
• OC’^2+OD’^2=OF^2
• (3-n)^2+(4+m)^2=(6-r)^2
• (3+3r-3)^2+(4- 2r+2)^2=(6-r)^2
• (3r)^2+(6-2r)^2=(6-r)^2
• 9r^2+36-24r+4r^2=36-12r+r^2
• 12r^2-48r+36=0
• r^2-4r+3=0
• Delta=16-4×3=4
• r=(4+-2)/2
• r=3 or r=1
• If r=3 => m=-4 : cannot be negative
• Therefore r=1 => m=0 and n=0
• Therefore ODFC is a rectangle

CASE D : D in OD’ and C in OC’

• OD’FC’ is a rectangle=>
• OC’^2+OD’^2=OF^2
• (3+n)^2+(4+m)^2=(6-r)^2
• (3-3r+3)^2+(4- 2r+2)^2=(6-r)^2
• (6-3r)^2+(6-2r)^2=(6-r)^2
• 36-36r+9r^2+36-24r+4r^2=36-12r+r^2
• 12r^2-48r+36=0
• r^2-4r+3=0
• Delta=16-4×3=4
• r=(4+-2)/2
• r=3 or r=1
• If r=3 => m=-4 : cannot be negative
• Therefore r=1 => m=0 and n=0
• Therefore ODFC is a rectangle

Conclusion : In all cases ODFC is a rectangle and r=1

There are many colinear points in this drawing :