Proposed Problem
See also: Complete Problem 285, Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, April 25, 2009
Problem 285: Circular Sector 90 degrees, Semicircles, Circle, Tangent, Radius
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We can use Descarte's Theorem for 4 mutually tangent circles: 2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2
ReplyDeletewhere a,b.c & d are the circle curvatures given by a=-1/R (the curvature is negative for the outermost circle),b=2/R,c=3/R (by problem 284)and d=1/r. On substitution we get, d=2c or r=R/6
Ajit: ajitathle@gmail.com
Simpler:
ReplyDeleteBuild semicircles with centres C and D.
Build a circle with center F' with radius R/6 and touching the first two.
Extent OF' with another R/6 to a point Z
But OF' = CD = 5/6R (OCF'D is a rectangle ; radius of the D semicircle is R/3)
Now OZ is equal to OB, therefore Z is on the O quarter circle (since the centers of the O circle and the F' circles and Z are on the same line)
M.
Even simpler is the following: Let AD = DE = a. (R/2+a)^2=(R-a)^2+(R/2)^2 giving a = R/3. Now, 2R/3 = OD = CF = R/2 + r or r = R/6
ReplyDeleteAjit, this is only valid if you can show tht DFCO is a rectangle, which you have not shown.
ReplyDeleteHow would you show that DFCO is a rectangle?
ReplyDeleteCenters and tangency points are colinear...
ReplyDeleteSee the drawing
ReplyDeleteR 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:
AD=2
OD=4
OF=6-r
DF=2+r
CF=3+r
Is ODFC a rectangle?
If it is true, then, obviously (DF=OC), 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=16-4x3=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 B : D’ in OD and C in 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=3-3r
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=16-4x3=4
r=(4+-2)/2
r=3 or r=1
If r=3 => m=4 flat triangle which is not
Therefore r=1 => m=0 and n=0
Therefore ODFC is a rectangle
CASE C : D in OD’ and C’ in 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=2-2r
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=3-3r
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=16-4x3=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’
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=2-2r
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=3-3r
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=16-4x3=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
Therefore r=R/6
Brilliant Harvey !
ReplyDeleteI think this is a problem which becomes much simpler using circle reflection or inversion. We use the largest circle as the reflection circle, the two R/2 circles become two parallel lines, and the two R/3 circles become circles of the same radii R as the big circle (which is left unchanged of course). Then, the centers of the circles involved in the inverted plane form an isosceles triangle of base 2R, side R+r' (r' being the radius of the inverted small circle), and altitude R-r', from where we obtain r'=R/4. With very little algebra we invert back that value to get r=R/6.
ReplyDeleteThis is synthetic geometry in my view, although using a more sophisticated tool than Euclid's elements.
DA=Radius of green semi-circle=R/2-r
ReplyDeleteOD=R-(R/2-r)=R/2+r
OC=R/2
CD=R/2+(R/2-r)=R-r
OD^2+OC^2=CD^2
(R/2+r)^2+(R/2)^2=(R-r)^2
R^2/4+Rr+r^2+R^2/4=R^2-2Rr+r^2
3Rr=R^2/2
R=6r
r=R/6