Proposed Problem
Click the figure below to see the complete problem 336 about Two equal circles, a Common Tangent and a Square.
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Complete Problem 336
Level: High School, SAT Prep, College geometry
Friday, August 7, 2009
Problem 336. Two equal circles, a Common Tangent and a Square
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Extend FM to meet AC in N. Also extend DA & EB to meet the two circles in D' & E' respectively.
ReplyDeleteNow, AN = r - x/2 & NM = r - x. From triangle MAN, (r-x/2)^2+(r-x)^2=r^2 from which we get x=2r/5 or x=2r. The former is the side of square FGHM while the latter is the side of DEE'D'
vihaan: vihaanup@gmail.com
Extend MH so that it meets the radii AD and BE at points P and Q
ReplyDeleteLet PM = HQ = y
Then we have two equations:
(r-x)^2 + y^2 = r^2 ..... (1)
and
2y + x = 2r ..........(2)
Eliminate y between the two equations to get a quadratic which resolves to
x = (2/5)*r or x = 2r.
x = 2r is absurd.
So x = (2/5)*r is the answer
which will
See diagram here.
ReplyDeleteLet O middle of FG and extend EB to intersect circle B in K
∆OGH and ∆OEK are SAS similar and DE tangent to B in E => ∠OHG = ∠HKE = ∠OEH
Hence ∆OGH and ∆HGE are similar => EG = 2x and OE = r = 5x/2 QED
See the drawing
ReplyDeleteIn ΔDMF : DM^2=DF^2+FM^2
(1) DM^2= (r-x/2)^2+x^2
In ΔAMD : AM^2=AD^2+DM^2 -2xAD (Proposition 13 Euclide Book II)
r^2= r^2+ DM^2 -2xr
(2) DM^2=2xr
(1) and (2) 2xr=(r-x/2)^2+x^2
2xr=r^2-xr+x^2/4+x^2
3xr=r^2+5x^2/4
Dividing by x^2 (x not equal to 0) =>3r/x=5/4+(r/x)^2
Let y=r/x : y^2-3y+5/4=0
Delta = 4 and y=5/2 or y=1/2
Therefore x=2r/5 or x=2r
Since the only definition of M and H are
M belongs to the square and circle A,
and H belongs to the square and Circle B
Therefore, there are 2 different solutions:
x=2r/5 (blue square)
or x=2r (green square)
(r-x)^2+(r-x/2)^2=r^2
ReplyDeleter^2-2rx+x^2+r^2-rx+x^2/4=r^2
r^2-3rx+5x^2/4=0
4r^2-12rx+5x^2=0
(2r-5x)(2r-x)=0
x=2r (rej.) or x=2r/5