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

Labels:
circle,
common tangent,
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'

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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

ReplyDeletedrawingIn 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)