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Equal Tangents (Posted on 2010-10-03) Difficulty: 3 of 5
C(UV) denotes the circle with diameter UV.

T(P,QR) denotes the tangential distance |PS|,
where point P lies outside C(QR), point S
lies on C(QR), and PS is tangent to C(QR).

Let A, B, C, and D be distinct, collinear
points in that order.

Construct a point E on line AD such that
|EF| = T(E,AB) = T(E,CD) = |EG|





















See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution No Subject | Comment 3 of 11 |

Denote   O1E by b  and O2E by c

b^2-r1^2=c^2-r2^2

b^2-c^2=r1^2-r2^2

(b-c)*O1O2=(r1-r2)*BC

(b-c)=((r1-r2)*BC)/O1O2

all the quantities  on the right side are known:

r1-r2    difference of the radii

O1O2     distance between the circles' centers

BC        =O1O2- (R1+R2)

SO: 

b-c  IS CONSTRUCTABLE (several ways , inter alia similar triangles    or intersecting chords in a circle etc).

b-c and b+c  define point E.

 

Very nice problem!!

Edited on October 3, 2010, 3:42 pm
  Posted by Ady TZIDON on 2010-10-03 15:40:05

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