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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.)
Some Thoughts Algebraic solution | Comment 1 of 11

Consider an x axis where A is at 0, and B,E,C, and D are at x values of B,E,C, and D.

With a little help from Pythagoras:
(E - B/2)^2 - (B/2)^2 = (C-E + (D-C)/2)^2 - ((D-C)/2)^2

Which leads to:
E = (D*C)/(D+C-B)

... although I think a compass and straight edge solution is what is being sought


  Posted by Larry on 2010-10-03 14:14:53
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