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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)

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Question Does this work? | Comment 5 of 11 |

I'm not sure whether this works:

1.  Draw the circles, connect their centres, O1 and O2, and extend that line to A and D.
2. Construct perpendicular radii from the two centres to O1' and O2'. Construct the lines O2'O1, crossing the O2 circle at  G, and O1'O2, crossing the second circle at F. Construct the radii O1F and O2G.
3. Construct O1F extended and O2Y extended, crossing at some point T. Now by constructing or copying, place an image of the diagram, rotated by 180 degrees, over the diagram, with O2 at O1, and vice versa. There will now be a quadrilateral, O1-T-O2-T' between the circles.
4.  Construct a line between T and T'; this will cross AD at E; lines drawn from E to F and G will be of equal length and form tangents to the circles at F and G.
I wasn't able to prove formally that this construction 'works', however Ady's expression b^2-c^2=r1^2-r2^2, may assist, as this is equivalent to (r1+r2)(r1-r2), and by construction, the angles TO1T'and TO2T' are equal to the sum of the angles at O1O1'F and O2O2'G, while the angles O1TO2 and O1T'O2 are clearly each ((90-a)+(90-b)).
Either a proof or a disproof would be greatly appreciated!

  Posted by broll on 2010-10-04 02:53:11
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