All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
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 Construction | Comment 4 of 11 |
This description may leave a little to be desired but I believe that it fulfils the requirements:

Let the centre of circle AB be A'.
Let the centre of circle CD be C'.

Construct square XYCB.

Join A' to X intersecting circle AC at F.
Join C' to Y intersecting circle CD at G.

Join F to G.

Construct a perpendicular bisector of FG to intersect line BC at E.

|FE|  =  |GE|.
  Posted by brianjn on 2010-10-04 02:20:35
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information