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

Home > Shapes > Geometry
The Real Cyclic Quad (Posted on 2014-08-23) Difficulty: 3 of 5
About a set of four concurrent circles (circles passing through a one common point) of same radius r,four of the common tangents are drawn to determine the circumscribing quadrilateral ABCD.Prove that ABCD itself is a cyclic quadrilateral.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 1.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Part of it. | Comment 1 of 2

It's certainly true if all the tangents are drawn exterior to the group of circles.

Assuming this is what is meant by 'circumscribing quadrilateral':

Call the centres of the circles O1,O2,O3,O4.

Then quadrilateral O1O2O3O4 is cyclic, because the one common point, p, is exactly one radius away from each of O1,O2,O3,and O4.

Now line O1O2, for example, connects two of the centres of the circles. The common tangent is parallel to this line, hence the points of tangency are each exactly one more perpendicular radius away from O1 and O2.

The result, after  the same procedure is completed for the remaining 3 sides of  O1O2O3O4,  is that the angles of the external quadrilateral ABCD are the same as those of O1O2O3O4.

Accordingly, ABCD is itself also a cyclic quadrilateral.

 - Over to you, Bractals, for the rest of it.


Edited on August 23, 2014, 11:34 pm
  Posted by broll on 2014-08-23 11:45:34

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information