perplexus.info :: Shapes : Hexagon Hinder 2

Hexagon Hinder 2 (Posted on 2016-05-06)

The hexagon PQRSTU is such that:

(1) All its vertices lie on a circle, and:
(2) PQ = RS = TU, and:
(3) the diagonals PS, QT, RU meet at a point.

N is the intersection of PS and RT

Find:

 RN/NT
---------
(PR/RT)²

See The Solution Submitted by K Sengupta

Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Observation

| Comment 1 of 2

If the hexagon is regular all 3 qualifications hold and the ratio is easily seen to be 1.

This isn't the same as proving the more general case, of course.

Posted by Jer on 2016-05-07 21:35:32

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