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        riddles >> medium >> Cyclic quadrilateral and orthogonal circle
        
(Message started by: O2009 on Oct 18^th, 2009, 9:59pm)

Title: Cyclic quadrilateral and orthogonal circle
Post by O2009 on Oct 18^th, 2009, 9:59pm

Let ABCD be a cyclic quadrilateral with no parallel sides. Let AB and CD intersect at M and let AD and BC intersect at N. Prove that the circle with MN as diameter is orthogonal to the circle through ABCD.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by humbern on Mar 23^rd, 2010, 8:26pm

I've been playing with this far too long, with no joy.

Cyclic quads, ortho circs, harmonic division.

Can the author or a kind soul help out?

...And explain how this is "easy"?

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by towr on Mar 24^th, 2010, 3:09am

Nice picture, what software did you use for that?

Anyway, I think by now we can safely assume this problem isn't easy. I'll move it to medium for now.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by humbern on Mar 24^th, 2010, 8:12am

The software is Cinderella. The frustration is mine. Any ideas out there?

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by Ronno on Apr 1^st, 2010, 6:10am

Yes I agree, this should be at least in medium. Having solved this before I'll give a hint.

In your diagram, notice that [hide]AC, BD and LT are concurrent[/hide]

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by humbern on Apr 2^nd, 2010, 2:17pm

Thank you!
And . . . cool!
And . . . I have to think about this....

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by malchar on Apr 2^nd, 2010, 10:24pm

What does it mean for circles to be orthogonal?

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by towr on Apr 3^rd, 2010, 3:11am

on 04/02/10 at 22:24:30, malchar wrote:

What does it mean for circles to be orthogonal?

It means that at the points where the circles cross, if you draw the tangents of the circles, those tangents are at 90 degree angle.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by humbern on May 17^th, 2010, 9:26am

Somebody please post an answer. This is killing me.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by Grimbal on May 19^th, 2010, 5:44am

I also don't know the answer.

But it seems that the line MN is the polar of the intersection of AC and BD.
http://mathworld.wolfram.com/InversePoints.html
I don't know if that helps.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by JohanC on May 21^st, 2010, 1:05pm

on 05/19/10 at 05:44:23, Grimbal wrote:

I also don't know the answer.

But it seems that the line MN is the polar of the intersection of AC and BD.
http://mathworld.wolfram.com/InversePoints.html
I don't know if that helps.

In that case, you might also be interested in
http://www.cut-the-knot.org/Curriculum/Geometry/SymmetryInCircle.shtml
and http://www.cut-the-knot.org/Curriculum/Geometry/InversionDemo.shtml
There are some nice applets to play with.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by Noke Lieu on May 23^rd, 2010, 9:04pm

... I struggle to check if I've got the cart before the horse, but taking a simplistic approach...

MTN is 90^o
Point Y is where TN bisects circle ABCD

extend PT throught to the other side of the circle ABCD to point Z
That also gives TYZ as 90^o

From there, angle MT(red circle tangent)= angle TZY... and so on.

(did have a diag, but can't seem to save it, sorry)

actually, I reckon I've stuffed up- This only follows if they're orthongonal.

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by Immanuel_Bonfils on May 27^th, 2010, 9:38pm

Grimbal went close!
Indeed, line MN is polar of the intersection of AC and BD, let's call it X..
But the best is that the line MK is polar of N wrt circumcircle O1, of the quadrilateral ABCD, so MK is perpendicular to ON, let's say at Y, and Y is a point of the circle O2 with diameter MN.

So from the condition for Y to be the inverse (or the image) of N wrt O1 we have

|OY|.|ON| = |OT|², where T is a point of O1 and |OT| is his radius.

But |OY|.|ON| = |OT|² is also the power of O wrt O2, so OT is the tangent to O2 taken from O . Done .

It's circle inverse geometry, but I think folks aren't much fan of, since I've post some time ago a problem in the subject http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1214764919 and nothing happened...

Title: Re: Cyclic quadrilateral and orthogonal circle
Post by humbern on Jun 1^st, 2010, 12:22pm

Thank you so much! This is actually really thrilling, although I'm still trying to put your reasoning into the simplest possible terms, given my purely recreational background. But I see the through-line. You might be interested to know that the forward to the recent biography of Donald Coxeter was written by Douglas Hofstadter (of Goedel, Escher, Bach fame), and he recalls his biggest aha! moment of math education as being taught about circle inversion. I'll be thinking about your "nothing happened" problem as soon as I can explain this current one to my co-workers, who have watched me banging my head...