Tangent + Bisector problem

Hi All,

I've been trying to solve this deceptively simple math problem for the last 2 days, but haven't had much luck yet. Any help/pointers much appreciated!

Here goes:

Given a triangle ABC, and a circle centered at A such that B & C are outside the circle. How can I find a point Q on the circle such that QM is tangential to the circle, and bisects angle BQC?

There's also a second such point on the other quadrant of the circle.

Replies to This Discussion

Permalink Reply by David Rutten on October 27, 2014 at 4:15pm

Just to be clear, both Q and M are up for grabs right? I probably won't have time to work on this though until next week, which is probably too late.

Permalink Reply by Suryansh Chandra on October 27, 2014 at 5:06pm

Yup.

Should've probably mentioned earlier.. M just happens to be the intersection of the angle bisector with BC. It can be anywhere on the line.

Next week is fine too if I still can't find a solution to this :)

if I find the solution I'll post here for sure!

Thanks a lot!

Permalink Reply by Pieter Segeren on October 27, 2014 at 5:07pm

Hi Suryansh (or is it Chandra?),
I have to admit I do not recognise a special situation here for which there is a rule. So here is an approach - of how I understood - using Galapagos (one side only for now). It found a solution with deviation of about 0.003 degrees. Maybe it could be a starting point.

Attachments:

Tangent&Bisector_PSG.gh, 23 KB

Permalink Reply by Suryansh Chandra on October 27, 2014 at 6:10pm

Hi Pieter,

Thanks for the reply.. I have a feeling there is an equation or a geometric way of building this. Galapagos would be good if the problem is small, but the diagram above is a small section of the problem: in my case this tangent-bisector needs to be found over 10,000 times for different ABC and radius every time..

Best,

Suryansh.

Permalink Reply by to] on October 27, 2014 at 5:14pm

Hi Suryansh,

how are you ? so see attached image for your question.

I think the bisector problem with tangent to the circle you can achieve with following approach.

01: bisector of CACB

02: draw line normal to the bisector and Extend the BC

03: new circle through AC and the new point

04: intersection with your circle is the tangent point

hope it works for you....

best to]

Permalink Reply by Suryansh Chandra on October 27, 2014 at 6:06pm

Hi Thomas,

Thanks for the solution -- its very interesting that you first create the isosceles equivalent of ABC and intersect its circumcircle with the original -- definitely very close. I gave it a shot and made a quick sketch, below screenshot and definition.. maybe I'm doing something wrong?

In any case, I owe you a beer! Let me know you're in London next :) I am going to dig further in this direction in the meantime..

PS. David.. I think there's a screenshot bug I also discovered :)

Attachments:

TangentBisector-01.gh, 14 KB

Permalink Reply by Suryansh Chandra on October 27, 2014 at 7:45pm

On further tinkering, I found that your method does work when you consider angles from Q to the points on the circumcircle, but not the original points B & C..

to be continued..

Permalink Reply by to] on October 29, 2014 at 8:58am

Hi,

For sure I will get in contact when I am around in London.

I worked a little bit further on it but still no success to come up with algorythm for that. I have a setup working for iterating through it till a certain threshold is hit but with 10000 elements this is not really an option.

What I know so far is that the intersection of the tangents are on the bisector .the circle and intersections are forming a conic. So quite curious who comes up with a solution as the regular triangle case is solved geometrical.

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