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Hello Everyone.
With referring to Werner's suggestion and Ttokoro's suggestion on solving with MathCad ( or Prime )
https://community.ptc.com/t5/Mathcad/Butterfly-problem/m-p/848009#M204472
I'm interesting in it :
Given:
Let ABC be a triangle with incircle ( O ), and let ( O ) touch BC, CA, AB at A', B', C' respectively. M is a
arbitrary-point move on B'C'. Lines BM and CA meet at B''. Lines CM and AB meet at C''. Lines A'M
and B''C'' meet at A''.
Prove :
A'' move on incircle ( O ) ( with M move on B'C'. )
Thanks in advance for your time and help.
Best Regards.
Loi.
Solved! Go to Solution.
Here is a brute force attack with Mathcad.
Good luck for trying to let the symbolics prove it! 😈
But as you can see we can at least demonstrate the property for specific numeric values ...
EDIT: modified screenshot and attached file! In the originals the radius was calculated wrongly
This characteristic property of a tangent quadrilateral should help:
https://www.cut-the-knot.org/Curriculum/Geometry/CircumQuadri.shtml
Here is a brute force attack with Mathcad.
Good luck for trying to let the symbolics prove it! 😈
But as you can see we can at least demonstrate the property for specific numeric values ...
EDIT: modified screenshot and attached file! In the originals the radius was calculated wrongly
Many, many thanks for your time and help, Werner.:
Best Regards.
Loi.
I thought you were looking for a general Mathcad proof for ALL possible triangles, not just for a few specific ones?
Without loss of generality we can use B(0/0) and C(1(0) (or C(10/0) as you have chosen).
So we have three variables which can be chosen at will - the coordinates of A (we can assume that they are positive, and the parameter lambda for chosing M (which of course can also be outside o the triangle as in your examples, just chose lambda smaller than 0 or greater than 1).
But you will have already noticed that the symbolic evaluation of the general expression in my sheet seems to run endless without any result (but then, maybe I was too impatient).
Best Regards.