Hi Benoit, nice idea to start a new thread to talk about this stuff. This 
is an answer to you too Theirry :)

The goals I have been working towards with this area of a triangle stuff 
are 1. supporting Heron <http://us.metamath.org/mpeuni/heron.html>'s 
theorem and 2. working towards Pick's theorem from the 100 list, however 
that is supposedly very hard 
<https://www.cl.cam.ac.uk/~jrh13/papers/pick.pdf> so it's a long way off. 

So far I have done areaquad <http://us.metamath.org/mpegif/areaquad.html>, 
to explain how it works look at the left image here 
<https://i.imgur.com/CvuWUQl.png>. It says given a quadrilateral with sides 
parallel to the y-axis the area is 1/2 x. (F + E - (D + C)) x. (B - A), 
where those are all just single real numbers, for example the top left 
vertex is at (A,F) e. RR^2. If you have the situation where one side is 0 
length, for example E = C, then this reduces to area = 1/2 x. (F - C) x. (B 
- A), which is 1/2 x. base x. height as expected. 

If you look at the right diagram you can see how any triangle can be 
expressed as two of these quads connected together. For this shape we get, 
I think, 1/2 x. (W - Y) x. (R - P). This is a slightly odd result and not 
how areas of triangles are usually expressed. My guess would be it would be 
good to take this result and then change it into a more familiar form.

I think the way area is expressed in Heron is quite natural, where the area 
of triangle is basically the cross product, |A||B|sin theta where A and B 
are two of the sides as vectors and theta is the angle between them.

Benoit I like your way of using barycentric coordinates to express the 
interior of the triangle, that is quite elegant. I am not sure how to fit 
it into this scheme though, if you look at U and V in areaquad 
<http://us.metamath.org/mpegif/areaquad.html> those are the upper and lower 
lines of the quad and defining things parallel to the y axis makes the area 
calculation much easier. I was imagining continuing to use that definition 
but with more pieces, having the left part and the right part of the 
triangle defined separately, however I am aware of how inelegant that is.

Thierry I hope I have answered some questions about what the final formula 
would look like. I am sorry I don't understand your triangle formulation 
here <https://groups.google.com/d/msg/metamath/DDN4noc0TYU/1bs81kSoAwAJ>, I 
think maybe you are using cross product somewhere? I like the idea of 
starting with just 3 complex numbers, that has a nice feel to it, and is 
the same as Heron.

In general I really like working collaboratively so I am grateful for all 
the help so far and if anyone wants to collaborate on moving these things 
forward I would like that, I am not so strong in metamath in general. :)

-- 
You received this message because you are subscribed to the Google Groups 
"Metamath" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
To view this discussion on the web visit 
https://groups.google.com/d/msgid/metamath/bc0b60d6-ea00-4847-8f02-b9fa96e349f6%40googlegroups.com.

Reply via email to