Centres - returns centres of a 2d triangle

[Avi]

Centres is a little function that returns Incentre, Orthocentre, Centroid or Circumcentre of a given 2d triangle.
It can also return Euler line's equation for a triangle.

The incenter is the intersection of the angle bisectors of the triangle. The centroid is the intersection of the lines from each vertex of the triangle to the middle of its opposite side. The circumcenter is the intersection of the perpendicular bisectors of the sides. The orthocenter is the intersection of the altitudes of the triangle.

Euler later proved that the centroid, circumcenter and orthocenter are collinear in any triangle. The line that these three points are on in a triangle is called the Euler Line. It is defined for every triangle except an equilateral triangle, where all the points coincide.

EXAMPLE

Code: Select all

;p:=centres([[4, 0], [-2, 4], [0, 6]], "o")
;msgbox % "Orthocentres - `n x=" p.1 "`n y=" p.2
;return

;-------------------- Function ----------------------
; r := i = incentres
; m = centroid
; c = circumcentre
; o = orthocentre
; leave r blank to return eqn of euler line

centres(z, r:=""){
	mx := (z.1.1+z.2.1+z.3.1)/3 , my := (z.1.2+z.2.2+z.3.2)/3 	;medians done
	s := ( c := sqrt( (z.2.1 - z.1.1)**2 + (z.2.2-z.1.2)**2 ) ) + ( a := sqrt( (z.3.1-z.2.1)**2 + (z.3.2-z.2.2)**2 ) )
		+ ( b := sqrt( (z.3.1-z.1.1)**2 + (z.3.2-z.1.2)**2 ) )
	ix := (a*z.1.1+b*z.2.1+c*z.3.1)/s , iy := (a*z.1.2+b*z.2.2+c*z.3.2)/s 	; incentres done
	midx_a := (z.3.1+z.2.1)/2 , midy_a := (z.3.2+z.2.2)/2 , slp_a := -1*(z.3.1-z.2.1)/(z.3.2-z.2.2) 	;perpendicular slope
	cc_a := midy_a-(slp_a*midx_a)  		; b = y - mx
	midx_b := (z.3.1+z.1.1)/2 , midy_b := (z.3.2+z.1.2)/2 , slp_b := -1*(z.3.1-z.1.1)/(z.3.2-z.1.2)
	cc_b := midy_b-(slp_b*midx_b) 		; b = y - mx
	cx := (cc_b-cc_a)/(slp_a-slp_b) , cy := cc_a + (slp_a*cx) 	; y := b+mx --- circumcentres done
	oc_a := z.1.2-(slp_a*z.1.1) 	; b = y - mx
	oc_b := z.2.2-(slp_b*z.2.1)
	ox := (oc_a-oc_b)/(slp_b-slp_a) , oy := oc_a + (slp_a*ox) 	; orthocentres done
	if r in m,i,c,o
		return [Round(%r%x), Round(%r%y)]
	else return "y=" Round(m:=(oy-cy)/(ox-cx)) "x+" Round(oy-m*ox)
}

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[Morpheus]

Just out of curiosity, assuming that mass is distributed equally across the surface of the triangle, which of the above centers (If any) would be the center of gravity?

[Alibaba]

I'm very sure it's the centroid...

[Avi]

It's not orthocentre and circumcentre as they are present at a vertex and side respectively in a right-angled triangle.
I am confused between incentre and centroid as both look to be candidates for C.O.M location when tested with few simple triangles (equilateral, isoceles, right-angled) . There should be proof with integration.

@Alibaba_ any reason why you are sure ?

[strobo]

One can parametrize
f(t, s) := a+(b-a)t+(c-b)s , for 0 <= t <= 1 and 0 <= s <= t
where a, b, c are the vertices of the triangle. If the triangle has nonvanishing 2 dimensional Volume, i.e. det(b-a, c-b) != 0, one can use f(t, s) to integrate over the triangle.
(1/Volume)*\int_0^1 dt \int_0^t ds |det(Df)|*f(t, s) == (a+b+c)/3
which is the center of mass (in case it is distributed as Morpheus said).