Description Notation Value
side AB a
stretch bond by length (AA') x
Find out
Bond length (OA) OA
stretch bond by length (AA') x1
Find out
AO 2*Ah/3
A'O A1O (AO+x)
Oh Ah/3
hh' hh1(x/2)
Oh' Oh1=1/2(x+a/√3)
A'h' A1h1:(3x+a√3)/2
A'B' A1B1(a+x√3)
perimeter     ABC periA=3a
area             ABC areaA=a2√3/4
perimeter    A'B'C' periA1
area             A'B'C' areaA1
periA/areaA pAA(4√3/a)
periA'/areaA' pA1A1(4√3)/(a+√3x)
periA/periA' pAA1[1/(1+x√3/a)]
areaA/areaA' aAA1 (pAA1*pAA1)
circumcircle radius  ABC crABC
incircle radius          ABC irABC
circumcircle radius  A'B'C' crA1B1C1
incircle radius          A'B'C' irA1B1C1
circumcircle area     ABC caABC
incircle area             ABC iaABC
circumcircle area    A'B'C' caA1B1C1
incircle area            A'B'C' iaA1B1C1
circumcircle radius/incircle radius : ABC rarABC (2)
circumcircle radius/incircle radius : A1B1C1 rarA1B1C1(2)
circumcircle radiusA'B'C'/circumcircle radiusABC crdivide (1+(x√3/a))
ij=jh=hi (the triangle hij is called the medial triangle & the sides are 1/2 of parallel side of ABC. so hi is parallel to & half of AB & so also others.) a/2
i'j'=j'h'=h'k' (a+x√3)/2=ij+x√3/2
AR (Ah/2)  
RO ( Ah/6)  
suppose, there is an atom at O and 1 atom each of the same element at A',B',C' . Now draw a perpendicular at O at 90 degree to the plane of the equilateral triangle i.e. XY plane and lift the atom to the point k. As point k moves upward from O,the medians A'h', B'i',C'j' are shrunk, say by x to Ah, Bi, Cj respectively. Then    
the height KO is : KO= √[x(2OB' -x)]  
KO ( max)=A1O when x=OB'  
KO (min)= zero when x=0  
Angle AKO  in degree of the AK with vertical:  cos-1(KO/OB')  
O(x,y,z) --
T(-a/3,y,z) --
S(a/3,y,z) -
T1 --
S1 -
K -
OT (a/3)  
OT1 [(a+√3x)/3]  
OS1 [(a+√3x)/3]  
T1S1  [2(a+√3x)/3]  
A(x,a/√3,z) -
A1(x, [a+√3x]/√3,z) -
h(x,-a/2√3,z) -
h1 -
B(-a/2 +x ,-a/2√3,z) -
B1(-[a+√3x]/2 +x, -[(a+√3x)/2√3],z) -
C(a/2 +x, -a/2√3,z) -
C1([a+√3x]/2 +x, -[(a+√3x)/2√3],z) -
i(Ax+Cx / 2 ,Ay+Cy / 2 , Az + Cz  /2 ) -
j(Ax+Bx / 2 ,Ay+By / 2 , Az + Bz  /2 ) -
i1(A1x+C1x / 2 ,A1y+C1y / 2 , A1z + C1z  /2 ) -
j1(A1x+B1x / 2 ,A1y+B1y / 2 , A1z + B1z  /2 ) -
A1C1 (a+√3x)  
A1B1 (a+√3x)  
in-center : point where bisectors of the 3 angles meet    
circum-center (O): point where perpendicular bisectors of 3 sides meet.    
centroid (G): point where 3 medians meet. G is between O & H. GH=2GO; OH=3GO;GH is the diameter of the ortho-centroidal circle. Mid point N between O & H is the center of nine point circle.    
Ortho-center (H): meeting point of perpendiculars from 3 vertex points to opposite sides    
circum-center(O), centroid(G), ortho-center(H) lie in one line which is called Euler Line.O lies on one side at half the distance to that of H lying at the other side of G. GO=HG/2 and HO=3GO.    
circum-center, centroid, ortho-center lie in one line which is called Euler Line. In equilateral triangles, all 4 points as above coincide. Normally, the in-center does not lie in the Euler line except isoceles triangle ( of course, in equilateral triangles, all 4 points coincide)    
in a triangle, if perimeter remains constant and keeping 1 side fixed, other 2 sides are varied so that their sum remains the same, the resultant locus will be an ellipse with the 2 extreme points of the fixed line as the foci of the ellipse.    
in a triangle perimeter remaining constant, equlateral triangle has maximum area.    
Euler Line Explanation    

It's surprising that these three points lie on a straight line. But you might see why from the picture.

Focus your attention on the centroid G. For each point, like A on one side of it, there is another, like A' on the other side of it but half as far away. On one side is B, the other B'; on one side C, the other C'. In fact, this correspondence sends the whole triangle ABC to the smaller, but similar, triangle A'B'C', called the medial triangle. The sides of the medial triangle A'B'C' are parallel and half the length of the sides of the original triangle ABC.

You can see from the figure that this correspondence sends the altitudes of the original triangle, which are AD, BE, and CF, to the altitudes of the medial triangle, which are A'D', B'E', and C'F'. Since the altitudes of the original triangle meet at the orthocenter H of the original triangle, the altitudes of the medial triangle will meet at its orthocenter H' which you can see in the figure is labelled O. Behold! This orthocenter O of the medial triangle is the circumcenter of the original triangle! Thus, this correspondence sends H to O, that is, H and O are on the opposite sides of the centroid G, and O is half as far away from G as H is.