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 





Ah 


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=a^{2}√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] 


OS(a/3) 


OS1 [(a+√3x)/3] 


TS(2a/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 ) 
 

Ah 


Bi 


Cj 


A1C1 (a+√3x) 


A1B1 (a+√3x) 


incenter : point where bisectors
of the 3 angles meet 


circumcenter (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 orthocentroidal circle. Mid point
N between O & H is the center of nine
point circle. 


Orthocenter (H): meeting point of perpendiculars from 3
vertex points to opposite sides 


circumcenter(O), centroid(G), orthocenter(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. 


circumcenter, centroid, orthocenter lie in one
line which is called Euler Line. In
equilateral triangles, all 4 points as above coincide. Normally, the
incenter 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.





