| 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=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] |
|
|
| 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) |
|
|
| 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.
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