UniversoFormulas
Work about the second lesson: Laboratory Notes
Consider an isosceles triangle. Let R be the radius of its circumscribed circle and r the radius of
its inscribed circle.
First section: Prove that the centers of both circles do not coincide
Second sectio: Calculate their distance in terms of r and R
First section:
The statement of the activity giveS me a isosceles triangle. If I use the Euler's line for the
geometry case, I will GET several results where the points could be situated along the line given
and it depends on whether the triangle where I will be working is an isosceles or equilateral.
The Euler´s line possesses the more characteristic points of a triangle, Orthocenter,
circumcenter and centroid regardless of the shape of the triangle given. It´s not important if
it is a equilateral, isosceles or scalene, all points are above the Euler´s line.
However, incenter point, the point where the three bisections of their inNer angles crossES, it
may or may not be on the Euler´s line no matter the type of angle.
If the incenter point is above the Euler´s line and it doesn´t coincide with the rest of the
important points then the triangle is isosceles.
If the incenter points is on the Euler´s line and it coincides with the the rest of the important
points then the triangle is equilateral.
I demonstrate this first section using argument to absurdity.
I suppose that the centter of both circles coincide, that is, I deny the initial proposition by
searching a contradiction.
If circumcenter and incenter points are the same, that is, they coincide in the plane then I know
that it must be an equilateral triangle by Euler´s line. As the shape of triangle given is an
isosceles, I arrive to the next contradiction: The triangle is both equilateral and isosceles.
Thus, the initial proposition is false.
Figure of isosceles triangle and equilateral triangle. The letters represent respectively O
orthocenter, G circumcenter, H centroid and I incenter.
As the shape of the left triangle given is an isosceles, the Euler´s line from this triangle has all
importans points, orthocenter, circumcenter, centroid and incenter above it.
The shape of triangle given is an isosceles,
Second section:
In order to obtain the distance between the two centers of the two different circumferences, in
this case, circumcenter and incenter, like a function of the terms r and R.
I will make use of from Euler´s line in the two dimensional Euclidean space.
The statement of the theorem says:
Let d be the distance between the incenter and circumcenter of a triangle
Then
2
d =R ( R−2 r )
Where R is the radius of circumcenter and and r incenter.
In order to prove the afored mentioned theorem, I am going to apply two lemmas for my
developing demonstration. I will begin with the following image where I show (a los cuales)
elements to which I will refer during (durante) the demonstration.
These elements are points, angles and lengths like radius.
The statement of the first lemma is
Let the incenter of △ABC be I
Let the circumcenter of △ABCbe O
Let OI be produced to the circumcircle at G and J
Let CI be produced to the circumcircle at P
Let F be the point where the incircle of △ABC meets BC
I’m given that:
the distance between the incenter and the circumcenter is d
The inradius is r
The circumradius is R
Then
IP⋅CI=(R+d)(R−d)
Thus, with the first lemma I get a mathematical expression which relates the inradius r,
circumradius R and distance between them d.
Inside the first lemma, it have used concepts like intersecting Chords Theorem to affirm what
lemma says.
As proof of Intersecting Chords Theorem is short I briefly explain how it is with other image.
Join A with B and C with D, as shown in the next
image:
Then I have:
∠AEB ≅ ∠DEG by (agente causa) two straight lines
make equal opposite angles
∠BAE≅ ∠CDE by Angles are equal in the same
segment of circle
It is well known that (es bien conocido) triangles with
two equal angles are similar and they have the next
congruence relation
△AEB∼△DEC
This relation indicates first and second triangle have
the same shape but in different position in the plane.
Thus,
AE DE
=
EB EC
AE∙ EC =DE ∙ EB
QED.
The statement of the second lemma is
Let the bisector of angle C of triangle △ABC be produced to the
circumcircle at P.
Let the I be the incenter of △ABC
By first lemma
AP=BP=IP
Substituting
IP ∙ CI=(R+d )(R−d)
By second lemma
IP=BP
And so
GI ∙ IJ =PB ∙ CI
Now using the extension of law of sines in △CPB
PB
=2 R
sin(∠ PCB)
And so
GI⋅IJ=2Rsin(∠PCB)⋅CI
By fourth Euclid´s common notions
∠PCB=∠ICF
And so
GI⋅IJ=2Rsin(∠ICF)⋅CI
We have that
IF=r
and by Radius at Right Angle to Tangent:
∠IFC is a right angle.
By the definition of sine
r
sin (∠ ICF)=
CI
and so
sin ( ∠ ICF ) ∙ CI =r
Substituting in (1)
GI ∙ IJ =2 Rr
( R+d ) ∙ ( R−d ) =2 Rr
2 2
R −d =2 Rr by difference of two squares
2 2
d =R −2 Rr
2
d =R ( R−2 r )
Thus, I have arrived to the object of this second section where it should be the mathematical
expression of the distance betweeen (no among) centers.
It apparently is the correct solution due to distance d is a function where its variables are the
radius r and R.
