deriving big formulas with derive and what happened then david sjöstrand sweden
Post on 31-Dec-2015
221 Views
Preview:
TRANSCRIPT
The incenter of a triangle• A triangle has the
vertices (x1, y1), (x2, y2) and (x3, y3).
• In 1992 I calculated the coordinates of the incenter as the intersection, (x, y), between two of the angle bisectors of the triangle. I received this result. incenter.dfw
Vector notation
• If we identify points, X, and vectors,
we can write the above formula
aA bB cCT
a b c
OX��������������
Concurrent lines
• The angle bisectors, the altitudes, the medians and the perpendicular bisectors are all concurrent.
• When are three lines passing the vertices of a triangle concurrent?
• D is a point on the line passing the points A and B.
• Then there are real numbers a and b such that
aA bBb B D a D A D
a b
Definition
A point D given by
divides the segment AB into two parts in the ratio a/b, counted from B.
a/b > 0 iff D lies between A and B.
If a/b < 0 iff D does not lie between A and B.
iff = if and only if
aA bBD
a b
• A and B are two points.
• Then
is a point on the line passing A and B.
If we call this line, line(A, B) we can write
aA bB
a b
( , )aA bB
line A Ba b
Theorem 1
The lines
are concurrent.
Their point of intersection is
and ,bB aA
line Cb a
, ,bB cC
line Ab c
,cC aA
line Bc a
aA bB cCT
a b c
This means that if we have this situation
then we have three concurrent lines having the mentioned point of intersection.
Proof:
Therefore
In the same way we can prove that that
Q.E.D.
( ),
( )
bB cCaA b c bB cCb c line A
a b c b c
,aA bB cC bB cC
line Aa b c b c
,aA bB cC aA cC
line Ba b c a c
,aA bB cC aA bB
line Ca b c a b
Medians
If you let a = b = c in Theorem 1 you receive the well known formula for the intersection point of the medians of a triangle.
3
aA bB cC aA aB aC A B C
a b c a a a
There is a converse of Theorem 1
Theorem 2
If the three lines line(A, A1), line(B, B1) and line(C, C1) are concurrent, there are three real numbers a, b and c, such that
1 ,bB cC
Ab c
1
aA cCB
a c
1andaA bB
Ca b
Proof
A1 is on line(B,C).Therefore there are real numbers b and c, such that A1 =
(bB + cC)/(b + c).
We also have
B1 = (c1C + a1A)/(c1 + a1) =
(c/c1 (c1C + a1A))/(c/c1(c1 + a1)) =
(cC + aA)/(c + a),
where a = ca1/c1.
Now line(A,A1), line(B,B1) and
line(C,(aA+bB)/(a + b)) are concurrent according to Theorem 1.
Therefore C1 = (aA + bB)/(a + b).
Q.E.D.
Corollary 1
line(A, A1), line(B, B1) and line(C, C1) are concurrent if and only if
1 1 1
1 1 1
1AC BA CB
C B AC B A
Proof
If line(A, A1), line(B, B1) and line(C, C1) are concurrent, we have the situation in the figure according to Theorem 2.
Now
1 1 1
1 1 1
1AC BA CB b c a
C B AC B A a b c
If 1 1 1
1 1 1
1AC BA CB
C B AC B A
1 1 1
1 1 1
1AC BA CB b c e
C B AC B A a d f
c
g
b ac eb e b c e b cd e
b aa f a b f a bd fd e
b c ag c
a b g
Altitudes – Orthocenter
We get
Therefore the altitudes of a triangle are concurrent
cos cos cos1
cos cos cos
c B a C b A
b C c A a B
The vertices of a triangle are the midpoints of a given triangle (medians.dfw)
Using ITERATES to find the midpoint of a triangle
top related