Deriving big formulas with Derive and what happened then
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Transcript of Deriving big formulas with Derive and what happened then
Deriving big formulas with Derive and what happened
then
David Sjöstrand
Sweden
How technology inspired me to learn more mathematics
David Sjöstrand
Sweden
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
I used the big formulas to plot inscribed circles in Excel.
INSC.XLS
• If I had neglected to make the below assignments I had received a much nicer result
A nicer result for the incenter
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
D divides the segment AB in the ratio -9/13 because -13(B - D) = 9(D-A)
Example
• 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
Thus we have the situation we have in Theorem 1
Therefore the lines are concurrent
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