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