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