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

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

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

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

Theorem 1 The lines are concurrent. Their point of intersection is

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.

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.

There is a converse of Theorem 1 Theorem 2 If the three lines line(A, A 1 ), line(B, B 1 ) and line(C, C 1 ) are concurrent, there are three real numbers a, b and c, such that

Proof A 1 is on line(B,C).Therefore there are real numbers b and c, such that A 1 = (bB + cC)/(b + c). We also have B 1 = (c 1 C + a 1 A)/(c 1 + a 1 ) = (c/c 1 (c 1 C + a 1 A))/(c/c1(c 1 + a 1 )) = (cC + aA)/(c + a), where a = ca 1 /c 1. Now line(A,A 1 ), line(B,B 1 ) and line(C,(aA+bB)/(a + b)) are concurrent according to Theorem 1. Therefore C 1 = (aA + bB)/(a + b). Q.E.D.

Corollary 1 line(A, A 1 ), line(B, B 1 ) and line(C, C 1 ) are concurrent if and only if

Proof If line(A, A 1 ), line(B, B 1 ) and line(C, C 1 ) are concurrent, we have the situation in the figure according to Theorem 2. Now

If

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

The vertices of a triangle are the midpoints of a given triangle (medians.dfw) Using ITERATES to find the midpoint of a triangle