1 / 27

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).

frederique-krish
Télécharger la présentation

Deriving big formulas with Derive and what happened then

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Deriving big formulas with Derive and what happened then David Sjöstrand Sweden

  2. How technology inspired me to learn more mathematics David Sjöstrand Sweden

  3. 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

  4. I used the big formulas to plot inscribed circles in Excel. INSC.XLS

  5. If I had neglected to make the below assignments I had received a much nicer result

  6. A nicer result for the incenter

  7. Vector notation • If we identify points, X, and vectors, we can write the above formula

  8. 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?

  9. D is a point on the line passing the points A and B. • Then there are real numbers a and b such that

  10. 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

  11. Example D divides the segment AB in the ratio -9/13 because -13(B - D) = 9(D-A)

  12. 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

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

  14. This means thatifwe have this situation then we have three concurrent lines having the mentioned point of intersection.

  15. Proof: Therefore In the same way we can prove that that Q.E.D.

  16. 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.

  17. 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

  18. 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.

  19. Corollary 1 line(A, A1), line(B, B1) and line(C, C1) are concurrent if and only if

  20. 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

  21. If

  22. Thus we have the situation we have in Theorem 1 Therefore the lines are concurrent

  23. Altitudes – Orthocenter We get Therefore the altitudes of a triangle are concurrent

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

More Related