Results 1 to 4 of 4

Math Help - Help with Feuerbach's circle proof using dilations

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    13

    Help with Feuerbach's circle proof using dilations

    I included the assignment as an attachment. I am trying to prove Feuerbach's Nine Point circle theorem using dilations. For the first question, I think that if {\delta}(O)=O' if I let {\delta}(Q)=Q' where Q and Q' are points on their respective circles, then I essentially have two triangles formed where side OQ has length r, radius of C(O, r) and side O'Q' with length r', radius of C(O',r'=kr). I am not sure if my reasoning is valid but if it is, how can I use this to deduce that Q' will always be on C(O', r').


    I really appreciate the help.
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    Quote Originally Posted by ordinalhigh View Post
    I included the assignment as an attachment. I am trying to prove Feuerbach's Nine Point circle theorem using dilations. For the first question, I think that if {\delta}(O)=O' if I let {\delta}(Q)=Q' where Q and Q' are points on their respective circles, then I essentially have two triangles formed where side OQ has length r, radius of C(O, r) and side O'Q' with length r', radius of C(O',r'=kr). I am not sure if my reasoning is valid but if it is, how can I use this to deduce that Q' will always be on C(O', r').


    I really appreciate the help.
    You practically have the answer already:
    You have an arbitrary circle with center O and radius r, and an arbitrary point Q on that circle. And you have defined a circle with center O' (the image of O) and radius kr. So all you need to know is if Q' (the image of Q) is on the second circle. That will be true if the distance from O' to Q' is kr.
    So you have defined two triangles XOQ and XO'Q'. Are they similar? If so, the ratio of the sides are all equal....

    I think you can take it from there. I think if you draw diagrams for the other three problems you'll be able to do them, too.

    Let me know if you have any more trouble.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    13
    Quote Originally Posted by hollywood View Post
    You practically have the answer already:
    You have an arbitrary circle with center O and radius r, and an arbitrary point Q on that circle. And you have defined a circle with center O' (the image of O) and radius kr. So all you need to know is if Q' (the image of Q) is on the second circle. That will be true if the distance from O' to Q' is kr.
    So you have defined two triangles XOQ and XO'Q'. Are they similar? If so, the ratio of the sides are all equal....

    I think you can take it from there. I think if you draw diagrams for the other three problems you'll be able to do them, too.

    Let me know if you have any more trouble.
    I think I understand it now. The similarity of the two triangles implies that the segment O'Q' is of length kr, so it must lie on the circle C(O',r').

    I think that for the second one, I can use midline property to show that the line BC of a triangle ABC is a perpendicular bisector of the extended altitude from A to the point on the circumcircle since the segment HA' will be perpendicular to BC.

    I think I have an idea of how the results will be used to prove the Nine point circle theorem but I am stuck on problem three. How can I show that the segments from the orthocenter to the diametral opposites of the points A, B, C, will bisect the sides of the triangle?
    Last edited by ordinalhigh; March 8th 2010 at 06:55 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    I finally have your result (too late, I guess).

    Consider the quadrilateral BHCA'.

    Since AA' is a diameter and B is on the circle, ABA' is a right angle.
    If you continue CH to a point X on AB, it is an altitude, so CX is perpendicular to AB.
    So CH is parallel to A'B.

    Since AA' is a diameter and C is on the circle, ACA' is a right angle.
    If you continue BH to a point Y on AC, it is an altitude, so BY is perpendicular to AC.
    So CA' is parallel to HB.

    Thus the quadrilateral BHCA' is a parallelogram, and the diagonals of a parallelogram bisect each other, so HA' and BC bisect each other.

    Are you taking geometry at a university? The assignment seems a little difficult for high school. What did you end up turning in? And what was your professor's solution?

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Applying dilations and reflection help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 9th 2010, 05:17 AM
  2. transformations, dilations help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 11th 2009, 05:45 PM
  3. another dilations question
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 14th 2008, 03:22 AM
  4. simple question about dilations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 13th 2008, 12:44 AM
  5. Dilations Help
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 18th 2007, 03:40 PM

Search Tags


/mathhelpforum @mathhelpforum