Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Euler's Line Parallel to a Triangle's Side

  1. Feb 6, 2006 #1
    Can Euler's line be parallel to any of a triangle's sides? I'm sure it can, but when is it and how could I prove this?
     
  2. jcsd
  3. Feb 6, 2006 #2
    I have a problem like Euler line is parallel to the side BC of a Tr.ABC. Now prove that tanB*tanC = 3. I think you will be good enough to help me. I had posted this back in homework column but no one could help me.
     
  4. Feb 7, 2006 #3
    I don't think I'll be able to help you, I'm still learning about Euler's line myself. Sorry.
     
  5. Feb 13, 2006 #4
    I'm sorry about bumping the thread, but I'd really like to know. Is there no one who could tell me?
     
  6. Feb 14, 2006 #5
    Same is my condition. I have never seen anything about Euler line. This is the first question where I have seen that for the first time.
    I think that 9-point circle theorem, and the relations in trigonometry when properly used should get the result. I have not tried. But even I don't know whether this works I want you also to try this way since we can collectively help mutually each other. Forget the help from experts who would wimply look at the question and give you the answer.

    Another way of finding answer is co-ordinate geometry. Ask me if you couldn't follow the any of the things I am talking.
    I have the determination to solve this question by ourselves. Hope even you have that mommentum much with you.
     
  7. Feb 14, 2006 #6
    Thank you for your interest, it seems either no one knows the answer or no one has the time or will to supply it. Replies are few, while views are in abundance.

    I've taken a look at the nine point circle, but it doesn't bring any inspiration. It's the first time I've looked into it, though, and only shallowly.

    I'm not sure what you mean by co-ordinate geometry, could you explain what that is?
     
  8. Feb 14, 2006 #7
    Take side BC on X axis and then express A as (c*cosB, c*sinB). Now express centroid and other points on Euler line in terms of A, B and C. We know that the slope of the line joining these points is 0 or they have the same Y-co-ordinate. This idea struck me today and I haven't tried it. I will post the results I get from that method by tommorow.
     
  9. Feb 15, 2006 #8
    Well I took a look at it. I attached an image I made. When Euler's line is parallel to BC, HD=GE=OK. That's as far as I get, I'm not sure where to go next.
     

    Attached Files:

  10. Feb 15, 2006 #9

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I am not familiar with the Euler line, could you please define it for me? Also, it would help if you were to define the points on your diagram. I see a bunch of lines, some of which I understand and others I don't, help me understand it.
     
  11. Feb 16, 2006 #10
    In the image I've drawn the altitudes, intersecting at Orthocenter H, the medians, intersecting at Centroid G and the perpendicular bisectors, intersecting at Circumcenter O. The points H, G and O, as well as some other special points, I believe, are collinear, the line through them is called Euler's line.
    I've also drawn a line through G, perpendicular to side BC, for the purpose of creating the last of the three line segments that are to be equal when Euler's line is parallel to side BC.
     
  12. Feb 16, 2006 #11
    What is the equation for H in a co-ordinate frame Nanaki?
    The triangle is ABC correcsponding to (c*cosB, c*sinB), (0,0), (a,0)
    I am interested to see Integral here. Hope you help us.
     
  13. Feb 16, 2006 #12
    I'm not sure what you mean, but the co-ordinates of H are (BH*cosB, BH*sinB), B being the angle withing triangle BDH.
     
  14. Feb 17, 2006 #13
    I want it in terms of A, B and C given. Did you understand what I meant by co-ordinate geometry.
    G = (a + c*cosB/3, c*sinB/3)
    Take O = (k,l) and we get k = a/2 and from the equating that it is equidistant from A and C, we get l. Now equate l with c*sinB/3. We get some relations which does not solve the problem. But I think Altitude might get us too front.
    For O,
    AO = BO = CO,
    (c*cosB - a/2)^2 + l^2 = (a/2)^2 + l^2 = (a - a/2)^2 + l^2
     
  15. Feb 17, 2006 #14
    I'm afraid I can't quite follow what you're saying.
     
  16. Feb 26, 2006 #15
    Are you still interested in finding out the solution?
     
  17. Feb 26, 2006 #16
    Yes I am. But I couldn't proceed much intensely from what I did till now.
    I have to cover a bit on slopes further to march to orthocentre. Probably you will have to wait atleast 2 weeks for a reply from me.
     
  18. Feb 26, 2006 #17
    Alright. I've been looking, but I really don't know what to look at.
     
  19. Mar 3, 2006 #18
    I've figured it out, with some help from a friend.

    I've attached an image to help explain.

    I know there are a lot of lines, but it should be understandable. Dark blue are the altitudes, light blue are the perpendicular bisectors, brown is one median, black is Euler's line.

    I've drawn Euler's line parallel to side AB. I've drawn a line through centroid G, perpendicular to side AB and parallel to altitude CC'. Now there are two similar triangles, namely CGH and GFX. Now we need a little bit of knowledge of the median: the centroid of a triangle always divides the medians in two parts with lengths that are proportional to eachother at 2:1 (not sure what it's called, hope you understand). I can prove this as well, but that's not the point right now. It means that the sides of CGH are twice as long as those of GFX. This means that CH=2GX; GX=C'H => CH=2C'H. So, Euler's line is parallel to a side when the altitude that's perpendicular to that side is divided into two parts with proportions 2:1. Or, in other words, taken vertex V and the intersection V' of the altitude from that vertex with the side, VH=2V'H.
     

    Attached Files:

  20. Mar 4, 2006 #19
    Can I proceed a bit more? I was waiting for you to understand some basics of co-ordinate geometry so that I will explain you something. Your condition is weak. So the problem is not over. What you said is a very trivial thing. The condition as I saidbefore should be based on the sides and angles of triangles and not on altitudes or medians. So read this as a continuation of what you said considering triangle with the three points A (c*cosB, c*sinB), B(0, 0) and C(a, 0). Let the three altitudes be AD, BE and CF and the orthocentre at H.

    Now D = (c*cosB, 0) and since Euler line is parallel and as per your condition, H = (c*cosB, 2/3c*sinB).

    Now a bit about slopes. If a line has slope tan(theta), the slope of a line perpendicular to it is -cot(theta). The slope of a line is tan(theta) if it makes an angle theta with the positive X-direction.

    Now AC makes an angle 180-C with the positive X-axis. So slope of BE by the above rule is -cot(189-C) = cotC.
    slope of BE = slope of BH =
    difference between Y-co-ordinates of B and H/diff in X-co-ordinates
    = 2/3c*sinB/c*cosB = 2/3tanB.

    As above cotC = 2/3tanB.
    Muliplying both sides by 3/2tanC we get tanB*tanC = 3/2
     
  21. Mar 4, 2006 #20
    Indeed you are right, and your post is very insightful. However, though finding the co-ordinates of H like this works in a triangle where BC is on the x-axis, does it also work on random triangles?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Euler's Line Parallel to a Triangle's Side
Loading...