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Math Help - Tough question - Coordinate geometry.

  1. #1
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    Tough question - Coordinate geometry.

    Six points are taken at a circle centered at the origin, such that the mean if it's abcissas is 4/3 and that of it's oordinates is 2/3. Line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k. Find h+k
    Can someone solve this?

    Thanks.
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  2. #2
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    Quote Originally Posted by cyanide911 View Post
    Six points are taken at a circle centered at the origin, such that the mean if it's abcissas is 4/3 and that of it's oordinates is 2/3. Line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k. Find h+k

    Can someone solve this?
    Thanks.
    Select any three points, calculate orthocenter[1] and centroid[1]; those two points define line[1].

    Use the other three points and calculate orthocenter[2] and centroid[2]; those two points define line[2].

    The intersection of those two lines will give you the (h,k) coordinates.

    The real difficulty of your problem is the generation of six coordinate points on a circle that average to (4/3,2/3).

    If you supply the coordinates for the six points, the remainder is easy.
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  3. #3
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    Yeah, that's what makes it tough.
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  4. #4
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    I believed that this thread had been covered/closed
    but I received a notice that it had been bumped.

    Attached are three images.
    From image 1, IF you have only 1 point, then that
    point (4/3, 2/3) occurs on the circle with
    R = sqrt( (4/3)^2 + (2/3)^2 ).

    The line, from the orgin to the point, is defined
    by y = x/2

    ---
    For two points:
    (see image 2)

    (1) to be equidistant from the origin or center of the circle
    &
    (2) the sum of the x_coordinates to average 4/3 = (x1+x2)/2
    &
    (3) the sum of the y-coordinates to average 2/3 = (y1+y2)/2

    They must be on a line perpendicular to y = x/2
    that passes through the point (4/3, 2/3) and be equidistant (distance w) from the point (4/3,2/3).

    That line is defined by y = -2x + 10/3

    With those two points (x1,y1) and (x2,y2) the Radius can be uniquely determined.

    ---
    for six points:
    (see image 3)

    (1) to be equidistant from the origin or center of the circle
    &
    (2) the sum of the x_coordinates to average 4/3 = (x1+x2+x3+x4+x5+x6)/6
    &
    (3) the sum of the y-coordinates to average 2/3 = (y1+y2+y3+y4+y5+y6)/6

    The points (3 through 6),
    must be on a line parallel to y = -2x + 10/3

    That parallel offset is dimension "v".

    (Note the possibility of a difficult situation if dimension v, causes (x4,y4) and (x5,y5) to exceed the length of radius determined above.)

    With that, you can find the necessary 6 coordinate points to construct an answer to the first post.

    Hope that helps.
    Attached Thumbnails Attached Thumbnails Tough question - Coordinate geometry.-rxy0001.jpg   Tough question - Coordinate geometry.-rxy0002.jpg   Tough question - Coordinate geometry.-rxy0003.jpg  
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  5. #5
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    http://www.mathhelpforum.com/math-he...lers-line.html

    If it weren't for the above reference I would not have associated this
    problem with the 9-point circle.

    a circle centered at the origin (=circumcenter),
    such that the mean if it's coordinates is (4/3, 2/3).
    A line joining the orthocenter of a triangle formed by any 3 points and centroid of a triangle formed by the remaining 3 pass through a fixed point h,k.
    Find h+k
    The distance from the circumcenter to the centroid of a triangle is twice the distance from the centroid to the 9 point circle point.

     \dfrac{3}{2} \times (4/3,2/3) = (2,1) = (h,k)

    so h+k = 3
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