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Math Help - stuck on trig

  1. #1
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    stuck on trig

    Could someone help me find the length in cm of the lines between the points in the picture attached.

    The 3 angles around point 6 are 120 degrees each. This is also the case for point 8 which is linked to point 2, 3 and 7.

    Now the 3 angles formed by the lines leaving point 7 are also 120 degrees.

    If the distance between point 1 and 2, 2 and 3, 3 and 4, 4 and 5 is 1 cm.

    5,6,4 forms an isosceles triangle and so does 2,8,3 triangle.

    What is the total distance of the lines?
    Last edited by Natasha1; May 12th 2006 at 12:49 AM.
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  2. #2
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    The distances of the lines between points 5,6,4 and 2,8,3 (which are 2 isoceles triangles) is 2 cm.

    This is how much I get as I have 4 sin 30 = 2

    I am missing the distances from points 1,7 and 6,7 and also 7,8 to get my total distance can someone help :-) please
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  3. #3
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    Quote Originally Posted by Natasha1
    Could someone help me find the length in cm of the lines between the points in the picture attached.

    The 3 angles around point 6 are 120 degrees each. This is also the case for point 8 which is linked to point 2, 3 and 7.

    Now the 3 angles formed by the lines leaving point 7 are also 120 degrees.

    If the distance between point 1 and 2, 2 and 3, 3 and 4, 4 and 5 is 1 cm.

    5,6,4 forms an isosceles triangle and so does 2,8,3 triangle.

    What is the total distance of the lines?
    I assume this is about the shortest route to connect the vertices of a regular pentagon.
    I have seen this topic posted here before. It is about the shortest roadway to connect points of convex polygons, or something like that. That there are points of intersections inside the polygon where the roads are always at 120 degrees with each other...

    So in your figure, intersection points 6,7,8 are those such points.
    The figure is symmetrical about the imaginary vertical line passing through line 1-7. So we need to find the lengths of lines 1-7, 7-6, 6-5 and 6-4 to solve the problem. Call those lines a, b, c and d, respectively.

    Given: regular pentagon, side = 1`cm each.

    Tools:
    ---Sum of interior angles of any n-gon: I = (n-2)(180deg).
    So, a triangle has 180deg total; a quadrilateral has 360deg total.

    ---Any interior angle of regular n-gon: i = (n-2)(180deg)/(n).
    So in a regular pentagon, an interior angle is 108deg.

    ---Law of Sines.
    ---Law of Cosines.

    Now, in your figure, we start by dropping a vertical line segment 1-7 or "a". At point 7, we branch into two lines, one of whom is 7-6 or b. From point 6, we branch into 6-5 or c, and 6-4 or d.

    "a" bisects the 108deg angle at point 1.
    Call the side 5-1 as e.
    In quadrilateral abce, angle 6-5-1 = 360 -54 -120 -120 = 66deg.

    Call side 4-5 as f.
    In triangle cdf,
    angle 6-5-4 = 108 -66 = 42deg.
    So, angle 6-4-5 = 180 -120 -42 = 18deg.
    (This triangle then is not isosceles.)
    By Law of Sines,
    f/sin(120deg) = c/sin(18deg) = d/sin(42deg)
    Since f=1cm,
    c = 1*sin(18deg)/sin(120deg) = 0.356822 cm. ---------***
    d = 1*sin(42deg)/sin(120deg) = 0.772645 cm. ---------***

    Draw line segment 1-6. Call it g.
    In triangle gce,
    By Law of Cosines,
    (g)^2 = (c)^2 +(e)^2 -2(c)(e)cos(66deg)
    (g)^2 = (0.356822)^2 +(1)^2 -2(0.356822)(1)cos(66deg)
    g = 0.914908 cm ---*
    By Law of Sines,
    c/sin(angle 5-1-6) = g/sin(66deg)
    (0.356822)/sin(angle 5-1-6) = (0.914908)/sin(66deg)
    So, angle 5-1-6 = arcsin[(0.356822)*sin(66deg)/(0.914908)] = 20.872561 degrees. ---*

    In triangle abg,
    g = 0.914908 cm.
    angle 6-1-7 = 54 -20.872561 = 33.127439 deg. -----------*
    angle 7-6-1 = 180 - 33.127439 -120 = 26.872561 deg. ----*
    By Law of Sines,
    (0.914908)/sin(120deg) = a/sin(26.872561deg) = b/sin(33.127439deg)
    So,
    a = (0.914908)*sin(26.872561deg)/sin(120deg) = 0.477521 cm. ------***
    b = (0.914908)*sin(33.127439deg)/sin(120deg) = 0.577350 cm. ------***

    Therefore, the total length of the shortest way is
    0.477521 +2(0.577350 +0.356822 +0.772645) = 3.891155 cm. -----answer.
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  4. #4
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    Quote Originally Posted by ticbol
    So in your figure, intersection points 6,7,8 are those such points.
    The figure is symmetrical about the imaginary vertical line passing through line 1-7.
    How do we know that? I think it is probably true for the network of
    shortest length, but I don't quite see how to prove it at present.


    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    How do we know that? I think it is probably true for the network of
    shortest length, but I don't quite see how to prove it at present.


    RonL
    How do I know? Easy. If you have a regular pentagon, which is symmetrical about a line from a vertex to the the midpoint of the opposite side, and you follow the rule/way of that thing about intersection points of 3 lines at 120 degrees from each other, going inside the pentagon by starting from that said vertex, then you will have a network that is also symmetrical about that same axis. Easy.
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  6. #6
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    Quote Originally Posted by ticbol
    How do I know? Easy. If you have a regular pentagon, which is symmetrical about a line from a vertex to the the midpoint of the opposite side, and you follow the rule/way of that thing about intersection points of 3 lines at 120 degrees from each other, going inside the pentagon by starting from that said vertex, then you will have a network that is also symmetrical about that same axis. Easy.
    That is the way I would hand wave an explanation, but it does not convince
    me, all it convinces me of is: for one way of doing it so the network is of
    minimum length then the mirror image about the diameter of the
    ciumscribed circle though vertex 1 is also a minimal solution.

    RonL
    Last edited by CaptainBlack; May 12th 2006 at 05:40 AM.
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  7. #7
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    Quote Originally Posted by CaptainBlack
    That is the way I would hand wave an explanation, but it does not convince
    me, all it convinces me of is: for one way of doing it so the network is of
    minimum length then the mirror image about the diameter of the circumscribed circle though vertex 1 is also a minimal solution.

    RonL
    I am not convincing you. That is my solution. Show yours, and I will also comment on it.
    Last edited by ticbol; May 12th 2006 at 05:23 AM.
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  8. #8
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    Quote Originally Posted by ticbol
    I am not convincing you. That is my solution. Show yours, and I will also comment on it.
    I'm not denying that that is the solution, what I am claiming is that
    as far as I can see there is a gap in the argument that symmetry of the
    pentagon about an axis implies symmetry of the positions of the Steiner
    points about that axis rather than a symmetry of the objective function
    that is to be minimised by the point placement, about that axis.

    RonL
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  9. #9
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    Quote Originally Posted by CaptainBlack
    I'm not denying that that is the solution, what I am claiming is that
    as far as I can see there is a gap in the argument that symmetry of the
    pentagon about an axis implies symmetry of the positions of the Steiner
    points about that axis rather than a symmetry of the objective function
    that is to be minimised by the point placement, about that axis.

    RonL
    So Steiner points they are.
    About that symmetry of the network again. Again, follow Steiner. Drop a line along that axis of symmetry. Stop a some random point 7. Draw two rays that are 120 degrees to the right or to the left of that starting line/axis.
    On one of the two rays, proceed, then stop at some random point 6. Align this point six, and also the point 7, so that the next two rays fromt point 6 will hit points 4 and 5. There you have the left half of the network.
    Now go to the right half. Do the same thing---stopping at some point 8, drawing two rays from point 8, adjust, adjust until these last two rays hit corners 2 and 3.
    Is there a possibility that the right half of the network not a mirror image of the left half, relative to the said axis of symmetry of the regular pentagon?
    None.
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  10. #10
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    Quote Originally Posted by ticbol
    So Steiner points they are.
    About that symmetry of the network again. Again, follow Steiner. Drop a line along that axis of symmetry. Stop a some random point 7. Draw two rays that are 120 degrees to the right or to the left of that starting line/axis.
    On one of the two rays, proceed, then stop at some random point 6. Align this point six, and also the point 7, so that the next two rays fromt point 6 will hit points 4 and 5. There you have the left half of the network.
    Now go to the right half. Do the same thing---stopping at some point 8, drawing two rays from point 8, adjust, adjust until these last two rays hit corners 2 and 3.
    Is there a possibility that the right half of the network not a mirror image of the left half, relative to the said axis of symmetry of the regular pentagon?
    None.
    No but your construction has already presupposed the point I am
    suggesting needs justification:

    Drop a line along that axis of symmetry
    This is the step that I think needs justification.

    RonL
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  11. #11
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    Quote Originally Posted by CaptainBlack
    No but your construction has already presupposed the point I am
    suggesting needs justification:



    This is the step that I think needs justification.

    RonL
    I see. So why not experiment. Try starting with a line that is not along any axis of symmerty of the regular pentagon. Then see if you can get a network that is shorter than that found when the network starts along an axis of symmetry.
    My solution was based on Natasha's drawing where the line segment 1-7 appeared to be perpendicular to the side 3-4, thereby placing 1-7 along an axis of symmetry.
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  12. #12
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    Quote Originally Posted by ticbol
    I see. So why not experiment. Try starting with a line that is not along any axis of symmerty of the regular pentagon. Then see if you can get a network that is shorter than that found when the network starts along an axis of symmetry.
    My solution was based on Natasha's drawing where the line segment 1-7 appeared to be perpendicular to the side 3-4, thereby placing 1-7 along an axis of symmetry.
    Experiment is not proof, as I have repeatedly said I believe that your
    solution is correct. If that is so no experiment will produce a shorter network.

    My contribution to this discussion is now at an end.

    Ron:
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  13. #13
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    Thanks for all this discussion it has immensely helped me.

    I am still struggling to show the same for the hexagon though. An illustration is given in the picture attached
    Last edited by Natasha1; May 12th 2006 at 07:46 AM.
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