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Math Help - Triangle split in two

  1. #1
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    Smile Triangle split in two

    Code:
    A
    .
    |..
    |. ..
    | . . .
    | .  .  .
    |  .  .   .
    |  .   .    .
    |   .   .     .
    |____...........
    E    D    C    B
    Note that only the dotted lines are of direct interest. The filled line AED is for reference only (creating right angle E and height AE).

    ABD is a great triangle consisting of two minor triangles ABC and ACD.
    Angle BAD is split in two equal parts (BAC = CAD = BAD/2) by the line AC.
    Angle BAD is known (the angle at the top, but note that angle DAE is not known),
    Length of AC is known (the dividing center line).
    Length of BD is known (the base of the great triangle).
    We also know that angles ADB and ACB obtruse (>90) if that helps.

    How do I solve for the rest of the angles and lengths of the figure?

    If I had known AB or AD instead of AC, then it would have been quite easy. Now it is maybe much more tricky than it seems. Or am I missing something obvious?

    Grateful in advance!
    Last edited by Optiminimal; May 14th 2007 at 04:17 AM. Reason: Typo
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  2. #2
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    Oh, no comments yet? I would need help from a talented eye to take a look at this. It's the old schoolbook "a triangle with three knowns", with a twist.

    There must exist some useful relationship between AB/AC and AC/AD, and also BC/CD, depending on the value of the angle BAD.
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  3. #3
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    I could offer an additional piece of information, however I don't think it is necessary and possibly it adds no useful information. It is also not very easy to explain:
    Code:
    A
    .
    |..
    |. ..
    | . . .
    | .  .  .
    |  .  .   .
    |  .   .    .
    |   .   .     .
    |____...........
    E    D,,,,C,,,,B
          ,,,,,,,,,
            ,,,,,
           F ,,
    The area filled with comma signs represent a half circle.

    Now we also know that line BD is the diameter of a circle on a plane which is at a right angle to the line AE. So we know that the radius R = BD/2. The center of the circle lies somewhere on the line CB. Note that C is not the center of the circle.

    Point F lies on the perimeter of the circle. From C extends a line to F. This line is of course shorter than the radius. CF is at a right angle to BD.

    The meaning of point F, as it relates to point C, is that we know that the angle CAF is the greatest angle which could be formed between line BD and the perimeter of the circle, such that the base (corresponding to CF) is at a right angle at line BD.

    If point A was straight above the center of the circle, the angle CAF would be maximized iff C was the center of the circle and CF = radius. But as the case now is, point A has moved to the side and this CAF is maximized when C is offset from the center (and F must follow so that line CF remains at a right angle to line BD).
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  4. #4
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    Looks like you have a more complex problem that you try to solve. It would be helpfull if you would post the inital problem. Are there exact measures for the known sides and angle?
    What I would try to do for solving your triangle, is using sine and cosine laws, in triangle ACD, respectively BCD, and maybe ADC.
    Numbers would definitely help to see the solution easily.
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  5. #5
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    Hello,

    I don't know if this helps a little bit further:

    An angle bisector divides a side of a triangle into two distances. The ratio of these 2 distances is the same as the ratio of the 2 adjacent sides of the triangle.

    With your triangle:
    Code:
     AD     DC
    ---- = ----
     AB     CB
    In my opinion your problem can't be solved because you need 3 independent pieces(?) of a triangle to calculate all distances and angles of the triangle.
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  6. #6
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    Thank you for the answers!
    earboth:
    You might be right that it cannot be solved analytically. Although it looks like it's temptingly close...

    I've just gotten the advice to use a numerical method: Choose values of the angle EAD and take advantage of the right angle triangles thus formed to calculate the known sides (AC and BD), using also the other known angles. Then change the guessed angle EAD until it fits with what is known.

    If I do this for many combinations of values on lengths AC, BD and angle BAD, would it be possible to derive an analytical formula from the resulting data? I mean, I could just fit a curve to it and then use the parameters of that curve to immediately find, say, length AB given data in a particular situation.

    alinailiescu:
    I need a method to solve it for any possible combination of numeric values. If there is no analytical solution then I need a numerical one. If you have any comments on the proposal I described above, they are very welcome.

    And yes, this is a part of a bigger puzzle. There is a picture taken of the circle. On the picture there is not a circle but an ellipse because of the distortions of perspective. C is at the center of the ellipse. What I'm getting at is to derive the position of the center of the circle from the data given from the photo.
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  7. #7
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    Look how I think that this problem can be solved. I did not find the final solution, but it is possible
    Attached Thumbnails Attached Thumbnails Triangle split in two-save0015.jpg  
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  8. #8
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    Cont.
    Attached Thumbnails Attached Thumbnails Triangle split in two-save0016.jpg  
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  9. #9
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    Thumbs up

    Thank you very very much, alinailiescu!

    I typed your eq. 5 into Quickmath and got four solutions for d:

    d=
    [IMG][/IMG]

    The four solutions being given by variations in plus/minus in front of the square roots.

    I will soon make a new post describing the bigger problem, since I have encountered other issues as well. Math applied on the reality is rarely as simple as it first looks.

    Again, thanks very much!
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  10. #10
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    Post

    I have now made a post of the "grand problem" here:
    Grand Problem
    It should be clear how the partial problem in this thread relates to it.
    North-South corresponds to BD.
    Delta Z corresponds to BC-(BD/2).
    Camera position corresponds to A.

    Unfortunately, I do not really know AC, corresponding to distance Camera-Center of ellipse. I thought that I would have that information, but I do not directly.

    At least I managed to make som nice 3D-geometry illustrations, didn't I!

    PS
    alinailiescu, your username sounds quite Romanian. My mother-in-law is from Iasu! I just thought that dropping that information would help motivate you to take a look at my problems again, since I know that Romanians are quite patriotic AND often very good at math...
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