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Math Help - Hyperbolic Triangle

  1. #1
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    Hyperbolic Triangle

    We are presented with the problem:

    The hyperbolic lines l, m and n defined by:

    l={zEH | abs(z-1)=2}
    m={zEH | Re(z)=2}
    n={zEH | abs(z-3)=2rt3}

    form a hyperbolic triangle with vertices: A:= m n n, B:= l n n and C:= l n m.

    (i) Find the complex numbers which represent A, B and C
    (ii) Draw a diagram to scale of the hyperbolic lines l, m and n. Label their points of intersection and their intersections with the real axis with the appopriate letters and complex numbers.
    (iii) Two of the interior angles are denoted: x at B and y at C. Give exact expressions for cos(y), sin(y) and cot(x). Calculate the angles x and y to 3dp.
    (iv) Let a denote the length of the h-line segment BC and b denote the length of the h-line segment AC. Calculate a and b exactly in terms of natural logarithms and also approximately, to 4dp.
    (v) Using the exact answers in (iii) and (iv), verify exactly, for the given hyperbolic triangle, the identity: cos(y)cosh(a)=sinh(a)coth(b)-sin(y)cot(x).

    My apologies if this is in the wrong forum as this is university Geometry, but there was not a university Geometry forum appropriate to put it in. It is not pre-university but it is is Geometry. I'm not seeking full-blown solutions just a little kickme in the right direction. This is my weakest module and I really don't quite understand it. There are no course materials, reading lists etc, and the teaching staff do not accept questions or queries about the work. So I am asking here. Thanks if anyone can give me a few pointers.

    Lumus
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  2. #2
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    Quote Originally Posted by LumusRedfoot View Post
    We are presented with the problem:

    The hyperbolic lines l, m and n defined by:

    l={zEH | abs(z-1)=2}
    m={zEH | Re(z)=2}
    n={zEH | abs(z-3)=2rt3}

    form a hyperbolic triangle with vertices: A:= m n n, B:= l n n and C:= l n m.

    (i) Find the complex numbers which represent A, B and C
    (ii) Draw a diagram to scale of the hyperbolic lines l, m and n. Label their points of intersection and their intersections with the real axis with the appopriate letters and complex numbers.
    (iii) Two of the interior angles are denoted: x at B and y at C. Give exact expressions for cos(y), sin(y) and cot(x). Calculate the angles x and y to 3dp.
    (iv) Let a denote the length of the h-line segment BC and b denote the length of the h-line segment AC. Calculate a and b exactly in terms of natural logarithms and also approximately, to 4dp.
    (v) Using the exact answers in (iii) and (iv), verify exactly, for the given hyperbolic triangle, the identity: cos(y)cosh(a)=sinh(a)coth(b)-sin(y)cot(x).

    My apologies if this is in the wrong forum as this is university Geometry, but there was not a university Geometry forum appropriate to put it in. It is not pre-university but it is is Geometry. I'm not seeking full-blown solutions just a little kickme in the right direction. This is my weakest module and I really don't quite understand it. There are no course materials, reading lists etc, and the teaching staff do not accept questions or queries about the work. So I am asking here. Thanks if anyone can give me a few pointers.

    Lumus
    I assume that you have no trouble with (ii), drawing the diagram like this:



    You should also have no difficulty using Pythagoras to check that A = (2,√(11)), B = (0,√3), C = (2,√3).

    For (iii), the conformal property of the hyperbolic space tells you that the hyperbolic angles are the same as the Cartesian angles. You can get those by calculus from the Cartesian equations of the circles, finding the slope of the tangents to the circles at B and C, and using some trigonometry.

    For (iv), use the formula for the hyperbolic length given here (together with the formula giving the inverse cosh in terms of a natural logarithm).

    Edit. In fact, you don't need calculus for (iii). Just use the fact that the tangent is perpendicular to the radius. You can read off the slope of the radius directly from the diagram.
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    What about (i)?
    How do I find the complex numbers? And how do they relate to the coordinates?

    From that diagram, I see that these are circles? I wouldn't have thought they were circles.

    Please don't assume that I don't have a problem with any of the parts or I would not have posted them.
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  4. #4
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    Quote Originally Posted by LumusRedfoot View Post
    What about (i)?
    How do I find the complex numbers? And how do they relate to the coordinates?

    From that diagram, I see that these are circles? I wouldn't have thought they were circles.

    Please don't assume that I don't have a problem with any of the parts or I would not have posted them.
    Hmm, I certainly did assume that if you are taking a university course in hyperbolic geometry then you would be familiar with some properties of complex numbers. For example, the point (a,b) in the plane corresponds to the complex number a+ib. Also, the equation |z-1| = 2 means that the distance from z to 1 is 2, so that z lies on a circle of radius 2 centred at 1. (And similarly |z-3| = 2√3 meams that z is on a circle of radius 2√3 centred at 3.)

    For (i), I gave the coordinates of A, B and C. I didn't notice that the question asked for the complex numbers which represent them. In terms of complex numbers, A = 2+√(11)i, B = √3i and C = 2+√3i.

    I don't want to sound dismissive if you are struggling with this material. It will be very hard for you to cope with this advanced course on hyperbolic geometry unless you are comfortable about the more elementary properties of complex numbers. Please feel free to post problems here. The more specific you can be about exactly where your difficulties lie, the easier it will be for us to help.
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    I'm completely fine with complex numbers, I just don't know how it relates to this question, in terms of what I'm being asked to do. If it was set up as an equation I could solve it but I don't know how to intepret the wording.

    Also I should note that knowledge of complex numbers (whasoever) is not a prerequisite for this module.

    For example, the point (a,b) in the plane corresponds to the complex number a+ib. Also, the equation |z-1| = 2 means that the distance from z to 1 is 2, so that z lies on a circle of radius 2 centred at 1. (And similarly |z-3| = 2√3 meams that z is on a circle of radius 2√3 centred at 3.)
    These are not things which have been taught on the course. I didn't know these. I mean it isn't stated that this is in the complex plane. I'm familiar with the complex plane/argand diagram etc. Just not really the hyperbolic stuff.

    Is part (i) basically just then asking to find the point at which the two given lines intereset? Would I tackle this just as in with the cartesian plane? I just don't know how to set up this equation and what I am looking for.

    For (i), I gave the coordinates of A, B and C. I didn't notice that the question asked for the complex numbers which represent them. In terms of complex numbers, A = 2+√(11)i, B = √3i and C = 2+√3i.
    Thank you very much for giving me the answers. It is much appreciated. If you could confirm what I asked above it would give me more confidence to solve such a problem on my own.
    Last edited by LumusRedfoot; April 26th 2011 at 12:39 AM.
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    I am trying to edit my post but for some reason there is no edit button like usual.

    What I would like to know is how you are working out the root 11. How do I find the imaginary part of A, given the real part is 2?
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    Quote Originally Posted by LumusRedfoot View Post
    What I would like to know is how you are working out the root 11. How do I find the imaginary part of A, given the real part is 2?
    Pythagoras. The circle centred at 3 has radius 2√3. So the square of the distance from A to the real axis is (2√3)^2 – 1^2 = 12 – 1 = 11.
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    Where are you getting -1^2 from? Please are you skipping a step here that I am supposed to assume?

    What point are you actually looking for, what is 'A'?

    I get that the real part is has to be 2 in both parts, so A is where the two lines meet, but how are you finding the imaginary part. Also what is the imaginart part of m, only the real part is given so I guess you have to figure that out too?
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    Quote Originally Posted by LumusRedfoot View Post
    Where are you getting -1^2 from? Please are you skipping a step here that I am supposed to assume?

    What point are you actually looking for, what is 'A'?

    I get that the real part is has to be 2 in both parts, so A is where the two lines meet, but how are you finding the imaginary part. Also what is the imaginart part of m, only the real part is given so I guess you have to figure that out too?
    A is as defined in your original post. It is the point where the vertical line m (given by Re(z)=2) meets the hyperbolic line n (which is a circle of radius 2√3 centred at 3). If you look at the diagram in my comment #2 above, you should be able to see how to apply Pythagoras' theorem to the right-angled triangle with vertices at A and the points 2 and 3 on the real axis.
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    Well, I can't see anywhere where a -1^2 comes from, that's not in the question... also I asked what is the imaginary part of m because only the real part is given.
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  11. #11
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    Quote Originally Posted by Opalg View Post
    A is as defined in your original post. It is the point where the vertical line m (given by Re(z)=2) meets the hyperbolic line n (which is a circle of radius 2√3 centred at 3). If you look at the diagram in my comment #2 above, you should be able to see how to apply Pythagoras' theorem to the right-angled triangle with vertices at A and the points 2 and 3 on the real axis.
    Quote Originally Posted by LumusRedfoot View Post
    Well, I can't see anywhere where a -1^2 comes from, that's not in the question... also I asked what is the imaginary part of m because only the real part is given.
    It seems to be a more medical problem ...

    see attachment
    Attached Files Attached Files
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    That's not how you're supposed to work it out though. As you can see, that is part (i), before part (ii) where you have to draw it. So there should be a non-graphical way of working it out.

    How would I set up the equation to find it? Not just jumping into things like right angled triangles, because that is not the intended method and it is not what is supposed to be done.

    So, can we try to solve part (i) WITHOUT the diagram please. Its just meant to be some kind of simultaneous equation.
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    Ok I kind of get it a bit better now.

    But I have question: How come there isn't also a 'negative A', as in one flipped in the x axis? Both positive and negative would meet the conditions...

    also what is 'z' (in terms of i.e. Re(Z)=2, I know z is representative of complex numbers, but for 'A' for example, would you note it Re(A)? as in z is like the general point on the line).

    Also how are we finding B? i note that abs(z-1)=2 and abs(z-3)=2rt3 must be met but how do we solve this.
    you're supposed to be able to do part i WITHOUT part ii. i.e before drawing a diagram. so can you please help me by giving the method and i will calculate
    Last edited by LumusRedfoot; May 1st 2011 at 07:58 AM.
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  14. #14
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    Quote Originally Posted by LumusRedfoot View Post
    Ok I kind of get it a bit better now.

    ...

    Also how are we finding B? i note that abs(z-1)=2 and abs(z-3)=2rt3 must be met but how do we solve this.
    you're supposed to be able to do part i WITHOUT part ii. i.e before drawing a diagram. so can you please help me by giving the method and i will calculate
    Let's start from the beginning:

    1. You know that a complex number is expressed as: z = a + bi

    with |z| = \sqrt{a^2+b^2}~\implies~z^2=a^2+b^2

    2. The point B ist determined by:

    \left|\begin{array}{l}|z-1|=2 \\ |z-3|=2\sqrt{3}\end{array}\right. \implies \left|\begin{array}{l}|a+bi-1|=2 \\ |a+bi-3|=2\sqrt{3}\end{array}\right.

    3. Using the 2nd formula of #1 you'll get:

    \left|\begin{array}{l}(a-1)^2+b^2 = 4\\ (a-3)^2+b^2=12\end{array}\right.

    Solve for (a, b)
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    hello
    that's very helpful, thank you for answering what i asked

    what i don't get from that is how you can just pull the -1 and -3 into the square with the 'a' part? don't you have to use a^2 + b^2 + 2ab if you square the whole thing? i mean, how does it being z-3 or z-1 make the formula differ to if its just z? in terms of the squaring?

    edit: youre really good thanks
    its because im only 14 i was moved ahead i think a bit too fast, I coped with the a level just fine but this is a bit hard for me.
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