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Math Help - Hyperbolic Triangle

  1. #16
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    Quote Originally Posted by LumusRedfoot View Post
    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...
    Although the question does not explicitly say so, it is usually assumed that hyperbolic space consists of the upper half of the complex plane. So in this problem I think it's safe to ignore the lower half of the plane.

    Quote Originally Posted by LumusRedfoot View Post
    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?
    To calculate |z|^2, you have to take the real part of z and the imaginary part of z, square them both, and then add. If z = a+ib (where a and b are both real) then z-1 = a+ib-1 = (a-1) + ib. The real part is a-1 and the imaginary part is b. So |z-1|^2 = (a-1)^2+b^2.

    Quote Originally Posted by LumusRedfoot View Post
    its because im only 14 i was moved ahead i think a bit too fast.
    Good grief, I think you're right. This is very advanced material, and it looks as though this course assumes a lot of previous experience of working with complex numbers and their geometry.
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  2. #17
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    Quote Originally Posted by Opalg View Post
    Although the question does not explicitly say so, it is usually assumed that hyperbolic space consists of the upper half of the complex plane. So in this problem I think it's safe to ignore the lower half of the plane.


    To calculate |z|^2, you have to take the real part of z and the imaginary part of z, square them both, and then add. If z = a+ib (where a and b are both real) then z-1 = a+ib-1 = (a-1) + ib. The real part is a-1 and the imaginary part is b. So |z-1|^2 = (a-1)^2+b^2.


    Good grief, I think you're right. This is very advanced material, and it looks as though this course assumes a lot of previous experience of working with complex numbers and their geometry.
    ah, right, so the 'a-1' part is just the real part of a 'new' compelx number? and we then treat that as if it were the original 'a'. THEN square that bracket? i can solve it from there on and i got the answer as described. thanks a lot.

    i might need more help with the next parts when I get to it but I'm going to get on with what I can do, thanks!
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  3. #18
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    please help me with part iii

    i don't know how to deal with the curvy angles.
    what is the formula for a hyperbolic angle? i looked on wiki and a few math sites but couldn't really find it. can someone explain in laymans terms please!
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  4. #19
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    Quote Originally Posted by LumusRedfoot View Post
    please help me with part iii

    i don't know how to deal with the curvy angles.
    what is the formula for a hyperbolic angle? i looked on wiki and a few math sites but couldn't really find it. can someone explain in laymans terms please!
    The nice thing about hyperbolic angles is that they are the same as regular (euclidean) angles. So x is the angle between the two circles at the point B. I will show in detail how to find cot(x), as required for part (iii). Then you can do something similar for the angle y at the point C.

    The angle between two circles at a point where they intersect is equal to the angle between the tangents to the circles at that point. So you need to find the slopes of the tangents to the two circles at B.

    A tangent to a circle is perpendicular to the radius at that point, so we'll start by finding the slopes of the radii at B.

    For the smaller circle, its centre is at the point (1,0). And B is the point (0,√3). The radius is the line joining (1,0) and (0,√3), and its slope is √3. The tangent is perpendicular to the radius, so its slope is 1/√3. This means that the tangent makes an angle \arctan(1/\sqrt3) = \pi/6 with the x-axis.

    For the larger circle, its centre is at the point (3,0). The radius in this case is the line joining (3,0) and (0,√3), and its slope is √3/3 = 1/√3. The tangent is perpendicular to the radius, so its slope is √3. This means that the tangent makes an angle \arctan(\sqrt3) = \pi/3 with the x-axis.

    Therefore the angle between the two tangents is \pi/3-\pi/6 = \pi/6 radians. Finally, \tan(x) = \tan(\pi/6) = 1/\sqrt3 and so \cot(x) = 1/\tan(x) = \sqrt3.
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  5. #20
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    can you please go over the method of how to work out 'A' for part (i) as i don't understand the method used to find 'B', using the non-graphical method
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  6. #21
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    do you need to do a differentiation on the circle equations? to find the angle of the tangents.

    how does the SLOPE (GRADIENT?)help you to find the ANGLES? i can work out the two tangers

    i understand what you are doing subtracting the smaller angle from the bigger angle using the xaxis for both to isolate the x angle. but i still dont know where your tan expressions are coming from. also is there a way of finding x straight away without doing the subtracting? (just so i know)

    also does a tangent have just a slope or is it a line equation that i need? do i need to do the m formula?

    (are u just using that the slopes forms triangles with the xaxis and using trig
    can you not just use the slopes of the radiuses themselves instead of the tangents and use that the opposite angle is equal. because you can make the triangles with the x axis and the radiuses to B and use the triangle with the y axis to isolate the angle with the two circle centres atB, it is equal to x?)

    i can work out what to do with te angle at c because it isnt like you perscribed with the two cricles. it is circle and one vertical line so i dont know what to do. i mean surely its differrent right. the hyperbolic angle being the same with the two cricles is because they both fold away form each toher? so with a circle and a line its not going to be the same...=/

    i really can't figure out what to do at point C...
    with B you have the two circles and you can use the TWO tangets to find the answer. C is made with a circle and a straight line
    Last edited by LumusRedfoot; May 6th 2011 at 04:21 AM. Reason: i really can't figure out what to do at point C...
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  7. #22
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    i habe the same problem at point C i have worked out at point B but as was stated what do you do when you are using the Re(z) =2 line?
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  8. #23
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    also can you show a graphical model of this calculation for reference to the workings
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  9. #24
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    Quote Originally Posted by LumusRedfoot View Post
    do you need to do a differentiation on the circle equations? to find the angle of the tangents. You could find the slope of the tangents using calculus, by differentiating the equations of the circles, but the method using trigonometry is easier and quicker.

    how does the SLOPE (GRADIENT?)help you to find the ANGLES? i can work out the two tangents The connection between the slope of a line and the angle that it makes with the x-axis is that if the line has slope m, and the angle it makes with the axis is θ, then m = tan(θ). (In fact, that is probably why the tan function got its name: it tells you the slope of a tangent.)

    i understand what you are doing subtracting the smaller angle from the bigger angle using the xaxis for both to isolate the x angle. but i still dont know where your tan expressions are coming from. also is there a way of finding x straight away without doing the subtracting? (just so i know) I don't think there is a way of finding this angle except by subtraction.

    also does a tangent have just a slope or is it a line equation that i need? do i need to do the m formula? You don't need the complete equation of the tangent, just its slope.

    (are u just using that the slopes forms triangles with the xaxis and using trig Yes.
    can you not just use the slopes of the radiuses themselves instead of the tangents and use that the opposite angle is equal. because you can make the triangles with the x axis and the radiuses to B and use the triangle with the y axis to isolate the angle with the two circle centres at B, it is equal to x?) That is an excellent suggestion. The angle between the tangents is the same as the angle between the radiuses, which is easier to calculate.

    i can work out what to do with the angle at c because it isnt like you perscribed with the two cricles. it is circle and one vertical line so i dont know what to do. i mean surely its differrent right. the hyperbolic angle being the same with the two cricles is because they both fold away form each toher? so with a circle and a line its not going to be the same...=/ In fact, it's much easier in this case, because one of the curves is already a straight line, so you don't have to bother finding out its tangent. You even know the angle that this line makes with the x-axis, namely a right angle. So all you need to find is the angle between the vertical line and the tangent to the circle at C.
    ..
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  10. #25
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    opalg on that last point

    i thought the angle between the tangets coudl be used because the two circles 'bend away' from each other thus taking the tangets either side makes the angle equal to the angle with the two tangets?

    do you not need to compensate for the fact thatyoure looking for the angle with a line one one side and a circle on the other? such that it wouldnt be equal to the angle between the line and the tanget?

    but using this method i make it -1/root 3? what do you think? am i doing it right
    or you can see that the tangent at B and C are equal just one negative and one positive cos they are oppositie again... just either side of the circle l

    its backwards so the angle is negative? do we just make it positive?>?

    if you do it backwards pi/2 - (- pi/6), if you do it forwards, pi/2 - pi/6? different results. can you have a negative angle. it doesnt seem right. i made the anser to y = pi/3. and then just take cos and sin for that as the answers. but the answers are in numbers not expressions. is that alright is that ok.

    nilding who r u


    edit please now help me with part 4. i have found two formulas... oen usus integrals, one does not use integrals. am i supposed to be looking at one with integrals or one without. for the hyperbolic arc length...

    is the length of a, arccosh(5/3), can someone verify, and is the ln form, ln(3). just ln(3)? as the answee?
    i am only 14 why am i doing this my mates are dont even know what a quadratic is

    edit also as for AC, is it not just root 11 - root 3? its just a line, you can get the length by reading it off the coordinates just like that...?
    Last edited by LumusRedfoot; May 8th 2011 at 06:04 AM.
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  11. #26
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    Quote Originally Posted by LumusRedfoot View Post
    i thought the angle between the tangents could be used because the two circles 'bend away' from each other thus taking the tangents either side makes the angle equal to the angle with the two tangents? It doesn't matter which way they are bending, it's the angle at the actual point of intersection that matters. That is why you can replace the curves by their tangents at that point.

    do you not need to compensate for the fact that youre looking for the angle with a line one one side and a circle on the other? such that it wouldnt be equal to the angle between the line and the tangent? No, you don't need to "compensate", you just need the angle between the line and the tangent.

    but using this method i make it -1/root 3? what do you think? am i doing it right
    or you can see that the tangent at B and C are equal just one negative and one positive cos they are opposite again... just either side of the circle l Yes, that is the smart way to do it. The two points B and C are symmetrically placed on opposite sides of the circle.

    its backwards so the angle is negative? do we just make it positive?>? Just make it positive. It is the size of the angle that you are looking for, not its orientation.

    if you do it backwards pi/2 - (- pi/6), if you do it forwards, pi/2 - pi/6? different results. can you have a negative angle. it doesnt seem right. i made the anser to y = pi/3. and then just take cos and sin for that as the answers. but the answers are in numbers not expressions. is that alright is that ok. Yes, the angle at C is pi/3.
    Quote Originally Posted by LumusRedfoot View Post
    nilding who r u
    There are three of you (LumusRedfoot, Neilding and Alexg42 from this other thread) who are all working on this problem. I suggest that you should get in touch with each other. You can use the private message feature in this forum to do that. In the long run, you will find it more useful to share ideas with each other than to keep coming for help here.

    Quote Originally Posted by LumusRedfoot View Post
    please now help me with part 4. i have found two formulas... oen usus integrals, one does not use integrals. am i supposed to be looking at one with integrals or one without. for the hyperbolic arc length...

    is the length of a, arccosh(5/3), can someone verify, and is the ln form, ln(3). just ln(3)? as the answee? Yes! ln(3) is correct.
    i am only 14 why am i doing this my mates are dont even know what a quadratic is You're either a genius or a maniac. Maybe both.

    edit also as for AC, is it not just root 11 - root 3? its just a line, you can get the length by reading it off the coordinates just like that...?
    I use the integral method. You have to parametrise the path by some function \sigma(t)\; (a\leqslant t\leqslant b). Then the hyperbolic length is given by the integral \int_a^b\frac{|\sigma'(t)|}{\text{Im}(\sigma(t))}d  t. For the circular arc BC, you can parametrise it by \sigma(t) = 1+2e^{it}\ (\pi/3\leqslant t\leqslant 2\pi/3). For the line AC you can use \sigma(t) = 2+it\ (\sqrt3\leqslant t\leqslant\sqrt{11}). But note that the hyperbolic length is not the same as the ordinary length. So it will not be equal to \sqrt{11}-\sqrt3. You have to get it by feeding that function into the integral formula.

    Spoiler:
    The (hyperbolic) length of AC is \ln(\sqrt{11/3}).
    Last edited by Opalg; May 8th 2011 at 08:40 AM. Reason: tidied up LaTeX
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