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Let $A=(0,112), B=(0,126), C=(98,112)$ be points in the hyperbolic upper half plane H. Find the hyperbolic distances $d_h(A,B), d_h(A,C), d_h(B,C)$. Every answer should be in the form of a logarithm.

So I worked out just $BC$ since I am sure $AB$ and $AC$ work out the same procedure.

To find $BC$:

we know $B=(0,126)$ and $C=(98,112)$

I found the midpoint=$(49,119)$ and I also found the reciprocal slope=$7$. Using this information, I found the x intercept to be $x=(32,0)$

Using the distance formula, I found CX=BX=130

and since the hyperbolic distance=$\ln\frac{\csc(B)-\cot(B)}{\csc(A)-\cot(A)}$, I obtained:

$$\ln\frac{\frac{130}{126}-\frac{-32}{126}}{\frac{130}{112}-\frac{-32}{112}}$$

I just want anyone to confirm if this looks right.

EDIT:

From above:

$B=(X_1,Y_1)=(0,126)$

$C=(X_2,Y_2)=(98,112)$

$X=(X_3,Y_3)=(32,0)$

$CX=BX=130$

$$D(B,C)=\ln\frac{\frac{B_X}{Y_1}-\frac{X_1-X_3}{Y_1}}{\frac{CX}{Y_2}-\frac{X_2-X_3}{Y_2}}$$

Plugging in the values:

$$D(B,C)=\ln\frac{\frac{130}{126}-\frac{-32}{126}}{\frac{130}{112}-\frac{66}{112}}$$

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Not sure about your formula at all, and your "midpoint" is not on the hyperbolic line $BC$ (remember the hyperbolic line is a half circle)

https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model gives another formula:

$$ \operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arcosh} \left( 1 + \frac{ {(x_2 - x_1)}^2 + {(y_2 - y_1)}^2 }{ 2 y_1 y_2 } \right) $$

then according to https://en.wikipedia.org/wiki/Hyperbolic_function#Inverse_functions_as_logarithms

$$ \operatorname {arcosh} (x) = \ln \left(x + \sqrt{x^{2} - 1} \right) $$

so if you get the same answer via this method there must be something right about it.

if you get a different answer there is something wrong somewhere.

Good luck

Added later:

I did some calculating and I think your formula was wrong, you should multiply the values not divide them, but maybe nicer is to formulate them as:

$$ \ln \left( \frac{(1+ \cos(A) ) (1+ \cos(B) )}{\sin(A)\sin(B)} \right) $$

Or even more nicely as

$$ \ln \left( \frac{1+ \cos(A) )}{\sin(A)} \right) + \ln \left( \frac{1+ \cos(B) )}{\sin(B)} \right) $$

being the sum of the distances A - top of the line and B - top of the line.

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  • $\begingroup$ Turns out I made a mistake in the above where one of the 32 is suppose to be a 66. After I corrected, I did your method and I got the SAME exact answer except my answer is negative. I shall look over and see where I made a mistake. $\endgroup$ – mika Apr 13 '15 at 4:51
  • $\begingroup$ Never mind, I realize my mistake. I shall write down your method too, its super useful when I do calculations at home and I can use a calculator. $\endgroup$ – mika Apr 13 '15 at 4:58
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    $\begingroup$ @mika i added a bit to my answer, i think your formula was wrong to start with , where did you get it from? (please give reference) you also always needed to take the absolute value , distances are never negative. $\endgroup$ – Willemien Apr 13 '15 at 6:20
  • $\begingroup$ I got it from Stahl's A Gateway to Modern Geometry: The Poincare Half-Plane. I shall update above if you are interested. $\endgroup$ – mika Apr 13 '15 at 18:41
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    $\begingroup$ No need to do so, but do check your formula's (and make them understandable) they have a meaning .(see them as the sum or difference of two seperate formula's) $\endgroup$ – Willemien Apr 14 '15 at 6:09

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