# Frustrating geometry question

I have been battling with this problem for some time now but I am still stuck. Please would some kind soul help me out?

So I have a Riemann sphere with 2 point $X,Y$ on it. (The Riemann sphere is just a unit sphere right?) The spherical distance between $X,Y$ is $l$. The projection point is equidistant from $X,Y$. I have to find the stereographic projection of $X,Y$, where we are told that the projections lie on the real axis.

My argument: the angle $X$ makes with the axis normal to the plane $\mathbb C$ would be $l\over 2$. So the "projection line" radiating from the projection point, passing through $X$ and intersecting $\mathbb R$ makes an angle $l\over 4$ with the vertical. So the distance of the projection from the origin $0\in\mathbb C$ would be $2\tan({l\over 4})$.

The given answer: The distance of the projection from the origin $0\in\mathbb C$ is $\tan({l\over 4})$

I don't understand why there isn't a factor of $2$ since if the radius of the sphere is $1$, surely the diameter has length $2$??

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Since you asked yourself "the RS I'd just the unit sphere right?", it may help you to have a precise definition. –  john w. Feb 8 '12 at 15:33
@johnw.: You are right, I have double-checked and my memory seems to have served me right -- it is indeed a unit sphere. But I my problem persists! –  queer Feb 8 '12 at 16:08
I'm having trouble understanding the problem but this might help: your reasoning for having 2 is that the diameter is 2. But you are computing the distance from 0 in $\mathbf{C}$ to the point on RS. So the radius is the only relevant number? i. e. Why do you want the diameter factor? –  john w. Feb 8 '12 at 19:06
@johnw.: I would have uploaded a sketch but I don't have enough reputation to do so. Why would I be finding the distance of 0 in $\mathbb C$ to the RS? Surely the projection lies on $\mathbb C$ in particular the real axis (as given) so I would be looking for the distance of the projected point (on C) to the origin (on C)...? –  queer Feb 8 '12 at 20:15
How are you doing the stereographic projection? There are two definitions that I've seen used in different books. One is that you put the sphere on top of the plane so that the south pole sits at the origin. The other is that you put the sphere so that the plane intersects the sphere at the equator. A bit of geometry will tell you that these two definitions differ in the projected radius by a factor of 2. So perhaps you should double check which stereographic projection you are using, and which projection the book is using. –  Willie Wong Feb 9 '12 at 9:12