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Does $\exists$ on the hyperbolic plane, a convex quadrilateral $Q$ and a convex pentagon $P$ with the same angle sum? I found this question to be rather interesting.

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Yes. Both $0.{}$ –  Will Jagy Apr 27 '13 at 5:50
    
Could you provide an example? –  Richard Carpenter Apr 27 '13 at 5:59

1 Answer 1

If you know about ideal polygon then this is easy because every ideal polygon has angle sum zero.

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I thought of that but is an ideal polygon convex? –  Richard Carpenter Apr 27 '13 at 6:05
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If boundaries of ideal polygon are geodesics then t is convex. This is true in general i.e. if you have a polygon with geodesic boundary then it is convex. P.S.: Here convexity means in hyperbolic sense not in euclidean sense i.e. any two point in the polygon can be joined by an hyperbolic geodesic which lies in that polygon. –  tessellation Apr 27 '13 at 6:09
    
I see. I have one more question. Does the quadrilateral or the pentagon have all right interior angles? I would assume it cant be the quadrilateral since that would imply a rectangle which does not exist. –  Richard Carpenter Apr 27 '13 at 6:14
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Theorem:Gauss-Bonnet: An n-gon with angles $\alpha_1,\alpha_2,...,\alpha_n$ has area $(n-2)\pi-(\alpha_1,\alpha_2,...,\alpha_n).$ Now if a quadrilateral has all interior angle $\pi/2$ then its area is zero which in not possible. But for a pentagon this case is possible because an ideal pentagon ha all angles 0 and if we start to draw smaller pentagon the interior angles will became small up-to angle $3\pi/5$ which is greater than $\pi/2.$ Hence there will be a polygon yu desired. This logic applies for all polygon with more than 4 sides. –  tessellation Apr 27 '13 at 6:40
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Yes I am. Actually if you take a regular pentagon of area $\pi/2$ which will exists by the above argument its all angles will be $\pi/2$. –  tessellation Apr 27 '13 at 6:52

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