In euclidean or hyperbolic geometry, there will be a quadrilateral with sides of length (r,s,t,u) if and only if the sum of any three exceeds the fourth. Algebraically, this is equivalent to the condition B(r,s,t,u) = (r+s+t-u)(s+t+u-r)(t+u+r-s)(u+r+s-t) > 0. This follows since, if we order the lengths so that u is the largest, then only the first factor could be negative.
The function B appears in Brahmagupta's formula for the area of a cyclic quadrilateral
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In general, a given list will correspond to infinitely many quadrilaterals. If we choose any positive number x so that there are triangles with sides (r,s,x) and (x,t,u), then we can place such triangles so that the sides of length x correspond. In general, it will be possible to vary x slightly, so we get the claimed infinity of solutions. A more interesting question arises if we ask for a convex cyclic quadrilateral.
In euclidean geometry, the condition is just that B(r,s,t,u) > 0. In other words, we can
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It is not obvious that such x exist when B(r,s,t,u) > 0 Here is a proof |
In hyperbolic geometry, the situation is more interesting. We shall present two proofs. One follows on from earlier work, and is really the analogue of a euclidean proof. The second uses a new and rather strange mapping from part of the disk to the plane.
a theorem on cyclic hyperbolic quadrilaterals
Of course the condition is equivalent to the existence of convex cyclic euclidean
The mapping approach leads to further hybrid theorems in a systematic way. Before looking further at polygons, we must discuss convexity.
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