Answer to Question #4126 in Geometry for john allen
Let R1 and R2 be the corresponding radii of circles, and O, T be the centers of the circles.
Notice that it is implicitly assumed that assuming that each of the AOB and CTD is less of equal to pi.
Then we have a isosceles triangles AOB and CTD such that
For definiteness assume that
We claim that the arc A^B intersected by AB is greater than C^D, the arc intersected
(*) A^B > C^D
A^B = R1 angle(AOB),
C^D = R2 angle(CTD).
So instead of (*) we have to prove that
(**) angle(AOB) > angle(CTD).
Moreover, since each of these angles is less of equal to pi, and so
angle(AOB)/2, angle(CTD)/2 is less of equal to pi/2,
it suffices to show that
(***) tan(AOB/2) > tan(CTD/2).
From triangle AOB we get that
tan(AOB/2) = AB/(2*R1)
tan(CTD/2) = CD/(2*R2)
Since AB=CD, and R1<We obtain (***), and therefore (*).
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