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Circle has no role. I cannot replicate these numbers. When each circle passes through the centre of the other circle, so the circle centres are apart, a rhombus of two equilateral triangles fits within the lens. The area of this rhombus is so when , contradicting My calculations give me that distance gets you Conversely, to get lens area I find we need --Lambiam07:38, 21 December 2022 (UTC)[reply]
Yes, circle has no role. Let be the angle at X (or Y) between the lines to the junctions of ABD and of CEFG. The lens has area , where , and the diagram says (D+G) has area 0.03, so . That gives (in radians). The length from X (or Y) to the midpoint of XY is , and twice this gives . My trig is rusty and I may have made some errors, but I think the principle and the order of magnitude are right. If were 0.23, we'd get and . Certes (talk) 13:31, 21 December 2022 (UTC)[reply]
Okay, well this is interesting. Yes, in the example, circle C has no role here. Someone said "When each circle passes through the centre of the other circle," but I do not think that is probable. Here were my steps:
Circles , , and have centers , , and and radii . The centers form and intersect such that , ,
Simplifying:
oh I see this was my problem, I think... I distributed the 2 onto
Update: well, here i am checking my own work again. i found another error. i deleted some values of d within . I should know better than to ask for help right away... i can do this... but the doubt is strong in this one
no that's wrong too, ... i got it mixed up in again...
You can shift two equal-sized circles such that each passes through the other's centre; I used this special case merely to easily establish bounds that were violated by a purported solution.
The function is transcendental, and one cannot hope to solve the equation by a combination of algebraic and trigonometric manipulations for rational values of except when is an integer, in which case the equation is solved by --Lambiam09:01, 22 December 2022 (UTC)[reply]