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Question: Formulae derived in Geometric expressions is it correct ?

Question:Formulae derived in Geometric expressions is it correct ?

janhardo 490
geometry circle
June 25 2020
1

Could it be correct ?


 

If L= 1.158R  then half of smallest circle is covered  to proof

 

I derived a formulue for this in GX

 

 

 

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restart;
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abs(((((((L)^(2)*(-1))+((R)^(2)*4)))^(1/2)*L*(-1/2))+(arctan((((((L)^(2)*(-1))+((R)^(2)*4)))^(1/2)*(L)^((-1))))*(L)^(2))+(arctan((((((L)^(2)*(-1))+((R)^(2)*4)))^(1/2)*(((L)^(2)+((R)^(2)*(-2))))^((-1))*L))*(R)^(2)*(-1))));

abs(-(1/2)*(-L^2+4*R^2)^(1/2)*L+arctan((-L^2+4*R^2)^(1/2)/L)*L^2-arctan((-L^2+4*R^2)^(1/2)*L/(L^2-2*R^2))*R^2)

 (1)
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verg:=%;

abs(-(1/2)*(-L^2+4*R^2)^(1/2)*L+arctan((-L^2+4*R^2)^(1/2)/L)*L^2-arctan((-L^2+4*R^2)^(1/2)*L/(L^2-2*R^2))*R^2)

 (2)
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subs(R=1,verg);#  

abs(-(1/2)*(-L^2+4)^(1/2)*L+arctan((-L^2+4)^(1/2)/L)*L^2-arctan((-L^2+4)^(1/2)*L/(L^2-2)))

 (3)
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verg2:=%

abs(-(1/2)*(-L^2+4)^(1/2)*L+arctan((-L^2+4)^(1/2)/L)*L^2-arctan((-L^2+4)^(1/2)*L/(L^2-2)))

 (4)
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solve(verg2 = Pi,L); # area half small circle with R=1 => 2*Pi* R^2/2 = Pi ,

Warning, solutions may have been lost

  

L must be expressed in R ?

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