# Calabi's Triangle

Author:
ortollj

## Solution for the Angle Values:

#  we need C_5 position being on the X axis # so we need y(C_5)=y(C_4)-c*cos(α)  with y(C_4)=y(T_1) - c sin(α)=0 # y(C_5)=y(T_1) - c sin(α)-c*cos(α)=0 # y(C_5)=L tan(α) - c sin(α)-c*cos(α)=0 # c= size length of the squares as we have tan(α)=c/(L-c/2) so # c = L tan(α) / (1 + (tan(α) / 2)) # y(C_5)=L tan(α) - c (sin(α)+ cos(α))=0 # y(C_5)=L tan(α) - (L tan(α) / (1 + (tan(α) / 2))) (sin(α)+ cos(α))=0 # y(C_5)=1 -  (1/ (1 + (tan(α) / 2))) (sin(α)+ cos(α))=0 # 1 =  1/(1 + (tan(α) / 2))  (sin(α)+ cos(α)) # (1 + (tan(α) / 2)) =    (sin(α)+ cos(α)) # 2 + tan(α)  =    2 (sin(α)+ cos(α)) # 2 + sin(α)/cos(α)  =    2 (sin(α)+ cos(α)) # 2 + sin(α)/(1-sin(α)^2)^0.5  =    2 (sin(α)+ (1-sin(α)^2)^0.5) # 2(1-sin(α)^2)^0.5 + sin(α)  =    2 (sin(α) (1-sin(α)^2)^0.5 + (1-sin(α)^2)) # we set x= sin(α) # 2(1-x^2)^0.5 + x  =    2 (x (1-x^2)^0.5 + (1-x^2)) ok solution x_0=0.631109 x_1=-0.958053 # # 2 + tan(α)  =    2 (sin(α)+ cos(α)) # we set t=tan(α/2) sin(α)=2*t/(1+t^2) and cos(α)=(1-t^2)/(1+t^2) and tan(α)=2*t/(1-t^2) # 2 + 2*t/(1-t^2)  =    (4*t/(1+t^2) + 2*(1-t^2)/(1+t^2)) # (2*(1-t^2) + 2*t)*(1+t^2) =   (1-t^2)* (4*t + 2*(1-t^2)) # -2*t^4+2*t^3+2*t+2 = 2*t^4-4*t^3-4*t^2+4*t+2 # -4*t^4 + 6*t^3 + 4*t^2 - 2*t = 0 # -4*t^3 + 6*t^2 + 4*t - 2 = 0