Batman w geometrii

Wzór (wersja graficzna)

Wzór (wersja komputerowa)

(x/7)^2*SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\*SQRT(ABS(y+3*SQRT(33)/7)/(y+3*SQRT(33)/7))-1=0 ABS(x/2)-((3*SQRT(33)-7)/112)*x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0 9*SQRT(ABS((ABS(x)-1)*(ABS(x)-0.75))/((1-ABS(x))*(ABS(x)-0.75)))-8*ABS(x)-y=0 3*ABS(x)+0.75*SQRT(ABS((ABS(x)-0.75)*(ABS(x)-0.5))/((0.75-ABS(x))*(ABS(x)-0.5)))-y=0 2.25*SQRT(ABS((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x)))-y=0 (6*SQRT(10))/7+(1.5-0.5*ABS(x))*SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6*SQRT(10))/14)*SQRT(4-(ABS(x)-1)^2)-y=0

Źródło: http://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real