To demonstrate how this works, let’s start off with a simple problem; the Delian problem:Doubling the volume of the cube is equivalent to doubling the cubic root of the side of the cube, which is equivalent to constructing the ?2. To do so, we could start with two parabolas that share a common vertex (at the origin of an (x, y) plane and have perpendicular axes (the x and y axes). This is a simple and a specific example.We consider the parabolas with the equations:p1:y2=2ax and p2:x2=2by for a=2bSince those parabolas intersect at two points (share the vertex and intersect at a point with positive coordinates) they could only have one real common tangent. This tangent cannot be parallel to either axis:t:y=cx+dWe mark the point: P1x1, y1=p1tThus the tangent is described with the equation:yy1=ax+ax1 y=ay1x+ax1y1Thus:c=ay1 and d=ax1y1 y1=acand x1=dc a2c2=2adc a=2cdMark the point: P2x2, y2=p2txx2=by+by2 y=x2bx-y2Therefore:c=x2b and d= -y2 x2=bc and y2= -d b2c2= -2bd d= -bc22Thus we find:a=2cd and d= -bc22 a= -bc3 c= -3abSince a=2bc= -32c is the slope of the tangent line, as it is negative it’s not parallel to either axes of the parabolasTo construct this quality: c=32, we should construct a right triangle with the side of unit length and hypotenuse over the tangent line t, the second side will be equal to 32Slope c=yx=y’x’, choosing x’=1 c=y’=32 , the length of the hypotenuse would be irrelevant.Using origami, we can fold the common tangent of the two parabolas with on crease, then we can fold the unit length from any point on the tangent on the x-axis and taking the unit length at any degree from the tangent. Then we fold the tangent on itself to find the third vertex of the triangle.p1:(y-n)2=2a(x-m) and x2=2byThe equation describing any common tangent of p1 and p2 is t:y=cx+dAgain we assume P1x1, y1=p1t(y-n)(y1-n)=a(x-m)+a(x1-m) y=ay1-nx+n+ax1-2amy1-nThus:c=ay1-n and d=n+ax1-2amy1-n y1=a + nccand x1=d – n c+2mAndp1:(y-n)2=2a(x-m) a2c2=2a (d – nc+m) a=2c (d-n+cm)And for the point on the other parabola P2x2, y2=p2t, we get for txx2=by+by2 y=x2bx-y2Leading us eventually to d= -bc22Substituting for d, we obtaina=2c (d-n+cm) a=2c (-bc22-n+cm) a= -bc3+mc2-nc bc3-2mc2+2nc+a=0Thus the slope of t:y=cx+d is one of the solutions to the equation:c3-2mbc2+2nbc+ab=0Given: x3+px2+qx+c=0, one can fold the roots by the following method:p=-2m, q=2n, and r=am= -p2, n=q2, and a=rThat way, we would only need to find coordinates of the focus of the p1: P1(-p2+r2, q2)and the directrix of the the parabola p1 which represented with the equation is;l1:x= -p2-r2(NB; this equation also describes the line carrying the side of the right triangle, drawn in the end, that carried the constructed slope)These would be the focus and directrix of p1, for p2the focus P2(0, 12)and the directrix l2is described by the equation: y= -12. Using the beloch fold, the crease would yield the common tangents.