In the decision of a task the opportunity of construction of a circle, equal on the area is shown a square, that is “the quadrature of a circle” that has enabled to solve “a quadrature of a circle” with accuracy on eight signs on standard number π is solved, and to express length of a circle a direct piece

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If calculation of a task of a message on a lot of signs the result of size of the side of a square will be equal … 1,7724538968686925718887244115238…

THE DECISION

About a circle of radius OR (fig. 1) which size of taken as a unit length, we shall describe the correct eight-final star Q formed from two equal squares, one of which square ABCD. … S_ABCD = (AB)² = (2OR) ²

Any side of a square will cut off from each rectangular top of other square on a triangle, one of which PCP₁. From here … S_Q = S_ABCD + 4S_PCP1

In radius CR from each rectangular top of figure Q we shall describe arches on its sides, and points of crossing of the sides and arches it is combinable straight

lines. In triangle PCP₁ of such straight line will be KK₁. Being crossed with a diagonal of a square, straight line KK₁ forms point R₁. In figure Q each acting rectangular triangle equal to triangle PCP₁, will share on two equal figures, a triangle and a trapeze what triangle KCK₁ and trapeze PKK₁P₁ are. If to remove in figure Q all eight equally acting rectangular triangles, one of which triangle KCK₁ we shall receive figure T - "gear" with acting trapezes on the area of the equal area of square ABCD …

S_T = S_Q – 8S_KCK1 = (S_ABCD + 4S_PCP1) – 8× ½ S_PCP1 =S_ABCD

S_T = S_ABCD

* * *

On fig. 2 which represents a fragment fig. 1, center O K. We shall receive triangle OKR₁ in which we shall carry out median ON. In radius OR1 we shall carry out an arch, which will cut off from median ON piece MN, and from hypotenuse OK - piece LK. …

We bring calculation of the received is combinable with point pieces …

OR = 1

OC = OR× √2 = 1 × 1,4142135…

RC = OC – OR = 1,4142135… - 1 = 0,4142135…

KK₁ = RC × √2 = 0,4142135…× 1,4142135…= 0,5857863…

RR₁ = RC – (KK₁/2) = 0,4142135…- 0,2928931…= 0,1213204…

OR₁ = OR + RR₁ = 1 + 0,1213204…= 1,1213204…

OK² = OR₁² + (KK₁/2)² = 1,1213204...² + 0,2928931…² = 1,1589416…²

LK = OK – OR₁ = 1,1589416… - 1,1213204… = 0,0376212…

ON² = OR₁² + (KK₁/4)² = 1,1213204…² + 0,1464465…² = 1,130843…²

MN = ON – OR₁ = 1,130843… 1,1213204… = 0,0095226…

Radius of a circle equal to square ABCD we shall accept conditionally for OR_x. We find its arithmetic size from equality of the areas of a conditional circle with radius OR_x and square ABCD …

π × OR_X² = (2OR)² 3,1415926…× OR_X² = 4

OR_X = 1,1283791…

Conditional point R_x we shall arrange any way on piece KK₁ and it is combinable its dotted straight line with center O. We shall receive conditional rectangular triangle OR₁R_x. Arithmetic size of conditional cathetus R₁R_x we shall receive from the decision …

R₁R_X² = OR_X² – OR₁² = 1,1283791…² – 1,1213204…² = 0,1260164…²

We shall receive the same size from a proportion made of sizes of pieces, before received vectorially …

[(OR+MN) - LK] / (OR+MN) = RR₁ / R₁R_X

R₁R_X = [(OR+MN) × RR₁] / [(OR+MN) – LK] =[(1+0,0095226…)×0,1213204…] / [(1+0,0095226...) –0,0376212…] = = 0,1260164…

Arithmetic size R₁R_x we shall express a geometrical piece. Pieces MN and LK we shall transfer on diagonal OC in radiuses ON and OK. Piece MN will be postponed from point R₁ up to point E, and piece LK from point R₁ up to point F. Then piece OR we shall put on continuation of diagonal OC so that the beginning of piece OR was point E, and the end - point O₁. Of point F we shall construct a perpendicular to OO₁ on which we shall postpone size of piece RR₁, from point F up to point F₁. Through points O₁ and F₁ we shall carry out a straight line before crossing from

straight line KK₁ in point R₂. Thus, conditional size R1Rx was expressed by geometrical piece R₁R₂. Received point R₂ it is combinable a straight line with center O. We shall receive radius OR₂ of a circle equal on the area to square ABCD …

OR₂² = OR₁² + R₁R₂² =1,1213204…²+ 0,1260164…²= 1,1283791 … ²

If to accept a square equal on the area to a circle with radius OR for conditional square A_xB_xC_xD_x we shall receive a proportion …

S_OR / S_OR2 = S_AõBõCõDõ / S_ABCD or … OR / OR₂ = ½ A_XB_X / ½ AB

Which we shall put in system of coordinates (fig. 3) to express conditional size ½ A_xB_x a geometrical piece. The left part of a proportion we shall put on an axis àáñöèññ, right - on an axis of ordinates. Points M (OR₂; ½ AB) and O give a beam in which àáñöèññîé OR it will be reflected new point M₁ which projection to an axis of ordinates, will vectorially reflect ½ the sides of required square A₁B₁C₁D₁, equal on the area to a circle of radius OR …

½ A₁B₁ = (OR× ½ AB) / OR₂

½ A₁B₁ = 1 / 1,1283791…= 0,8862269…

A₁B₁ = 0,8862269…× 2 = 1,7724538… If calculation of a task of a message on a lot of signs the side of square A₁B₁ will be equal 1,7724538968686925718887244115238…, and S_A1B1C1D1 = 3,1415928165250138836954861078059…

LENGTH OF THE CIRCLE

The finding of the side of square A₁B₁C₁D₁ enables to express L _OR - length of a circle of a circle of radius OR a direct piece. We shall make a proportion …

OR / OR₂ = L _OR / P_A1B1C1D1 or … OR / OR₂ = ¼ L _OR / A₁B₁

Which we shall put in system of coordinates (fig. 4). The left part of a proportion we shall put on an axis àáñöèññ, right - on an axis of ordinates.

Points M (OR₂; A₁B₁) and O give a beam on which àáñöèññîé OR new point M₁ which projection to an axis of ordinates will vectorially reflect a direct piece ¼ L _{OR …} is formed

¼ L_OR = (OR× A₁B₁) / OR₂ = 1,7724538… / 1,1283791…= 1,5707963…

½ L_OR = 1,5707963…× 2 = 3,1415926…

L_OR = 1,5707963…× 4 = 6,2831852…

***

¼ L_OR = 3,1415928165250138836954861078045…

In turn ¼ of length of a circumference of a circle of radius OR₂ too it is expressed by direct piece A₁B₁.

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