People clearly realize that 1 times a regular hexagon is a regular hexagon, 2 times a regular hexagon is a regular polygon, 3 times a regular hexagon is a regular polygon, and ... n times is called a regular hexagon? Polygon (abbreviated as regular n-polygon). "The ratio of the perimeter of a regular N-shape to the diagonal of its center (3. 14 15926 ... ratio 1) is called the positive N-edge ratio". (n is an infinite natural number of 0, 1, 2, 3…, which cannot be lost or ignored).

Because n is an infinite natural number, the positive n-edge ratio (3. 14 15926 ... so-called π value) is also an infinite infinite number.

When the diameter of a circle overlaps the diagonal of a regular N polygon passing through the center point, although the diameter and diagonal length are equal. But the perimeters of the two do not overlap, but are similar, close, close or equal or unequal to each other. The reason is that the points on any straight line are infinite, so the points on the perimeter of the inscribed regular N-polygon will never completely coincide with the points on the circle.

If the inscribed regular n-sided polygon is separated from the circle, the positive n-sided rate is still positive n-sided rate and pi is still pi.

Positive n-edge rate is not equal to pi; Pi is not equal to the positive n edge rate.

Because pi refers to the ratio of circumference to diameter, their ratio is 6+2√3 to 3; The ratio of positive N-sides refers to the ratio of the perimeter of positive N-sides to the diagonal line passing through the center, and their ratio is 3.1415926 ...1.

For this reason, the perimeter formula 2πR of regular N-polygon only replaces the perimeter formula and is not equal to the perimeter; The area formula πR of regular N-polygon? It only replaces the formula of circular area and is not equal to circular area.

Objectively speaking, a circle is a circle and a regular N-polygon is a regular N-polygon. When the regular N polygon is circumscribed by a circle, the circumference of the circle inscribed with the regular N polygon is calculated by formula 2πR, and the circumference must be smaller than the circumference of the circle. When the circular sleeve circumscribes the regular N polygon, the area formula πR of the regular N polygon circumscribed by the circle is used. To calculate the area, the area must be larger than the circular area (note: in fact, πR? It is the reciprocal product transformation of the circumscribed n-side area of a circle and the rectangular area, not the reciprocal product transformation of the circular area and the rectangular area).

When we take the same positive N edge rate value for this π, we will give the formulas 2πR and πR? Yes: π will deviate from the formula πR if it satisfies the formula 2πR? ; If π is to satisfy the formula πR? , it will deviate from the contradiction of formula 2 π r.

According to Einstein's "theory of relativity", it is concluded that "matter and matter combine into a whole (solid, liquid, vapor) called an object; The size of an object surrounded by space includes the number of unit cubes called volumes. The combination of immaterial and immaterial polymerization into a complete vacuum is called space; The size of the space surrounded by objects is called volume. "

Because the difference between objects and space is the difference between matter and immaterial, the universe is made up of matter and immaterial, and objects and space occupy nature together.

Therefore, all objects and all spaces in the world are born relative. When they are at rest, there is no problem of "objects occupying space or space occupying objects". Only when objects and space exchange positions with an object volume and a space volume, and objects and space interact, will "objects occupy space and space occupy objects" appear. Because the volume of an object is relative to the volume of space, so is the volume. Both are indispensable, otherwise the object cannot be moved or carried.

Because the minimum limit of volume relative to volume is zero (that is, geometric point refers to zero volume or zero volume, zero area or empty product, zero length or zero distance); However, the volume of the object and the volume of the space are infinitely non-zero, that is, the volume or volume, area or empty product, length or distance are all greater than zero. There are no maximum and minimum values, and there is no size limit.

So every infinitesimal of a body, a surface and a line in infinite bisection geometry is still an infinitesimal. Infinite infinitesimal is infinite infinitesimal, and infinite infinitesimal is not equal to the minimum limit zero.

The above is the tip of the iceberg of positive and negative geometry and limit theory in relativity.

Therefore, in the past, it was a misunderstanding that people converted the equal parts and equal products of a circular surface into rectangular surfaces. That is, the circular area s is not equal to the rectangular area πR? Exactly: "circular area s=7(d/3)?" (d stands for diameter). π takes 3. 14 15926 ... It is not the ratio of the circumference to the diameter of a circle. To be precise, it is the ratio of the perimeter of a regular N-polygon to the diagonal line passing through the center.

Then, why do you say, "The area of a circle is equal to seven times the square of one third of its diameter"?

This must start with softening equal area deformation.

For example, a cuboid plasticine with a length of 7 meters, a width of 1 m and a height of 1 m has a rectangular area of 7 square meters above or below. When 7 cubic meters of cuboid plasticine becomes a cylinder with a height of 1 m, the area of its upper and lower bottom circles will still be 7 square meters. That is, a rectangular area of 7 square meters softens into a circular area of 7 square meters. If 1 unit length is represented by a, then the area of a 7-square-meter circle is 7a? . For this reason, any circular area s can be regarded as 7a? .

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Every square on the chessboard is an A? Let's analyze it: seven aces? Soften the equal product into a circle (Figure-1). Is the circle area 7a? ; Circular area 7a? Re-softening the equal product into (Figure -2)H-shaped area is also 7a? ; On the H-shape (Figure-2), add two A's? Is it a big square with an area of 9a? ; These three numbers are called (the last three numbers). The size of their respective areas varies with.

One chess piece is a point, seven chess pieces are seven points, and the diameter q of the point is called the point diameter. There is a point in the middle and six points on the periphery, which are arranged tangentially around a circle to form a circular outline (Figure -4). The circumscribed circle of the outline has an area of S and a diameter of 3q. Then an H-shaped contour is formed by tangency of seven points (Figure -5), and the circumscribed H-shaped area of the contour is 7Q? ; Finally, nine points are tangent to form a square outline (Figure -6), and the circumscribed square area of the outline is 9Q? . These three figures are called (the next three figures), and their respective circumscribed areas all vary with the size of the point diameter q.

The above six figures are not difficult to see:

(Figure-1) Circular area 7a? And (fig. 2) the H-shaped region 7a? Are they all (Figure -3) with a large square area of 9a? Seven out of nine, (Figure -4) the circumscribed circle is the inscribed circle of the circumscribed square (Figure -6).

From the top and bottom of the six graphs: Because of the first group, (graph-1) circle is similar to (graph -4) circumscribed circle; In the second group, (Figure -2)H shape is similar to (Figure -5) circumscribed H shape; In the third group, (Figure -3) the big square is similar to the circumscribed square (Figure -6). Therefore, whether the area and area of each group with similar shapes are equal is related to A and Q; Or whether a and q are equal is related to the area and area of each group of similar shapes.

When a=Q, it is obvious that the similar shapes of the second group and the third group are: A and Q are equal, and the areas of similar shapes are equal (7a? =7Q？ 、9a？ =9Q？ ); Or an area of similar shape is equal to the area (7a? =7Q？ 、9a？ =9Q？ ), a and q are equal.

But is the first group of similar shapes equal to A and Q, and the area is equal to the area?

This needs to be proved by data reasoning:

It is known (Figure -4) that the circumscribed circle area S is 63 square centimeters, and A and Q are equal. At this time (Figure -4), this circular area of 63 square centimeters locks the corresponding areas of the two (the lower three figures) and (the upper three figures).

Because A is equal to Q, the circle of 63 square centimeters (Figure -4) is both the inscribed circle of a square (Figure -6) and the inscribed circle of a big square (Figure -3). For this reason, the inscribed circle areas of (Figure -6) and (Figure -3) are also 63 square centimeters respectively.

Because (Figure -3) the big square can be used as the circumscribed circle of a circle of 63 square centimeters, it is based on the fact that the side length 3a of the big square is equal to the diameter 3Q of the inscribed circle (the diameter 3Q of the inscribed circle is generated according to the circle area of 63 square centimeters).

Therefore, any size of the inscribed circle area (Figure -3) will change the size of the side length 3a of the big square (Figure -3), so that the side length 3a is not equal to the diameter 3Q of the circle with 63 square centimeters, and the big square (Figure -3) cannot be used as the circumscribed circle of the circle with 63 square centimeters.

If (Figure -3) the inscribed circle area is greater than 63 square centimeters, then (Figure -2) 7a? H-shape and (Figure -3)9a? A large square will be sensitive to large strain (7a? & gt7Q？ 、9a？ & gt9Q？ )。 Display 9a? The big square of the square extends outward, beyond the known inscribed circle of 63 square centimeters), resulting in the side length 3a being greater than the diameter 3Q, which violates that A equals Q.

If the inscribed circle area in (Figure -3) is less than 63 square centimeters, then (Figure -2) 7a? H-shape and (Figure -3)9a? Large square versus small strain (7a? & lt7Q？ 、9a？ & lt9Q？ )。 Display 9a? The inward contraction of the big square will also break away from the known inscribed circle of 63 square centimeters, resulting in the side length 3a being smaller than the diameter 3Q, which also violates that A equals Q.

Therefore, only when the inscribed circle area (Figure -3) is equal to the circumscribed circle area of 63 square centimeters (Figure -4) can it be 7a? =7Q？ 、9a？ =9Q？ Do 9a? This big square is used as the circumscribed circle of a circle of 63 square centimeters. At the same time, the side length 3a of the big square is also equal to the diameter 3Q of the inscribed circle, keeping a and q equal. So (Figure -3) the size of the big square is determined according to the known inscribed circle of 63 square centimeters.

This shows that this is true for circles of any size. When the circle (Figure-1) overlaps the inscribed circle of 63 square centimeters (Figure -3).

If (Figure-1) the circular area is 7a? More than 63 square centimeters, then (Figure -2) 7a? H-shape and (Figure -3)9a? A large square will be sensitive to large strain (7a? & gt7Q？ 、9a？ & gt9Q？ )。 Display 9a? The big square "A" expands outward and breaks away from the inscribed circle of 63 square centimeters, resulting in that the side length 3a is greater than the diameter 3Q, and A is also greater than Q.

If (Figure-1) the circular area is 7a? Less than 63 square centimeters, then (Figure -2) 7a? H-shape and (Figure -3)9a? Large square versus small strain (7a? & lt7Q？ 、9a？ & lt9Q？ )。 Display 9a? The big square of the square will shrink inward, and it will also break away from the known inscribed circle of 63 square centimeters, resulting in that the side length 3a is less than the diameter 3Q and a is less than Q.

Therefore, only (Figure-1) circular area 7a? It is equal to (Figure-3) the inscribed circle area of 63 square centimeters, so it can be 7a? =7Q？ 、9a？ =9Q？ Do 9a? This big square is used as the circumscribed circle of a circle of 63 square centimeters. At the same time, the side length 3a of a square is also equal to the diameter 3Q of a circle of 63 square centimeters, and keep a equal to Q. Then (Figure-1) the area of the circle 7a? The size of is determined by the area of the inscribed circle (Figure -3).

Confirmed: (Figure-1) Circular area 7a? It is equal to (Figure -4) the circumscribed circle area S. It also shows that "the area of a circle is 7/9 of its circumscribed circle area".

Because the circular area S=7a? So a=√s/7. That is, if the inscribed circle area of a square (Figure -3) is 7 square centimeters, then a=√7/7= 1 centimeter. If the inscribed circle area of a square (Figure -3) is 28 square centimeters, then a=√28/7=2 centimeters. If the inscribed circle area of a square (Figure -3) is 63 square centimeters, then a=√63/7=3 centimeters.

The above proves that the first group of similar shapes are the same: A and Q are equal, and the areas of similar shapes are equal.

From this, three groups of similar shapes are deduced: the area and area of each group of similar shapes are equal, and A and Q are equal; Or a and q are equal, and the area and area of each group of similar shapes are equal.

At the same time, we also found an axiom: "If the area of a circle is 7a? , then its circumscribed circle area is 9a? " .

According to the axiom, the theorem is deduced: "The area of a circle is equal to seven times the square of one third of its diameter".

Formula for the area of a circle: ∵s=7a? . d=3a。

∴s=7(d/3)？ Give a gift to the 70th anniversary of the National Day!

HPFYKG？ An illiterate mathematical discovery? Dongjingui 201June 27, 4