# How to cut a rectangle into 3 rectangles

Table of Contents:

### Problem 5.

Octagon and square. Most of the previously proposed puzzles are fairly easy to solve using the two-way overlay method. However, this method does not always lead to the desired result. For example, converting a regular octagon to a square is probably easier to do through trial and error, using the symmetry of the octagon, than by overlapping two stripes. You can use another method for solving cutting problems. the mosaic method. The essence of this very intuitive and simple method, which makes it possible to solve very difficult problems of transforming regular polygons, is the superposition of two suitable (usually semi-regular) mosaics. So, how to transform a regular hexagon into a square so that the number of cut parts is the smallest?

### Task 2.

Rectangle 1 x 3 and transform to square symmetrically. How to symmetrically cut a rectangle with sides 1 x 3 into 9 pieces to make a square out of them?

### Problem 4.

Two six-pointed stars and a hexagon. Cut a regular hexagon into the smallest number of pieces so as to make two six-pointed stars.

### Problem 3.

Greek cross and square. Cut the Greek cross so that the pieces can be folded into a square. a) four straight lines using central symmetry; b) two straight lines.

## Geometric Logic Problems

### Problem 1.

4 x 9 **rectangle** and square. How to cut a 4 x 9 rectangle into the minimum number of pieces to make a square?

## Problem solutions.

Since the areas of the rectangle and the square must be equal, the side of the square is 6. Therefore, we set aside 6 on the large sides of the **rectangle** and make a stepped cut, which can be guessed about, remembering the central symmetry of the rectangle. It is easy to make sure that for arbitrary sides of the **rectangle**, the stepped cut will not allow making an equal square.

First, we cut 2 squares along the diagonals and place 4 of the resulting triangles around the remaining square, attaching parts of their hypotenuses to its sides. We mentally connect the points of right angles to each other with lines and along them we make the last 4 cuts, cutting off small obtuse triangles from right-angled triangles, for which, as follows from the figure, there are just corresponding places.

## how to split a rectangle

I have four latitude and longitude values that create a rectangle (area). Now I want to split the rectangle into grids.

So I define latlong (11.20804, 122.22839), (11.20804, 122.52914), (11.06926, 122.52914), (11.06926, 122.22839) after splitting the grids I want to give each grid an id and want the latitude and longitude values of each grid_id.I from all forces tried to split the rectangle into grids and assign them grid_id. Any help would be appreciated

### 1 answer

I have a given size of the view width (for example 450 which I got from viewWidth = rectOftable.size.width;) where rectOftable is a rectangle with the size (width is height) of the content view. I have now created an NSAttributedString that contains a string that can have.

In 2d space, there is a **rectangle** and a circle that overlap each other. As far as possible I am calculating the smallest distance (depth) that I need to separate the circle and rectangle?

Something like that. You will need to play with it and decide if you are ok if there is no perfect square at the bottom. This means a starting point, not a complete solution.

I just read that you need grid ids, so guess like 1,1, so you have to tweak this to suit your needs.

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I have a given view width (like 450 which I got from viewWidth = rectOftable.size.width;) where rectOftable is a rectangle with size (width is height).

In 2d space, there is a **rectangle** and a circle that overlap each other. As far as possible I am calculating the smallest distance (depth) that I need to separate the circle and **rectangle**?

I have rectangle R1, with width w1, height h1. I am given a smaller rectangle R2 (w2, h2), the origin of which coincides with R1, that is, (0,0). How can I divide the remaining space in two.

I have a **rectangle** with coordinates A, B, C, D, which respectively have coordinates: A- (3.9) B- (6.9) C- (6.6) D- (3.6) where A- top left coordinate, A D is bottom left.

I am trying to split a rectangle with specific coordinates into 8 smaller **rectangles** (two columns and four rows). Input data, for example, will be: rec = [(0, 0), (0, 330), (200, 330).

I am trying to split a rectangle in such a way that it divides equally. I also don’t know if the svg rectangle can be split? This is the svg code:

I want to code a video platform and I have a problem and I cannot come up with a solution right now. I want to split a rectangle into equal parts Came up with a solution for a square, but I cannot.

## prompt

Three rectangles are not a lot, so you can iterate over the cases of their arrangement in a square and check if in each case the rectangles can be similar.

## Decision

If we draw a little of the division of the square into three rectangles in order to understand how they can be located in it at all, then rather quickly you can come to the conclusion that there are only two different cases (up to the rotations of the square). Indeed, three, two or one rectangles can adjoin the upper side of the square. If there are three of them, then the configuration shown in Fig. 1 on the left. If two, then is the configuration shown on the right in this figure. If only one **rectangle** adjoins the upper side, then the other two are located below it, and their common side is either horizontal (and then this is the same as the first configuration) or vertical (then this is the same as the second configuration).

### Splitting Rectangles into Equal Parts (1.G.3)

About the first configuration, it is immediately clear that all three **rectangles** are equal to each other: by condition they should be similar, but from the location it turns out that they are equal*about*major sides.

Let’s deal with the second configuration. We assume *orientation* rectangle the direction of its longer side (it is clear that we have here only elongated **rectangles**, in which one side is longer than the other). How the top two **rectangles** can be oriented?

They cannot both be vertical (as in Fig. 1), because then they will be equal (b*about*The larger sides coincide), and therefore the ratio of the larger side to the smaller side is less than 2 (since the smaller side is equal to half the side of the square, and the larger side is not more than the whole side of the square). And the lower rectangle will have this ratio greater than 2. Hence, it cannot be similar to the upper.

They can be both horizontal (Fig. 2, left). Then the two upper **rectangles** are again equal and it is easy to calculate that in order for all three rectangles to be similar, it is necessary that the sides of each relate to each other as 3: 2.

### 02c Dividing **rectangles** into fifths

Finally, could it be that one of the top rectangles is horizontal and the other is vertical? Let’s check. This situation is depicted in Figure 2 on the right. Let us introduce designations as in this figure. Considering the similarity of rectangles, we find:

Since the sides of the square are equal, we get equalities:

Right equality allows you to express *y*:

### 02b Dividing **rectangles** into thirds

after which the equation on the left is obtained

This cubic equation has one real root \ (\ rho \ approx1,3247 \ ldots \), so this is the case. In total, there are three ways to cut a square into similar **rectangles**.

## A task

How many ways can you cut a square into three **rectangles**, each of which is similar to the other two? Recall that two **rectangles** are similar if the sides of the first are related to each other in the same way as the sides of the second. Methods that differ only in the rotation or reflection of the square are considered one.

## Afterword

Since for cubic equations, formulas are known that give exact solutions, then you can be sure that there is a root and it is one. In radicals, this number is written as follows:

It can also be written as an endless sequence of nested radicals:

Interestingly, this number has its own “name”: the Dutch architect (and part-time monk) Hans van der Laan named it *plastic number* (plastic number). Van der Laan did not create very many buildings and they were mainly churches, but his theoretical work carried a certain weight. In particular, he developed a theory of harmonious relationships between building elements, in which the plastic number played a central role.

Fig. 3. Buildings designed by Hans van der Laan. *Left*: Benedictine monastery in Tumelilla, Sweden. *On right*: interior of an abbey in Maastricht, Netherlands. Photo from divisare.com

This name, according to his idea, reflected the fact that this number can be given geometric “shapes”. We got acquainted with one example of such a form in the task. Another example arises like this. Suppose you have an unlimited supply of boxes (rectangular parallelepipeds) of different sizes with whole side lengths. Let’s start with a 1 × 1 × 1 box, attach another such box to it on the side. you get a 2 × 1 × 1 box. Let’s attach the same to it in front to get a 2 × 2 × 1 box. Attach a 2 × 2 × 2 box to it from below to make a 2 × 2 × 3 box. Then you need to continue like this: put new boxes alternately from the side, front, bottom, and choose their size so that the two dimensions (these are the dimensions of the face to which the next box is attached) coincide with the dimensions of the current box, and the third dimension was the same as the changed measurement two “moves” before. The first steps are shown in Figure 4. For example, the fifth “move” on the right is a 2 × 2 × 3 box and its “length” (measured along the arrows in this figure) is 2, because two moves before that the box had a “width” equal to 2 (this is the right box in the top row).

Fig. 4. Construction of a “plastic” box. Drawing from article V. W. De Spinadel, A. R. Buitrago Towards van der Laan’s Plastic Number in the Plane

If you continue this process, then the size of the boxes will naturally increase. But the ratio of their sides (“adjacent” in length, as shown in Fig. 4) will tend to the final limit, which is the plastic number.

The idea behind the justification is as follows. Note that the sizes of the boxes are triples of adjacent numbers from the sequence 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16. If we denote *n*-th member of this sequence *Pn*, then at *n* 3 the equality holds *Pn* = *Pn*−2 *Pn*−3. precisely, it is this linear recurrence that defines this sequence, which is called the Padovan sequence. It turns out that it is possible to express the general term of a recurrent sequence in terms of the roots of its characteristic polynomial. You can familiarize yourself with this topic in more detail at the indicated links, now it is only important that for a given sequence the characteristic polynomial is: \ (x ^ 3-x-1 \), and its real root, as we know, is a plastic number *ρ*. Therefore, by the way, the sequence of powers of this number 1, *ρ*, *ρ* 2. *ρ* 3 satisfies the same recurrence relation (this observation actually gives rise to the method of expressing a member of a sequence in terms of the roots of a polynomial). This polynomial also has two complex roots. If we denote them by *q* and *s*, then for some constants *a*, *b*, *c* equality *Pn* = *aρ n* *bq n* *cs n* will be true for all natural *n*. But since complex roots *q* and *s* modulo less than 1, their degrees tend to zero with increasing *n*.

In this sense, the plastic number for the Padovan sequence is the same as the other (and much more famous) “architectural” number. the golden ratio. for the Fibonacci sequence (and the silver section for the Pell numbers).

You can read more about the properties of the plastic number in the article V. W. De Spinadel, A. R. Buitrago Towards van der Laan’s Plastic Number in the Plane.