### How to cut a square into 7 triangles

## Geometric problems (for cutting)

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The Development of Children’s Brain Control Functions: Useful Tips and Exercises for Teachers

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Exercise: Geometric Problems (for cutting)

### The basics on a Speed square

Developing the creative abilities of students

development of attention, memory, independent and teamwork skills

developing mental imagination, ingenuity, and “wit”

Today’s geometric problems (for cutting) will be related to one seemingly simple geometric figure.

The main merit of the square is its use as a convenient unit of area. Indeed, squares are very convenient for mashing flat areas, but, say, circles do not do so without holes and overlaps. Often mathematicians say “squaring” instead of “finding the area”.

Thus, the problem of finding the area of a circle is called the quadrature circle problem. The square is the protagonist in Pythagoras’ theorem.

Cut a square piece of paper into 20 equal triangles and make 5 equal squares.

You want to divide a cross made up of five squares into parts that make up a single square.

A square contains 16 cells. Divide the square into two equal parts so that the line of the cut runs along the sides of the squares.

Cut a 7×7 square into five pieces and rearrange them to make three squares: 2×2, 3×3, and 6×6.

Cut the square into 4 parts of equal shape and size so that each part has exactly one shaded square.

Dividing a square into smaller squares of the same area is very simple: just draw a grid of equally spaced straight lines parallel to the sides of the **square**. The number of squares obtained would be a square, yes, yes! This is why the product of two identical numbers was called a square. Is it possible to divide a square into several squares which do not have identical numbers??

This question remained unsolved for a long time. Many even eminent mathematicians believed that such a partitioning was impossible. But in 1939, a partitioning of the square into 55 different squares was constructed. In 1940, two ways were found to dissect a square into 28 different squares, then into 26 squares, and in 1948, a dissection into 24 different squares was obtained. In 1978 it was found the partition of 21 different squares and it was proved that the partition into a smaller number of different squares can no longer be found.

## Problems for cutting 7

Here are a couple more cutting problems. The cutting problems develop ideas about symmetry, logic and imagination, quick thinking and independent thinking.

Problem 1. Cut the square in the drawing into four equal parts, so that each would contain three shaded cells.

Task 2. Cut a 4 by 9 cell rectangle into two equal parts so that it could be made into a square.

Problem 3. Cut the rectangle 4 by 8 cells into two equal parts so that later it could be made into a square.

It is impossible. The total number of cells is 32. This number is not a complete **square**, so such a rectangle cannot be made into a square with a side with a whole number of cells.

Problem 4. From a rectangle of 10 by 7 cells cut out 6 center cells. Cut the obtained figure into two parts so that it would be possible to make a square from them.

## An inventive problem: where did the extra empty square come from?

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In fact, this is not an optical illusion, but an interesting problem. The squares of the painted over figures are of course equal (32 squares), but what is visually observed as triangles 135 is not, in fact, triangles with different squares (S135 = 32.5 squares). That is, the error disguised in the problem condition is that the initial figure is named a triangle (in fact, it is a concave 4-corner). This is clearly seen in the diagram below. the “hypotenuse” of the upper and lower figures pass through different points: (8,3) above and (5,2) below. The secret is in the properties of the blue and red triangles. This is easy to check by calculation.

The ratio of the lengths of the corresponding sides of the blue and red triangles are not equal to each other (2/3 and 5/8), so these triangles are not similar, and therefore have different angles at the corresponding tops. Let us call the first figure, which is a concave quadrilateral, and the second figure, which is a concave octagon, pseudo-triangles. If the lower sides of these pseudotriangles are parallel, then the hypotenuses in both pseudotriangles are actually broken lines (the upper figure creates an inward bend and the lower figure creates an outward bend). If we overlay the upper and lower figures 135 on each other, a parallelogram is formed between their “hypotenuse” and contains the “extra” area. This parallelogram is shown in the figure-scheme in correct proportions.

The acute angle in this parallelogram is arcctg46 0°118.2. The minute hand of a fully functional watch advances to this angle in 12.45 seconds. This is the value by which the obtuse angle in the parallelogram under consideration differs from the unconcave angle. Visually so insignificant difference is invisible, but it is well visible in the animation.

According to Martin Gardner, the task was invented by New York amateur illusionist Paul Curry in 1953. However, the principle behind it was already known in the 1860s. You may notice that the lengths of the sides of the figures in this problem (2, 3, 5, 8, 13) are consecutive Fibonacci numbers.

## How to make a square out of 3 triangles

The methods of fast patchwork significantly simplify and accelerate the work of skilled worker. Over the years, needlewomen have invented a huge number of different ways to simplify your work. About the most popular, and we will tell.

In the Anglo-American school of quilting all measurements are in inches ( 2,5 cm). If a square is a whole block unit, then triangles in quilting are divided into half-square triangles, and quarters of a whole square (quarter square triangles). The difference is in the direction of the grain of the thread. Therefore, some triangles are cut in two **square** triangles and others in four triangles.

When using the quick method of sewing a triangle square, it is important to properly calculate the size of the blank and the allowances.

Half square triangles: Cut a square the same size as the side of the triangle, with 2 seam allowances of 6mm and a diagonal seam allowance of 8.4mm. The size of the seam allowance does not depend on the size of the **triangles** and squares, it depends on the size of the seam allowance. If you are used to using a different seam allowance: 7mm or 5mm. then the value of the additional allowance is calculated by the formula : (allowance x allowance). Calculate the square root of the result and divide it by 2. (Rationale: Since according to the Pythagoras formula the square of the hypotenuse is equal to the sum of the squares of the cathetuses, and in this case the hypotenuse is two allowances, then the cathetus is the extra allowance, we take the hypotenuse, take the square root of it and divide by two).

Example: You need 2 parts of two triangles, a square of 5x5cm. You use an allowance of 6 mm (0.6 cm) Cut two squares of size 7 x 7 cm (5 0.6 0.6 0.84). Fold them right sides together. For convenience, you can mark the diagonal along which you will be stitching with a marker or chalk. Put two stitches along the diagonal (at a distance of 6 mm from the diagonal). Make a diagonal cut, between the stitches. Unfold the pieces and fold out the seam allowances.

If you need more than two composite squares. You can not cut the fabric into squares, and use a strip width of a few squares or a large square (see the diagram below).

Quarter square triangles: Cut a square of the same size as the hypotenuse (longest) of the triangle 2 allowance for a 6mm seam 2 allowance for a diagonal seam 8.4mm. Example: You need 2 pieces of four triangles square 10 x 10 cm (and an allowance of 0.6 cm). (see. (See assembly diagram below.) You will need to cut two squares of size 10 (0.6 2) (0.84 2) = 12.9 cm. т.е. 12.9 x 12.9 cm. Fold the squares face-to-face, mark the diagonals, and scythe the stitching lines. Sew scythe line seams along one diagonal. Cut along the diagonal of the squares in half. Unfold the parts and iron out. Fold the resulting squares face side by face, combining mowing line seams. Sew scythe stitches along the second diagonal. Cut the squares in half along this diagonal. Unfold the pieces and iron out. Create two squares of 4 **triangles** each.

Quick triangles half a square. Quick method of assembling triangles into squares.

Layout of the square pattern for this method.

Cut out squares of two colors. Fold them face-to-face. Place two stitches along the diagonal (6mm from the diagonal).

Make a diagonal cut. Unfold the pieces and iron out the seam allowances

Squares of two triangles several squares in a row.

A square of four triangles (quarter square triangles): a quarter-square Note, here the left-hand thread runs along the diagonal of the future **square** (along the hypotenuse of the triangle)

Cut out the square. Sew seams along sides of **square** marked in red. Cut the square diagonally. Open and seal details.

Make squares with a valley thread along the diagonal of the square.

In the same pattern, stitch along the blue. or along the green lines, and by cutting the **square** along the diagonals, you will get compound triangles of quarter-squares with the valley along the short side (cathetus) of the triangle.

Example of assembling a 3/3 square of half-squares and quarter-squares.

Squares of 4 triangles (quarter-squares).

Cut out squares, fold over, mark with scythe fishing line seams and scythe fishing line cut. Stitch. Cut.

Iron out parts. Fold them face-to-face, combining mowing line seams. Mark the slash line of the seams and the slash line of the cut. Stitch. Cut.

Create two squares with bows (a square of quarter-squares).

I would like to introduce another flat geometric constructor, which can be called Pythagoras-2.

A 10×10 cm square is dissected as shown, resulting in 9 geometric shapes: 4 large **triangles**, 2 small ones, one middle one, a **square** and a rectangle.

Before working with the patterns, the kids do several tasks with specific shapes.

Task 1. Take 2 large triangles and a square. Make: a rectangle, a triangle, and 2 different quadrangles, one of which is a trapezoid.

Task 2. Take 2 small triangles and the middle one. Make: a square, a triangle, a rectangle, and 2 different quadrilaterals, one of which is a trapezoid.

Assignment 3. Take 2 small triangles, a medium triangle, and a large triangle. Make: a square, a triangle, a rectangle, and 2 different quadrilaterals, one of which is a trapezium.

Next, the children build different patterns, gradually moving on to unbroken patterns.

The following constructor is also author’s work and by analogy with Tangram I call it a Tregram, because it is obtained by cutting an equilateral triangle. Play exercises with this constructor can be conducted for filling in and composing planar images from sets of geometric shapes. An equilateral triangle made of cardboard (side length 20 cm, each side divided into 5 equal parts of 4 cm) is cut into 10 pieces, as shown in the figure.

The result is 4 small triangles, 2 rhombuses, a trapezoid, a parallelogram, a large triangle and a hexagon.

__In the first step__ Children get acquainted with all parts of the constructor, making them out of **triangles** and other small shapes:

By attaching two triangles to each other, children make a rhombus. 2. By attaching another triangle to the rhombus, the children get a trapezoid, which can also be made with three triangles. 3. By attaching another triangle to the trapezoid, the children obtain a parallelogram. The same figure can also be made from other small shapes. 4. Then children draw their own conclusions about which shapes they can put together by putting the small shapes on the big ones.

__In the second stage__ children fill in the interior space of the figures-silhouettes on the sheets, using all parts of the construction set.

__At the third stage__ children make plane images from exploded patterns with a gradual transition to partially exploded patterns.

__At the fourth stage__ children model images according to their own ideas.

By the type of Nikitin’s cubes, I made another flat constructor. This set consists of 15 squares of 5×5 cm:

8 squares are painted half diagonally;

__At the first stage__ of play tasks, we use only 4 squares, painted half of them, and make all the images from them.

__At the second stage__ children make images out of 9 squares, using the entire set of.

Goal. To teach children to make geometric figures from a certain number of sticks, using the method of fitting one figure, taken as a basis, to another.

Material: The children have counting sticks on their desks, the board, and chalk for this and the next lesson.

Course of work. 1. The teacher asks the children to count out 5 sticks each, check and put them in front of them. Then he says, “Tell me how many sticks it takes to make a triangle, each side of which will be equal to one stick. How many sticks would be needed to make two of these triangles? You have only five sticks, but they must also make two equal triangles. Think about how to do this, and make it up.”.

When most of the children have completed the task, the teacher asks them to tell how to make two equal triangles out of five sticks. Draws the children’s attention to the fact that they can do the task in different ways. How to do this should be sketched out. When explaining, use the expression “attached to one triangle the other from below” (left, etc.).д.), and in explaining the solution of the problem use also the expression “attached to one triangle the other using only 2 sticks”.

Make 2 equal squares out of 7 sticks (the teacher clarifies in advance what geometric figure can be made out of 4 sticks). Gives the task: count 7 sticks and think how to make 2 equal squares on the table from them.

After completing the task, consider different ways of fitting one square to the other, the teacher sketches them on the board.

Analyzing questions: “How did he make 2 equal squares out of 7 sticks?? What made first, what then? How many sticks make 1 square?? How many sticks it took to make the second square? How many sticks it took to make 2 equal squares?”

Goal. Making shapes by fitting. To see and show the new, resulting figure by making up; to use the expression: “fitting one figure to another,” to think about practical actions.

Course of work. The instructor invites the children to recall the shapes they made, using the method of fitting. Tells them what they will do today to learn how to make new, more complex shapes. Gives tasks:

Count 7 sticks and think about how to make 3 equal triangles from them.

After completing the task, the teacher invites all the children to make 3 triangles in a row so that a new quadrilateral figure is formed This solution is sketched on the board with chalk by the children. The teacher asks them to show 3 separate **triangles**, a quadrilateral and a triangle (2 pieces), a quadrilateral.

* Figs. 2 Making shapes from triangles*

Make 4 equal triangles from 9 sticks. Think about how to do it, talk about how to do it, then do the task.

After that, the teacher invites the children to draw the figures they made with chalk on the board and tell about the sequence of the task.

Analysis questions: “How to make 4 equal triangles out of 9 sticks? Which of the triangles made up first? What are the resulting shapes and how many?”

The teacher, clarifying children’s answers, says, “You can start with any triangle and then add others to it on the right or left, above or below.

Goal. Encourage children to find their own ways of making up the pieces based on thinking about the solution beforehand.

Progress. The teacher asks the children questions: “Of how many sticks can you make a square, each side of which is equal to one stick? 2 squares? (of 8 and 7). How you will make 2 squares out of 7 sticks?”

Count 10 sticks and make 3 equal squares from them. To think about how to make up and tell.

As they progress, the teacher challenges several children to sketch the shapes they make on the board and tell the sequence of making. Invites all children to make a figure out of 3 equal squares arranged in a row, horizontally. He draws one on the board and says, “Look at the board. Here is a drawing of how this problem can be solved in different ways. You can add another square to one and then a third. (Shows.) And you can make a rectangle out of 8 sticks, then divide it into 3 equal squares with 2 sticks.”. (Shows.) Then asks questions: “What shapes have turned out and how many? How many rectangles are obtained? Find and show them.”.

Make a square and two equal triangles from five sticks. First tell and then make up.

When doing this task, children usually make the mistake of making two triangles by the learned method of fitting, resulting in a quadrilateral. Therefore, the teacher draws the children’s attention to the condition of the problem, the need to make a square, and offers leading questions: “How many sticks do we need to make a square?? Since you have sticks? Is it possible to make by fitting 1 triangle to another? How to make? Which shapes should we start with??” After completing the task, the children explain how they did: they must make a square and divide it with 1 stick into 2 equal triangles.

Objective. Exercise the children’s ability to express a guessing solution.

How to do it. 1. Make a square and 4 triangles from 9 sticks. To think about and tell how to make up. (Several children make assumptions.)

If the children have difficulty, the teacher advises: “Remember how you made a square and two triangles out of five sticks. Think and guess how to do the task. Whoever is the first to solve the problem will draw the resulting figure on the board.

After completing and sketching the answer, the teacher invites all children to make the same shapes for themselves

* Rice. 3 Making shapes out of triangles*

What kind of geometric shapes are formed?? How many triangles, squares, and quadrilaterals? How to make up? What is the most convenient, fastest way to make?”

Make two squares out of 10 sticks: a small square and a big square.

Make 5 triangles from 9 sticks.

If necessary, during the second and third tasks, the tutor gives leading questions, advice: “Think first, then compose. Do not repeat mistakes, look for a new way of solving. Does the problem talk about the size of the triangles? This is a problem for ingenuity, you have to think, guess how to solve the problem.

So, in the initial period of teaching 5-year-old children to solve simple wit problems, they independently, mostly by practicing with sticks, look for a way to solve. In order to develop their ability to plan their course of thought, children should be invited to express preliminary reasoning or combine it with practical probes, explain the method and way of solving.

Several types of solving problems of the first group are possible. Having mastered the way of arranging figures under the condition of common sides, children very easily and quickly give 2-3 variants of the solution. Each figure differs from the previous one in its spatial position. At the same time children master the way of construction of the given figures by dividing the received geometrical figure into several (a quadrangle or a square into 2 triangles, a rectangle into 3 squares).

With 5-6 year-old children, the more complicated shape shuffling problems should begin with those where a certain number of sticks need to be removed to change the shape, and the simplest ones for shuffling the sticks.

The process of children’s search for solutions of problems of the second and third groups is much more difficult, than of the first group. For this purpose it is necessary to remember and comprehend the character of transformation and the result (what figures should turn out and how many) and constantly during search of the decision to correlate it with the expected or already carried out changes. In the process of solving it is necessary to visually and mentally analyze the problem, and to imagine possible changes in the figure.

Thus, in the process of solving tasks children must master such thinking operations of task analysis, as a result of which it is possible to imagine various transformations, check them, then, discarding incorrect ones, look for and try new ways of solving. Training should be aimed at developing children’s ability to think through moves mentally, to solve a problem fully or partially in their mind, and to limit practical trial.

In what order should children 5-6 years old be offered the second and third groups of problems on ingenuity??

* Fig. 4*

* Fig. 5*

* Fig. 6*

* Fig. 7*

* Fig. 8*

* Fig. 9*

* Fig. 10*

* Fig. 11*

* Rice. 12*

* Fig. 13*

Characteristic of these and other similar ingenuity problems is that the transformation needed for solving them changes the number of squares the given figure is made of (Zadachas 2, 5, etc.).), changing their size (problems 6, 7), and modifying shapes, for example, transforming squares into rectangles in problem 1.

During the lessons, in order to guide the search activity of children, the tutor uses various techniques to foster in them a positive attitude to a long persistent search, but at the same time, a quick response, rejection of the developed way of searching. Children’s interest is sustained by a desire to succeed, which requires active thought work.

**How to cut a square into 7 triangles**

Prove that a 10×10 checkerboard cannot be cut along the grid lines into 1×4 rectangles. (Solutions by D.Ю. Kuznetsov.)

Solution 1 Divide the board into 2×2 squares and color them in staggered order (fig.1). Note that any 1×4 rectangle contains an equal number (2 each) of black and white cells, but for a given coloring board, there are 52 black cells and 48 white cells, t.е. not equal. So, cutting the 10×10 board into 1×4 rectangles is impossible.

Solution 2 Let’s color the board diagonally in 4 colors (Fig.2). Note that any rectangle contains one cell of each of the four colors, but in the given colouring there are 25 cells of the 1st and 3rd colors, 26 cells of the 2nd and 24 cells of the 4th, t.е. not equally. So, it is impossible to cut a 10×10 board into 1×4 rectangles.

The bottom right and left corner squares are cut out of the chessboard. Can the resulting figure be dissected into 1×2 dominoes?? What if you cut out the bottom right and the top left?

Can a 6×6 board be cut into dominoes so that there are exactly 11 horizontal? (Horizontal coloring in two colors.)

Color the picture in the four colors so that the neighboring parts are painted in different colors. Can we do with three colors?? (See the following. Activity 6: Coloring a Geographic Map. grade 5-6).

In a 4×4 square, the cells of the left half are painted black and the rest are painted white. In a single operation, you are allowed to recolor all the cells inside any rectangle to the opposite color. How do you get a chess piece from the original coloring in three operations?

Some grasshoppers are sitting on a straight line, and the distance between neighboring grasshoppers. are the same. Every minute, one of them jumps to a point symmetric to it relative to another grasshopper. Can Sasha the grasshopper be in the place where his neighbor Lyosha was sitting in the beginning??

a) Is it possible to cut a chessboard into pieces consisting of 4 squares in the shape of the letter “T”??

b) Is it possible to cut a 10×10 chessboard into such pieces?

Is it possible to divide an 8×8 square with a corner cut off into 1×3 rectangles?

Is it possible to cut a 10×10 board into four-cell pieces in the shape of the letter “G”?? (Horizontal coloring in two colors.)

An 8×8 board is cut into 2×1 dominoes. Can there be 15 vertical and 17 horizontal dominoes?

The triangle is cut into triangles (25 pieces) as shown. Juke can walk around the triangle, moving between neighboring (side by side) **triangles**. What is the maximum number of triangles a Juke can go through if it has been in each of them no more than once??

What is the greatest number of rhombuses, each made of two equilateral triangles with side 1, that can be cut out of an equilateral triangle with side 5 (see the diagram below. Figure. of the previous problem).

A triangular castle is divided into 100 identical triangular halls. In the middle of each wall is a door. How many halls can a person who does not want to visit more than once visit??

## Master class Origami Divide the square into equal parts Paper

Dividing the square into equal parts. This is always just a preparatory stage for folding. However, without certain skills, just he can be quite difficult, especially if the number of parts, is a simple number:3, 5, 7, as well as 9. About this in detail.

Let’s mark the middle of the upper side. To do this we will make a small tack.

Fold the corner of the **square** to the middle of the opposite side.

In this case the intersection point of the side opposite to this corner and the side adjacent to it, divides the side in the ratio 1:2. So with only the folds we have found one third of the side of the square.

Arranging the square. Tie on the left side is 1/3 of it.

Using this tack we form a fold. In doing so, it should be parallel to the upper and lower sides.

Turn the sheet on the opposite side.

We fold the obtained rectangle in half.

Thus we obtain three parallel folds. They divide the square into three equal parts.

We will divide the square into five equal parts.

Let’s mark with a tack the middle of the side.

Make a fold that passes through the bottom left corner of the square and our mark at the same time. The bottom right corner is located horizontally 2/5 of the right edge.

We divide the resulting segment in half. The width of the bent strip is 1/5.

Straighten the sheet. Now divide the rest of the sheet into four equal parts.

Fold the left side to the intended vertical fold. Thus, divide this gap in half.

Unfold the sheet. It remains for each of the wide strips to divide in half again.

Fold the left side to the side fold identified in the previous step.

It remains, divide the last sector. To do this, align the right side line with the leftmost vertical fold.

Straighten the sheet. The division into five equal parts is completed.

In order to divide the sheet into seven equal parts, you must first divide it into five, as described above.

We do the fold, at which the bottom right corner coincides with the second mark on the right.

Spreading the sheet. The point on the right side, which was formed thanks to this fold. This is 3/7 from the top edge or 4/7 from the bottom edge.

Combine the bottom right corner with the point obtained on the right side. Make a fold that is parallel to the top and bottom sides.

Fold the bottom side to the horizontal fold. The width of this strip will be 1/7th of the side.

Make the above fold “mountain” and combine it with the mark, obtained above, which divided the side of the 3/7 and 4/7.

Align the upper side with the fold obtained in the previous step.

Flatten the sheet. It remains to divide each of the top two rectangles in half more.

Align the top side with the fold obtained in the previous step.

Align the top side with the lowest horizontal fold.

Unfold the sheet. Our square is horizontally divided into seven equal parts.

In order to divide the sheet into nine equal parts, you must first divide it into three, as described above.

We make a fold where the bottom right corner aligns with the first mark on the right.

The point on the right side will divide it by 4/9 (top) and 5/9 (bottom). Further division into equal parts can be different. Below is one way to complete the division of the square into equal parts.

Thanks to the point obtained on the right side, we make a fold parallel to the upper and lower edge. The difference by which the lower part will be wider than the upper part. and that’s 1/9.

Flip to the opposite side.

Fold back the top layer of paper. The fold should coincide with the edge of the bottom layer.

Turn it back to the opposite side. Unfold the sheet.

The resulting fold in the previous step is combined with the line, which is obtained by using a tack.

Align the top edge with the same line. We have something like a basic shape “door”. Now each of the four wide rectangles is left to divide into two more.

We straighten the sheet. Our horizontal square is divided into nine equal parts.

## Olympiad, logic, and fun math problems. Cutting Tasks

Mathematics tutors and teachers of various extracurricular activities and circles are offered a selection of entertaining and developing geometric cutting problems. The tutor’s goal in using such problems in his classes is not only to interest the student in interesting and spectacular combinations of cells and shapes, but also to form a sense of lines, angles, and shapes. The set of problems is mainly designed for children in grades 4-6, although I cannot exclude their use even with high school students. Exercises require high and steady concentration from students, and are perfect for developing and training visual memory. Recommended for math tutors preparing students for entrance exams to math schools and classes that have special requirements for a child’s level of independent thinking and creativity. The level of the problems corresponds to the level of the entrance Olympiads of the Lyceum Second School (Second Mathematics School), the Junior Faculty of Mechanics and Mathematics of Moscow State University, the Kurchatov School, etc.

Math Tutor’s Note: Some solutions to problems, which you can see by clicking on the appropriate pointer, show only one possible example of a cut. I’m quite willing to admit that you might have some other correct combination, so don’t be afraid of that. Check your mouse solution carefully, and if it satisfies the condition, then feel free to take on the next problem.

1) Try to cut the figure shown in the picture into 3 equal shaped pieces:

Math Tutor Hint: The small shapes are very similar to the letter T View Math Tutor Solution

2) Now cut this figure into four equal parts:

Math Tutor Hint: It’s easy to guess that the small pieces will consist of 3 squares, and there aren’t many pieces of three squares. There are only two kinds: a corner and a rectangle 1×3. View math tutor solution:

3) Cut this figure into 5 pieces of equal shape:

Math Tutor Hint: Find the number of squares that make up each such figure. These figures look like the letter G. View math tutor solution

4) Now you need to cut a figure of ten cells into 4 unequal rectangles (or squares).

Directions from the math tutor: Pick a rectangle and then try inscribing three more squares into the remaining squares. If this does not work, change the first rectangle and try again. View Math Tutor Solution

5) The problem becomes more complicated: you need to cut the figure into 4 different shapes (not necessarily rectangles).

### How can you cut a square into equal triangles? | Nathan Dalaklis

Math tutor tip: First draw all kinds of different shapes separately (there will be more than four) and repeat the brute-force method as in the previous problem. View the math tutor’s solution:

6) Cut this figure into 5 figures of four cells of different shapes so that each of them has only one green cell.

Math Tutor Hint: Try starting the cut from the top edge of a given figure and you will immediately see how to proceed. View the math tutor’s solution:

7) Based on the previous problem. Find the total number of differently shaped figures consisting of exactly four cells? The figures can be twisted and turned, but they cannot be lifted from the surface on which they lie. That is, the two figures given will not be considered equal because they cannot be made from each other by rotation.

Math Tutor Hint: Examine the solution to the previous problem and try to imagine the different positions of these shapes as they rotate. It is easy to guess that the answer to our problem will be a number 5 or more. (Actually, even more than six.). There are a total of 7 types of the described shapes. View the math tutor’s solution

### Cutting squares into equal-area triangles

8) Cut a **square** of 16 cells into 4 equally shaped pieces so that each of the four pieces has exactly one green cell.

Math tutor tip: The kind of small shapes are not a square or rectangle, or even a corner of four cells. So what kind of shapes should we try to cut? View Math Tutor Solution

9) Cut the pictured figure into two pieces so that the resulting pieces can be made into a square.

Hint from the math tutor: There are 16 cells in all, so the square must be about the size of 4×4. Also, somehow you need to fill in the box in the middle. How to do this? Maybe some kind of shift? Then, since the length of the rectangle is equal to an odd number of squares, the cut should not be made with a vertical cut, but with a broken scythe line. So that the top is cut on one side of the middle cell and the bottom is cut on the other. View Math Tutor Solution

10) Cut a rectangle of size 4×9 into two pieces so that the result can be made into a square.

Math Tutor Hint: A rectangle has a total of 36 cells. Therefore, the square will turn out to be 6×6. Since the long side has nine cells, three of these cells must be cut off. How this cut will go further? View Math Tutor Solution

11) The five-cell cross shown in the picture needs to be cut (you can cut the cells themselves) into pieces that would make a square.

Math Tutor Hint: It is clear that no matter how we cut along the lines of the cells, we won’t get a square, since there are only 5 cells. This is the only problem in which you are allowed to cut outside the cells. However, it is still a good idea to keep them as a reference point. For example, it is worth noting that we somehow need to remove the depressions that we have, namely, in the inner corners of our cross. How to do it? For example, by cutting some protruding triangles from the outer corners of the cross. View Math Tutor’s solution: Comment on the solution: cut as shown in the picture and insert the blue triangles into the empty areas shown by the purple triangles.

Alexander Kolpakov. Tutor in Mathematics Moscow, Strogino.

Cool site! Thanks for the most interesting problems with answers on the whole Internet!

## Subdivisions into various similar triangles

And what triangle can be dissected into triangles similar to it, among which there are no equals? It turns out that any nonequilateral. Before explaining the solution, let us remind you that in similar triangles, the ratios of the corresponding sides are equal. The generalized Thales theorem will help to construct the desired partition: parallel lines cut off proportional segments on the sides of the angle.

Consider the triangle *ABC*, in which *BC* / *AC* = *k* 1. Applying to the triangle *ABC* **triangles** 1, 2, 3, 4, and 5 We obtain a triangle divided into 6 unequal similar triangles.

The triangles *ABC*, 1, 2, 3, 4 are all different because each next in *k* times greater than the previous one.

But **triangles** 4 and 5 can be equal if *k* *k* 3 = *k* 4. Then we complete triangles 6 and 7, and replace triangle 5 by triangle 8. **Triangles** 7 and 8 are not equal because *k* 6 ≠ *k* *k* 3 *k* 5. For if *k* *k* 3 = *k* 4. then *k* 6 = *k* 2 (*k* *k* 3 ) = *k* 3 *k* 5 *k* *k* 3 *k* 5.

a) Let arbitrary triangle *ABC*. Let us draw the middle line *MN* parallel to the side of *AB*, and in the resulting triangle *CMN* let’s lower the height *CD*. In addition, let us lower on the line *MN* perpendiculars *AK* и *BL*. Then it is easy to see that ∆*AKM* = ∆*CDM* и ∆*BLN* = ∆*CDN* as right triangles that have an equal pair of sides and a pair of angles.

Hence the method of cutting this triangle and then rearranging the pieces. Exactly, let us make cuts along the segments *MN* и *CD*. Then rearrange the triangles *CDM* и *CDN* in the place of the triangles *AKM* и *BLN* respectively, as shown in Fig. 2. We get a rectangle *AKLB*, as required in the problem.

Note that this method will not work if one of the angles *CAB* or *CBA*. obtuse. This is because in this case the height *CD* does not lie inside the triangle *CMN*. But this is not too bad: if we draw the middle line parallel to the longest side of the original triangle, we will drop the height from the obtuse angle in the cut-off triangle, and it will necessarily lie inside the triangle.

b) Let the given rectangle *ABCD*, sides of which *AD* и *AB* are equal to *a* и *b* respectively, whereby *a* *b*. Then the area of the square we want to get as a result must be equal to *ab*. Hence, the length of the side of the square is √*ab*, which is less than *AD*, but more than *AB*.

Let us construct the square *APQR*, equal to the required one, so that the point *B* lies on the segment *AP*, and the point *R*. on the segment *AD*. Let *PD* intersects the segments *BC* и *QR* at the points of *M* и *N* respectively. Then it is easy to see that the triangles *PBM*, *PAD* и *NRD* are similar, and in addition, *BP* = (√*ab*. *b*) и *RD* = (*a*. √*ab*). So,

Therefore, ∆*PBM* = ∆*NRD* on the two sides and the angle between them. Also from here it is not difficult to derive the equations *PQ* = *MC* и *NQ* = *CD*, and therefore ∆*PQN* = ∆*MCD* also on the two sides and the angle between them.

From all the above reasoning the method of cutting. Exactly, we first lay down on the sides of *AD* и *BC* the segments *AR* и *CM*, whose lengths are √*ab* (on how to draw segments of the form √*ab*, see. the problem “Right polygons”. box in the “Solution” section). Next, reconstruct the perpendicular to the segment *AD* at the point *R*. Now all that remains is to cut off the triangles *MCD* и *NRD* and rearrange them as shown in Fig. 3.

Note that this method requires that the point *M* is inside the segment *BK * (otherwise not the entire triangle *NRD* is contained inside the rectangle *ABCD*). That is, it is necessary that

If this condition is not satisfied, we must first make this rectangle wider and less long. To do this, just cut it in half and rearrange the pieces as shown in Fig. 4. It is clear that after performing this operation, the ratio of the larger side to the smaller side will decrease by a factor of four. So, by doing it a sufficiently large number of times, we end up with a rectangle to which the dissection from Fig. 3.

c) Consider the two given squares *ABCD* и *DPQR*, Place them next to each other so that they overlap on the side *CD* of the smaller square and have a common vertex *D*. Let us assume that *PD* = *a* и *AB* = *b*, where, as we have already noted, *a* *b*. Then on the side of *DR* of the larger square we can consider such a point *M*, that *MR* = *AB*. According to the Pythagoras theorem.

Let the lines passing through the points of *B* и *Q* parallel to the lines *MQ* и *BM* intersect at *N*. Then the quadrilateral *BMQN* is a parallelogram, and since all sides are equal, it is a rhombus. But ∆*BAM* = ∆*MRQ* on three sides, whence it follows (given that the angles *BAM* и *MRQ* lines) that. Thus, *BMQN*. square. And since its area is equal to (*a* 2 *b* 2 ), then this is exactly the **square** we need to get.

In order to proceed to the cut, it remains to note that ∆*BAM* = ∆*MRQ* = ∆*BCN* = ∆*NPQ*. After that, what needs to be done becomes obvious: we need to cut the triangles *BAM* и *MRQ* and rearrange them as shown in Fig. 5.