Introductory speech of the teacher:
A small historical note: Many scientists have been fond of cutting tasks since ancient times. Solutions to many simple cutting problems were found by the ancient Greeks and Chinese, but the first systematic treatise on this topic belongs to the pen of Abul-Vef. Geometers took seriously the task of cutting shapes into the smallest possible number of pieces and then constructing another shape in the early 20th century. The famous puzzle founder Henry E. Dudeny was one of the founders of this section.
Nowadays, puzzle lovers are fond of solving cutting problems before because there is no universal method for solving such problems, and everyone who undertakes to solve them can fully demonstrate their ingenuity, intuition and ability to creative thinking. (In the lesson, we will indicate only one of the possible examples of cutting. It can be assumed that the students can get some other correct combination - do not be afraid of this).
This lesson is supposed to be held in the form of a practical lesson. Divide the participants of the circle into groups of 2-3 people. Provide each of the groups with figures prepared in advance by the teacher. Students have a ruler (with divisions), pencil, scissors. It is allowed to make only straight cuts with scissors. Having cut some shape into parts, it is necessary to compose another shape from the same parts.
Cutting tasks:
1). Try cutting the figure shown in the figure into 3 equal parts:
Hint: The small shapes are very similar to the letter T.
2). Now cut this shape into 4 equal parts:
Hint: It is easy to guess that small figures will consist of 3 cells, and there are not many figures of three cells. There are only two types of them: a corner and a rectangle.
3). Divide the figure into two equal parts, and from the resulting parts, fold the chessboard.
Hint: Suggest starting the task from the second part, how to get a chessboard. Recall the shape of the chessboard (square). Count the available number of cells in length and width. (Remind that there should be 8 cells).
4). Try cutting the cheese into eight equal pieces with three strokes of the knife.
Hint: Try cutting the cheese lengthways.
Tasks for independent solution:
1). Cut a square out of paper and do the following:
· Cut into such 4 parts, of which you can make two equal smaller squares.
· Cut into five pieces — four isosceles triangles and one square — and fold them to make three squares.
a) Cut an arbitrary triangle into several pieces so that a rectangle can be folded out of them.
b) Cut an arbitrary rectangle into several pieces so that a square can be folded out of them.
c) Cut two arbitrary squares into several pieces so that one large square can be folded out of them.
Hint 1
b) First, make a rectangle out of an arbitrary rectangle, the ratio of the larger side to the smaller one does not exceed four.
c) Use the Pythagorean theorem.
Hint 2
a) Draw the height or center line.
b) Place a rectangle on the square you want to make and draw a "diagonal".
c) Attach the squares to each other, on the side of the larger square measure a segment equal to the length of the smaller square, then connect it to the "opposite" vertices of each of the squares (see Fig. 1).
Solution
a) Let an arbitrary triangle be given ABC... Let's draw the middle line MN parallel to the side AB, and in the resulting triangle CMN lower the height CD... In addition, let us drop on the straight line MN perpendiculars AK and BL... Then it is easy to see that ∆ AKM = ∆CDM and ∆ BLN = ∆CDN as right-angled triangles, in which the corresponding pair of sides and a pair of angles are equal.
Hence the method of cutting a given triangle and then rearranging the pieces. Namely, let's make cuts along the segments MN and CD... After that, we shift the triangles CDM and CDN in place of triangles AKM and BLN respectively, as shown in fig. 2. We got a rectangle AKLB as required in the task.
Note that this method will not work if one of the corners CAB or CBA- stupid. This is due to the fact that in this case the height CD does not lie inside the triangle CMN... But this is not too scary: if we draw the middle line parallel to the longest side of the original triangle, then in the cut-off triangle we will lower the height from an obtuse angle, and it will necessarily lie inside the triangle.
b) Let a rectangle be given ABCD whose sides AD and AB are equal a and b respectively, and a > b... Then the area of the square that we want to get in the end should be equal to ab... Therefore, the side length of the square is √ ab which is less than AD but more than AB.
Let's build a square APQR, equal to the sought one, so that the point B lay on the segment AP and point R- on the segment AD... Let be PD intersects segments BC and QR in points M and N respectively. Then it is easy to see that the triangles PBM, PAD and NRD are similar, and besides, BP = (√ab – b) and RD = (a – √ab). Means,
Therefore, ∆ PBM = ∆NRD on both sides and the corner between them. It is also easy to deduce the equalities from here PQ = MC and NQ = CD, and hence ∆ PQN = ∆MCD also on both sides and the corner between them.
The cutting method follows from all the above reasoning. Precisely, first we lay aside on the sides AD and BC segments AR and CM whose lengths are equal to √ ab(on how to construct segments of the form √ ab, see the task “Regular polygons” - the sidebar in the “Solution” section). Next, we restore the perpendicular to the segment AD at the point R... Now all that remains is to cut off the triangles MCD and NRD and rearrange them as shown in fig. 3.
Note that in order to use this method, it is required that the point M ended up inside the segment BK(otherwise not the whole triangle NRD contained within a rectangle ABCD). That is, it is necessary that
If this condition is not met, then first you need to make this rectangle wider and shorter. To do this, it is enough to cut it in half and transfer the pieces as shown in fig. 4. It is clear that after such an operation, the ratio of the larger side to the smaller side will decrease by four times. So, doing it a sufficiently large number of times, in the end we will get a rectangle to which the cutting from Fig. 3.
c) Consider two given squares ABCD and DPQR by attaching them to each other so that they intersect on the side CD smaller square and had a common vertex D... We will assume that PD = a and AB = b, and, as we have already noted, a > b... Then on the side DR larger square, you can consider such a point M, what Mr = AB... By the Pythagorean theorem.
Let the lines passing through the points B and Q parallel straight MQ and BM respectively, intersect at the point N... Then the quadrilateral BMQN is a parallelogram, and since all sides of it are equal, it is a rhombus. But ∆ BAM = ∆MRQ on three sides, whence it follows (given that the angles BAM and MRQ straight lines) that. Thus, BMQN- square. And since its area is ( a 2 + b 2), then this is exactly the square that we need to get.
In order to proceed to cutting, it remains to note that ∆ BAM = ∆MRQ = ∆BCN = ∆NPQ... After that, what needs to be done becomes obvious: it is necessary to cut off the triangles. BAM and MRQ and rearrange them as shown in fig. 5.
Afterword
Having solved the proposed problems, the reader may well think about the following question: when, in general, one given polygon can be cut by straight lines into a finite number of such pieces, from which another given polygon is composed? With a little reflection, he will understand that it is at least necessary that the areas of these polygons are equal. Thus, the original question turns into the following: is it true that if two polygons have the same area, then one of them can be cut into pieces that add up the second (this property of two polygons is called scissor-congruence)? It turns out that this is really so, and this is what the Boyai-Gervin theorem, proved in the 30s of the XIX century, tells us. More precisely, its wording is as follows.
Boyai-Gerwin theorem. Two polygons are equal in size if and only if they are equally congruent.
The idea behind the proof of this remarkable result is as follows. First, we will not prove the assertion of the theorem itself, but the fact that each of the two given equal-sized polygons can be cut into pieces that add up to a square of the same area. To do this, first we divide each of the polygons into triangles (such a partition is called triangulation). And then we turn each triangle into a square (for example, using the method described in points a) and b) of this problem). It remains to add one large from a large number of small squares - we can do this thanks to point c).
A similar question for polyhedra constitutes one of the famous problems of David Hilbert (the third), presented by him in a lecture at the II International Congress of Mathematicians in Paris in 1900. Characteristically, the answer turned out to be negative. The consideration of two such simplest polyhedra as a cube and a regular tetrahedron shows that neither of them can be cut into a finite number of parts so that the other is composed of them. And this is not accidental - such cutting simply does not exist.
The solution to the third problem of Hilbert was obtained by one of his students - Max Dehn - already in 1901. Dehn discovered an invariant quantity that did not change when polyhedra were cut into pieces and new shapes were added from them. However, this value turned out to be different for some polyhedra (in particular, a cube and a regular tetrahedron). The latter circumstance clearly indicates the fact that these polyhedra are not scissor-congruent.
Objective 1: A rectangle, the sides of which are expressed in integers, can be cut into figures of the form (the side of the cell in the figure is equal to one). Prove that it can be cut into 1 × 5 rectangles.
(D. ~ Karpov)
Solution: The area of this rectangle is divided entirely by the area of the specified figure, that is, by 5. The area of the rectangle is equal to the product of the lengths of the sides. Since the lengths of the sides are integers and 5 is a prime number, the length of one of the sides must be divisible by 5. Divide this side and the opposite side into segments of length 5, and the other two sides into segments of length 1, after which we connect the corresponding points on opposite sides with straight lines. Objective 2: Solve the system of equations in real numbers(A. ~ Khrabrov)
Solution: Answer: the system has a unique solution: a = b = c = d = 0. Adding the two equations of the system, we obtain the equation 8a² + 9b² + 7c² + 4d² = 16ab + 8cd From the inequalities 2ab ≤ a² + b² and 2cd ≤ c² + d² it follows, that the right-hand side of this equation is not greater than the left-hand side, and equality can be achieved only if b = 0, c = 0, a = b and c = d. Hence, the only possible solution to this system is a = b = c = d = 0.The second option is solved in a similar way.
Objective 3: In rhombus ABCD, points E and F are taken on sides AB and BC, respectively, such that CF / BF = BE / AE = 1994. It turned out that DE = DF. Find the EDF angle.
From the condition of the problem (in both versions) it follows that BE = CF. Set aside on the side AB the segment AK equal to BE. Triangles ADK and CDF are equal in two sides and angle (AD = CD, AK = CF, ∠ DAK = ∠ DCF). Hence, DK = DF = DE, that is, the triangle DKE is isosceles. In particular, the angles DKE and DEK at its base are equal. Therefore, triangles ADK and BDE are equal (on two sides and angle: AK = BE, DK = DE, ∠ DKA = ∠ DEB). Hence AD = BD, that is, triangle ABD is equilateral. Therefore, ∠ BAD = 60, ∠ ABC = 120.
(A. ~ Khrabrov)
Solution: Let team A beat team B in a match with these rules (possibly securing the victory ahead of schedule). This means that for any conceivable outcome of the remaining (unscored) penalties, the result of team A would be higher than that of team B. Imagine that the teams continued to shoot the penalty after the end of the match and shot all the remaining penalties, and team A did not score any more goals and Team B never missed again. In this case, the total number of goals scored by A will still remain more than those scored by B (this is what the words "early victory" mean). How much more can it be? Only by 1 or 2. Indeed, if the difference were more than two, then the victory of team A would have become inevitable even earlier, before the last pair of penalties was taken.Further, we note that during the continuation of the match we are considering, exactly half of all the blows hit the goal. Thus, of all 129 pairs of shots, exactly half of them hit the goal, that is, exactly 129. These 129 goals are divided between A and B so that A has 1 or 2 more. This uniquely determines the number of goals scored by team A - 65.
Task 5: Solve the equation in natural numbers:(D. ~ Karpov)
Solution: This equation has a unique solution: x = 2, y = 1, z = 2 (in both versions). The fact that it is a solution follows from the general identity a² + (2a + 1) = (a + 1) ² \, applied in the first variant to a = 105, and in the second to a = 201.There are no other solutions, since if z> 2, then the right-hand side of the equation is divisible by 8, and the left-hand side is not, since 105 x can give, when divided by 8, only a remainder of 1, and 211 y - only remainders of 1 and 3. It remains to note that that for z = 1 there are no solutions either, and for z = 2 the values y = 1 and x = 2 are uniquely determined.
In the understanding of mathematics tutors and teachers of various electives and circles, a selection of entertaining and developing geometric cutting problems is offered. The purpose of using such tasks by the tutor in his classes is not only to interest the student with interesting and effective combinations of cells and figures, but also to form in him a sense of lines, angles and shapes. The set of tasks is focused mainly on children in grades 4-6, although it is possible to use it even with high school students. The exercises require students to have a high and steady concentration of attention and are great for developing and training visual memory. Recommended for mathematics tutors preparing students for entrance exams to mathematics schools and classes that have special requirements for the level of independent thinking and creativity of the child. The level of the tasks corresponds to the level of the entrance olympiads to the “second school” lyceum (second mathematical school), the small Mechmatist of Moscow State University, the Kurchatov school, etc.
Math Tutor's Note:
In some solutions to problems, which you can view by clicking on the corresponding index, only one of the possible examples of cutting is indicated. I fully admit that you may end up with some other correct combination - no need to be afraid of this. Check carefully the solution of your mind and if it satisfies the condition, then feel free to tackle the next problem.
1) Try to cut the figure shown in the figure into 3 equal parts:
: Small shapes are very similar to the letter T
2) Now cut this shape into 4 equal parts:
Math Tutor's Tip: It is easy to guess that small figures will consist of 3 cells, and there are not so many figures of three cells. There are only two types of them: a corner and a rectangle 1 × 3.
3) Cut this shape into 5 equal parts:
Find the number of cells that each such figure consists of. These figures are similar to the letter G.
4) And now you need to cut a figure of ten cells into 4 unequal each other rectangle (or square).
Math Tutor Specification: Select some rectangle, and then try to write three more in the remaining cells. If it doesn't work, then change the first rectangle and try again.
5) The task becomes more complicated: you need to cut the figure into 4 different in shape figurines (not necessarily rectangles).
Math Tutor's Tip: First, draw separately all kinds of shapes of different shapes (there will be more than four) and repeat the method of iterating over the options as in the previous task.
:
6) Cut this figure into 5 figures of four cells of different shapes so that only one green cell is painted over in each of them.
Math Tutor's Tip: Try to start cutting from the top edge of the given shape and you will immediately understand how to proceed.
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7) Based on the previous problem. Find how many figures of various shapes, consisting of exactly four cells, are there? The figures can be twisted, rotated, but you cannot lift the table (from its surface), on which it lies. That is, the two given figures will not be considered equal, since they cannot be obtained from each other by turning.
Math Tutor's Tip: Study the solution to the previous problem and try to imagine the different positions of these figures when turning. It is not hard to guess that the answer in our problem will be the number 5 or more. (In fact, even more than six). There are 7 types of the described figures in total.
8) Cut a square of 16 cells into 4 equal-shaped pieces so that each of the four pieces contains exactly one green cell.
Math Tutor's Tip: The appearance of the small figures is not a square or rectangle, or even a corner of four cells. So what shapes should you try to cut into?
9) Cut the depicted figure into two parts so that the resulting parts can be folded into a square.
Math Tutor's Tip: There are 16 cells in total, which means that the square will be 4 × 4 in size. And somehow you need to fill the window in the middle. How to do it? Could it be some kind of shift? Then, since the length of the rectangle is equal to the odd number of cells, the cutting should be done not with a vertical cut, but along a broken line. So that the upper part is cut off on one side of the middle cells, and the lower part on the other.
10) Cut a 4 × 9 rectangle in two so that you can add a square as a result.
Math Tutor's Tip: There are 36 cells in the rectangle. Therefore, the square will be 6x6. Since the long side consists of nine cells, three of them must be cut off. How will this cut go further?
11) A cross of five cells, shown in the figure, needs to be cut (you can cut the cells themselves) into parts from which a square could be folded.
Math Tutor's Tip: It is clear that no matter how we cut the cells along the lines, we will not get a square, since there are only 5 cells. This is the only problem in which it is allowed to cut not in cells... However, it would still be good to leave them as a guide. for example, it is worth noting that we somehow need to remove the indentations that we have - namely, in the inner corners of our cross. How would you do it? For example, cutting off some protruding triangles from the outer corners of the cross ...
