Olympiad, logical and entertaining tasks in mathematics. Tasks for cutting
a) Cut the arbitrary triangle into several pieces so that the rectangle can be folded from them.
b) Cut the arbitrary rectangle into several pieces so that the square can be folded.
c) Cut two arbitrary squares into several pieces so that one large square can be folded.
Tip 1.
b) First, make up such a rectangle from an arbitrary rectangle, the ratio of the main side of which does not exceed four.
c) Use the Pythagore theorem.
Tip 2.
a) spend height or middle line.
b) Verify the rectangle to the square, which should turn out, and swipe "Diagonal".
c) Apply the squares to each other, on the side of the larger square, measure the segment equal to the length of a smaller square, then connect it with the "opposite" vertices of each square (see Fig. 1).
Decision
a) let it be given an arbitrary triangle ABC. Cut the middle line MN. Parallel side AB, and in the resulting triangle CMN. Lower the height CD. In addition, lowered direct MN. Perpendiculary AK and BL.. Then it is easy to see that Δ AKM. = ∆CDM. and Δ. BLN. = ∆CDN. as rectangular triangleswhich are equal to the corresponding pair of parties and steam corners.
Hence the method of cutting this triangle and subsequent shifting pieces. It is, we will draw cuts by segments MN. and CD. After that, put triangles CDM. and CDN. in place of triangles AKM. and BLN. Accordingly, as shown in Fig. 2. We got a rectangle AKLB.Just 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 height CD does not lie inside the triangle CMN.. But this is not too scary: if you spend the middle line in parallel the longest side of the original triangle, then in the cut-off triangle we will lower the height of a stupid angle, and it will definitely lie inside the triangle.
b) let him give a rectangle Abcd.whose asges AD and AB equal a. and b. accordingly, and a. > b.. Then the square of the square we want to get in the end should be equal aB. Consequently, the length of the side of the square is √ aBWhat is less than ADbut more than AB.
Let's build a square Apqr.equal to the desired, so that the point B. lying on the cut Ap., and point R. - On the cut AD. Let be Pd. Crossing segments BC. and QR At points M. and N. respectively. Then it's easy to see that triangles PBM., Pad and NRD. like, and besides BP. = (√aB – b.) I. Rd. = (a. – √aB). It means
Consequently, Δ. PBM. = ∆NRD. On two sides and the corner between them. Also from here it is easy to withdraw equality PQ. = MC. and NQ. = CDSo, Δ PQN. = ∆MCD. Also in two sides and the corner between them.
Of all the above reasoning, the cutting method follows. It is, first we postpone on the sides AD and BC. Segments AR and Cm., the lengths of which are equal √ aB (on how to build segments of the form √ aBFor the task "Right polygons" - insert in the section "Decision"). Next, we restore perpendicular to the segment AD At point R.. Now only cut triangles MCD. and NRD. And shifting them as shown in Fig. 3.
Note that in order for this method to be used, it is required to take the point M. It turned out inside the segment Bk. (otherwise not the whole triangle NRD.connected inside rectangle Abcd.). That is necessary to
If this condition is not performed, then you first need to make this rectangle are wider and less long. To do this, it is enough to cut it in half and shifting pieces as shown in Fig. 4. It is clear that after this operation, the ratio of the main side to the smaller will decrease four times. So, doing it enough big number Once, in the end, we get a rectangle to which cutting with rice applies. 3.
c) Consider two square data Abcd. and Dpqr., attaching them to each other so that they intersect on the side CD smaller square and had a total vertex D.. We assume that Pd. = a. and AB = b., moreover, as we have already noted, a. > b.. Then side Dr. larger square can be considered such a point M., what Mr. = AB. According to Pythagore's theorem.
Let direct passing through the points B. and Q. Parallel direct MQ. and BM. Accordingly, intersect at the point N.. Then quadrigal Bmqn. is a parallelogram, and since he has all the parties equal, then this is a rhombus. But Δ. BAM. = ∆Mrq According to three parties, from where it follows (considering that the angles BAM. and Mrq straight) that. In this way, Bmqn. - Square. And since its area is equal ( a. 2 + b. 2), then this is the square that we need to get.
In order to proceed to cut, it remains to notice that Δ BAM. = ∆Mrq = ∆BCN. = ∆NPQ.. After that, what needs to be done becomes obvious: it is necessary to cut triangles BAM. and Mrq And shifting them as shown in Fig. five.
Afterword
Slowing offered tasks, the reader, it is quite possible, thinks about such a question: And when can one polygon can be cut with straight lines to a finite number of such pieces, of which another polygon is developing? Slightly reflection, it will understand that at least it is necessary that the area of \u200b\u200bthese polygons are equal. Thus, the source 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, of which the second is developing (this property of two polygons is called equivalent)? It turns out that this is true, and this is telling us the theorem of Boyii-Gervin, proven in the 1930s of the XIX century. More precisely, its wording is consisting of.
Boiii-Guerin theorem. Two polygons areometric if and only if they are equivalent.
The idea of \u200b\u200bevidence of this wonderful result is as follows. First, we will not prove the approval of the theorem, but the fact that each of the two data is equal to the polygons can be cut into pieces from which the square of the same area is folded. To do this, first we break each of the polygons on triangles (such a partition is called triangulation). And then each triangle will turn into a square (for example, with the help of the method described in paragraphs a) and b) of this task). It remains to be folded from a large number of small squares one big - we can do this thanks to the point B).
A similar question for polyhedra is one of the famous problems of David Hilbert (third), submitted to them in the report at the II International Congress of Mathematics in Paris in 1900. It is characteristic that the answer to him turned out to be negative. Already considering two such simple polyhedra, as a cube and the correct tetrahedron, shows that none of them turn out to be cut into a finite number of parts so that the other is different. And this is not by chance - there simply does not exist.
The decision of the third problem of Hilbert was obtained by one of his students - Max Den - already in 1901. Den found an invariant value that did not change when cutting polyhedra to pieces and folding from them new figures. However, this value was different for some polyhedra (in particular, Cuba and the correct tetrahedron). The latter circumstance explicitly indicates the fact that these polyhedra are not equivalent.
Task 1: The rectangle, the sides of which are expressed by integers, can be cut on the form figures (the side of the cell in the figure is equal to one). Prove that it can be cut to 1 × 5 rectangles.
(D. ~ Karpov)
Decision: The area of \u200b\u200bthis rectangle is divided by a focus on the area of \u200b\u200bthe specified figure, that is, by 5. The area of \u200b\u200bthe rectangle is equal to the product of the lengths of the sides. Since the lengths of the parties are integers, and 5 - a simple number, then the length of one of the parties should be divided by 5. We divide this side and the opposite length of length 5, and the other two sides - on the length of length 1, then connect the corresponding points on the opposite sides by straight lines. Task 2: Decide in real numbers system of equations(A. ~ Khrabrov)
Decision: Answer: The system has a single solution: a \u003d b \u003d c \u003d d \u003d 0. Having folded the two equations of the system, we obtain equation 8a² + 9b² + 7c² + 4d² \u003d 16ab + 8cd from inequality 2ab ≤ a² + B² and 2cd ≤ C² + D² should be That the right side of this equation is no more than the left, and the equality can be achieved only if B \u003d 0, c \u003d 0, a \u003d b and c \u003d d. So the only one possible solution This system is a \u003d b \u003d c \u003d d \u003d 0.The second option is solved similarly.
Task 3: In the ABCD Rombe on the sides of AB and BC, respectively, the points E and F are taken, such as CF / BF \u003d BE / AE \u003d 1994. It turned out that de \u003d df. Find the angle of EDF.
From the condition of the problem (in both options) it follows that be \u003d cf. I will postpone on the AB side AK, equal to BE. Adk and CDF triangles are equal to two sides and angle (ad \u003d CD, AK \u003d CF, ∠ dak \u003d ∠ DCF). So, DK \u003d DF \u003d DE, that is, the DKE triangle is a challenge. In particular, the angles of DKE and DEK are equal when it is founded. Consequently, the ADK and BDE triangles are equal (on two sides and angle: ak \u003d be, dk \u003d de, ∠ dka \u003d ∠ DEB). Hence the AD \u003d BD, that is, the ABD triangle is equilateral. Consequently, ∠ Bad \u003d 60, ∠ ABC \u003d 120.
(A. ~ Khrabrov)
Decision: Let the team and defeating the team b in a match with such rules (perhaps, ensuring his victory ahead of schedule). This means that with any thought outcome of the remaining (induce) penalty, the result of the team A would be higher than B. teams imagine that the teams continued to break through the penalty after the end of the match and struck all the remaining penalty, and the team and did not scored any more ball And the team would have never missed anymore. At the same time, the total number of goals scored and will still be larger than those scored b (this is exactly what the words "Armchair Victory" mean). How much can it be more? Only on 1 or on 2. Indeed, if the difference turned out to be more than two, the victory of the team would be inevitable even earlier, before the breaking of the last panel of the penalty.Next, we note that with the continuation of the match in the gate, exactly half of all blows came to the gate. Thus, and from all 129 pairs of shocks in the gate got exactly half, that is, exactly 129. These 129 goals are divided between a and b so that at a 1 or 2 more. This uniquely determines the number of goals scored by the A - 65 team.
Task 5: Decide equation in natural numbers:(D. ~ Karpov)
Decision: This equation has a single solution: x \u003d 2, y \u003d 1, z \u003d 2 (in both options). The fact that it is a solution follows from the general identity A² + (2a + 1) \u003d (A + 1) ² ², used in the first embodiment to A \u003d 105, and in the second - to a \u003d 201.There are no other solutions, since if Z\u003e 2, then the right side of the equation is divided into 8, and the left - no, since 105 x can give only the residue 1, and 211 y to be 0 only - only residues 1 and 3. It remains to be noted, That at z \u003d 1 solutions is also not, and at z \u003d 2, the values \u200b\u200by \u003d 1 and x \u003d 2 are uniquely defined.
Teacher's introductory word:
Small historical reference: Many scientists from ancient times were fond of the tasks for cutting. Solutions of many simple tasks Ancient Greeks were found on the cut, but the first systematic treatise on this topic belongs to Peru Abul-Vefa. Geometers seriously engaged in solving tasks for cutting figures to the smallest number of parts and the subsequent construction of another figure at the beginning of the 20th century. One of the founders of this section was the famous founder of Henry E. Dyudeni puzzles.
Nowadays, puzzle lovers are fond of solving problems before because the universal method of solving such tasks does not exist, and everyone who takes them to decide can fully show their smelting, intuition and the ability to creative thinking. (In the class we will indicate only one of the possible examples of cutting. It can be assumed that students may turn out some other correct combination - it is not necessary to fear it).
This lesson is supposed to be held in the form of practical classes. Break the participants a mug into groups of 2-3 people. Each of the groups will be provided in advance by a teacher's figures. Students have a ruler (with divisions), pencil, scissors. It is allowed to produce with scissors only straight cuts. By cutting some kind of figure to pieces, you need to make another figure from the same parts.
Cutting tasks:
1). Try to cut the figure shown in the figure to 3 equal in the form of part:
Tip: Little figures are very similar to the letter T.
2). Currente this figure to 4 equal in the form of part:
Hint: It is easy to guess that small figures will consist of 3 cells, and three cell figures are not so much. There are only two types: a corner and a rectangle.
3). Divide the shape into two identical parts, and fold the chessboard from the parts received.
Tip: Suggest to start performing a task from the second part, how to get a chessboard. Remember what form has a chessboard (square). Calculate the existing number of cells in length in width. (Remind that cells should be 8).
4). Try three knife movements to cut the cheese on eight equal pieces.
Hint: Try cut cheese along.
Tasks for self solutions:
1). Cut the square of the paper and do the following:
· Cut into such 4 parts from which two equal smaller squares can be made.
· Cut on five parts - four isceived triangles and one square - and fold them so that it turns out three squares.
In the nimination of tutors in mathematics and teachers of various electives and circles, a selection of entertaining and developing geometric problems for cutting is offered. The purpose of using such tasks to use such tasks in its classes is not only to interest the student in interesting and effective combinations of cells and figures, but also to form a sense of lines, angles and forms. The task set is mainly oriented on children 4-6 classes, although its use is not excluded even with high school students. Exercises require students with high and device concentrations and are well suited for the development and training of visual memory. Recommended for tutors of mathematics engaged in the preparation of students to entrance exams In mathematical schools and classes that prevent special requirements for the level of independent thinking and the creative abilities of the child. The level of tasks corresponds to the level of introductory olympiad in the Lyceum "Second School" (second mathematical school), Male Mehmat MSU, Kurchatov School, etc.
Note Tutor in Mathematics:
In some solutions to tasks that you can view clicking on the appropriate pointer, only one of the possible samples of cutting is specified. I fully admit that you can get some other faithful combination - do not have to be afraid of it. Check carefully the solution of your soap and if it satisfies the condition, then boldly take on the following task.
1) Try cutting the figure of 3 equal in the figure in the figure:
: Little figures are very similar to the letter T
2) Cut now this figure on 4 equal in the form of part:
Mathematics tip: It is easy to guess that small figures will consist of 3 cells, and three cell figures are not so much. There are only two types: a corner and a 1 × 3 rectangle.
3) Cut this figure to 5 equal in the form of parts:
Find the number of cells from which each such figure consists. These figures are similar to the letter G.
4) And now you need to cut a figure of ten cells to 4 unequal A friend of the rectangle (or square).
Indication of the tutor in mathematics: Highlight some rectangle, and then in the remaining cells, try to enter three more. If it does not work, then change the first rectangle and try again.
5) The task is complicated: the figure should be cut on 4 different in form figures (not necessarily on rectangles).
Mathematics tip: Draw first separate all kinds of figures of different shapes (There will be more than four) and repeat the method of overlooking the variants as in the previous task.
:
6) Cut this figure on 5 figures of four cells of different shapes so that only one green cell is painted in each of them.
Tip Tutor in Mathematics: Try to start cut from the top edge of this figure and you will immediately understand how to act.
:
7) based on the previous task. Find how many figures have different shapes consisting exactly of four cells? The figures can be twisted, rotated, but you can not lift the spike (from its surface) on which it lies. That is, the two above figures will not be considered equal, as they cannot be obtained from each other by turning.
Tip Tutor in Mathematics: Examine the solution of the previous task and try to imagine the various positions of these figures when you turn. It is easy to guess that the answer in our task will be the number 5 or more. (In fact, even more than six). In total, there are 7 types of figures described.
8) Cut the square of 16 cells by 4 equal in the form of the part so that in each of the four parts there was exactly one green cell.
Mathematics tip: The form of small figures is not a square and not a rectangle, and not even a corner of four cells. So what are the figures to try to cut?
9) Figure the figure, cut into two parts in such a way that the square can be folded from the parts obtained.
Matematsky Tutor Tutor: In total, in the figure of 16 cells - it means that the square will be size 4 × 4. And somehow you need to fill the window in the middle. How to do it? Maybe some shift? Then because the length of the rectangle is equal to odd cells, the cutting must be carried out by a vertical cut, but by a broken line. So that the upper part is cut off from one side from the middle cells, and the lower on the other.
10) Cut the rectangle size of 4 × 9 into two parts with such a calculation so that the square can be folded as a result of them.
Mathematics tip: Total in rectangle 36 cells. Therefore, the square will be 6 × 6. So the ka mountain side consists of nine cells, then three of them need to be cut off. How will this incision go further?
11) Captive from five cells shown in the figure is required to cut (you can cut the cells themselves) to such parts from which the square could be folded.
Mathematics tip: It is clear that as if we do not cut the cells on the lines - I will not get a square, since the cells are only 5. This is the only task in which it is allowed to cut not by cells. However, they will still be well left as a reference point. For example, it is worth noting that we somehow need to remove the deepening, which we have - namely, in the inner corners of our cross. How to do it? For example, cut off some discover triangles from the external corners of the cross ...