Finding the area by nodes. Start in science
This topic will be of interest to students in grades 10-11 in preparation for the exam. The Peak formula can be used when calculating the area of a figure depicted on checkered paper (this task is proposed in the USE test and measuring materials).
During the classes
"The subject of mathematics is so serious
which is useful not to miss the opportunity
make it a little fun"
(B. Pascal)
Teacher: There are tasks that are unusual and do not look like tasks from school textbooks? Yes, these are tasks on checkered paper. Such tasks exist in control and measurement USE materials. What is the peculiarity of such problems, what methods and techniques are used to solve problems on checkered paper? In this lesson, we will explore tasks on checkered paper related to finding the area of the depicted figure, and learn how to calculate the areas of polygons drawn on a checkered sheet.
Teacher: The object of the study will be tasks on checkered paper.
The subject of our study will be problems for calculating the area of polygons on checkered paper.
And the goal of the study will be the Peak formula.
B - the number of integer points inside the polygon
Г - the number of integer points on the border of the polygon
This is a handy formula that can be used to calculate the area of any polygon without self-intersections with the vertices at the knots of the checkered paper.
Who is Peak? Peak Georg Aleksandrov (1859-1943) - Austrian mathematician. Discovered the formula in 1899.
Teacher: Let us formulate a hypothesis: the area of the figure calculated by the Pick formula is equal to the area of the figure calculated by the geometry formulas.
When solving problems on checkered paper, we need geometric imagination and fairly simple information that we know:
The area of a rectangle is equal to the product of the adjacent sides.
Square right triangle is equal to half the product of the sides forming a right angle.
Teacher: Grid nodes are points where grid lines intersect.
The interior nodes of the polygon are blue. The nodes on the borders of the polygon are brown.
We will consider only such polygons, all of whose vertices lie at the nodes of the checkered paper.
Teacher: Let's do some research for a triangle. First, we calculate the area of the triangle using the Peak formula.
AT + G/2 − 1 , where AT G— the number of integer points on the border of the polygon.
B = 34, G = 15,
AT + G/2 − 1 = 34 + 15 :2 − 1 = 40, 5 Answer: 40.5
Teacher: Now we calculate the area of the triangle using the geometry formulas. The area of any triangle drawn on checkered paper can be easily calculated by representing it as the sum or difference of the areas of right-angled triangles and rectangles, the sides of which go along the grid lines passing through the vertices of the drawn triangle. Students do the calculations in their notebooks. Then they check their results with the calculations on the board.
Teacher: Comparing the results of the studies, draw a conclusion. We found that the area of the figure calculated using the Peak formula is equal to the area of the figure calculated using the geometry formulas. So the hypothesis turned out to be correct.
Next, the teacher suggests calculating the area of \u200b\u200b"his" arbitrary polygon using the geometry formulas and the Pick formula and comparing the results. You can “play” with the Peak formula on the site of mathematical studies.
At the end of the article, one of the papers on the topic "Calculation of the area of an arbitrary polygon using the Pick formula" is proposed.
More pExample:
The area of a polygon with integer vertices is AT + G/2 − 1 , where AT is the number of integer points inside the polygon, and G is the number of integer points on the boundary of the polygon.
B = 10, G = 6,
AT + G/2 − 1 = 10 + 6 :2 − 1 = 12 ANSWER: 12
Teacher: I suggest that you solve the following tasks:
Answer: 12
Answer: 13
Answer: 9
Answer: 11.5
Answer: 4
|
Calculating the area of a figure.
Pick Method
The work of a student of class 5B MBOU secondary school No. 23 in Irkutsk
Balsukova Alexandra
Head: Khodyreva T.G.
2014
Calculating the area of a figure. Pick method
Object of study : tasks on checkered paper
Subject of study : problems for calculating the area of a polygon on checkered paper, methods and techniques for solving them.
Research methods Keywords: comparison, generalization, analogy, study of literature and Internet resources, information analysis.
Purpose of the study:
choose the main, interesting, understandable information
Analyze and organize the information received
Find various methods and techniques for solving problems on checkered paper
check area formulas geometric shapes using Pick's formula
Create an electronic presentation of the work to present the collected material
Geometry is the most powerful tool for the refinement of our mental faculties and enables us to think and reason correctly.
(G. Galileo)
Relevance of the topic
Passion for mathematics often begins with thinking about a problem. So, when studying the topic "Areas of polygons", the question arises whether there are tasks that are different from the tasks considered in the textbook. Such tasks include tasks on checkered paper. What is the peculiarity of such tasks, are there any special methods and techniques for solving problems on checkered paper. In a math class, the teacher introduced us to an interesting method for calculating polygons. I began to study the literature, Internet resources on this topic. It would seem that fascinating things can be found on a checkered plane, that is, on an endless piece of paper, lined into identical squares. It turns out that the tasks associated with checkered paper are quite diverse. I learned how to calculate the areas of polygons drawn on a checkered piece of paper. For many tasks on paper in a cage there is no general rule for solving, specific methods and techniques. This is their property that determines their value for the development of not a specific educational skill or skill, but in general the ability to think, reflect, analyze, look for analogies, that is, these tasks develop thinking skills in their broadest sense.
And I also learned that such tasks are considered in the control - measuring materials GIA and USE. Therefore, I consider the study of this material useful for its application not only in the future educational process, but also for solving non-standard Olympiad problems.
2.The concept of area
Square- numerical characteristic of a two-dimensional geometric figure, showing the size of this figure. Historically, area calculation was called . A figure that has area is called squaring .
The area of a flat figure in terms of geometry
1. Square- the measure of a flat figure in relation to the standard figure, which is a square with a side equal to one length.
2. Square- a numerical characteristic attributed to flat figures of a certain class (for example, polygons). The area of a square with a side equal to a unit of length, taken equal to a unit of area
3. Square- a positive value, the numerical value of which has the following properties:
Equal figures have equal areas;
If the figure is divided into parts that are simple figures(that is, those that can be divided into a finite number of flat triangles), then the area of \u200b\u200bthis figure is equal to the sum of the areas of its parts;
The area of a square with a side equal to the unit of measurement is equal to one.
Thus, we can conclude that the area is not a specific value, but only gives some conditional characteristic of a flat figure. To find the area of an arbitrary figure, it is necessary to determine how many squares with a side equal to one length, it contains. For example, let's take a rectangle in which a square centimeter fits exactly 6 times. This means that the area of the rectangle is 6 cm2.
The choice of the area of a square with a side equal to the unit of measurement as the minimum unit of measurement for all areas is not accidental. This is the result of an agreement between people that arose in the course of "natural" centuries-old selection. In addition, there were other proposals for a unit of measure. So, for example, it was proposed to take the area of an equilateral triangle as such a unit (i.e., any flat figure could be represented as a “sum” of a certain number equilateral triangles), which would lead to a change in the numerical representation of areas.
Thus, formulas for calculating areas appeared in mathematics and were not immediately realized by a person - this many scientists living in different eras and different countries. (Wrong formulas did not find a place in science and went into oblivion). The true formulas were supplemented, corrected and substantiated over thousands of years, until they reached us in their modern form.
Of course area measurement consists in comparing the area of a given figure with the area of a figure taken as a unit of measurement. As a result of the comparison, a certain number is obtained - the numerical value of the area of \u200b\u200bthe given figure. This number shows how many times the area of a given figure is greater (or less) than the area of the figure, taken as a unit of area.
T Thus, we can conclude that the area is an artificial quantity, historically introduced by man to measure some property of a flat figure. The need to enter such a value was due to the growing need to know how large this or that territory is, how much grain is needed to sow a field or calculate the floor surface area for decorating ornamental tiles.
Peak Formula
To estimate the area of a polygon on checkered paper, it is enough to calculate how many cells this polygon covers (we take the area of \u200b\u200bthe cell as a unit). More precisely, ifS is the area of the polygon, B is the number of cells that lie entirely inside the polygon, and G is the number of cells that have an interior. We will consider only such polygons, all the vertices of which lie at the nodes of the checkered paper - in those where the grid lines of the polygon intersect at least one common point.
The area of any triangle drawn on checkered paper can be easily calculated by representing it as the sum or difference of the areas of right-angled triangles and rectangles whose sides follow the grid lines passing through the vertices of the drawn triangle.
To calculate the area of such a polygon, you can use the following theorem:
Theorem . Let - the number of integer points inside the polygon, - the number of integer points on its boundary, - its area. ThenPick's formula:
Example. For the polygon in the figureL = 7 (red dots), 9 (green dots), soS = 7+ 9/2 -1 = 10,5 square units.
Pick's theorem- classic result and .
The area of a triangle with vertices at the nodes and containing no nodes either inside or on the sides (except for the vertices) is equal to 1/2. This fact.
3. History
Pick's formula was discovered by the Austrian mathematician Georg Alexander (1859-1942) in . At the age of 16, Georg finished school and entered. At the age of 20 he received the right to teach physics and mathematics. In 1884 Peak went to to . There he met another student of Klein,. Later, in 1885, he returned towhere he spent the rest of his scientific career.
Georg Pick was friends with Einstein. Pick and Einstein not only shared scientific interests, but were also passionate about music. Pick, who played in a quartet that consisted of university professors, introduced Einstein to the scientific and musical societies of Prague.
The circle of mathematical interests of Peak was extremely wide. In particular, they are more than 50 scientific works. Pick's theorem, discovered by him in 1899, was widely known for calculating the area of a polygon. In Germany, this theorem is included in school textbooks.
4.Applications of Pick's formula
The Pick formula is used not only to calculate the areas of polygons, but also to solve many problems of the Olympiad level.
Some examples of using Pick's formula when solving problems:
1) The chess king went around the board of 8 × 8 cells, having visited each
home field exactly once and with the last move returning to the original
field. A broken line connecting in series the centers of the fields that
the king passed, has no self-intersections. What area can
limit this broken line? (The side of the cell is 1.)
It immediately follows from the Pick formula that the area bounded by the lo-
mana is 64/2 − 1 = 31; here the lattice nodes are the centers 64
fields and, by assumption, they all lie on the boundary of the polygon. So
Thus, although there are quite a lot of such "trajectories" of the king, but all of them
limit polygons of equal areas.
Tasks from the control and measuring materials of the GIA and the Unified State Examination
Task B3
Find the area of the figure depicted on checkered paper with a cell size of 1 cm 1 cm (see Fig.). Give your answer in square centimeters.
4.Conclusion
In the process of research, I studied reference, popular science literature. I learned that the problem of finding the area of a polygon with vertices at the nodes of the grid inspired the Austrian mathematician Pick in 1899 to prove the wonderful Pick formula.
As a result of my work, I expanded my knowledge of solving problems on checkered paper, determined for myself the classification of the problems under study, and became convinced of their diversity.
I learned how to calculate the areas of polygons drawn on a checkered sheet. The considered tasks have a different level of difficulty - from simple to Olympiad. Everyone can find among them tasks of a feasible level of complexity, starting from which, it will be possible to move on to solving more difficult ones.
I came to the conclusion that the topic that interested me is quite multifaceted, the tasks on checkered paper are diverse, the methods and techniques for solving them are also diverse. Therefore, our I decided to continue working in this direction.
5. Literature used:
1. N. B. Vasil’ev, “Around the Pick formula,” Kvant. - 1974. - No. 12
2. Kokse Prasolov VV Tasks in planimetry. - M.: MTsNMO, 2006.t e r G.S.M. Introduction to geometry. - M.: Nauka, 1966
3. Roslova L.O., Sharygin I.F. Measurements. - M.: Ed. "Open World", 2005.
Internet resources:
:
Feedback on work
“Calculation of the areas of plane figures. Pick Method"
Consideration of this topic will improve cognitive activity a student who later on in geometry lessons will begin to see the harmony of the drawing and will cease to perceive geometry (and mathematics in general) as a boring science.
Reviewed by math teacher
Khodyreva Tatyana Georgievna
Starkova Kristina, 8B grade student
The paper considers Pick's theorem and its proof.
The problems of finding the area of polygons are considered
Download:
Preview:
DEPARTMENT OF GENERAL AND VOCATIONAL EDUCATION
ADMINISTRATION OF THE TCHAIKOVSKY MUNICIPAL DISTRICT
PERM REGION
VI MUNICIPAL RESEARCH CONFERENCE
STUDENTS
Municipal Autonomous General Educational Institution
"medium comprehensive school No. 11"
SECTION: MATHEMATICS
Application of Pick's Formula
Student of 8 "B" class
MAOU secondary school №11Tchaikovsky
Leader: Batueva L, N.,
Mathematics teacher MAOU secondary school №11
Tchaikovsky
year 2012
I. Introduction……………………………………………………. 2
II. Peak Formula
2.1.Grids.Knots…………………………………………….4
2.2.Triangulation of a polygon…………………………5
2.3. Proof of Pick's theorem………………………6
2.4 Studying the areas of polygons…………9
2.5. Conclusion…………………………………………………..12
III. Geometric problems with practical content ... 13
IV. Conclusion………………………………………………..14
V. List of used literature………………………..16
- Introduction
Passion for mathematics often begins with thinking about a problem. So, when studying the topic "Areas of polygons", the question arose whether there were tasks that were different from the tasks considered in geometry textbooks. These are tasks on checkered paper. We had questions: what is the peculiarity of such problems, are there special methods and techniques for solving problems on checkered paper. Seeing such tasks in the control and measuring materials of the Unified State Exam and the State Academic Examination, I decided to definitely investigate the tasks on checkered paper related to finding the area of the depicted figure.
I began to study the literature, Internet resources on this topic. It would seem that what is fascinating can be found on a checkered plane, that is, on an endless piece of paper drawn into identical squares? Don't judge hastily. It turns out that the tasks associated with checkered paper are quite diverse. I learned how to calculate the areas of polygons drawn on a checkered piece of paper. For many tasks on paper in a cage there is no general rule for solving, specific methods and techniques. This is their property that determines their value for the development of not a specific educational skill or skill, but in general the ability to think, reflect, analyze, look for analogies, that is, these tasks develop thinking skills in their broadest sense.
We defined:
Object of study: tasks on checkered paper
Subject of study: problems for calculating the area of a polygon on checkered paper, methods and techniques for solving them.
Research methods: modeling, comparison, generalization, analogy, study of literary and Internet resources, analysis and classification of information.
- Purpose of the study:Derive and test formulas for calculating the areas of geometric shapes using the Peak formula
To achieve this goal, we propose to solve the following tasks:
- Select the necessary literature
- Select material for research, choose the main, interesting, understandable information
- Analyze and organize the information received
- Find different methods and techniques for solving problems on checkered paper
- Create an electronic presentation of the work to present the collected material to classmates
the variety of tasks on paper in a box, their "entertainment", the lack of general rules and methods of solving cause difficulties for schoolchildren when considering them
- Hypothesis:. The area of the figure calculated by the Pick formula is equal to the area of the figure calculated by the planimetry formula.
When solving problems on checkered paper, we need geometric imagination and fairly simple geometric information that is known to everyone.
II. Peak Formula
2.1. Lattices. Knots.
Consider on the plane two families of parallel lines dividing the plane into equal squares; the set of all points of intersection of these lines is called a point lattice or simply a lattice, and the points themselves are called lattice nodes.
Internal nodes of a polygon - red.
Knots on the faces of a polygon - blue.
To estimate the area of a polygon on checkered paper, it is enough to calculate how many cells this polygon covers (we take the area of \u200b\u200bthe cell as a unit). More precisely, if S is the area of the polygon, B is the number of cells that lie entirely inside the polygon, and G is the number of cells that have at least one common point with the interior of the polygon.
We will consider only such polygons, all of whose vertices lie at the nodes of the checkered paper - in those where the grid lines intersect.
The area of any triangle drawn on checkered paper can be easily calculated by representing it as the sum or difference of the areas of right-angled triangles and rectangles whose sides follow the grid lines passing through the vertices of the drawn triangle.
2.2 Triangulation of a polygon
Any polygon with vertices at the grid nodes can be triangulated - divided into "simple" triangles.
Let some polygon and some finite set be given on the plane To points lying inside the polygon and on its boundary (moreover, all the vertices of the polygon belong to the set TO ).
Triangulation with vertices To is called a partition of a given polygon into triangles with vertices in the set To such that each point in To serves as a vertex for each of those triangulation triangles to which this point belongs (that is, points from To do not fall inside or on the sides of the triangles, fig. 1.37).
Rice. 1.37
Theorem 2. a) Any n -gon can be cut by diagonals into triangles, and the number of triangles will be equal to n – 2 (this partition is a triangulation with vertices in vertices n-gon).
Consider a non-degenerate simple integer polygon (that is, it is connected - any two of its points can be connected by a continuous curve that is entirely contained in it, and all its vertices have integer coordinates, its boundary is a connected polyline without self-intersections, and it has a non-zero area) .
To calculate the area of such a polygon, you can use the following theorem:
2.3. Proof of Pick's theorem.
Let B be the number of integer points inside the polygon, Г be the number of integer points on its boundary,- its area. Then Pick's formula: S=V+G2-1
Example. For the polygon in the figure B=23 (yellow dots), D=7, (blue dots, let's not forget the vertices!), sosquare units.
First, note that Pick's formula is true for the unit square. Indeed, in this case we have B=0, D=4 and.
Consider a rectangle with sides lying on the lattice lines. Let the lengths of its sides be equal and . We have in this case, B=(a-1)(b-1) , G=2a+2b, then by the Pick formula,
Consider now a right triangle with legs lying on the coordinate axes. Such a triangle is obtained from a rectangle with sides and , considered in the previous case, by cutting it diagonally. Let them lie on the diagonalinteger points. Then for this case B \u003d a-1) b-1, 2 G \u003d G \u003d 2a + 2b 2 +c-1 and we get that4) Now consider an arbitrary triangle. It can be obtained by cutting off several right-angled triangles and, possibly, a rectangle from a rectangle (see pictures). Since Pick's formula is true for both a rectangle and a right-angled triangle, we get that it will also be true for an arbitrary triangle.
It remains to take the last step: move from triangles to polygons. Any polygon can be divided into triangles (for example, by diagonals). Therefore, we just need to prove that when adding any triangle to an arbitrary polygon, Pick's formula remains true. Let the polygon and triangle have a common side. Let's assume that forPick's formula is valid, we will prove that it will be true for the polygon obtained from adding . Since and have a common side, then all integer points lying on this side, except for two vertices, become interior points of the new polygon. Vertices will be boundary points. Let us denote the number of common points by and get B=MT=BM+BT+c-2 - the number of internal integer points of the new polygon, Г=Г(М)+Г(T)-2(s-2)-2 - the number of boundary points of the new polygon. From these equalities we get: BM+BT+c-2 , G=G(M)+G(T)-2(s-2)-2 . Since we have assumed that the theorem is true for and for separately, then S(MT)+S(M)+S(T)=(B(M)+ GM2 -1)+B(T)+ GT2 -1)=(B(M)+ B(T))+( GM2+HT2)-2 =G(MT)-(c-2)+ B(MT) +2(c-2)+22 -2= G(MT)+ B(MT)2-1 .Thus, the Pick formula is proved.
2.4 Study of areas of polygons.
2) On checkered paper with cells measuring 1 cm x 1 cm is depicted triangle. Find its area in square centimeters. | ||
Picture | According to the geometry formula | According to Pick's formula |
S=12ah Str.ABD=1/2 AD ∙ BD=1/2 ∙ 2 ∙ 1=1 Str.BDC=1/2 DC ∙ BD=1/2 ∙ 3 ∙ 1=1.5 Str.ABC=Str.BDC-Str.ABD= 1,5-1=0,5 | S= V+G2-1 G=3 ;V=0. S=0+3/2-1=0.5 |
3) A square is depicted on checkered paper with cells measuring 1 cm x 1 cm. Find its area in square centimeters. | ||
Picture | According to the geometry formula | According to Pick's formula |
S=a∙b Sq.KMNE=7 ∙ 7=49 Str.AKB=1/2 ∙ KB ∙ AK=1/2 ∙ 4 ∙ 4=8 Str.AKB=Str.DCE=8 Str.AND= 1/2 ∙ ND ∙ AN=1/2 ∙ 3 ∙ 3=4.5 Str.AND=Str.BMC=4.5 Spr.= Sq.KMNE- Str.AKB- Str.DCE- Str.AND- Str.BMC=49-8-8-4.5-4.5=24 | S= V+G2-1 D=14; W=19. S=18+14/2-1=24 |
4) On checkered paper with cells measuring 1 cm x 1 cm is depicted | ||
Picture | According to the geometry formula | According to Pick's formula |
S1= 12a∙b=1/2∙7∙1= 3.5 S2= 12a∙b=1/2∙7∙2=7 S3= 12a∙b=1/2∙4∙1=2 S4= 12a∙b=1/2∙5∙1=2.5 S5=a²=1²=1 Sq.= a²=7²=49 S=49-3.5-7-2-2.5-1=32cm² | S= V+G2-1 D=5; V=31. S=31+ 42 -1=32cm² |
|
5) On checkered paper with cells measuring 1 cm x 1 cm four square. Find its area in square centimeters. | ||
S= a b a=36+36=62 b=9+9=32 S \u003d 62 ∙ 32 \u003d 36 cm 2 | S= V+G2-1 D=18, V=28 S=28+ 182 -1=36cm 2 |
|
6) On checkered paper with cells measuring 1 cm x 1 cm is depicted four square. Find its area in square centimeters | ||
S1= 12a∙b=1/2∙3∙3=4.5 S2= 12a∙b=1/2∙6∙6=18 S3= 12a∙b=1/2∙3∙3=4.5 S=4.5+18+4.5=27 cm² | S= V+G2-1 D=18; W=28. S=28+ 182 -1=36cm² |
|
7) On checkered paper with cells measuring 1 cm x 1 cm is depicted four square. Find its area in square centimeters | ||
S1= 12a∙b=1/2∙3∙3=4.5 S2= 12a∙b=1/2∙6∙6=18 S3= 12a∙b=1/2∙3∙3=4.5 S4= 12a∙b=1/2∙6∙6=18 Sq.=9²=81cm² S=81-4.5-18-4.5-18=36cm² | S= V+G2-1 D=18; W=28. S=28+ 182 -1=36cm² |
8) On checkered paper with cells measuring 1 cm x 1 cm is depicted four square. Find its area in square centimeters | ||
Picture | According to the geometry formula | According to Pick's formula |
S1= 12a∙b=1/2∙2∙4=4 S2= 12ah =1/2 ∙ 4 ∙ 4=8 S3= 12ah =1/2 ∙ 8 ∙ 2=8 S4= 12ah =1/2 ∙ 4 ∙ 1=2 Spr.= a∙ b=6 ∙ 8=48 S5=48-4-8-8-2=24 cm² | S= G+V2-1 D=16; W=17. S=17+ 162 -1=24 cm² |
Conclusion
- Comparing the results in the tables and proving Pick's theorem, I came to the conclusion that the area of the figure calculated using the Pick formula is equal to the area of the figure calculated using the derived planimetry formula
So my hypothesis turned out to be correct.
III.Geometric problems with practical content.
The Pick formula will also help us to solve geometric problems with practical content.
Task 9 . Find the area of the forest (in m²) depicted on a plan with a square grid of 1 × 1 (cm) on a scale of 1 cm - 200 m (Fig. 10)
Solution.
Rice. 10 V \u003d 8, G \u003d 7. S \u003d 8 + 7/2 - 1 \u003d 10.5 (cm²)
1 cm² - 200² m²; S = 40,000 10.5 = 420,000 (m²)
Answer: 420,000 m²
Task 10 . Find the area of the field (in m²) depicted on a plan with a square grid of 1 × 1 (cm) on a scale of 1 cm - 200 m. (Fig. 11)
Solution. Let's find S the area of the quadrilateral depicted on checkered paper using the Peak formula: S = B + - 1
V \u003d 7, D \u003d 4. S \u003d 7 + 4/2 - 1 \u003d 8 (cm²)
Rice. 11 1 cm² - 200² m²; S = 40000 8 = 320,000 (m²)
Answer: 320,000 m²
Conclusion
In the process of research, I studied reference, popular science literature, learned how to work in the Notebook program. I found out that
The problem of finding the area of a polygon with vertices at the nodes of the grid inspired the Austrian mathematician Pick in 1899 to prove the wonderful Pick formula.
As a result of my work, I expanded my knowledge of solving problems on checkered paper, determined for myself the classification of the problems under study, and became convinced of their diversity.
I learned to calculate the areas of polygons drawn on a checkered sheet. The tasks considered have a different level of difficulty - from simple to Olympiad. Everyone can find among them tasks of a feasible level of complexity, starting from which, it will be possible to move on to solving more difficult ones.
I came to the conclusion that the topic that interested me is quite multifaceted, the tasks on checkered paper are diverse, the methods and techniques for solving them are also diverse. Therefore, our I decided to continue working in this direction.
Literature
1. Geometry on checkered paper. Small MEHMAT MSU.
2. Zharkovskaya N. M., Riss E. A. Checkered paper geometry. Pick's formula // Mathematics, 2009, no. 17, p. 24-25.
3. Tasks open bank tasks in mathematics FIPI, 2010 - 2011
4.V.V.Vavilov, A.V.Ustinov. Polygons on lattices. M.MTsNMO, 2006.
5. Thematic studies.etudes.ru
6. L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev and others. Geometry. 7-9 classes. M. Enlightenment, 2010
There is a wonderful formula that allows you to count polygon area on the coordinate grid almost without errors. It's not even a formula, it's real theorem. At first glance, it may seem complicated. But it is enough to solve a couple of tasks - and you will understand how cool this feature is. So go ahead!
Let's start with a new definition:
A coordinate stack node is any point that lies at the intersection of the vertical and horizontal lines of this grid.
Designation:
In the first picture, the nodes are not marked at all. The second one has 4 nodes. Finally, in the third picture, all 16 nodes are marked.
What does this have to do with problem B5? The fact is that the vertices of the polygon in such problems always lie at the nodes of the grid. As a consequence, the following theorem works for them:
Theorem. Consider a polygon on a coordinate grid whose vertices lie at the nodes of this grid. Then the area of the polygon is:
where n is the number of nodes inside the given polygon, k is the number of nodes that lie on its boundary (boundary nodes).
As an example, consider an ordinary triangle on a coordinate grid and try to mark the internal and boundary nodes.
The first picture shows an ordinary triangle. On the second picture, its internal nodes are marked, the number of which is n = 10. On the third picture, the nodes lying on the border are marked, there are k = 6 of them in total.
Perhaps many readers do not understand how to count the numbers n and k. Start with internal nodes. Everything is obvious here: we paint over the triangle with a pencil and see how many nodes are shaded.
With boundary nodes, it's a little more complicated. polygon border - closed broken line, which intersects the coordinate grid at many points. The easiest way is to mark some "starting" point, and then go around the rest.
Boundary nodes will be only those points on the polyline at which they simultaneously intersect three lines:
- Actually, a broken line;
- Horizontal grid line;
- vertical line.
Let's see how it all works in real problems.
A task. Find the area of a triangle if the cell size is 1 x 1 cm:
First, let's mark the nodes that lie inside the triangle, as well as on its border:
It turns out that there is only one internal node: n = 1. There are six boundary nodes: three coincide with triangle vertices, and three more lie on the sides. Total k = 6.
Now we calculate the area using the formula:
That's all! Problem solved.
A task. Find the area of a quadrangle drawn on checkered paper with a cell size of 1 cm by 1 cm. Give your answer in square centimeters.
Again, we mark the internal and boundary nodes. There are n = 2 internal nodes. Boundary nodes: k = 7, of which 4 are vertices of the quadrilateral, and 3 more lie on the sides.
It remains to substitute the numbers n and k in the area formula:
Pay attention to the last example. This task was actually proposed for diagnostic work in 2012. If you work according to the standard scheme, you will have to do a lot of additional constructions. And by the method of knots, everything is solved almost orally.
Important note on areas
But the formula is not everything. Let's rewrite the formula a little, bringing the terms on the right side to common denominator . We get:
The numbers n and k are the number of nodes, they are always integers. So the whole numerator is also an integer. We divide it by 2, which implies an important fact:
The area is always expressed whole number or fraction. Moreover, at the end of the fraction there is always “five tenths”: 10.5; 17.5 etc.
Thus, the area in problem B5 is always expressed as an integer or a fraction of the form ***.5. If the answer is different, it means that a mistake has been made somewhere. Keep this in mind when you take the real exam in mathematics!
Draw a polygon on checkered paper. For example, such as shown in Figure 1.
Let's try to calculate its area now. How to do it? Probably the easiest way is to break it into right-angled triangles and rectangles, the areas of which are already easy to calculate and add up the results. The method I used is simple, but very cumbersome, and besides, it is not suitable for all polygons.
Consider a non-degenerate simple integer polygon (that is, it is connected - any two of its points can be connected by a continuous curve that is entirely contained in it, and all its vertices have integer coordinates, its boundary is a connected polyline without self-intersections, and it has a non-zero square). To calculate the area of such a polygon, you can use the following theorem:
Pick's theorem. Let be the number of integer points inside the polygon, be the number of integer points on its boundary, and be its area. Then Pick's formula:
Example. For the polygon in Figure 1 (yellow dots), (blue dots, don't forget the vertices!), so square units.
Proof of Pick's theorem. First, note that Pick's formula is true for the unit square. Indeed, in this case we have
Consider a rectangle with sides lying on the lattice lines. Let the lengths of its sides be equal to and. We have in this case and, according to the Pick formula,
Consider now a right triangle with legs lying on the coordinate axes. Such a triangle is obtained from a rectangle with sides and, considered in the previous case, by cutting it diagonally. Let integer points lie on the diagonal. Then for this case we get that
Now consider an arbitrary triangle. It can be obtained by cutting off several right-angled triangles and, possibly, a rectangle from a rectangle (see figures 2 and 3). Since Pick's formula is true for both a rectangle and a right-angled triangle, we get that it will also be true for an arbitrary triangle.
It remains to take the last step: move from triangles to polygons. Any polygon can be divided into triangles (for example, by diagonals). Therefore, we just need to prove that when adding any triangle to an arbitrary polygon, Pick's formula remains true.
Let the polygon and the triangle have a common side. Assume that Pick's formula is valid for, and we will prove that it will also be true for the polygon obtained from the addition. Since and have a common side, all integer points lying on this side, except for two vertices, become interior points of the new polygon. Vertices will be boundary points. Let us denote the number of common points by and get
The number of interior integer points of the new polygon,
The number of boundary points for the new polygon.
From these equalities we get
Since we assumed that the theorem is true for and for separately, then
Thus, the Pick formula is proved.
This formula was discovered by the Austrian mathematician Peak Georg Aleksandrov (1859 - 1943) in 1899. In addition to this formula, Georg Pick discovered the Pick, Pick - Julia, Pick - Nevalina theorems, proved the Schwarz - Pick inequality. AT Appendix 1 you can see the non-standard tasks I have considered for applying the Pick formula.