If we proceed from the relationship between the desired and the data of the problem, then the condition of the problem for the construction can be expressed analytically.

The analytical expression of the construction problem in the form of an equation, and its solutions in the form of the roots of this equation, help to find a geometric solution, as well as to determine with what tools it can be performed.

The solution of problems by the algebraic method is reduced to the construction:

• average proportional to two given segments x = 4ab
• the fourth proportional to the three given segments, expressed

. „ bc

pressed by the formula x \u003d -;

By the algebraic sum of these segments x = a±b, x-a + b-c + d,

x = 3a±2b etc.; _

By formulas like x = 1a + b.

The algebraic method for solving geometric construction problems is as follows:

• 1) unknown quantities appearing in the condition of the problem are denoted by the letters x, y, z etc.;
• 2) compose equations that relate these unknowns with the values ​​given in the problem a, b, c, ...;
• 3) solve the formulated equations;
• 4) examine the responses received;
• 5) perform the required construction.

Before moving on to solving construction problems by the algebraic method, let us consider the construction of some segments given by the ratios between the lengths of other segments.

1. Sometimes in geometric construction problems the ratio of two quantities is given in the form a:b; a 3: b 3; a 4: b 4 etc.

Let us show that any of these ratios can be replaced by the ratio of two segments.

Problem 6.47. Construct a segment given by the ratio a p: b p, where

p e N.

Solution

Draw two arbitrary mutually perpendicular lines KL and MN(Fig. 6.52) and denote by the letter O the point of their intersection. On straight lines KL and MN from the point O we postpone the segments OA and oa x, respectively equal to given segments Kommersant and a. By connecting points A and A b restore at the point A g perpendicular to AA X KL at some point L 2 . At point A 2 we restore the perpendicular to A 2 A 1 and continue it until it intersects with the line MN at point A 3, etc.

Let us determine the value of each of the following ratios: OA x: OA; OA 2 : OA x; OA 3: OA g etc.

Since right triangles OAA x, OA, A 2, OA^A 3 ,... are similar, that means:

oa, and ,

By construction, - L = -, and therefore, by virtue of equalities (*), we obtain OA b

Let us determine the value of the ratio It will not change if we

each of its terms is divided by the same amount OA b and therefore

t t OA, a O Ap a OA 2 a OA b

But from the equalities -- = - and -- = - we see that -- = - and -= -.

OAOA x b oa, Kommersant oa, a

By virtue of the last two equalities, we can rewrite equality (**) as follows:

OA 2 _ a 2 OA ~ b 2 "

We can find other relations by analogous reasoning.

2. Consider the problem of constructing the mean proportional to two given segments, i.e. segment -jab.

Problem 6.48. Construct the mean proportional of the segments a and b. Solution

On one straight line, we lay down successively the segments AC = a and SW = b(Fig. 6.53)

Rice. 6.53

On the segment AB how to construct a circle ofiC, 1 on the diameter.

At point C, we restore the perpendicular to the line AB.

We have NC = -jab. Really, AANB- rectangular.

According to the well-known theorem AACN similar ANCB, which means where

NC 2 = AC CB, or in other notation NC 2=ab. Finally we have NC-Jab.

3. When solving construction problems, it is very often necessary to build a segment that is the fourth proportional to three given segments. Let's consider a solution to this problem.

Problem 6.49. Given three segments a, b, s. Build such a line X,

a c what - = -.

Solution

Take any angle O. On one side of the angle, set aside the segments OA = a and OS = s, and on the other - a segment OB-b(Fig. 6.54)

Draw a line through point C R || AB. She will cross the beam OV at the point D. Let's prove that OD- desired segment X. triangles OAB

and OCD are similar. Therefore, i.e. OD = x.

Rice. 6.54

In a particular case, this problem allows us to divide a segment into P equal parts. Let's denote this segment as b. Take any segment With, let it go a - ps(Fig. 6.55).

Rice. 6.55

„ ac b b 1 .

Since - = -, then x =- with = - c = - b. oh a ps p

4. Consider a more complex ratio of segments.

Problem 6.50. Construct a segment given by the ratio 2 [a: 2/b, where p e N.

Solution

Let us assume that the ratio of quantities is given in the form [a: -Jb, where a and Kommersant- segment data.

To determine those two segments, the ratio of which is equal to Va: Vb, we proceed as follows.

On an arbitrary straight line from a selected point To Let's put aside two segments in succession: KN-a and NM = b(Fig. 6.56)

Rice. 6.56

On the segment KM, as on the diameter, construct a semicircle KRM.

At the point N let's restore the perpendicular NN" to the segment KM. Straight NN" crosses the arc KRM at some point L.

Connecting the dot LsKim. Segments KL and LM- desired, i.e.

Indeed, we have -=--. But A KLM similar to A LMN, but-

KL LN KL 2 LN 2

this-=-and, therefore, -=--, but from the last equality

LM NM LM NM 2

KN LN 2 KL 2 KN „

and equality-=-- it follows that-- =-. Extracting square

NM NM 2 LM 2 NM

the root of both parts of the last equality, we find:

To get two segments whose ratio is [a: yfb, you must first construct such two segments type of, the ratio of

ryh is determined by the equality - = -j=, and then through the same

construction find segments R and q, which are defined by the equality p_yfm h Vn

Similar constructions can be used to find segments whose ratio is equal to 2 fa: 2 yх 2 + h h 2, then from equality (*) we obtain

Construction. 1. We build a segment y \u003d yj (2h b) 2 -h a 2(Fig. 6.61).

Rice. 6.61

2. Building x = ^^-(Fig. 6.62).

Rice. 6.62

3. Finally, we build the desired isosceles triangle ABC by reason AC \u003d 2hi height DB = hb(Fig. 6.63).

Rice. 6.63

Proof. It is necessary to prove that in the constructed isosceles triangle ABC heights BD-hb and AE- h a . The first equality is obvious, and the validity of the second follows from the reversibility of all the formulas given in the analysis. _

Study. We notice that the segment y \u003d yl (.2h b) 2 -h 2 can be constructed only if (2/i b) 2 -h a 2>0, or 2 h b > h a .

Under this condition, it is possible to construct a segment x and, consequently, the required triangle ABC. Since two isosceles triangles with equal bases and equal heights are equal, the problem has a unique solution.

Comment. The problem admits a simpler solution in another way. If through a dot D draw a line parallel to the height AE and the crossing side Sun at the point F, then a triangle DFB can be built on the leg 0.5 h a and hypotenuse h b , which will lead to the construction of the desired triangle.

Problem 6.57. Through a given point outside the circle BUT draw a secant that would be divided by this circle in this respect.

Solution

Analysis. Let's say that the problem is solved: secant AL satisfies the condition of the problem (Fig. 6.64). Draw a secant from point A AU, passing through the center O of the given circle. Since the point A is given to us, it means that we know the segments AD and AS. Denote by the letter x the length of the segment AK. If we draw secants from point A, which is outside the circle, then the product of the entire secant by its outer part is a constant value, and therefore

Rice. 6.64

From the drawing we see that AL = x + LK.

l shs.. ph

And since by condition x: LK=m : P, those. bk =- that means AL=x+ - -

= -(t + p). t

Therefore, the equality (*) will take the following form: x-(m + n) = AD? AU, where

Construction. 1. Based on the formula (**), by a well-known construction, we determine the segment x.

• 2. From a point BUT make a notch on this circle To with a radius equal to the found x.
• 3. Connecting the dots BUT and To and continuing this line, we get the desired secant.

Note that we have not given the reasoning that takes place in solving this problem at the stages of proof and research (we leave the reader to carry out these stages on his own).

Problem 6.58. Find a point outside the given circle such that the tangent drawn from it to this circle is half the secant drawn from the same point through the center.

Solution

Analysis(Fig. 6.65). We denote by the letter x the distance to the desired point from the center O of the circle. As is known, AB 2 -DA? AC(1) but DA = x - z (2), AC = x + z(3) and, therefore, AB 2\u003d (x - g) (x + g) \u003d x 2 - g 2 and AB \u003d 1x 2 -g 2 (4).

Rice. 6.65

Since according to the condition AC \u003d 2AB, then from formulas (3) and (4) we have x + r - \u003d 21x 2 - r 2, whence x 2 + 2rx + r 2 \u003d 4x 2 - 4r 2, or Zx 2 - 2rx - 5r 2 \u003d 0. Therefore,

those. x a \u003d - g and x 2 \u003d - g.

In this problem, x cannot be a negative value, and therefore we discard the second root.

Construction. Let's continue one of the diameters (CD) given circle

and on it we set aside from the point D line segment D.A., equal -g (DA = AO - OD = 5 2 3

Г - г = -г (6)).

Dot BUT- desired.

Proof. AC = x + z = -z + z, those. ac=-g (7).

.- /2 8 4 AC

From formulas (1), (6), (7) we find: AB=y/DA-AC=J-r--r=-r=

which confirms the correctness of the construction made (we invite the reader to carry out the research stage independently).

1. General remarks on the solution of problems by the algebraic method.

2. Tasks for movement.

3. Tasks for work.

4. Tasks for mixtures and percentages.

Using the algebraic method to find an arithmetic way to solve text problems.

1. When solving problems by the algebraic method, the desired quantities or other quantities, knowing which it is possible to determine the desired ones, are denoted by letters (usually x, y,z). All independent relationships between data and unknown quantities, which are either directly formulated in the condition (in verbal form), or follow from the meaning of the problem (for example, the physical laws that the quantities under consideration obey), or follow from the condition and some reasoning, are written in form of equality of inequalities. In the general case, these relations form a certain mixed system. In special cases, this system may not contain inequalities or equations, or it may consist of only one equation or inequality.

The solution of problems by the algebraic method is not subject to any single, sufficiently universal scheme. Therefore, any indication relating to all tasks is of the most general nature. The tasks that arise in solving practical and theoretical issues have their own individual characteristics. Therefore, their study and solution are of the most diverse nature.

Let us dwell on solving problems whose mathematical model is given by an equation with one unknown.

Recall that the activity for solving the problem consists of four stages. Work at the first stage (analysis of the content of the problem) does not depend on the chosen solution method and has no fundamental differences. At the second stage (when searching for a way to solve the problem and drawing up a plan for solving it), in the case of using the algebraic method of solution, the following are carried out: the choice of the main relation for compiling the equation; the choice of the unknown and the introduction of a designation for it; expression of the quantities included in the main ratio, through the unknown and the data. The third stage (implementation of the plan for solving the problem) involves the compilation of an equation and its solution. The fourth stage (checking the solution of the problem) is carried out in the standard way.

Usually when writing equations with one unknown X adhere to the following two rules.

rule I . One of these quantities is expressed in terms of the unknown X and other data (that is, an equation is drawn up in which one part contains a given value, and the other contains the same value, expressed by X and other given quantities).

rule II . For the same quantity, two algebraic expressions are compiled, which are then equated to each other.

Outwardly, it seems that the first rule is simpler than the second.

In the first case, it is always required to compose one algebraic expression, and in the second, two. However, there are often problems in which it is more convenient to make two algebraic expressions for the same quantity than to choose an already known one and make one expression for it.

The process of solving text problems in an algebraic way is performed according to the following algorithm:

1. First, choose the ratio on the basis of which the equation will be drawn up. If the problem contains more than two ratios, then the ratio that establishes some connection between all unknowns should be taken as the basis for compiling the equation.

Then the unknown is chosen, which is denoted by the corresponding letter.

All unknown quantities included in the ratio chosen for compiling the equation must be expressed in terms of the chosen unknown, based on the rest of the ratios included in the problem, except for the main one.

4. From these three operations, the compilation of an equation directly follows as the design of a verbal record with the help of mathematical symbols.

The central place among the listed operations is occupied by the choice of the main relation for compiling equations. The considered examples show that the choice of the main ratio is decisive in the formulation of equations, introduces logical harmony into the sometimes vague verbal text of the problem, gives confidence in the orientation and protects against chaotic actions for expressing all the quantities included in the problem through the data and the desired ones.

The algebraic method of solving problems is of great practical importance. With its help, they solve a wide variety of tasks from the field of technology, agriculture, and everyday life. Already in high school, equations are used by students in the study of physics, chemistry, and astronomy. Where arithmetic turns out to be powerless or, at best, requires extremely cumbersome reasoning, there the algebraic method leads easily and quickly to the answer. And even in the so-called "typical" arithmetic problems, relatively easily solved by arithmetic, the algebraic solution, as a rule, is both shorter and more natural.

The algebraic method of solving problems makes it easy to show that some problems that differ from each other only in the plot have not only the same relationships between the data and the desired values, but also lead to typical reasoning through which these relationships are established. Such problems give only different specific interpretations of the same mathematical reasoning, the same relationships, that is, they have the same mathematical model.

2. The group of tasks for movement includes tasks that talk about three quantities: paths (s), speed ( v) and time ( t). As a rule, they are talking about uniform rectilinear motion, when the speed is constant in magnitude and direction. In this case, all three quantities are related by the following relation: S = vt. For example, if the speed of a cyclist is 12 km/h, then in 1.5 hours he will travel 12 km/h  1.5 h = 18 km. There are problems in which uniformly accelerated rectilinear motion is considered, that is, motion with constant acceleration (a). Distance traveled s in this case is calculated by the formula: S = v 0 t + at 2 /2, where v 0 – initial speed. So, for 10 s of falling with an initial speed of 5 m / s and a free fall acceleration of 9.8 m 2 / s, the body will fly a distance equal to 5 m / s  10 s + 9.8 m 2 / s  10 2 s 2 / 2 = 50 m + 490 m = 540 m.

As already noted, in the course of solving text problems and, first of all, in problems related to movement, it is very useful to make an illustrative drawing (to build an auxiliary graphical model of the problem). The drawing should be done in such a way that it shows the dynamics of movement with all meetings, stops and turns. A well-designed drawing allows not only a deeper understanding of the content of the problem, but also facilitates the compilation of equations and inequalities. Examples of such drawings will be given below.

The following conventions are usually adopted in motion problems.

Unless specifically stated in the task, the movement in individual sections is considered uniform (whether it is movement in a straight line or in a circle).

Turns of moving bodies are considered instantaneous, that is, they occur without spending time; the speed also changes instantly.

This group of tasks, in turn, can be divided into tasks in which the movements of bodies are considered: 1) towards each other; 2) in one direction ("after"); 3) in opposite directions; 4) along a closed trajectory; 5) along the river.

If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 a), then when the bodies move towards each other, the time after which they will meet is equal to S/(v 1 + v 2).

2. If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 b), then when the bodies move in one direction ( v 1 > v 2) the time after which the first body overtakes the second is S/(v 1 – v 2).

3. If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 in), then, having set off simultaneously in opposite directions, the bodies will be in time t be at a distance S 1 = S + (v 1 + v 2 ) t.

Rice. 16

4. If the bodies move in one direction along a closed trajectory of length s with speeds v 1 and v 2 , the time after which the bodies will meet again (one body will overtake the other), leaving simultaneously from one point, is found by the formula t = S/(v 1 – v 2) provided that v 1 > v 2 .

This follows from the fact that with a simultaneous start along a closed trajectory in one direction, a body with a higher speed begins to catch up with a body with a lower speed. The first time it catches up with him, having traveled a distance of S more than another body. If it overtakes him for the second, third time, and so on, this means that it travels a distance of 2 S, by 3 S and so on more than another body.

If the bodies move in different directions along a closed path of length S with speeds v 1 and v 2 , the time after which they will meet, having departed simultaneously from one point, is found by the formula t = v(v 1 + v 2). In this case, immediately after the start of motion, a situation arises when the bodies begin to move towards each other.

5. If the body moves along the river, then its speed relative to the shore and is the sum of the speed of the body in still water v and the speed of the river w: and =v + w. If a body moves against the current of a river, then its speed is and =v – w. For example, if the speed of the boat v\u003d 12 km / h, and the speed of the river w \u003d 3 km / h, then in 3 hours the boat will sail along the river (12 km / h + 3 km / h)  3 hours = 45 km, and against the current - (12 km / h - 3 km / h)  3 hours = 27 km. It is believed that the speed of objects with zero speed in still water (raft, log, etc.) is equal to the speed of the river.

Let's look at a few examples.

Example.From one point in one direction every 20 min. cars are leaving. The second car travels at a speed of 60 km/h, and the speed of the first is 50% more than the speed of the second. Find the speed of the third car if it is known that it overtook the first car 5.5 hours later than the second.

Solution. Let x km/h be the speed of the third car. The speed of the first car is 50% greater than the speed of the second, so it is equal to

When moving in one direction, the meeting time is found as the ratio of the distance between objects to the difference in their speeds. First car in 40 min. (2/3 h) travels 90  (2/3) = 60 km. Therefore, the third one will overtake him (they will meet) in 60/( X– 90) hours. Second in 20 min. (1/3 h) travels 60  (1/3) = 20 km. This means that the third one will catch up with him (they will meet) in 20/( X- 60) hours (Fig. 17).

P
about the condition of the problem

Rice. 17

After simple transformations, we obtain a quadratic equation 11x 2 - 1730x + 63000 = 0, solving which we find

The check shows that the second root does not satisfy the condition of the problem, since in this case the third car will not catch up with other cars. Answer: The speed of the third car is 100 km/h.

Example The motor ship passed 96 km along the river, returned back and stood for some time under loading, spending 32 hours for all. The speed of the river is 2 km / h. Determine the speed of the ship in still water if the loading time is 37.5% of the time spent on the whole round trip.

Solution. Let x km/h be the speed of the ship in still water. Then ( X+ 2) km/h - its speed downstream; (X - 2) km/h - against the current; 96/( X+ 2) hours - the time of movement with the flow; 96/( X- 2) hours - the time of movement against the current. Since 37.5% of the total time the ship was under loading, the net time of movement is 62.5%  32/100% = 20 (hours). Therefore, according to the condition of the problem, we have the equation:

Transforming it, we get: 24( X – 2 + X + 2) = 5(X + 2)(X – 2) => 5X 2 – 4X– 20 = 0. Having solved the quadratic equation, we find: X 1 = 10; X 2 = -0.4. The second root does not satisfy the condition of the problem.

Answer: 10 km/h is the speed of the ship in still water.

Example. Car drove way out of town BUT to city C through city AT Without stops. Distance AB, equal to 120 km, he traveled at a constant speed 1 hour faster than the distance sun, equal to 90 km. Determine the average speed of the car from the city BUT to city C, if it is known that the speed on the section AB 30 km/h more speed on the site Sun.

Solution. Let X km / h - the speed of the car on the site Sun.

Then ( X+ 30) km/h – speed on the section AB, 120/(X+ 30) h, 90/ X h is the time the car travels AB and Sun respectively.

Therefore, according to the condition of the problem, we have the equation:

.

Let's transform it:

120X+ 1(X + 30)X = 90(X + 30) => X 2 + 60X – 2700 = 0.

Solving the quadratic equation, we find: X 1 = 30, X 2 = -90. The second root does not satisfy the condition of the problem. So the speed in the section Sun equal to 30 km/h, on the section AB - 60 km/h It follows that the distance AB the car traveled in 2 hours (120 km: 60 km/h = 2 hours), and the distance Sun - in 3 hours (90 km: 30 km/h = 3 hours), so the whole distance AC he traveled in 5 hours (3 hours + 2 hours = 5 hours). Then the average speed of movement on the site AU, the length of which is 210 km, is equal to 210 km: 5 hours \u003d 42 km / h.

Answer: 42 km / h - the average speed of the car on the site AS.

The group of tasks for work includes tasks that talk about three quantities: work BUT, time t, during which work is performed, productivity R - work done per unit of time. These three quantities are related by the equation BUT = Rt. Tasks for work also include tasks related to filling and emptying tanks (vessels, tanks, pools, etc.) using pipes, pumps and other devices. In this case, the volume of pumped water is considered as the work done.

Tasks for work, generally speaking, can be attributed to the group of tasks for movement, since in tasks of this type it can be considered that all work or the total volume of the reservoir plays the role of distance, and the productivity of objects that do work is similar to the speed of movement. However, according to the plot, these tasks naturally differ, and some tasks for work have their own specific methods of solving. So, in those tasks in which the amount of work performed is not specified, all work is taken as a unit.

Example. Two teams had to complete the order in 12 days. After 8 days of joint work, the first team received another task, so the second team finished the order for another 7 days. In how many days could each of the teams complete the order, working separately?

Solution. Let the first brigade complete the task for X days, the second brigade - for y days. Let's take all the work as a unit. Then 1/ X - productivity of the first brigade, a 1/ y – second. Since two teams must complete the order in 12 days, we get the first equation 12(1/ X + 1/at) = 1.

From the second condition it follows that the second team worked 15 days, and the first - only 8 days. So the second equation is:

8/X+ 15/at= 1.

Thus, we have a system:

Subtracting the first equation from the second equation, we get:

21/y = 1 => y= 21.

Then 12/ X + 12/21 = 1 => 12/X – = 3/7 => x = 28.

Answer: the first brigade will complete the order in 28 days, the second in 21 days.

Example. Worker BUT and working AT can complete the job in 12 days BUT and working FROM– in 9 days, working AT and working C - in 12 days. How many days will it take them to complete the job, working together?

Solution. Let the worker BUT can do the job for X days, working AT- per at days, working FROM- per z days. Let's take all the work as a unit. Then 1/ x, 1/y and 1/ z – worker productivity A, B and FROM respectively. Using the condition of the problem, we arrive at the following system of equations presented in the table.

Table 1

Having transformed the equations, we have a system of three equations with three unknowns:

Adding the equations of the system term by term, we get:

or

The sum is the joint productivity of the workers, so the time in which they complete all the work will be equal to

Answer: 7.2 days.

Example. Two pipes are laid in the pool - supply and discharge, and through the first pipe the pool is filled for 2 hours longer than through the second pipe the water is poured out of the pool. When the pool was one-third full, both pipes were opened, and the pool turned out to be empty after 8 hours. How many hours can the pool fill through one first pipe and how many hours can a full pool drain through one second pipe?

Solution. Let V m 3 - the volume of the pool, X m 3 / h - the performance of the supply pipe, at m 3 / h - outlet. Then V/ x hours - the time required for the supply pipe to fill the pool, V/ y hours - the time required by the outlet pipe to drain the pool. According to the task V/ x – V/ y = 2.

Since the productivity of the outlet pipe is greater than the productivity of the filling pipe, when both pipes are turned on, the pool will be drained and one third of the pool will dry out in time (V/3)/(y – x), which, according to the condition of the problem, is equal to 8 hours. So, the condition of the problem can be written as a system of two equations with three unknowns:

The task is to find V/ x and V/ y. Let us single out a combination of unknowns in the equations V/ x and V/ y, writing the system as:

Introducing new unknowns V/ x= a and V/ y = b, we get the following system:

Substituting into the second equation the expression a= b + 2, we have an equation for b:

deciding which we find b 1 = 6, b 2 = -eight. The condition of the problem is satisfied by the first root 6, = 6 (p.). From the first equation of the last system we find a= 8 (h), that is, the first pipe fills the pool in 8 hours.

Answer: through the first pipe the pool will be filled in 8 hours, through the second pipe the pool will be drained after 6 hours.

Example. One tractor team has to plow 240 hectares, and the other 35% more than the first. The first brigade, plowing 3 ha less than the second brigade every day, finished work 2 days earlier than the second brigade. How many hectares did each brigade plow daily?

Solution. Let's find 35% of 240 ha: 240 ha  35% / 100% = 84 ha.

Consequently, the second team had to plow 240 ha + 84 ha = 324 ha. Let the first brigade plow daily X ha. Then the second brigade plowed daily ( X+ 3) ha; 240/ X– working hours of the first brigade; 324/( X+ 3) - the time of the second brigade. According to the condition of the problem, the first team finished work 2 days earlier than the second, so we have the equation

which after transformations can be written as follows:

324X – 240X - 720 = 2x 2 + 6x=> 2x 2 - 78x + 720 = 0 => x 2 - 39x + 360 = 0.

Having solved the quadratic equation, we find x 1 \u003d 24, x 2 \u003d 15. This is the norm of the first brigade.

Consequently, the second brigade plowed 27 ha and 18 ha, respectively, per day. Both solutions satisfy the condition of the problem.

Answer: 24 hectares per day were plowed by the first brigade, 27 hectares by the second; 15 hectares per day were plowed by the first brigade, 18 hectares by the second.

Example. In May, two workshops produced 1080 parts. In June, the first shop increased the output of parts by 15%, and the second increased the output of parts by 12%, so both shops produced 1224 parts. How many parts did each shop produce in June?

Solution. Let X parts were made in May by the first workshop, at details - the second. Since 1080 parts were manufactured in May, according to the condition of the problem, we have the equation x + y = 1080.

Find 15% off X:

So, at 0.15 X parts increased the output of the first workshop, therefore, in June it produced x + 0,15 X = 1,15 x details. Similarly, we find that the second shop in June produced 1.12 y details. So the second equation will look like: 1.15 x + 1,12 at= 1224. Thus, we have the system:

from which we find x = 480, y= 600. Consequently, in June the workshops produced 552 parts and 672 parts, respectively.

Answer: the first workshop produced 552 parts, the second - 672 parts.

4. The group of tasks on mixtures and percentages includes tasks in which we are talking about mixing various substances in certain proportions, as well as tasks on percentages.

Tasks for concentration and percentage

Let's clarify some concepts. Let there be a mixture of P various substances (components) BUT 1 BUT 2 , ..., BUT n respectively, the volumes of which are equal V 1 , V 2 , ..., V n . Mix volume V 0 consists of the volumes of pure components: V 0 = V 1 + V 2 + ... + V n .

Volume concentration substances BUT i (i = 1, 2, ..., P) in the mixture is called the quantity c i, calculated by the formula:

Volume percentage of substance A i (i = 1, 2, ..., P) in the mixture is called the quantity p i , calculated by formula R i = With i ,  100%. Concentrations With 1, With 2 , ..., With n, which are dimensionless quantities, are related by the equality With 1 + with 2 + ... + with n = 1, and the relations

show what part of the total volume of the mixture is the volume of the individual components.

If the percentage is known i-th component, then its concentration is found by the formula:

that is Pi – is the concentration i th substance in the mixture, expressed as a percentage. For example, if the percentage of a substance is 70%, then its corresponding concentration is 0.7. Conversely, if the concentration is 0.33, then the percentage is 33%. So the sum R 1 + p 2 + …+ p n = 100%. If concentrations are known With 1 , With 2 , ..., With n components that make up this mixture of volume V 0 , then the corresponding volumes of the components are found by the formulas:

The concepts weight (mass) concentralization components of the mixture and the corresponding percentages. They are defined as the ratio of the weight (mass) of a pure substance BUT i , in the alloy to the weight (mass) of the entire alloy. What concentration, volume or weight, is involved in a particular problem is always clear from its conditions.

There are tasks in which it is necessary to recalculate the volume concentration to the weight concentration or vice versa. In order to do this, it is necessary to know the density (specific gravity) of the components that make up the solution or alloy. Consider, for example, a two-component mixture with volume concentrations of the components With 1 and With 2 (With 1 + with 2 = 1) and the specific gravity of the components d 1 and d 2 . The mass of the mixture can be found by the formula:

wherein V 1 and V 2 – the volumes of the components that make up the mixture. The weight concentrations of the components are found from the equalities:

which determine the relationship of these quantities with volumetric concentrations.

As a rule, in the texts of such problems one and the same repeated condition occurs: from two or more mixtures containing components A 1 , A 2 , BUT 3 , ..., BUT n , a new mixture is compiled by mixing the original mixtures, taken in a certain proportion. In this case, it is required to find in what ratio the components BUT 1, BUT 2 , BUT 3 , ..., BUT n enter the resulting mixture. To solve this problem, it is convenient to introduce into consideration the volume or weight amount of each mixture, as well as the concentrations of its constituent components BUT 1, BUT 2 , BUT 3 , ..., BUT n . With the help of concentrations, it is necessary to “split” each mixture into separate components, and then, in the manner specified in the condition of the problem, compose a new mixture. In this case, it is easy to calculate how much of each component is included in the resulting mixture, as well as the total amount of this mixture. After that, the concentrations of the components are determined BUT 1, BUT 2 , BUT 3 , ..., BUT n in the new mix.

Example.There are two pieces of copper-zinc alloy with copper percentage of 80% and 30% respectively. In what ratio should these alloys be taken in order, by melting the pieces taken together, to obtain an alloy containing 60% copper?

Solution. Let the first alloy be taken X kg, and the second - at kg. By condition, the concentration of copper in the first alloy is 80/100 = 0.8, in the second - 30/100 = 0.3 (it is clear that we are talking about weight concentrations), which means that in the first alloy 0.8 X kg of copper and (1 - 0.8) X = 0,2X kg of zinc, in the second - 0.3 at kg of copper and (1 - 0.3) y = 0,7at kg of zinc. The amount of copper in the resulting alloy is (0.8  X + 0,3  y) kg, and the mass of this alloy will be (x + y) kg. Therefore, the new concentration of copper in the alloy, according to the definition, is equal to

According to the condition of the problem, this concentration should be equal to 0.6. Therefore, we get the equation:

This equation contains two unknowns X and y. However, according to the condition of the problem, it is not required to determine the quantities themselves X and y, but only their attitude. After simple transformations, we get

Answer: alloys must be taken in a ratio of 3: 2.

Example.There are two solutions of sulfuric acid in water: the first is 40%, the second is 60%. These two solutions were mixed, after which 5 kg of pure water was added and a 20% solution was obtained. If, instead of 5 kg of pure water, 5 kg of an 80% solution were added, then a 70% solution would be obtained. How many were 40% and 60% solutions?

Solution. Let X kg is the mass of the first solution, at kg - the second. Then the mass of a 20% solution ( X + at+ 5) kg. Since in X kg 40% solution contains 0.4 X kg of acid at kg of 60% solution contains 0.6 y kg of acid, and (x + y + 5) kg of 20% solution contains 0.2( X + y + 5) kg of acid, then by condition we have the first equation 0.4 X + 0,6y = 0,2(X +y + 5).

If, instead of 5 kg of water, add 5 kg of an 80% solution, you get a solution with a mass (x + y+ 5) kg, in which there will be (0.4 X + 0,6at+ 0.8  5) kg of acid, which will be 70% of (x + y+ 5) kg.

The main methods for solving geometric problems: geometric - the required statement is derived using logical reasoning from a number of well-known theorems; algebraic - the desired geometric value is calculated on the basis of various dependencies between elements geometric shapes directly or with the help of equations; combined - at some stages the solution is carried out by a geometric method, and at others by an algebraic one.

Triangles Signs of equality of triangles, right triangles. Properties and signs of an isosceles triangle. Problem 1. The median AM of the triangle ABC is equal to the segment VM. Prove that one of the angles of triangle ABC is equal to the sum of the other two angles. Problem 2. Segments AB and CD intersect at their common midpoint O. Points K 1 are marked on AC and BD such that AK=BK 1. Prove that a) OK=OK 1, b) point O lies on the line KK 1. Problem 3 (sign of an isosceles triangle). If the bisector of a triangle is the median, then the triangle is isosceles.

Task 4 (sign of a right-angled triangle by the median). Prove that if the median of a triangle is equal to half of the side to which it is drawn, then the triangle is right-angled. Problem 5 (property of the median of a right triangle). Prove that in a right triangle the median drawn to the hypotenuse is equal to half of it. Problem 6. Prove that in a right triangle with unequal legs the bisector right angle bisects the angle between height and median drawn from the same vertex. Problem 7. The median and height of a triangle drawn from one vertex divide this angle into three equal parts. Prove that the triangle is a right triangle.

Area properties. Areas of polygons Corollary from the triangle area theorem. If the heights of two triangles are equal, then their areas are related as bases. Theorem on the ratio of the areas of triangles with equal angles. If the angle of one triangle is equal to the angle of another triangle, then the areas of these triangles are related as the products of the sides containing equal angles.

Cevian intersection point theorems Theorem. In any triangle, the medians intersect at one point (centroid, center of gravity) and are divided by this point in a ratio of 2: 1, counting from the top. Properties of the median: 1. The median divides the triangle into two equal ones, that is, having the same area. 2. Three medians divide the triangle into six equal ones. 3. The segments connecting the centroid with the vertices of the triangle divide the triangle into three equal parts.

One of the main methods for solving problems in which the medians of a triangle are involved is the "doubling the median" method. Complete the triangle to a parallelogram and use the theorem on the sum of the squares of its diagonals. Problem 8. Find the ratio of the sum of squares of the medians of a triangle to the sum of the squares of all its sides.

Bisector property of an interior angle of a triangle. The bisector of the interior angle of a triangle divides the opposite side into parts proportional to its enclosing sides. Theorem. In any triangle, the bisectors intersect at one point (and the center), which is the center of the circle inscribed in it. Note: Obviously, the centroid and the center of a triangle always lie inside it.

. Solution. B A 1 1) In triangle ABC AA 1 is the bisector of angle A, therefore AB: AC = BA 1: CA 1 = BA 1: (BC - BA 1) I or C А B 1 2) In triangle ABA 1 BI is the bisector of angle B, so AI: IA 1 = BA: BA 1 or

Theorem on the perpendicular bisector to a segment. Each point of the perpendicular bisector to a segment is equidistant from the ends of this segment. Conversely, each point equidistant from the ends of the segment lies on the perpendicular bisector to it. Theorem. The perpendicular bisectors to the sides of a triangle intersect at one point, which is the center of the circle circumscribed about it. Theorem. In any triangle, the heights intersect at one point (orthocenter of the triangle). Question. Where is the orthocenter of acute, right, obtuse triangles?

Solution. B 1) Triangle BC 1 H is right-angled, and C 1 H 2) Triangle BC 1 C is right-angled, and A B 1 C

Using cast formulas. Where is the note. If one of the angles is obtuse, then in (*) the corresponding cosine must be taken modulo.

Interesting are the problems of finding the distance from an arbitrary vertex of a triangle to one of its remarkable points. First, we solve the problem of finding the distance from the vertex to the orthocenter. Problem 11. Heights BB 1 and CC 1 are omitted in the triangle ABC. Find the length of the segment HB, where H is the intersection point of the heights. B 1) triangle BC 1 Н is right-angled, and Solution. C 1 H 2) triangle BC 1 C is a right triangle, and A B 1 C

Problem 12. Find the distance from vertex B of triangle ABC to the orthocenter, if Solution. By the law of cosines Then

Problem 13. At angles A and B of triangle ABC (A

Problem 14. Which of the vertices of the triangle is closer to the center? Solution. C D I A Let I be the center, the point of intersection of the bisectors of the triangle ABC. Let us use the fact that the larger angle lies against the larger side of the triangle. If AB > BC then A

Problem 15. Which of the heights of the triangle is the smallest? Solution. C B 1 A 1 H A C 1 Let H be the intersection point of the altitudes of triangle ABC. If AC B. A circle with diameter BC passes through the points C 1 and B 1. B Considering that the smaller of the two chords is the one on which the smaller inscribed angle rests, we get that CC 1

Problem 16. The segment AH is the height of the triangle ABC. From vertices B and C, perpendiculars BB 1 and CC 1 are drawn to the line passing through point A. Prove that triangles ABC and HB 1 C 1 are similar. Find the area of ​​the triangle HB 1 C 1 if the area of ​​the triangle ABC is equal to S, and AC: HC 1 =5: 3. Proof. Since the triangles ANS and ACC 1 are rectangular, the points H and C 1 A lie on a circle with diameter AC. C 1 B B 1 H Similarly, points B 1 and H lie on a circle with diameter AB. With triangle ACC 1

Hence, Since (1) and (2) hold, And then the triangles ABC and HB 1 C 1 are similar. C 1 Similarity coefficient B B 1 H C means

Problem 17. Let the points A 1, B 1, C 1 in an acute-angled triangle ABC be the bases of the heights. Prove that the point H - the intersection of the heights of triangle ABC is the intersection point of the bisectors of the triangle A 1 B 1 C 1. Solution. On sides AC and BC B of triangle ABC, as on C 1 A diameters, we construct circles. H Points A 1, B 1, C 1 belong to these circles. 1 A B 1 C Therefore B 1 C 1 C = B 1 BC, as angles based on the same circular arc. B 1 BC = CAA 1, as angles with mutually perpendicular sides.

CAA 1 = CC 1 A 1 as angles based on the same circular arc. Therefore, B 1 C 1 C = CC 1 A 1, i.e. C 1 C is the bisector of the angle B 1 C 1 A 1. Similarly, it is shown that AA 1 and BB 1 are the bisectors of the angles B 1 A 1 C 1 and A 1 B 1 C 1. B C 1 A 1 H A B 1 C Independently investigate the cases of a right and obtuse triangle.

Integration of algebraic and geometric methods in problem solving

One of the urgent problems of school mathematical education at the present stage is the problem of the integration of mathematical knowledge, the formation of holistic ideas of students about mathematics as a science. The solution of this problem is especially important for the basic school, where two mathematical disciplines are studied: algebra and geometry.

The concept of "integration" [lat. integratio - restoration, replenishment; integer - whole] is interpreted as restoration, unification into a whole of any parts, elements; as a state of connectedness into a whole of separate differentiated parts, and also as a process leading to such a state. In education, integration is often understood as mutual influence, interpenetration and interconnection of the content of various academic disciplines.

Since the main activity of students in teaching mathematics is problem solving, it is advisable to integrate algebra and geometry along the line of their methods. The algebraic method (in relation to elementary mathematics) is interpreted as a method consisting in the use of letters and literal expressions, which are transformed according to certain rules. It is also called the method of literal calculations.

The geometric method is characterized as a method that comes from visual representations. The essential features of this concept are geometric (visual) representations and the laws of geometry, which reflect the properties of geometric shapes.

If we take the system of knowledge on which the method is based as the basis for the classification of algebraic and geometric methods, then we get the following methods.

1. Algebraic: method of identical transformations; method of equations and inequalities; functional method; vector method; coordinate method.

2. Geometric(we confine ourselves to planimetry): length method; triangle method; method of parallel lines; method of relations between the sides and angles of a triangle; method of quadrilaterals; area method; triangle similarity method; trigonometric method (a method based on the ratios between the sides and angles of a triangle, expressed through trigonometric functions); method of geometric transformations; graphical method (although this method is studied in the course of algebra, it is based on the use of geometric representations of functions and the laws of geometry associated with them).

We will assume that each method consists of certain techniques, and each technique consists of actions. Under the integration of algebraic and geometric methods, we mean the process of combining these methods or linking their techniques into one method.

In the field of teaching problem solving, the integration of methods involves the parallel (in one lesson) solution of a problem by different methods (algebraic and geometric) or the solution of an algebraic problem by a geometric method, and a geometric problem by an algebraic method. The means of integration can be special blocks of tasks, which include both algebraic and geometric problems. Let's give examples.

7th grade

Here you can use text problems from the algebra course and geometric problems solved by the method of equations.

Task 1. One elevator had twice as much grain as the other. 750 tons of grain were taken out of the first elevator, 350 tons were brought to the second elevator, after which the grain became equal in both elevators. How much grain was originally in each elevator?

To solve this problem, we use the method of equations and inequalities from algebra and the method of lengths from geometry, based on the properties of the length of a segment.

Algebraic Method. Let x tons of grain were originally in the second elevator, then 2x tons of grain were originally in the first elevator; (2x – 750) tons of grain remained in the first elevator, and (x + 350) tons of grain became in the second elevator. Since the grain in both elevators became equal, we can make an equation

2x - 750 = x + 350, hence x = 1100, 2x = 2 1100 = 2200.

Answer: 2200 tons of grain was in the first elevator and 1100 tons - in the second.

geometric method. We solve this problem using a line chart. A line chart is usually a segment or several segments, the lengths of which correspond to the numerical values ​​of the quantity in question. We solve the problem in stages.

1st stage. Construction of a line diagram. After reading the text of the problem, the students discuss the following questions (the teacher's help is possible).

1. How many situations are considered in the problem?

[Two: initial and final.]

2. From what situation should you start building a line chart?

[You can start building from the first situation and move from it to the second, or you can
first build a line diagram of the final situation and go from it to
initial. Consider the first option for constructing a line chart.]

3. What is the line diagram of the initial situation?

[Two segments, one of which is twice as long as the other. The first segment represents
the amount of grain in the first elevator, and the second - in the second elevator.]

After that, students build a diagram of the initial situation. Then the discussion continues.

4. How to move on the diagram from the first situation to the second?

[It is necessary to subtract from the first segment the segment that conventionally depicts 750 tons, and
add a segment representing 350 tons to the second segment.]

5. Are these segments taken arbitrarily?

[No, it should be taken into account that the newly obtained segments must
to be equal, since the grain in both elevators became equal.]

After performing actions with segments, students receive a diagram of the final situation. The first stage of work on the task ends with the designation of segments and the design of entries on the drawing.

2nd stage. Solution of the resulting geometric problem. The constructed line diagram turns an algebraic problem into a geometric one, the solution of which is based on using the properties of the segment length, namely:

1) equal segments have equal lengths; a smaller segment has a shorter length;
2) if a point divides a segment into two segments, then the length of the entire segment is equal to the sum of the lengths of these two segments.

Students write the solution in a geometric language, using the notation of segments, and the result is translated into natural language. In this case, this translation is carried out automatically due to the transfer of terminology (3rd stage). First, a detailed record of the solution should be made, indicating what each segment represents. Gradually, you can move on to a short note, as some facts are visible in the figure.

Let us give a detailed record of the solution of Problem 1.

Solution. 1st stage. Let segment AB represent the amount of grain in the first elevator (Fig. 1), then the segment will represent the amount of grain in the second elevator.

AB = 2CD - initial distribution of grain between elevators. 750 tons of grain were taken out of the first elevator, and 350 tons were brought to the second elevator, so we subtract from segment AB segment BK, conditionally representing 750 tons, and add segment DE, representing 350 tons, to segment CD.

2nd stage. Method I. CD = AF = FB (by construction),

FB = FK + KB = 350 + 750 = 1100, so CD = 1100, AB = 1100 2 = 2200.

3rd stage. Answer: in the first elevator there were 2200 tons of grain, in the second 1100 tons.

Students can make a short note of the solution to the problem, for example, it could be like this.

Solution. AB = 2CD - initial distribution of grain between two elevators; BK=750, DE=350.

AK = CE - final distribution of grain between elevators.

CD = AF = FB (by construction), FB = 350 + 750 = 1100, then

CD = 1100, AB = 1100 2 = 2200.

Answer: 2200 tons, 1100 tons.

The line diagram allows you to make various equations for the problem that students cannot write down without a drawing, that is, it becomes possible to solve the problem in algebraic ways in different ways. Let's take a look at some of them.

Method II. Let AK = CE = x, then, since AB = 2CD, we get x + 750 = 2(x - 350),

whence x = 1450, CD = 1450 - 350 = 1100, AB = 1100 2 = 2200.

Answer: 2200 tons, 1100 tons.

Method III. Let CD = x, then AB = 2x. Since AK = CE, we have 2x - 750 = x + 350

(The same equation is obtained when solving the problem without a diagram.)

A line chart allows not only to solve a problem without an equation, but often the answer can be "seen" right on the drawing.

Task 2. One garden plot has five times as many raspberry bushes as another. After 22 bushes were transplanted from the first section to the second, then in both sections of the raspberry bushes it became equally divided. How many raspberry bushes were on each plot?

Solution. 1st stage. Let segment AB represent the number of raspberry bushes in the first area, and segment CD the number of raspberry bushes in the second area (Fig. 2). AB and 5CD - initial distribution of raspberry bushes between plots.

Since both sections of the raspberry bushes became even, we divide the segment BE in half (BF = FE) and subtract the segment BF from the segment AB, and add the segment DK to the segment CD (DK = BF). AF = CK - final distribution of raspberry bushes between plots.

2nd stage. According to the condition, 22 bushes were transplanted from the first site to the second, which means BF = 22 = 2CD, then CD = 11, AB = 5CD = 5 11 = 55.

Answer: on the first site there were 55 raspberry bushes, on the second 11 bushes.

One of the advantages of using the geometric method in solving the problems considered is its clarity. Building a line diagram and moving from one of its states to another allows students to better perceive the situations described in the problem and, therefore, helps to find ways to solve it. Sometimes the answer is almost obvious on the drawing, this makes it possible to use a line diagram to check the solution to the problem, which is performed by an algebraic method without a drawing.

At the motivational stage of the formation of the geometric method, it is advisable to propose solving the problem by two methods: algebraic and geometric. The task should be selected in such a way that its solution using a line diagram is more rational than a solution without a drawing. Let us give an example of solving one of these problems.

Task 3. The first tank contains four times more liquid than the second. When 10 liters of liquid were poured from the first tank into the second, it turned out that what was left in the first became in the second tank. How many liters of liquid were in each tank originally?

Solution. Algebraic Method. We bring to the equation

where x l is the initial amount of liquid in the second tank.

Solving this equation, we find x = 10, then

4x = 4 10 = 40.

So, in the first tank there were 40 liters, and in the second 10 liters.

Geometric method. Let's build a line diagram of the initial distribution of liquid between two tanks. Let the segment AB represent the amount of liquid (l) in the first tank (Fig. 3), then the segment CD will represent the amount of liquid (l) in the second tank (construction can be started from segment CD). AB = 4CD - initial liquid distribution between two tanks.

The process of pouring liquid from one tank to another will be displayed as subtracting a certain segment from segment AB and adding it to segment CD. To find out the length of the segment that should be subtracted from segment AB, it is necessary to note the following: in the first and second tanks there were 5 parts of liquid, and in the first tank there were 4 parts, and in the second 1 part.

After transfusion, the total amount of liquid (5 parts) did not change, but in the second tank it became 2 parts, and in the first 3 parts. This means that the segment BE must be subtracted from the segment AB (BE = CD), and the segment DK (DK = BE) must be added to the segment CD, then , which corresponds to the transfusion of liquid. Therefore BE = 10, then

AB=40, CD=BE=10.

So, in the first tank there were 40 liters of liquid, and in the second 10 liters.

After solving the problem, you should compare both methods of solution with students, identify the advantages and disadvantages of each of them.

It should be noted that with the help of line charts, problems are solved in which the ratios of values ​​\u200b\u200bof quantities are given (less, more, by, in, the same) and one or more situations are considered.

Text problems, in which one of the quantities is the product of the other two, allow us to integrate the area method based on the properties of the area, and the method of equations and inequalities. Let's give examples.

Task 4. The team of lumberjacks daily exceeded the norm by 16 m 3, so it completed the weekly norm (six working days) in four days. How many cubic meters of timber did the brigade harvest per day?

Solution. Algebraic Method. We come to the equation

where x m 3 is the daily norm of the brigade according to the plan.

Geometric method. Since the problem considers the product of two quantities (A = pn), for clarity, we represent it in the form of a two-dimensional diagram. A two-dimensional diagram is the area of ​​one or more rectangles, the sides of which represent the numerical values ​​of the quantities under consideration (p and n), and the area of ​​the rectangle represents their product (S = A).

The solution of the problem, as in the case of a linear (one-dimensional) diagram, takes place in three stages:

1) building a two-dimensional diagram, that is, translating the problem into the language of segments and areas of figures;
2) solving the resulting geometric problem by compiling an equation based on the use of the properties of the area of ​​polygonal figures;
3) translation of the received answer from the geometric language into natural language.

1st stage. It is implemented during the analysis of the task text. Students answer the following questions.

1. Is it possible to build a two-dimensional diagram according to the condition of the problem?

[It is possible, since one of the quantities (the weekly norm of the brigade) is equal to
the product of the other two: the daily norm of the brigade and the number of days.]

2. What is a 2D chart?

[A rectangle one side of which defines
brigade daily allowance, and the other - the number of days.]

3. How many rectangles should be built?

[Two, their areas will determine the weekly rate of the brigade
according to the plan and actually completed work in four days.]

4. What can be said about the areas of these rectangles?

[They are equal, since completed in four
day's work is equal to the week's rate.]

Then the students, with the help of the teacher, complete the construction. The base and height of the first rectangle are taken arbitrarily, the second rectangle is equal to the first one, and their bases are segments lying on the same ray with a common origin (Fig. 4). The first stage ends with the designation of rectangles and the design of entries in the drawing.

At the beginning of learning the geometric method, a detailed record is kept of what the length, width and area of ​​each rectangle means, that is, the task is translated into geometric language.

2nd stage. The stage begins with the consideration of the areas of the resulting rectangles and the establishment of relationships between them (equalities, inequalities). The students are asked the question: name the rectangles with equal areas. The corresponding entry is:

S ABCD \u003d S AMNK \u003d S, S 1 \u003d S 2, since S 1 + S 3 \u003d S 2 + S 3.

Among the students there may be those who will complete the drawing with a large inaccuracy, that is, in the drawing, the BMNE and KECD rectangles will obviously not be equal in size. It should be brought to their attention and note that the lines KB and CN should be parallel.

Using the condition S 1 \u003d S 2, an equation is drawn up. Let us give an approximate notation of the solution of Problem 4 by the geometric method.

Solution. Let S ABCD determine the weekly norm for a team of lumberjacks. AB - productivity (m 3) of the brigade per day according to the plan; AD - number of days; S AMNK - the amount of work done by the team in four days.

SAMNK=SABCD=S;

S 1 \u003d S 2, since S 1 + S 3 \u003d S 2 + S 3.

S 1 \u003d 2KE, S 2 \u003d 16 4 \u003d 64,

means 2KE = 64, then KE = 32.

AB = KE = 32, AM = AB + BM = 32 + 16 = 48.

Answer: the brigade harvested 48 m 3 of forest per day.

With the help of a two-dimensional diagram and geometric relationships, in particular the equal area of ​​the rectangles ABCD and AMNK, another equation can be made. If AB = x, then we get

(the same equation is obtained when solving the problem without a drawing).

Task 5. The plant had to complete the order for the production of cars in 15 days. But already two days before the deadline, the plant not only fulfilled the plan, but also produced six more cars in excess of the plan, since it produced two cars every day in excess of the plan. How many cars was the plant supposed to produce according to the plan?

The peculiarity of solving this problem by the geometric method, in comparison with the solution of the previous problem, is that the areas S 1 and S 2 (see Fig. 4) are not equal, since, according to the condition, the plant not only fulfilled the plan, but also produced in excess of the plan six more cars. Students should keep this in mind both when constructing a drawing and when compiling an equation.

Solution. Let AB represent the production capacity of the plant per day according to the plan (Fig. 5). AD - order completion date according to the plan. Then S ABCD determines the entire order for the production of cars, AM represents the number of cars that the plant produced daily, AP is the lead time, and S AMNP corresponds to the number of cars that the plant produced in 13 days.

According to the condition, the plant produced six cars over the plan, so we have

S 1 + S 3 + 6 \u003d S 3 + S 2 or S 1 + 6 \u003d S 2,

but S 2 = 2 13 = 26, hence S 1 + 6 = 26, whence S 1 = 20. On the other hand, S 1 = 2AB, therefore 2AB = 20, then AB = 10, S ABCD = AB 15 = 10 15 = 150.

Answer: the plant was supposed to produce 150 cars according to the plan.

Geometric problems can also serve as a means of integrating methods in the 7th grade. Let's give examples.

Task 6. Point A divides the segment CD in half, and point B divides it into unequal parts. Prove that the area of ​​a rectangle with dimensions CB and BD is equal to the difference between the areas of squares with sides AD and AB

Solution. Let CD = x, BD = y. Then

Therefore, to solve the problem, it is necessary to prove the identity

As you can see, the method of areas and the method of identical transformations are involved in solving this problem.

Task 7. AP = PQ = QR = RB = BC, AB = AC (Fig. 7). Find angle A.

Solution. Let Р A = x, then Р 1 = Р A = x. P 2 \u003d 2x (as the outer corner of the triangle APQ), P 4 \u003d P 2 \u003d 2x.

P 3 \u003d 180 ° - (P 2 + P 4) \u003d 180 ° - 4x,

P 5 \u003d 180 ° - (P 1 + P 3) \u003d 3x,

P 6 = P 5 = 3x. P 7 = P B - P 6, but

that's why

Since P 8 \u003d P C, then P C + P 8 + P 7 \u003d 2P C + P 7 \u003d 180 °, or

Solving this equation, we get that x = 20°.

Answer: P A \u003d 20 °.

When solving this problem, the method of triangles and the method of equations and inequalities were used. Similar problems are found in geometry textbooks.

Iterative Algebraic Methods for Image Reconstruction

graduate work

4.1 Algebraic method

Let the function f(x) = f(x, y) describe some density distribution in some selected section of the object. The main task of computational tomography is to reconstruct the function f(x) from a set of experimentally obtained projections:

which are linear integrals of the desired distribution along the lines L:. Here is the scanning angle, is the delta function.

In practice, as a rule, projections are not set for all values ​​and, but only for a finite number of them. There are a number of practical problems for which the number of discretizations by 0 is very limited (from 3 to 5). Problems of this type are related to the problems of low-angle tomography and are among the most difficult to solve. The task can be formulated as follows: for a given finite set of projections of a function of two variables, obtain the best estimate of this function.

Let us formulate a general statement of the problem of restoring a solution to problem (4.1) using algebraic methods and construct an iterative algorithm for restoring such problems. The use of algebraic methods is fundamentally different from the method of integral transformations, since it involves the discretization of the image before the start of the restoration algorithm. The construction of a discrete model of the image reconstruction problem can be described as follows.

Let it be required to restore a two-dimensional function f(x)=f(x,y) defined in the region D R2. Assume that the recovery area D is enclosed in a square K, which is divided into n equal small squares called elises. Let us number all elises from 1 to n. At the same time, we will accept the main restriction, which is that the restored function f(x) takes a constant value fj inside the jth eliza, i.e., we replace the function f (x) with the discretized expression

if (x) j-th elise;

otherwise. (4.3)

Assume that we are given a set of linear continuous functionals that represent the direct Radon transform along a set of some lines:

Then is the projection of the function f(x) along the ray Li.

Applying operators to equality (4.2) and taking into account their continuity and linearity, we obtain a system of linear algebraic equations

where, i = 1, ..., m; j = 1, ..., n.

If the family of basis functions (bj) is given by formula (4.3), then

The length of the intersection of the i-th ray with the j-th elise.

We denote the coefficient matrix A=(), the image vector f=(f1, f2, ..., fn), the projection vector R=(R1, R1, ..., Rt). Then the solution of the problem is reduced to solving a system of linear algebraic equations of the form

In this case, the vector R is known with some error.

It should be noted that the form of system (4.5) depends on the specific choice of the system of basis functions bi and the set of functionals Ri. There are other ways to choose the partition grid of the domain D (and, hence, the basis functions bi). The functionals are chosen not only in the form (4.4), but also taking into account the actual length of the rays and using piecewise constant functions. Moreover, the statement of the problem does not depend on the geometry of the rays and is easily formulated for the three-dimensional case.

4.2 Using interlining operators

In this subsection, we consider a new method for representing an approximate solution of the problem of flat computed tomography (CT) in the form of piecewise constant functions. The method has a higher accuracy than the classical method for solving a plane RKT problem using piecewise constant functions.

partitioning E2 into quadrilaterals. Let us introduce the following notation.

The operator О1 is the operator of approximation of f(x,y) by piecewise constant functions in x. If y=const, then it is found from the condition of the best approximation f(x,y) in the band, yE. Similarly, the operator О2 is the operator of approximation of f(x, y) by piecewise constant functions in y.

If x=const, then j(x) is found from the best fit f(x,y) in the band, xE.

We introduce the following operators:

We find the values ​​from the condition of the best approximation of f by the number f(оij, ij) in

Lemma 3.1 Let a function, r=1,2 or and be a function with bounded variation. Then the operators Onm have the properties

Proof. Properties (3.25) and (3.26) follow from the fact that

Property (3.27) follows from the fact that

Properties (3.29) hold for all differentiable functions and for continuous functions with bounded variation.

Lemma 1 is proved.

Corollary 1. For and for continuous functions with bounded variation, we obtain the following error estimate.

Corollary 2. Replacing functions by piecewise constant functions of one variable with the same error estimate

we get the operator

Get values ​​for gi (x)

Get values ​​for Gi (y)

with the following properties:

Corollary 3. Operator

has the following properties:

If, r=1,2 or u is a function with limited variation, then

Proof. For the error, we can write the equality

This implies the inequality

Applying the estimates 3 and 4 to the right side of the obtained expression, we arrive at the estimate (3.42).

Corollary 3 is proved.

If m=n, then the operator has an error (it uses constants); approximation by the operator has an error. That is, the operator (it uses constants) has the same error as the operator:

The following paragraphs highlight the advantages of this method.

Number of unknowns

The use of interlination of functions in the construction of an approximate solution, namely, the representation of an approximate solution in the form:

led to the appearance of 2n3+n2 constants, which are unknown. Hence the operator uses O(n3) constant-unknowns. The operator has an error.

Using the operator - the classical representation of the approximate solution - leads to the appearance of n4 constants, which are unknown. Hence the operator uses O(n4) constant-unknowns. The operator has an error.

Summarizing the above, we conclude that using the operator requires finding O(n3) unknowns, while using the operator requires finding O(n4) unknowns to approximate the solution with the same error.

Therefore, the use of the operator gives significant advantages in terms of the number of arithmetic operations, since in order to achieve the same accuracy, it is necessary to solve a system of linear algebraic equations of a lower dimension.

To illustrate this fact, we present the following table:

Table 1

	Unknown	Unknown	Error

Comparisons show that to achieve the same accuracy, when using the operator, you can take fewer equations. For example, for n=9, the number of unknowns in the classical method is 4 times greater.

Due to the fact that the system must be overdetermined, and for n = 9 unknowns 1539 (for the case with interlination) and 6561 (for the classical method), and the number of equations should be taken more than the number of unknowns, it is clear that in the method with interlination these equations will be smaller.

A computational experiment carried out using the developed algorithms and programs confirmed these statements.

Region discretization

The use of schemes for solving the problem of flat computed tomography, based on the use of and determines the discretization of the area.

For - an irregular grid: a breakdown into squares with a side and rectangles with sides, and, elongated along the axis Ox and Oy, respectively. Grid nodes are located at the centers of squares and rectangles.

For - a regular grid: a breakdown into squares with a side. Grid nodes are located in the centers of the squares.

The positive effect of applying the operator is achieved due to a different arrangement of nodes, which causes a connection between the following relation:

Which coincide with the nodes located in the centers of the corresponding square, vertical and horizontal rectangles.

For these points, since at these centers, then we have exact solutions.

Hence, the approximate solution constructed with the help is an interpolation formula. With its help, the value of the function is calculated at any points of the region D, other than those indicated, in which there is an exact match

Concerning exact match in the specified centers. Means,

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