The proportion of outward bound exercises in "practice one practice" is110. It is the highest level of basic exercise, variant exercise and outward bound exercise. The purpose of writing expansion exercises is to further understand the important content and cultivate the ability of mathematical thinking and problem solving. Completing outward bound training requires rich mathematical literacy, which can be further improved and developed.
In order to promote the development of children's thinking ability in an orderly and effective way and reflect the charm of life mathematics, we might as well expand this math game problem and design it as a special research class, which can be used to lead junior high school students to carry out a training trip to cultivate mathematics literacy, tear up their thinking ability in special research, experience the pleasure of exploring "secrets", feel the comfort brought by mathematical modeling thought and understand the philosophy of life contained in mathematics.
First, from simple to complex, step by step logical reasoning.
Inspire children to think in an orderly way through the following examples, experience the process from simple to complex, and form a gradual reasoning habit.
Question 1: use three numbers, 3, 4 and 5, to form a one-digit number and a two-digit number at will, and find their product. The party with the most products wins.
Children don't need a little time to get the correct answer quickly. The formula of the maximum product is 5×43=2 15.
Question 2: Use the four numbers 2, 3, 4 and 5 to form two random two-digit numbers, and find their product. The party with the most products wins.
Students can focus on learning 53×42 or 52×43. Because children already have the experience of "writing the maximum number with several numbers", they will use it unconsciously. Through calculation, we can get the formula that 52×43=2236 is the maximum product.
Question 3: Use the five numbers 1, 2, 3, 4 and 5 to form a two-digit number and a three-digit number at will, and find their product. The party with the most products wins.
With the first two questions as the basis, children will be psychologically prepared to get the biggest product. On this basis, lead the children to compare the differences between question 3 and question 2. Let the children feel it, too Actually, the third question is whether to add the new number "1" after 52 or 43 on the basis of the second question. Then get the following formula:
52 1×43=22403,
52×43 1=224 12。
Therefore, it is easy to get the formula that 52×43 1=224 12 is the maximum product.
From the surface of the three problems, it follows a process from simple to complex. However, due to the orderliness, gradualness and gradualness of logical reasoning, children have fully experienced the advanced thinking and gained insights in the process of reasoning. Let children get better training in mathematics literacy.
Second, from numbers and words, hit floor is mathematically abstract.
Mathematical language is the smartest language in the world. Representing numbers with letters is a qualitative leap in students' thinking. Letters represent numbers and have a strong sense of symbols. Symbol consciousness is an important manifestation of students' mathematical abstraction, which can express more general mathematical principles and the essence and beauty contained in real life.
So we will refer to 1, 2, 3, 4, 5 mentioned above, and use the letters A, B, C, D, E in the order of 5, 4, 3, 2, 1. Numbers are represented by letters, and the size of the original number represented by each letter has no substantive significance. The rest is the relationship between them. A>b>c>d>e.
At this time, we can use letters to represent these numbers to study the above problems again.
Question 1: It is known that 9 ≥ A > B & gtC≥ 1, how to form a one-digit number and a two-digit number with three numbers of A, B and C, so as to maximize the product.
After the operation, it is not difficult to find that only two combinations have become our focus. That is, the product of two digits of A and C and the product of two digits of B and C and A. In order to explore these two combinations, which combination has the largest product? The difference method can be introduced step by step to compare numbers.
For example, because of 5>3, 5-3 >; 0, represented by letters: If a>b, then A-B > 0。 Conversely, if a-b >; 0, then a> B.
A two-digit number composed of A and C can be expressed as 10a+c, and a two-digit number composed of B and C can be expressed as10B+C. Therefore, we can guide students to list and calculate the following formulas:
( 10b+c)×a-( 10a+c)×b
= 10ab+ac- 10ab-bc
=ac-bc
=(a-b)c
& gt0
Therefore, the product of b and c and a (10b+c)×a is the largest. That is, bc×a is the largest (bc stands for two digits).
Question 2: It is known that 9 ≥ A > B>c & gtD≥ 1, how to use four numbers, A, B, C and D, to form two two-digit numbers to maximize the product.
On the basis of the previous question, there should be two combinations that can be focused on. The formulas are: (10a+c)×( 10b+d) and (10a+d)×( 10b+c). Can guide students to calculate:
( 10a+d)×( 10 b+c)-( 10a+c)×( 10 b+d)
= 100 ab+ 10ac+ 10bd+CD- 100 ab- 10ad- 10bc-CD
= 10a (DC)+10b (DC)
= 10(c-d)(a-b)
& gt0
Therefore, the largest combination of products is:
(10a+d)×( 10b+c), that is, AD× BC (both AD and BC represent a two-digit number).
Question 3: It is known that 9 ≥ A > B>c>d>e & gtF≥ 1, how to form two three-digit numbers with six numbers of A, B, C, D, E and F to maximize the product.
On the basis of the second question, there should be two combinations to focus on. The formulas are: (100a+10d+f) × (10b+10c+e) and (100a+10d+e )× (/) Can guide students to calculate:
( 100 a+ 10d+f)×( 100 b+ 10c+e)-( 100 a+ 10d+e)×( 100 b+ 10c+f)
= 10000 ab+ 1000 AC+ 100 AE+ 1000 BD+ 100 CD+ 10de+ 100 BF+ 10cf+ef- 1000 ab- 100 AC- 100
= 100a (English-French)+10d (English-French)+100b (French-English)+10c (French-English)
= 100 (A-B) (English+French)+10 (D-C) (English)
= 10 (English-French) [10 (A-B)+(D-C)]
& gt0
So the maximum combination of products is: (100a+10d+f) × (100b+10c+e), that is, ADF× BCE (both ADF and BCE are three digits).
Third, by analogy, the modeling idea reached its peak.
For the convenience of expression, recall that we use 6, 5, 4, 3, 2 and 1 to represent a, b, c, d, e and f.
Using three numbers to form a one-digit number and a two-digit number, when the product is maximized, we find that the result is: 6× 54;
Using four numbers to form two two-digit numbers, the product is maximized, and the result is: 63 × 54;
When five numbers make up a two-digit number and a three-digit number to maximize the product, the result is: 63 × 542;
When six numbers form two three-digit numbers, the product is maximized and the result is: 63 1×542.
Through the above examples, children can be guided to observe carefully, think positively and find out the rules:
1. Use numbers from to in turn;
2. Every time a new number is added, it is necessary to judge which of the original two numbers is bigger and which is smaller. Add the new number to the end of the smaller number in the original number to form a number with one more bit.
In fact, this is a minimalist model to solve this problem. Add from the largest to the smallest, and add the newly added number to the end of the smaller number at one time to solve the problem.
It also contains profound philosophy of life. In order to get the maximum product, that is, the optimal solution, we should give up the big and take the small when making the choice of addend. "Follow the small" is for the ultimate "maximum". It seems that every temporary retreat is for the convenience of sprinting across the gap in front.
So how to verify the correctness of this model? We use A and B to represent the original two numbers, A is greater than B, now add a number n, and adding N at the end of A or B can make their product bigger?
According to the model obtained above, it should be that the product added at the end of B will be larger. So according to the difference method, the inference is as follows:
( 10B+n)×A-( 10A+n)×B
= 10AB+An- 10AB-Bn
=(A-B)×n
& gt0
The conclusion is more general and universal by using letters to represent specific numbers for reasoning and authentication. At the same time, it also shows that the problem-solving model we have gradually obtained through concrete analysis is correct and has the value and significance of popularization.
If students are interested, our model can be further promoted. The premise we discussed earlier is decimal number. You can also extend decimal to hexadecimal or even n-decimal. If it is hexadecimal, we need 16 characters, representing 0 to 15. For example, we can use these characters, 0 123456789ABCDEF, where a stands for 10 and f stands for 15. Of course, the number of digits in a number is no longer called one, ten or hundred. Should be 16? Bit, 16? Bit, 16? Bit, 16? Bit, and so on. Similarly, if it is n-ary, then we need n characters representing from 0 to n- 1. The number on the right is n? Bit，n？ Bit，n？ Bit，n？ a little ...
As you all know, quadrilateral is a closed figure surrounded by four end-to-end line segments. We can call it a midpoint quadrilateral. Take the midpoint of each of the four sides of the quadrilateral and connect the four midpoints in turn to form a new quadrilateral.
Explore the midpoint quadrilateral of quadrilateral, generally starting from the square. Students can find that the midpoint quadrangle of a square is still a square through operation. Only the direction is off, and the area is half of the original square area.
Then we can explore the midpoint quadrilateral of the rectangle. Through operation, we can find that the midpoint quadrilateral of a rectangle is a diamond. The diamond inscribed with the midpoint quadrangle is a rectangle. Each midpoint quadrangle is half the area of the upper quadrangle.
The parallelogram inscribed with the midpoint is a new parallelogram. And the area is also half of the original parallelogram. The new parallelogram is still a parallelogram, and its area is 1/2 of the upper parallelogram and 1/4 of the uppermost parallelogram. The direction is consistent with the uppermost parallelogram.
The inscribed midpoint quadrilateral of isosceles trapezoid is a diamond. The inscribed midpoint quadrilateral of a common trapezoid is a parallelogram.
So far, we can find that all quadrangles, whether square, rectangle, diamond, parallelogram or trapezoid, are parallelograms. Of course, students also know that squares, rectangles and diamonds are special parallelograms. Just because squares, rectangles and diamonds are special parallelograms, they are more conducive to students' observation and imagination, and they can better understand logical reasoning thinking and thinking analysis methods from special to general.
On the basis of students' confirmation that parallelogram and trapezoid inscribed midpoint quadrangle must be parallelograms, guide students to ask: What kind of figure is an ordinary quadrangle inscribed midpoint quadrangle? Through hands-on operation and exploration, we can further experience and explore new discoveries.
Through drawing practice and careful observation, we can find that any quadrilateral inscribed with the midpoint is a parallelogram. So how to make students understand that the midpoint quadrilateral must be a parallelogram?
Here is an introduction to the understanding of the median line in the triangle. Take the midpoint of two sides of a triangle and connect them to get a line segment called the center line of the triangle. What are the properties of the midline of a triangle? We can use two identical triangles plus a middle line, and one of them is turned upside down to form a parallelogram. Through observation, it is not difficult to find that the two median lines of two spliced triangles are exactly on the same straight line. And are parallel and equal to the upper and lower bottom edges. So we can know from the feeling that the center line of the triangle is parallel to the bottom and equal to half of the bottom.
After confirming that the midline of the triangle is parallel and equal to half of the base. Back to the quadrilateral figure whose midpoint is inscribed in the ordinary quadrilateral, at this time we add a diagonal line to the original quadrilateral. Then hide the lower part and observe the upper part, we can find that the connecting line between the midpoints of the two sides of the triangle in the upper part is actually the midline of the triangle. So this middle line, it will be parallel to the diagonal of the original quadrangle just added, and equal to half of this diagonal. Similarly, hide the upper part and only observe the remaining triangles in the lower part. You can also find a triangle below. The center line of the triangle will also be equal to half of the diagonal and parallel to it. In this way, if you put the top and bottom together, you can easily find that the upper midline and the lower midline are parallel and equal to each other. These two median lines are the upper and lower sides of the quadrangle inscribed with the midpoint of the original quadrangle. Similarly, the left and right sides of the quadrilateral inscribed with the midpoint are parallel and equal. So we can confirm that the midpoint quadrilateral is a parallelogram.
Parallelogram, trapezoid and arbitrary quadrilateral, and their inscribed quadrilaterals are all parallelograms. From this point of view, parallelogram is a kind of belonging of quadrilateral. At this time, we can guide the children to go further. Parallelogram, trapezoid and arbitrary quadrilateral like this are called convex quadrilateral.
A convex quadrilateral is a quadrilateral with no angle greater than 180. If any side of a quadrilateral extends to both sides, the other sides are on the same side of the extended straight line. This quadrilateral is called convex quadrilateral.
The quadrilateral inscribed with the midpoint of the convex quadrilateral is a parallelogram, so what about the corresponding concave quadrilateral?
Then show the concave quadrilateral, and let the students operate it, find out the midpoint of the four sides, and connect the four midpoints to get a new quadrilateral. Students can find that one side of this new quadrilateral is outside the original quadrilateral. The new quadrilateral seems to be a parallelogram. So is it a parallelogram? The answer is yes, so how to guide children to think logically and prove?
We can observe and find a concave quadrilateral, in which both sides are concave. Because of this, a line is exposed outside the original quadrangle. At this time, we guide children to connect the two side endpoints of two concave line segments. You can find two triangles nested inside and outside. In two triangles nested inside and outside, their bases are the same, that is, the line segments just connected. The midline of these two triangles is a set of opposite sides of a quadrilateral inscribed with the midpoint. Based on the understanding of the midline of the previous triangle. It can be understood that this group of opposite sides are parallel and equal.
In addition, if the other two vertices of the original quadrangle are connected to get another diagonal line, we can also draw the conclusion that another group of opposite sides are parallel and equal.
So far, any quadrilateral of convex quadrilateral and concave quadrilateral is a parallelogram.
So what is the area of any quadrilateral inscribed with the midpoint quadrilateral?
The area of the inscribed midpoint of the parallelogram is half that of the original parallelogram. The proof is simple. You only need to connect four points to get eight small triangles, among which they are equal to each other. It can be proved that the area of the midpoint quadrilateral is half that of the original parallelogram.
How to prove that the area of the midpoint quadrangle of an ordinary quadrangle is half that of the original quadrangle?
You can connect a set of diagonal lines to make the points of the original quadrilateral. In this way, the original quadrilateral is divided into two triangles, and the midpoint quadrilateral is also divided into two parallelograms. At this time, we can explore the area relationship between a triangle and its internal parallelogram. Can be found through observation and reasoning. The base of parallelogram is the center line of triangle, which is half of the base of triangle, and the height of parallelogram is half of the height of triangle. According to the formula of parallelogram and triangle area, parallelogram is half of triangle area. So is the other side. Finally, it can be obtained that the area of the midpoint quadrangle is half that of the original quadrangle.
The same is true of the concave quadrilateral principle. The area of the quadrangle inscribed with the midpoint is still half that of the original quadrangle. The method is diagonal, and the area relationship between triangle and built-in parallelogram can be proved by the nature of the midline. Finally, the midpoint quadrangle is half the area of the original quadrangle.
The most intriguing thing is that the inscribed midpoint quadrilateral of any convex-concave quadrilateral is a parallelogram, and its area is half that of the original quadrilateral. Maybe the parallelogram is a beautiful quadrilateral. Why is there such a strong sense of unity and belonging?
In fact, in the mathematical world, seemingly complex mathematical imagination often has inherent unity and attribution. This requires students to have strong exploration spirit, eager exploration interest, correct exploration methods, rigorous exploration thinking and shocking life experience, and mathematics will become the most enjoyable knowledge.
The calculation of geometric figure area runs through the primary school learning stage. The study of geometry knowledge contributes to the development of students' spatial imagination ability, the cultivation of students' intuitive imagination ability and the cultivation of students' logical thinking ability from special to general.
Geometry area learning starts with the square area. At first, we knew the unit of area. When the side length of a square is one meter, we say that the area of the square is 1㎡. In the same way, know one square decimeter and one square centimeter.
Then learn the area formulas of squares and rectangles through measurement, that is, how many unit squares are there in a row of small squares and how many rows are there in a * * *. Through the summation of the same addend, we can get the formula of multiplication operation by understanding the multiplication operation.
On the basis of students' full understanding of square and rectangular area calculation. The calculation of parallelogram area is introduced in the textbook. The understanding of parallelogram area is to convert it into rectangle by cutting. Then use the rectangular area calculation formula and multiply the length by the width to calculate the area. In the process of cutting and transforming parallelogram graphics, students fully experience that the original size and area of parallelogram have not changed. It's just its shape that changes. When it becomes a rectangle, the width of the rectangle is one of the heights in the original parallelogram. It can also guide children to intuitively understand the length of incision during the cutting process. This is the key and difficult point in the whole transformation process.
When students understand the essence of transformation and the essential relationship between the height of parallelogram and the width of rectangle. The calculation of parallelogram area will be easily solved by students.
With the basic knowledge of parallelogram area calculation, triangle area calculation is much easier. Is nothing more than splitting in two. There is no discussion here.
When students have a deep understanding of the calculation of parallelogram area. The calculation of trapezoidal area is in front of students.
In fact, before students learn the area of parallelogram. The textbook fully strengthens students' understanding of parallelogram representation. For example, a parallelogram is cut with scissors to become two triangles, or a trapezoid and a triangle, or two parallelograms, or two trapezoids, and so on.
Before learning the formula of trapezoidal area. You can use this part as an import part. Especially the understanding and comprehension of dividing a parallelogram into two identical trapeziums. At the same time, the parallelogram can be divided into two identical triangles as a part of thinking guidance. Let the children find out through guiding tips. A parallelogram can be divided into two identical triangles. Thus, the calculation formula of triangle area is obtained. So is it possible to combine two identical trapezoids into a figure, which is the known figure of the area calculation formula we have learned?
Let the students operate by prompting and guiding. Students should gradually realize that two identical trapezoids can form a parallelogram by inverting one of them. So it is concluded that the area of trapezoid is actually half that of parallelogram. Only the bottom of this parallelogram becomes the sum of the upper bottom and the lower bottom. We can get the formula of trapezoidal area, S= (upper bottom+lower bottom) × height ÷2.
Understanding of the formula for calculating trapezoidal area. At this time, the trapezoid is replaced by a physical figure. For example, a pile of wood, 9 at the bottom, 7 at the second, 5 at the third and 3 at the fourth. How many pieces of wood do you need at this time? How can I get it? From the appearance of the graph, it can be expressed as 3+5+7+9.
When we guide children back to the angle of trapezoidal area calculation. Students can use their imagination to form a trapezoid by borrowing a group of identical stakes. Therefore, the calculation formula of wood root number can be obtained: S=(3+9)×4÷2.
Then comparing this formula with the trapezoid area formula, we can find that the root number of the lowest wood is equivalent to the bottom, the root number of the top wood is equivalent to the upper bottom, and the number of layers of wood is equivalent to the height of the trapezoid. In other words, the formula for calculating the trapezoidal area is consistent with the formula for calculating the number of trapezoidal stacked wood.
Of course, we students may find that it is not easy to add up directly with the trapezoidal area calculation formula. At this time, the teacher can guide students to realize that it is much easier to calculate with formulas when there are more layers of wooden stakes.
When students understand the value of formulas, we can guide them to abstract mathematics and throw away the basis of figures and objects. Direct presentation: 1+3+5+7+9+...+99
Of course, students can imagine with the help of trapezoidal graphic model. Gradually get the calculation formula: S=( 1+99)×50÷2. Here, the understanding of 50 in the formula is the most important and difficult. In trapezoid, it is the height of trapezoid. It's a pile of layers of wood. At this time, it should be the number of addends, that is, the number of terms.
Finally, students can understand that the difference between two adjacent items like 1, 3, 5, 7...99 is an arithmetic number, which is called arithmetic progression. To sum all numbers is to sum arithmetic progression. Namely: S= (the first item+the last item) × the number of items ÷2.
Taking the throne by intelligence is a puzzle game. There are eleven flags in a row of wooden troughs, and the last one is red. The rules of the game are: two people take 1 ~ 2 tablets each time in turn. Whoever can win the last red flag "throne" wins.
At first glance, it seems to be related to luck. In fact, it contains mysteries and rules to follow. May wish to guide students from less to more, step by step, and explore the laws contained in it.
First, guide students to understand that the rules of this game are: 1. Two people take turns; 2. Take 1 ~ 2 capsules each time; 3. Take the last one to win.
Secondly, lead students into the game and explore the inside story.
It is not difficult to find that when there is one or two pieces, the person who takes them first will definitely win. When the number of pieces is three, the last player wins. There are two strategies: first, the first winner takes one, and the second winner wins with two, that is, 1+2 type; Second, the first winner gets two, and the last winner gets one, which means typing 2+ 1.
At this time, let the students operate and experience repeatedly: one or two, the first one will win; Third, the latter will win.
Subsequently, from less to more, additional signs were added. When there are four flags, which one wins, the first one or the last one? Let the students practice the operation, experience it in the operation and draw a conclusion: the first winner takes one, and then there are three left. At this time, the last winner is the first winner of this competition. Similarly, when there are five pieces, it can be concluded that the first piece has taken two pieces, and then there are three pieces left. At this time, the last one, that is, the first one in this competition, won. The difference between these two strategies is that the first one needs one or two. The purpose is the same, there are three left.
After the experiment, when there are six pieces. Let the students repeatedly realize that the latter will win through the operation experiment. There are also two strategies: first, the first winner takes one, the second winner takes two, and the remaining three, and the strategy of repeating three wins; Second, when the first winner takes two, the last winner takes one and there are three left. Repeat the strategy of three and you win.
By analogy, let the students guess whether the first player wins or the last player wins when there are seven, eight or nine pieces, and then verify their feelings.
Finally, let the students experience the one, two, three, four, five, six, seven, eight and nine flags. The situation is the same, but it is reproduced or repeated.
In fact, the core of this game is the strategy of holding children when the three flags are hoisted. That is to say, this game has an essential minimalist mode-it can be divisible by 3. When the divisor is 3, all nonzero natural numbers are divided into remainder 1 number, remainder 2 number and divisible number. If you want to win, you must try to take the remainder of 1, and never take the divisible number. The remainder of 2 is the adjustment number.
In this way, no matter whether there are 1 1 or additional pieces, if one side doesn't know the inside story, the one who knows the inside story will definitely wait for an opportunity to win.
The rules of the game are made by people. Once the rules of the game are changed, it is necessary to re-explore the game strategy. If it is stipulated that you can take one, two or three at a time, how can you win?
At this time, students can be guided to use the same inquiry methods and strategies to explore.
When the number of flags is 1-3, the first one wins, and when there are four flags, the last one wins. The strategies are:1+3,2+2,3+1. When there are five pieces, the first one wins. The strategies are: 1+ 1+3, 1+2+2, 1+3+ 1. In fact, the first winner takes a flag first, and when it is converted into four flags, the last winner (the first winner of this competition) will win.
Similarly, when there are six pieces, the first one wins, and the strategy is: 2+ 1+3, 2+2+2, 2+3+ 1. In fact, the first winner takes two flags first, and then when it becomes four flags, the last winner (that is, the first winner of this competition) wins.
When seven pieces are played, the first player wins. The strategies are: 3+ 1+3, 3+2+2, 3+3+ 1. In fact, the first place won three flags first, and when it was changed to four flags, the last place (the first place in this competition) won.
When there are eight pieces, the last piece wins, and the strategy is to use it twice according to the strategy of four pieces.
Similar to the last game, the core of this game is the strategy of taking children when there are four flags. That is to say, this game has an essential minimalist mode-it can be divisible by 4. When the divisor is 4, all nonzero natural numbers are divided into remainder 1 number, remainder 2 number, remainder 3 number and divisible number. If you want to win, you must try to take the remainder of 1, and never take the divisible number. The remainder of 2 and 3 is the adjustment number.
When students have a thorough understanding, the rules of the game can be further changed. Through experiments, analysis and reasoning, a broader and more universal model can be obtained.
The rules of the game are: two people take 1~n tablets each time in turn. Whoever can win the last red flag "throne" wins.
Winning strategy: the minimalist mode of this game-divisible by n+ 1 When the divisor is n+ 1, all non-zero natural numbers are divided into remainder 1, remainder 2, remainder 3 ... remainder n and divisible number. If you want to win, you must try to take the remainder 1, and never take the divisible number. Remainder 2, remainder 3 ... remainder n is the adjustment number.