The "Huajin Cup" Youth Mathematics Invitational Tournament is held every two years. This year is the second session. Which session was it in 2000?
The solution "held every other year" means held every two years. This year is 1988, and in 2000 there will be 2000- 1988= 12, so there will be 12÷2=6 conversations. This year is the second, so.
A: The eighth session was held in 2000.
Because the number of analysis and discussion is small, it can be quickly counted directly: 1988, 1990, 1992, 1994, 1996, 1998, 2008.
An inflatable life buoy (as shown in Figure 32). The radius of the great circle shown by the dotted line is 33 cm. The radius of the small circle shown by the solid line is 9 cm. Two ants start from point A at the same time and crawl along the big circle and the small circle at the same speed. Q: The ants in the small circle climbed several times and met the ants in the big circle for the first time.
Because the speed of two ants is the same, from the formula of distance-speed = time, we know that the ratio of time for ants to climb a circle on a big circle and a small circle should be equal to the ratio of the length of the circle, which is equal to the ratio of the radius, that is, 33:9.
When two ants met for the first time, how many times did the ants climb on the small circle? That is to find a minimum time, which is an integer multiple of the time for ants to crawl on the big circle and the small circle respectively. As can be seen from the above discussion, if we choose the appropriate time unit, the ants in the small circle can crawl 9 units, while the ants in the big circle can crawl 33 units. In this way, the problem boils down to the minimum of 9 and 33.
Answer: The ant on the small circle climbed 1 1 times and met the ant on the big circle again.
The key to analyzing and discussing this topic is to see that the essence of the problem is to find the least common multiple. Pay attention to observation and see the mathematics in life, which Professor Hua often inspires teenagers to do.
Figure 33 is a chessboard. Please calculate how many holes there are in this board.
There are many ways to solve this problem. Because of the time limit, the direct number is too late and easy to make mistakes. The following figure (Figure 34) shows a better algorithm. The chessboard is divided into a parallelogram and four small triangles, as shown in Figure 34. The number of chess holes in parallelogram is 9×9=9 1, and each small triangle has 10 chess holes. So the total number of chess holes.
A: * * * There are 12 1 chess holes.
Analyze and discuss the students who have played checkers. Have you counted the number of chess holes before? Interested students can make statistics in their spare time to see who has the best method.
There is a four-digit integer. Add a decimal point before one of its numbers and add it to this four-digit number to get 2000.438+0. Find this four-digit number
Solution 1 Because the result has two decimal places, the decimal point cannot be added before the single digit. If the decimal point is added before the ten digits, the result is one percent of the original four digits, and if the original four digits are added, the result of 2000.85438+0 should be 1.0 1 times of the original four digits, and the original four digits are 2000.5000.00000000005
Similarly, if the decimal point is added before the hundred digits, then the number 2000.85438+0 should be 1.005438+0 times of the original four digits, and the decimal point should be added before the thousand digits. The number of 2000.8 1 should be 1.000 1 times of the original four digits, but (2000.8 1 ÷ 1) and (2000.81÷)
A: This four-digit number is 198 1.
Scheme 2 notes that in the original four digits, 8, 1 must appear in order. The decimal point cannot be added before one digit; You can't add it before thousands, otherwise the original four digits can only be 8 100, which is 2008+0.
Whether the decimal point is added before the decimal point or before the hundred digits, the number obtained is greater than 1 less than 100. This number plus the original four digits is equal to 2000.8 1, so the original four digits must be less than 2000, but greater than 1900, indicating that its first two digits must be 1, 9.
The analysis and discussion scheme 1 uses accurate calculation, while scheme 2 relies on "judgment". Judgment also requires skill, based on careful analysis of problems.
What needs to be pointed out here is that we can't draw the conclusion that the decimal point is just before the ten digits as soon as we see that the number 2000.8 1 has two decimal places. Please think about why.
Figure 35 is black and white plaid. The side length of the big white square is 14 cm, and the side length of the small white square is 6 cm. Q: What percentage of the total area is white in this cloth?
The area of the unpacked grid is 9 times that of Figure 36, and the area of the white part of the grid is 9 times that of Figure 36. In this way, we only need to calculate the percentage of the area occupied by the white part in Figure 36. The calculation is simple:
A: The area of the white part of the grid is 58% of the total area.
The key to analyzing and discussing this topic is to see that the grid can be divided into nine squares, as shown in Figure 35. This essentially takes advantage of the "symmetry" of the grid: the grid pattern is made up of a pattern arranged repeatedly and orderly.
Symmetry is not only an important concept in mathematics, but also a basic law in nature. Therefore, "symmetry" and "asymmetry" have always been important topics in different fields, such as mathematics, physics, chemistry and even aesthetics. The famous mathematician H. Weil once wrote a book called Symmetry (with Chinese translation), which is very rich in content.
Fig. 37 is a subtraction formula of two three-digit numbers, and each box represents a number. Q: What is the continuous product of the numbers in these six boxes?
Figure 37
To solve the problem of subtracting two numbers, it is customary to consider single digits first. But if you look closely, you will find that the single digits of two digits are uncertain: these two digits are added 1 or subtracted 1 at the same time, and their difference remains unchanged. In this way, the continuous product of the numbers in six boxes will be uncertain unless the number in one box is 0, so the product will always be 0. This inspired us to try to find the box.
Of course, the first digit of two three digits is not 0, so the first digit of the subtree is at least 1, and the first digit of the minuend is at most 9. But because the first bit of the difference is 8, there is only one possibility, that is, the first bit of the minuend is 9, and the first bit of the subtree is 1.
So the subtraction on the second digit can't be borrowed. The second bit of the minuend is at most 9, and the second bit of the minuend is at least 0, and the difference between these two numbers is 9, so there is only one possibility: the second bit of the minuend is 9 and the second bit of the minuend is 0. So we have determined that the number of one of the six boxes must be 0.
Answer: The continuous product of the numbers in the six boxes is equal to 0.
To analyze and discuss this problem, it is not necessary to completely determine these three figures, nor can it be completely determined. For example, the minuend and the reduction can be (996, 102), (994, 100), (999, 105) and so on.
Some students will say: the answer to this question is a guess.
"Guess" is also a method in mathematics. There are many famous conjectures in mathematics, which have had an important influence on the development of mathematics. There are two points to emphasize here: first, the "conjecture" in mathematics is not an unfounded "conjecture", but an answer conceived by indexers after in-depth analysis or a large number of examples; That makes sense. For example, in the solution of this question, we found through analysis that if there is no 0 in the six boxes, then the answer to this question is not unique, so the guess answer is 0. There is no point in guessing that the answer is 100. Second, "guessing" is not equal to the answer, and guessing can only be a strictly proved answer. For example, the famous Goldbach conjecture has not been proved yet, so it is still being proved.
The side length of the square in Figure 38 is 2m, the radii of the four circles are all 1 m, and the center of the circle is the four vertices of the square. Q: What are the areas of this square and four circles?
The common part of each circle and square is a sector, and its area is a quarter of the area of the circle. So the area of the whole figure is equal to the area of the square plus the area of four three-quarters circles, and the area of four three-quarters circles is equal to three times the area of the circle. So the area of the whole figure is equal to the area of the square plus the area of the circle three times, that is to say,
2× 2+π×/kloc-0 /×/kloc-0 /× 3 ≈13.42 (square meter).
A: The area covered by this square and four circles is about 13.42 square meters.
There are seven bamboo poles in a row. The length of the first bamboo pole is 1 m, and the length of each other is half that of the last one.
Q: What is the total length of these seven bamboo poles?
Solution: Take a 2-meter-long bamboo pole and cut it in half, each half length is 1 meter. Take one as the first bamboo pole, cut the other in half and take one as the second bamboo pole. In this way, by the time the seventh bamboo pole is cut off, the length of the remaining bamboo pole will be
Therefore, the total length of seven bamboo poles is 2 meters minus the length of the rest, that is
This paper analyzes and discusses an ancient arithmetic problem in China, that is, "one foot pestle, half a day, is inexhaustible". In other words, there is a foot-long short stick, half of which can never be cut off every day. So, how much is left every day? How much is left on the seventh day?
It is more than enough to calculate the total length of seven bamboo poles with the above solution. But if you count each bamboo pole first and then add it up, it will be too late. Students might as well give it a try.
There are three line segments A, B and C. A is 2.12m long, B is 2.7 1 m long and C is 3.53m long. Taking them as the upper bottom, the lower bottom and the height, three different trapezoids can be made. Which trapezoid has the largest area?
First, pay attention to the solution. Trapezoidal area = (upper bottom+lower bottom) × height ÷2. But now we are comparing the areas of three trapezoids, so we can multiply their areas by 2, so we only need to compare the size of (upper bottom+lower bottom) × height. We use the law of multiplicative distribution:
The area of the first trapezoid is twice:
(2. 12+3.53)×2.7 1=2. 12×2. 17+3.53×2.7 1
The second one:
(2.7 1+3.53)×2. 12=2.7 1×2. 12+3.53×2. 12
The third one:
(2. 12+2.7 1)×3.53=2. 12×3.53+2.7 1×3.53
First, compare the first and the second. The first addend on the right side of the two formulas, one is 2. 12×2.7 1 and the other is 2.7 1×2. 12. These two products are equal by the multiplication commutation law. Therefore, only the size of the second addend needs to be compared. It's obviously three.
Similarly, if we compare the first and the third, we find that they all have edges, the second addend is equal, and the first addend is 2. 12× 2.7 1
To sum up, the third trapezoid has the largest area.
Answer: The third trapezoid has the largest area.
To analyze and discuss this topic, we should make full use of the knowledge of multiplication exchange law and multiplication distribution law we have learned, rather than directly calculating the area. Obviously, it takes a lot of time to directly calculate the area of three trapeziums.
There is an electronic clock that lights up every 9 minutes and rings every hour. At noon 12, the electronic clock rings and lights up. Q: When will it ring and light up again?
Solution: Because the electronic clock rings every hour, we only need to consider which light to turn on. From noon 12, the light will come on every 9 minutes. How many 9 minutes does it take to get to the hour? Because 1 hour =60 minutes, the question in other words is: how many times of 9 minutes is an integer multiple of 60 minutes? So the essence of the problem is clear: it requires 9 minutes and 60 minutes.
It is not difficult to calculate that the least common multiple of 9 and 60 is 180. In other words, the electronic clock will ring and light up again after 180 minutes from noon, that is, after 3 hours.
A: The next time the light rings, it will be 3 pm.
The analysis and discussion of such problems will be encountered everywhere in life. Can students give some more examples?
A deck of playing cards has four suits, and each suit has 13 cards. You can draw any card from it. Q: How many cards do you have to draw to ensure that four cards are of the same suit?
No matter how you draw cards, you must have four cards of the same suit.
Let's see if we can draw 12, and make sure there are four flush cards. Although sometimes there may be four or more flushes in 12 cards, there may be exactly three flushes in each suit, so there is no guarantee that there will be four flushes.
So, can you draw 13 cards at will to ensure that there are 4 flush cards? Let's say this. The evidence is as follows:
If not, then each suit can only have three cards at most, and the total number of cards in four suits can only be 12, which is contradictory to drawing 13 cards. So it is enough to draw 13 card.
This method of proof is called reduction to absurdity.
Answer: At least 13 cards must be drawn to ensure that the four cards are of the same suit.
Use the so-called "pigeonhole principle" to analyze and discuss this topic. For example, there are four drawers to hold 13 books, and at least one drawer should hold four books. This principle is also called "Pigeonhole Principle" or "Overlapping Principle".
Although pigeonhole principle is simple, he has many ingenious applications in mathematics. Interested students can read Chang Gengzhe's pigeonhole principle and Others.
There is a class of students going boating. They calculated that if there was an extra boat, each boat could only hold six people. If one boat is reduced, each boat can only take nine people. Q: How many students are there in this class?
Solution 1 suppose a boat is added first, so each boat just takes 6 people. Now if two boats are removed, there will be 6×2= 12 students without boats. Now there are just 9 students on each boat, that is to say, an increase of 9-6=3 students on each boat can just arrange the remaining 12 students, so there are still 12 students.
A: There are 36 people in this class.
Solution 2 According to the topic conditions, the number of students in the class is both a multiple of 6 and a multiple of 9, so it is a common multiple of 6 and 9. The least common multiple of 6 and 9 is 18. If the total number of people is 18, then each ship needs18 ÷ 6 = 6 people in 3 ships, and each ship needs 65438+9 people.
This paper analyzes and discusses many similar problems in ancient China, such as "chickens and rabbits in the same cage" and so on.
This problem can also be solved by a series of equations. Students might as well give it a try.
Four small animals changed their seats. At first, the mouse sat in the seat of 1, the monkey sat in No.2, the rabbit sat in No.3 and the kitten sat in No.4. After that, they kept changing seats. They exchanged seats in the upper and lower rows for the first time, the left and right rows after the second exchange, the upper and lower rows for the third time, and the left and right rows for the fourth time ... This situation has been going on. Q
The solution to this problem is, which seat does the rabbit sit in after the tenth seat exchange? Let's find out the changing law of rabbit seat according to the meaning of the question.
As can be seen from the arrow diagram in Figure 40, every time the rabbit's seat is exchanged, it will turn one square clockwise, and every four times, the rabbit's seat will turn back to its original position. Knowing this rule, the answer is not difficult to get. After the tenth seat exchange, the rabbit's seat should be the second seat.
Answer: After exchanging seats for the tenth time, the rabbit sat in the second seat.
The movement of "small animals changing seats" is called "displacement" in mathematics, and the seat change of rabbits is called "rotation". Displacement and rotation are important concepts in group theory, geometry and other branches of mathematics. Although this question is simple, there are many interesting reasons in it!
In order to deepen students' understanding, we still have two questions to think about. Please think about them.
(1) Find out the seat changes of the other three small animals. What are their similarities and differences?
(2) Change the question in the title to: "What small animal is sitting in the fourth seat after the tenth seat is exchanged?" Do you know how to do it? Think about it.
The four numbers 1, 9, 8 and 8 can be arranged into several four numbers divided by 1 1.
Solve what kind of numbers can be divisible by 1 1 A judgment rule is: compare the sum of odd digits with the sum of even digits. If the difference between them can be divisible by 1 1, then the given number can be divisible by 1 1, otherwise it can't.
Now it is required to be divisible by 1 1, so we can think of it this way: after adding 3 to this number, it can be divisible by1/. So we get the judgment rule of "a number divided by 1 1": add even numbers to get a sum number, otherwise it is not.
To arrange 1, 9, 8, 8 into a four-digit number divided by 1 1, these four digits can be divided into two groups, each with two digits. One group is regarded as thousands and ten, and their sum is a; The other group is hundred digits and single digits, and the sum of them plus three is B. We should group them appropriately so that they can be divisible by 1 1. There are only the following four grouping methods:
After verification, (1) grouping method meets the previous requirements:
A = 1+8, B = 9+8+3 = 20, and B-A = 1 1 can be divisible by1,but the other three groups are not satisfied.
According to the decision rule, we can also know that if a number is divided by the remainder 8 1 1, then any two numbers in odd positions are exchanged, or any two numbers in even positions are exchanged, and the new number is divided by 1 1, then in the above (1) group, 65438. Any one of 9 and 8 can be regarded as the hundredth place. * * * There are four possible arrangements: 1988, 1889, 89 18, 88 19.
Answer: It can be arranged as 4 numbers divided by 8 divided by 1 1.
Analyze and discuss the different four digits of 1, 9, 8, 8/2. If all these 12 numbers are listed, it will be too complicated to test their division residues separately. Therefore, when solving problems, we must first simplify the testing process as much as possible.
Figure 4 1 is a chess board, which consists of 19 horizontal lines and vertical lines. Q: How many squares on the Go board are the same as the small squares in Figure 42?
The solution requires the brain to calculate the number of small squares accurately and quickly.
Let's first find a representative point in the small square on the right, such as point E in the lower right corner. Then, according to the meaning of the question, put the small squares on the Go board and carefully observe where the E point should be. Through observation, it is not difficult to find that:
(1) Point E can only be on the grid point of square ABCD (including the boundary) in the lower right corner of the chessboard.
(2) Conversely, every grid point in the ABCD square in the lower right corner can be regarded as the point E of the small square, and can only be regarded as the point E of the small square.
In this way, "the number of small squares" becomes "the number of lattice points in the square ABCD". It is easy to see that the lattice point in the square ABCD is 10× 10= 100.
A: * * * Yes 100.
There are many solutions to analyze and discuss this topic, and the above solutions are ingenious and quick. It uses a simple truth, that is, "there are as many one-to-one correspondences".
One-to-one correspondence is an important basic concept in mathematics. It can be seen from this topic that as long as such a concept is clarified, it will play a great role.
Question: What's the difference if both graphs are rectangles?
For example, the big chessboard is 20×30, and the small chessboard 10× 15. How many rectangles are there in the big chessboard like the small chessboard?
Rematch problems and solutions
calculate
solve
There are three cards with numbers written on them (Figure 43). Take out one, two and three cards from them and arrange them in any order to get different one, two and three digits. Please write down all the prime numbers.
Solution We know that a natural number greater than 1 is called a prime number if it has no other divisor except 1 and itself.
Let's recall the judgment rule of divisibility by 3: if the sum of digits of a number can be divisible by 3, then this number can also be divisible by 3.
Because the numbers on the three cards are 1, 2, 3 respectively, the sum of these three numbers is 6, which can be divisible by 3, so any three digits arranged by these three numbers can be divisible by 3, so it can't be a prime number.
Let's look at the two cards. Because 1+2=3, by the same token, the two-digit number composed of 1 2 can also be divisible by 3, so it is not a prime number. In this way, the remaining two digits to be discussed are13,31,23,32, of which 65438.
Finally, there are three digits: 1, 2,3.1is not a prime number. Both 2 and 3 are prime numbers.
In short, this problem has five prime numbers: 2,3,13,23,31.
A: * * * has five prime numbers: 2,3,13,23,31.
The analysis and discussion of this problem mainly examines the degree of mastery of the concept of prime numbers and the basic law of divisibility (such as the characteristics of divisibility by 3). Of course, if you write out the figures made up of two cards and analyze them one by one, you can also work them out. But this is not desirable.
There are three square pools, large, medium and small, the inner sides of which are 6 meters, 3 meters and 2 meters respectively. If two piles of gravel are immersed in the water of small and medium-sized pools, the water levels of the two pools will be raised by 6 cm and 4 cm respectively. If these two piles of gravel are submerged in the water of the big pool, how many centimeters will the water level of the big pool increase?
The solution is to sink the gravel into the water, and the volume increased by the rising water level is equal to the volume of the gravel sunk.
Therefore, the volume of crushed stone sunk into the pool is
3m× 3m× 0.06m = 0.54m 3。
The volume of gravel sinking into the small pool is
2m× 2m× 0.04m = 0. 16m3。
The volume of these two piles of rubble is * * *
0.54 m3+0. 16 m3 = 0.7 m3。
All of them sank into a large pool, and the volume of the water surface of the large pool increased by 0.7 m 3. The bottom area of the big pool is
6m× 6m = 36m 2.
So the water level rose:
There are dozens of holes in a circle (less than 100), as shown in Figure 44. Like playing checkers, Xiao Ming starts from hole A and jumps counterclockwise every few holes, hoping to jump back to hole A after one lap. He first tried to jump every two holes, but only to the B hole. He tried to jump every four holes and only to the B hole. Finally, he tried to jump.
This solution assumes that the holes on the circle have been numbered as follows; The number of hole A is 1, and then the number is 2, 3, 4 ... The number of hole B is the number of holes on the circle.
Let's first look at the holes that Xiaoming jumps into every two holes. It is easy to see that it should be in 1, 4, 7, 10, ... That is to say, the number on the hole that Xiaoming jumped is a multiple of 3 plus 1. According to the meaning of the question, Xiao Ming finally jumped into the B hole, so the total number of holes is a multiple of 3 plus 65438+.
Similarly, jump every 4 holes and finally jump to hole B, which means that the total number of holes is a multiple of 5 plus1; Jump every 6 holes, and finally jump back to A, indicating that the total number of holes is a multiple of 7.
If you subtract 1 from the number of holes, this number is a multiple of 3 and 5, so it is a multiple of 15. This multiple of 15 plus 1 is equal to the number of holes and divisible by 7. Note that 15 divided by 7 is still 1, so 15. 6 times 15 is divisible by 7. We can also see that 15 (less than 7) plus other multiples of 1 cannot be divisible by 7, while 15× 7 = 105 is greater than 100 and greater than 7.
A: There are 9 1 holes in the circle.
The analysis and discussion of this problem is actually a special case of the following kind of problems. The general problem is: there is an unknown integer, and we can find it only by knowing the remainder obtained by dividing it by some integers. In Sun Tzu's Art of War, an ancient Chinese mathematical masterpiece, there have been general methods to solve this kind of problems. This method is generally called "China's Remainder Theorem" internationally. Professor Hua once wrote a booklet for junior high school students, "Starting with Sun Tzu's Divine Calculations", which introduced the ingenious methods to solve this problem in a simple way, and also led to some other interesting questions, which was very enlightening. This booklet has been selected into Selected Works of Huake Puwen (Shanghai Education Publishing House
Try to fill in 1, 2, 3, 4, 5, 6, 7 in the box in Figure 45, and use each number only once:
So any two of these three numbers are coprime. One of the three numbers has been filled in, which is 7 14.
Solution We know that if the greatest common divisor of two numbers is 1, then these two numbers are called prime numbers.
The three-digit number 7 14 that has been filled in is a composite number, and its prime factor is decomposed into
7 14=2×3×7× 17.
So any two of these three numbers are coprime. One of the three numbers has been filled in, which is 7 14.
It can be seen that in order to make the numbers in the bottom box and 7 14 mutually prime numbers, only 5 can be selected among the remaining unfilled numbers 2, 3, 5 and 6, that is, only 5 can be filled in the third line.
Now let's discuss how to fill in the three boxes in the second line with the numbers 2, 3 and 6.
Because the common divisor of any two even numbers is 2, they are not coprime. And 7 14 is even, so the three digits in the second line can't be even, that is, the digits can't be filled with 2 and 6, so the digits can only be 3. In this way, the three digits in the second line can only be 263 or 623. But 623 is divisible by 7, so 623 and 765438+.
Finally, look at the number 263. It can be seen from the test that the prime factors 2, 3, 7 and 17 of 7 14 are not factors of 263, so 7 14 and 263 are coprime. Obviously, 263 and 5 are also coprime. So the numbers 714,263 and 5 are one.
Answer: The filling method is:
Figure 47 is a road map. The number on each section of the road is the number of minutes that Xiao Wang needs to complete this section. How many minutes does it take Xiao Wang to walk from A to B at the earliest?
Solution 1 For convenience of description, we mark each intersection with letters, as shown in Figure 48 and Figure 49.
First, we will gradually simplify the road map.
The route from a to b must pass through DC and GC. There are two routes from A to C: ADC time consumption 14+ 13=27 (minutes); AGC takes 15+ 1 1=26 (minutes). We won't take the previous route, so we can delete the DC part. However, it should be noted that AD cannot be deleted, because there are other routes from A to B (such as AHB) that pass through AD and need further analysis.
There are also two routes from G to E: CCE 16 minutes and GIE 16 minutes. We can choose any route, such as choosing the former and erasing GIE. (We can also choose the latter and erase CE, but we can't erase GC because there are other routes passing through it. ) In this way, the road map is simplified to the shape of Figure 49.
In fig. 49, there are two routes from A to F, one passing through H demand 14+6+ 17=37 (minutes), and the other passing through G demand15+1+1.
In Figure 50, there are also two routes from C to B. By comparing the required time, one route passing through E can be erased. Finally, there is the most time-saving route (Figure 5 1), which requires15+1+65438+.
A: It will take 48 minutes at the earliest.
Scheme 2 should grasp the key point C. If the road from A to B passes through point C, then choose the most time-saving road from A to C, that is, AGC chooses the most time-saving road from C to B, that is, CFB. So the most time-saving way from A to B through C is the connection of these two roads, namely AGCFB. The total time is 48 minutes.
For the rest, just compare the road from A to B that doesn't pass through point C with the road of AGCFB to see which one is more time-saving.
There are only two roads that don't pass through point C: ①ADHFB, which takes 49 minutes; ②AGIEB, it also takes 49 minutes.
So it takes 48 minutes to get from A to B at the earliest.
The above analysis and discussion of simplified summation need not be drawn one by one. As long as you mark the road sections to be erased on the original map, you can quickly find the shortest route. Even the most complicated road map can be easily simplified. Figure 52 is a slightly more complicated road map, and the numbers in the figure have the same meaning as this question. Please try the above step-by-step simplification method to find out the shortest time from a to B.
This topic has a special name in applied mathematics, which is called "the shortest path problem". The shortest path problem is widely used in transportation, planning and many other aspects. In practical problems, the road map is often very complicated, and it is very difficult to find all the routes from A to B. Therefore, it is very necessary to use the interpolation method mentioned above.
The median line EF of trapezoidal ABCD is 15cm (see figure 53), ∠ ABC = ∠ AEF = 90, and g is a point on EF. If the area of triangle ABG is 1/5 of the area of trapezoid ABCD, what is the length of EG?
[Solution] The area of trapezoidal ABCD is equal to EF×AB, while the area of three-column ABC is equal to (1/2)EG×AB, so the area ratio of triangular ABG to trapezoidal ABCD is equal to (1/2)EG to EF. According to the subject conditions, the area of triangular ABG is 1/5 of the area of trapezoidal ABCD.
Answer: EG is 6 cm long.
[Analysis and discussion] This question assumes ∠ ABC = ∠ AEG = 90, which is actually redundant. The supplement only takes into account that primary school students may not have learned the nature of the midline. Interested students can consider how to do this problem if this condition is removed.
There are three piles of weights, the first pile is 3g, the second pile is 5g, and the third pile is 7g. Please take the least number of weights to make the total weight 130g. Write down the method: how much weight is needed, including 3g, 5g and 7g?
[Solution] In order to simplify the problem, let's first analyze the relationship between the three rows of weights. Obviously, a 3g password cracker plus a 7g weight is exactly equal to two 5g weights (both 10). Therefore, if two 5-gram weights are replaced by a 3-gram weight and a 7-gram weight, the number and total weight of the weights will remain unchanged. In this way, we will remain the same.
This comes down to the following two situations:
First, there is no 5g weight. Obviously, in order to minimize the number of weights, 3g weights should be taken as little as possible, and the total weight of 130g minus 3g weights should be a multiple of 7. After calculation, we can know that we should take 0, 1, 2, 3, 4, 5 3g weights, and the rest. Then130-3g× 6 =112g = 7g×16. So we can take 16 7 g weights and 6 3 g weights, making a total of 22 weights.
Second, one of the weights taken is 5g. Then the total weight of 3g and 7g is 130g-5g = 125g. Similar to the first case, it can be calculated that two 3g weights and 17 7g weights should be taken, so the total * * is 17+2.
Comparing the above two situations, we know that at least 20 weights can be taken, just like the last ten situations; 2 3g， 1 5g， 17 7g； Of course, you can also use two 5g weights instead of one 3g and 1 7g weight. For example, you can take five 5g and 15 7g weights.
Answer: Take at least 20 weights, as mentioned above.
[Analysis and discussion] In this problem, there are three numbers (that is, the number of three weights) that can be changed. The above solution is essentially to fix a number (the number of 5 grams of weights) first, so that the remaining number will change and be easier to handle. If all three numbers are changed, it will become very messy. Even if we find a way to take only 20 weights, it is hard to say why this is the least.
If the students want to do such an exercise, they might as well calculate the maximum weight they can take under the conditions of this topic.
There are five round flower beds with diameters of 3m, 4m, 5m, 8m and 9m respectively. Please divide these five flower beds into two groups and give them to two classes respectively, so that the areas managed by the two classes are as close as possible.
We know that the area of each circle is equal to the square of the diameter multiplied by (π/4). Now we should group five circles, and the total area of the two groups should be as close as possible. The closer the ratio of the total area of the two groups is, the better! Because the area of each circle has a factor (π/ 4), and all we care about is the ratio of the areas, we don't remove all the same factors, but simplify the problem as: divide five circles into two groups to make the diameters of the two groups of circles equal.