A breakthrough in the thinking of middle school students in mid-range questions
An effective mathematics test paper, the most discriminating questions are mid-range questions, and the distinguishing object is also middle-grade students, an excellent mid-range question can effectively control students' scores in different ranges, beyond it, it is medium and above, and the more it is, it is medium and below. In terms of the time it takes to solve the problem, excellent students can quickly find the solution idea and complete it in a standardized manner, while middle students need a long time to find it, and junior students have no clue and can only write a solution in the end.
The key to breaking through the mid-range questions lies in the interpretation of the conditions of the question, the first is to be able to interpret the conclusions that can be drawn from the conditions, the second is to find the correlation between these conclusions, and the third is to use these first-level conclusions as conditions for the next step of derivation.
Topic. As shown in the figure, the acute angle ABC is connected to the circle O, AD BC, be AC, om BC, and the vertical feet are D, E, and M respectively
1) If acb=60°, find the size of abo;
2) Is OMB similar to AEB?Why?
3) Determine whether the area of OBD and OAE is equaland give reasons.
Analysis: 1) ACB is the circumferential angle, and the arc AB it is opposite to the central angle AOB, so AOB=120°, plus OA=OB, in isosceles AAB, the bottom angle ABO=30°;
2) OMB and AEB are both right-angled triangles, so when judging their similarity, they already have an equal angle, and among the various methods for determining similarity, the two angle conditions are the easiest to find, so it is natural to think of finding the remaining pair of corresponding angles;
In the acute angle of OMB and AEB, BAE is the circumferential angle, the arc it is opposite is the arc BC, and the central angle of the arc BC is BOC, and the OC should be connected here, as shown in the following figure:
Since OM BC, and OB=OC, in isosceles BOC, we can use "three wires in one" to prove BOM=1 2 BOC, and BAE=1 2 BOC, so BOM= BAE, we can prove OMB AEB;
3) OBD and OAE are not special triangles, so their area first considers the most basic area formula, that is, half of the base times the height, so we need to make the high OF OAE, as shown in the figure below:
The area of obd is 1 2bd om, the area of oae is 1 2ae of, if bd·om=ae·of, the problem can be solved;
Direction 1: Can you find a pair of similar triangles that happen to contain the above 4 line segments on their corresponding sides?
The answer is no;Thinking direction 2: Among the 4 line segments, 2 are from the similar triangle just proved, OMB and AEB, the former contains the line segment OM, the latter contains the line segment AE, and more coincidentally, they are also corresponding edges, so in the remaining 2 line segments, BD and OF, are there in another pair of similar triangles?
The answer is yes;
There are many triangles where BD is located, and special triangles are preferred, that is, RT ABD, and the triangle where OF is located is also RT AOF
Similar to the previous question, they already have a pair of equal right angles, and in the remaining acute angles, abd=1 2 aoc, the reason is that the circumferential angle of the same arc is half of the central angle, and aof=1 2 aoc, the reason is the same as the previous problem, so it can be proved that aof abd, so we can get the proportional line segment of:bd=ao:ab, in the previous omb aeb, we get the proportional line segment om:ae=bo:ab, observe the right side of these two proportional equations, by ao=bo can know that they are equal, so we get a new proportional line segment of:bd=om:ae, which is converted to the product formula bd·om=ae·of, and multiply each side by 1 2 to get s obd=s oae
Problem solving: In the process of reading the question, the interpretation of each sentence condition needs to be explained thoroughly, for example, the first sentence "acute angle abc is connected to the circle O", you need to recognize that the three sides of the triangle are all strings in the circle, each string has a circumferential angle and a central angle, and its three inner angles in the circle "identity" are circumferential angles.
There are many vertical conditions and many right angles in the problem, so when constructing similar triangles, it is conventional thinking to give priority to right triangles. For example, when we were looking for the solution idea of the third problem, many students tried to use the method of cutting and filling, but after trying many times, they were never able to cut and make them into regular figures or graphs with a certain area, and this idea was crooked from the beginning.
Many middle school students have a problem, that is, they are smart, they think they have thought of a method that others can't think of, but they don't know math problems, the most particular thing is precisely the conventional method, down-to-earth is the best solution, this mentality remains the same, and it is extremely difficult to get out of the thinking dilemma of middle school students.
In fact, this question has already set up the "ladder" of thought, the second question is an obvious hint, in the actual process of solving the problem, there are many middle school students who can complete the second question, but they are basically trapped in the third question how to represent the area, when we read the question conditions, we should find that without giving any line segment length, it is already clear that this area is "not seek" or "not required", there is really no reason to spend more time on this dead end.
The best idea we expect is to start with a successful interpretation of the problem conditions, and when the students fail to solve the problem, we might as well ask them what they think, and find a way to break through the dilemma from the students' thinking, which is the guidance.
The guidance in the classroom cannot be said to be exactly the same as the guidance of problem-solving ideas, the former guidance is based on the teacher's understanding of the student's cognition to make the classroom behavior, which is actively imposed by the teacher to the student, while the latter is the corrective classroom behavior made after listening to the student's ideas, which is triggered by the student, according to the thinking of different students, the guidance is not the same, more personalized.
When the teacher finishes a topic in class, there must be a blank space, that is, the time for students to digest themselves, and in this process, do not wait passively, but pay attention to the guidance of thinking formation after solving the problem, simply put, it is to tell students that it is okay to think like this, and it is not okay to think like that, so as to prove it, and so on.
From the three steps of reading, solving, and reflecting, we can observe and understand the students, so as to understand the teaching.