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Mathematics problem solving in senior high school entrance examination
When it comes to the grand finale of mathematics, people usually think that it is comprehensive and difficult, and it is not difficult to get a few points or full marks. Whether the finale can be solved smoothly is one of the most critical factors to determine the score of the senior high school entrance examination. Therefore, studying the problem-solving skills of the finale has become a hot topic in junior high school mathematics teaching research.

Let's take an example to talk about how to decompress the shaft.

In the plane rectangular coordinate system, the side AB of the rectangular ABCD is 2, and the AD is 1. AB and AD are on the positive semi-axis of X axis and Y axis respectively, and point A coincides with the coordinate origin. Fold the rectangle so that the point A falls on the edge DC, and let A' be the corresponding point where the point A falls on the edge DC.

(1) When the rectangular ABCD is folded along the straight line y =-x+b,

(as shown in figure 1), find the coordinates of point A' and the value of b;

(2) When the rectangular ABCD is folded along the straight line y =kx +b,

① Find the coordinates of point A' (represented by K) and the relationship between K and B; ② If we divide the position of the straight line and rectangle where the crease is located into three situations as shown in Figure 2( 1), (2) and (3), please write down the value range of k in each case.

1. Read and review the questions carefully and dig up useful information.

This paper takes a rectangle as a model, organically combines axisymmetric transformation with coordinate plane through folding, and explores the coordinates of points, the relationship between variables and the range of independent variables. * * * is similar in that the symmetrical point A' of point A falls on DC after being folded, so determining the coordinates of point A' is the starting point of thinking. Look at Figure 2. What can we get from observing the three folding phenomena? Oh! Graphic language tells us that the straight line intersects the rectangle at three different positions, which reflects the various values of k when the position of the symmetrical point changes.

2. Establish organic connection and make reasonable judgment and reasoning.

No matter between (1) and (2)① or (2)②, the design of the three problems all embodies the idea from special to general, so there must be a connection in solving problems. Therefore, we should start with special circumstances first, and then solve all the problems.

As in (1), the method of finding the coordinates of a' and the value of b provides almost the same idea for solving (2)①.

For example, when the symmetry point A' of point A coincides with point C and point D, the value range of k can be determined by finding the value of k..

3. Choose the solution method according to the requirements of the topic

In order to understand (1), we might as well try to set the coordinates of point a' and the intersection of the straight line where the crease is located and the two coordinate axes to see if there is any breakthrough. Got it! It is easy to find that △DOA' is similar to △OFE. So we can get the coordinates of point A', and then we can easily get the value of B by using Pythagorean theorem in Rt△DEA'. In (2) (2), these discrete symmetrical points actually move continuously from C to D, and their essence is a dynamic graphic problem. We can grasp the invariants (here is the special point) and discuss the classification accordingly.

4. Seriously organize the language science and standardize the answers.

What is the basis for implementing the established problem-solving plan? Through expression and scientific and standardized answers, it is an important part of solving problems to clearly show your ideas to others and get a fair evaluation.

The following is the answer to this question for readers' reference.

(1) As shown in Figure 3, let a straight line Y = 1/2-X+B intersect with OD at point E, intersect with OB at point F, and connect A'O, then OE=b, OF=2b, and the coordinate of point A 'is (a, 1).

Because ∠DOA'+∠A'OF=90O, ∠OFE+∠A'OF=90O,

So ∠DOA'=∠OFE, so △DOA '∩△OFE.

So =, that is =. So a=

So the coordinate of point A' is (,1).

Connect A'E, then A'E=OE=b,

In Rt△DEA', according to Pythagorean theorem, A'E2=A'D2+DE2,

That is b2=()2+( 1-b)2, and b=.