Solution questions occupy a considerable proportion in the senior high school entrance examination, mainly composed of comprehensive questions. As far as the types of questions are concerned, they include calculation questions, proof questions and application questions. Their characteristics and examination functions determine the complexity of thinking and the diversity of problem-solving design. Generally speaking, the problem-solving design depends on the method of solving the problem, whether it is overall consideration or local association. The principles of determining the method are: familiarity principle and concreteness principle; Simplification principle, harmony principle, etc.

(1) To solve the comprehensive and final questions, we should grasp the following links:

1. Examination: This is the beginning and basis for solving the problem. It is necessary to comprehensively investigate all the conditions and answering requirements of the question type, so as to correctly and comprehensively understand the meaning of the question, grasp the characteristics and structure of the question type as a whole, and facilitate the selection of problem-solving methods and the design of problem-solving steps.

We should grasp the "three characteristics" in the thinking of examining questions, that is, to clarify the purpose, improve the accuracy and pay attention to the meaning. The practice of solving problems shows that conditional suggestion can recognize and inspire the means of solving problems, and the conclusion can predict and guide the direction of solving problems. Only by carefully examining the questions can we get as much information as possible from the questions themselves. Don't be afraid of slowness in this step. In fact, there is "quickness" in "slowness". The direction of solving the problem is clear and the means of solving the problem are reasonable and appropriate.

2. Seeking reasonable ideas and methods to solve problems: Breaking away from the model and striving for innovation are the remarkable characteristics of the mathematics test questions in the senior high school entrance examination in recent years, especially solving problems. Therefore, it is forbidden to apply mechanical mode to seek ideas and methods for solving problems, but to identify the conditions and conclusions of problems from different sides and angles, and to understand the relationship between conditions and conclusions, the relationship between geometric characteristics of figures and the number and structural characteristics of numbers and formulas. Carefully determine the ideas and methods to solve the problem. When thinking is blocked, we should adjust our thinking and methods in time, re-examine the meaning of the question, pay attention to excavating the implied conditions and internal relations, and prevent us from falling into a dead end and giving up easily.

(2) Problem analysis

1 type linear geometry synthesis problem

The common examination forms of this kind of questions are reasoning and calculation. For reasoning, the basic idea is analysis and synthesis, that is, from the conclusions that need to be proved, find the conditions that make them valid, and at the same time deduce some conclusions from the known conditions and try to connect them. For calculation, the basic idea is to use the quantitative relationship between geometric elements (such as edges and angles) combined with the idea of equations to deal with it.

For example 1 (Neijiang, Sichuan, 2007), as shown in figure 1, in the middle, the moving point (not coincident with point A and point C) is on the side and intersects with the point.

(1) When the area of is equal to the area of a quadrilateral, find the length;

(2) When the perimeter of is equal to the perimeter of a quadrilateral, find the length;

(3) Is there a point in the world that makes it an isosceles right triangle? If it does not exist, please briefly explain the reason; If it exists, the request is long.

Analysis: The area equation in (1) can be converted into "the area ratio with △ACB is 1: 2", because △ECF∽△ACB needs a long length and can be solved by the relationship between similarity ratio and area ratio. Because the sides corresponding to similar triangles are proportional, we can use proportional line segments to find the relationship between line segments in the second question.

Solution: (1) Because the area of △ECF is equal to the area of quadrilateral EABF, S △ ECF: S △ ACB = 1: 2, and because EF∨AB, so △ECF∽△ACB. So, because CA = 4, CE =.

(2) Let the length of CE be x, because △ECF∽△ACB, so. So CF= According to the perimeter equation, we can get:. Solution.

(3)△EFP is an isosceles right triangle, and there are two cases:

① As shown in Figure 2, assume ∠ EP = EF = 90, and EP = ef. From AB = 5, BC = 3, AC = 4, ∠ C = 90,

So the high CD on the hypotenuse AB of Rt△ACB = Let EP = EF = X, which is obtained from △ECF∽△ACB.

, that is, the solution, that is, ef =

When ∠ EF = FP = 90, EF = FP and ef =.

② As shown in Figure 3, if ∠ PE = PF = 90 and PE = PF, the distance from point P to EF is.

Let ef = x, from △ECF∽△ACB, we get

, that is, the solution, that is, ef =

To sum up, there is a little p on AB, which makes △EFP an isosceles right triangle, and ef = or ef =.

Special note: Because the hypotenuse is not specified in the isosceles right triangle, it is necessary to discuss the possible situation.

Tracking exercise 1 (Yantai, Shandong, 2007) As shown in Figure 4, in the isosceles trapezoid ABCD, AD∑BC, point E is a moving point on the line segment AD (E does not coincide with A and D), and G, F and H are the midpoint of BE, BC and CE respectively.

(1) Try to discuss the shape of quadrilateral EGFH and explain the reasons.

(2) When the point E moves to what position, the quadrilateral EGFH is a diamond? And prove it.

(3) If the diamond-shaped EGFH in (2) is a square, please explore the relationship between line segment EF and line segment BC to prove your conclusion.

Reference answer: 1, (1) quadrilateral EGFH is a parallelogram. Just specify GF//EH and GF = EH.

(2) When point E is the midpoint of AD, the quadrilateral EGFH is a diamond. BE=CE can be obtained by congruence, and then EG = EH can be obtained.

According to the fact that EGFH is square, it can be concluded that EGFH = eh, ∠ BEC = 90. Because g and h are the midpoint of BE and CE respectively, EB = EC.

Because f is the midpoint of BC,

Type 2. Synthesis problem of circle

The common form is the synthesis of reasoning and calculation, and the basic idea of solving is still analysis-synthesis. It should be noted that because of its comprehensiveness, it is often necessary to make full use of the previous conclusions when answering the latter questions, which will be simple and convenient.

Example 2 (Maoming, Guangdong Province, 2007) As shown in Figure 5, points A, B, C and D are four points on ⊙O with a diameter of AB, C is the midpoint of the bad arc, and AC and BD meet at point E, AE = 2, EC = 1.

(1) Verification: ∽.

(2) Try to explore whether the quadrilateral ABCD is trapezoidal. If yes, please give proof.

And find its area; If not, please explain why.

(3) Extend AB to H so that BH =OB .. Prove that CH is the tangent of ⊙ O. 。

Analysis: (1) Just prove it; (2) If we want to judge that it is trapezoidal, we only need to explain DC∨AB. It is noted that there are many quantitative relations in the known conditions, so it is considered to explain them from the perspective of equilateral: first find DC, and then explain that OBCD is a diamond; (3) To prove that "CH is the tangent of ⊙O", just prove ∝.

Solution: (1) Because c is the midpoint of an arc, because ∠DCE=∠ACD.

therefore ...

(2) The quadrilateral ABCD is a trapezoid.

Proof: The connection was made by (1). Because, therefore, it is known. Because its diameter is ⊙υO, so, so, so the quadrilateral OBCD is a diamond. So, so the quadrilateral ABCD is a trapezoid.

If c and CF are connected perpendicular to AB and OC at point f, then, so.

So CF=BC×sin60 = 1.5.

So ...

(3) It is proved that the quadrilateral OBCD obtained from (2) is rhombic when connecting OC intersection BD to point G,

So we also know that OB = BH, so BH is parallel and equal to CD, so the quadrilateral BHCD is a parallelogram, so, so, so CH is the tangent of ⊙ O. 。

Special note: in reasoning, sometimes it may be necessary to use calculation to help prove, such as DC∨AB in this question.

Tracking exercise 2.

(Mianyang, Sichuan, 2007) As shown in the figure, AB is ⊙O, ∠BAC = 60,

P is a point on OB, and AB crosses the vertical line of P and intersects the extension line of AC at Q,

The tangent CD passing through point C intersects PQ at D and connects OC.

(1) proves that △CDQ is an isosceles triangle;

(2) If △ CDQ △ COB, find the value of BP:PO.

Reference answer: 2( 1) is known as ∠ACB = 90, ∠ABC = 30, ∴ ∠Q = 30, ∠ BCO = ∠ ABC = 30.

∵ CD is the tangent of⊙ O, CO is the radius, ∴ CD⊥CO, ∴ ∠DCQ =30, ∴ ∠DCQ =∠Q,

So △CDQ is an isosceles triangle.

(2) Let the radius of ⊙O be 1, then AB = 2, OC = 1, AC = AB∕2 = 1, BC =.

∫△cdq?△cob，∴ CQ = BC =。 So AQ = AC+CQ = 1+,

Then AP = AQ∕2 =( 1+)∕2, ∴ BP = AB-AP = (3-) ∕ 2,

PO = AP-AO =( - 1)∕2,∴ BP:PO =。

Type 3. Algebraic (or Geometric) Synthesis Problems and Statistics (or Probability)

This kind of questions are generally knowledge series questions, so let's break them one by one.

Example 3. (Jiangxi, 2007) In a math activity, a figure as shown in the figure was drawn on the blackboard. Before the activity, the teacher wrote down one of the following four equations on four pieces of prepared paper:

① ② ③ ④

Xiao Ming closes his eyes, randomly selects one from four pieces of paper, and then randomly selects another from the remaining pieces of paper. Please answer the following two questions with pictures:

(1) When drawing ① and ②, can ① and ② be used as conditions for judging whether they are isosceles triangles? Tell me your reasons;

(2) Please use a tree diagram or table to represent all possible results (represented by serial numbers) of the equation extracted from two pieces of paper, and find the probability that an isosceles triangle cannot be formed under the condition of the equation extracted from two pieces of paper.

Analysis: (1) Just stated BE=CE, so consider proof. (2) If it is not necessarily true, it is not necessarily an isosceles triangle. Then the solution can be obtained according to the definition of probability.

Solution: (1) Yes. Reason: by,

,

get ..

It is an isosceles triangle.

(2) Tree diagram:

All possible outcomes (① ②) (① ③) (① ④) (② ①) (② ③) (② ④) (③ ①) (③ ②) (③ ②) (④ ①) (④ ②) (④ ③).

The equations on two pieces of paper have 12 equal possibility results, and four of them ((① ③), (③ ①), (② ④) and (④ ②)) cannot form an isosceles triangle, so the probability of not forming an isosceles triangle is.

Special note: there are two situations where ""cannot be obtained. One is that "edges and corners" can't get congruence, and the other is that you can only get similarity.

Tracking exercise 3. Shenyang, Liaoning, 2007. In the three geometric bodies A, B and C shown in the figure, the direction shown by the arrow is taken as their front faces, and the front views of the three geometric bodies A, B and C are respectively A 1, B 1 and C1; The left views are A2, B2 and C2 respectively; The top views are A3, B3 and C3 respectively.

(1) Please write down the names of A 1, A2, A3, B 1, B2, B3, C 1, C2 and C3 respectively;

(2) Xiao Gang first drew these nine views on nine cards with the same size and shape, put three cards A 1, A2 and A3 in pocket A, three cards B 1, B2 and B3 in pocket B, and three cards C 1, C2 and C3 in pocket C..

① Find out the probability that the graphic names on three cards randomly selected by Xiao Liang are all the same by completing the tree diagram below;

(2) Xiao Liang and Xiaogang play games. According to the rules of the game, Xiao Gang wins when only two of the three cards randomly selected by Xiao Liang have the same graphic names. When the graphic names on the three cards are completely different, Xiao Liang wins. Is this game fair to both sides? Why?

Solution: (1) A B C

(2)① Tree diagram:

Reference answer: 3( 1) is known as A 1, A2 is a rectangle and A3 is a circle; B 1, B2 and B3 are all rectangles;

C 1 is a triangle, C2 and C3 are rectangles.

(2)① Complete the tree diagram as follows:

It can be seen from the tree diagram that * * * has 27 possible results, among which there are 12 results with the same graphic names on three cards, and the probability of the same graphic names on three cards is 1227 = 49.

The game is unfair to both sides. According to ①, P (Xiao Gang wins) = 49. The probability that the graphic names on the three cards are completely different is 19, that is, p (Liang Xiaosheng) = 19. This game is unfair to both sides.

Type 4. Functions (Equations) in Graphs

This kind of problem usually needs to be solved by the idea of equations and functions. Specifically, the relationship is often established by the formula of line segment proportion or area, and then solved by solving the equation or using the properties of the function.

Example 4. (Linfen, Shanxi, 2007) As shown in the figure, it is known that the sides of a square and a square are sum, and their centers are on a straight line, which intersects a point. When the square moves to the left along the straight line at a speed of 1 unit per second, the square also starts to rotate clockwise every second, and their shapes and sizes remain unchanged during the movement.

(1) Before starting exercise;

(2) When two squares move in their own ways at the same time.

When the movement lasts for 3 seconds, the square stops rotating. At this time,

, ;

(3) When the square stops rotating, the time for the square to move to the left is seconds, and the area of the overlapping part of two squares is, and the functional expression between and is obtained.

Analysis: (1), so (2) when moving for 3 seconds, point A falls on the floor at this time, so AE= =0. (3) The overlapping part is a square, as long as its side length is represented by X, note that the side length is different in different situations, which needs to be discussed.

Solution: (1) 9. (2) 0,6.

(3) When the square stops moving and continues to move to the left, the shape of the overlapping part with the square is also square. The functional relationship between the area of the overlapping part and the square should be divided into four cases:

① As shown in figure 1, the functional relationship between when, and is.

② As shown in Figure 2, when 4≤x≤8, the functional relationship with is y = 8.

③ As shown in Figure 3, when 8

(4) When the functional relationship between, and is.

Special note: (1) This problem is also a transformation problem. When calculating and proving, we should grasp the invariant elements in the transformation (such as equal angles, equal sides, congruences of figures, etc.). ) to deal with. If there are multiple right angles, the quantitative relationship can be established from the perspectives of similarity, trigonometric function and pythagorean theorem. (2) The problem of piecewise function in graphic changes can be discussed separately from the perspective of graphic characteristics.

Tracking exercise 4 (Hebei, 2007) as shown in the figure, in the isosceles trapezoid ABCD, AD∑BC, AB=DC=50, AD=75, BC = 135. Point P starts from point B and moves at a uniform speed of 5 units per second along the dotted line BA-AD-DC to point C; Point q starts from point c and moves in the direction of CB at a constant speed of 3 units per second. After passing through point Q, the light QK⊥BC goes up, and the intersecting line segment CD-DA-AB starts to move at point E, and points P and Q start to move at the same time. When point P coincides with point C, point Q also stops. Set points p and q move for t seconds (t > 0).

(1) When point P reaches the end point C, find the value of t and point out the length of BQ at this time;

(2) When point P moves to AD, why does the value of t make P Q∑DC?

(3) Let the area of light QK sweeping the trapezoidal ABCD be s, and find out points E respectively.

Functional relationship between s and t when moving to CD and DA; (Don't write the range of T)

(4) Can △ PQE become a right triangle? If yes, write the range of t; If not, please explain why.

Reference answer:

4: (1) T = 35 (seconds), and the length BQ from point P to terminal C is 135- 105 = 30.

(2) If PQCD, AD∨BC, the quadrilateral PQCD is a parallelogram, so that PD=QC, from QC=3t, BA+AP=5t, 50+75-5t = 3t,

The solution is t= When t =, there is PQ∨DC.

(3)① When point E moves on the CD, s = s ⊿ QCE = QEQC = 6t2;

② When point E moves on DA, S= S trapezoid QCDE = (ed+QC) DH = =120t-600.

(4) △PQE can be a right triangle. When △ PQE is a right triangle, the range of t is 0 < t ≤ 25 and t

≠ or t = 35

Tracking exercise 5 (Yangzhou, Jiangsu, 2007) is shown in the figure, in the rectangle, cm, cm (). At the same time, the moving point starts from this point and moves along and at the speed of cm/s. When the lines are perpendicular and intersect, it stops moving when the point reaches the end point. Set the moving time to seconds.

(1) If cm, seconds, then _ _ _ _ _ cm.

(2) If cm, find the time and work, and find their similarity ratio;

(3) If there is a moment when the trapezoid and the trapezoid area are equal during the movement, the range of values to be obtained;

(4) Is there such a rectangle? Is there a moment when the trapezoid, trapezoid and trapezoid area are all equal during the movement? The value of, if it exists; If it does not exist, please explain why.

Reference answer: 5, (1), (2), similarity ratio.

(3)△AMP∽△ABN can get PM=,, simplified, and, 3 < a ≤ 6.

(4) If the area of the trapezoid is equal to the area of the trapezoid, substitute it to get (negative value).

Type 5. Parabolic figure

Generally speaking, most of these problems are final, and the basic idea of solving them is analysis and synthesis. In addition to using the core knowledge of algebra and geometry flexibly, we should also pay attention to the application of basic mathematical thinking methods such as classification, combination of numbers and shapes, and transformation.

Example 5 (Henan, 2007) As shown in the figure, a parabola with straight symmetry axis passes through point A (6 6,0) and point B (0 0,4).

(1) Find the parabolic analytical formula and vertex coordinates;

(2) Let point E (,) be the moving point on the parabola, located in the fourth quadrant, and the quadrilateral OEAF is a parallelogram with OA as the diagonal. Find the functional relationship of the area s sum of parallelogram OEAF, and write the range of independent variables.

① When the area of the parallelogram OEAF is 24, please judge whether the parallelogram OEAF is a diamond?

② Is there a point E that makes the parallelogram OEAF square? If it exists, find the coordinates of point e; If it does not exist, please explain why.

Analysis: (1) The analytical expression of parabola can be obtained by using the undetermined coefficient method, and (2) The functional relationship can be established by using the area formula of parallelogram OEAF. ① The key to judge whether OEAF is rhombic is to see whether the adjacent sides are equal from the known conditions, that is, the area relationship needs to be transformed into the line segment relationship. ② Assuming that there is a qualified E, consider first satisfying the condition of "making OEAF square" and then seeing whether other conditions can satisfy "parabola".

Solution: (1) From the axis of symmetry of parabola, the analytical formula can be set as. Substitute the coordinates of a and b into the above formula, the analytical formula of parabola is and the vertex is.

(2) Because the point is on the parabola, located in the fourth quadrant, and the coordinates are appropriate,

So y < 0, that is, -y > 0, and -y represents the distance from point E to OA. Because OA is diagonal,

So ...

Because the abscissas of parabola and X-axis focus are x 1= 1 and x2=6, respectively. And point e is in the fourth quadrant, and the ordinate of point e is less than 0, so the abscissa of point e is 1

The value range of. Yes 1 < < 6.

(1) According to the meaning of the question, when S = 24, point E is solved.

There are two, E 1(3, -4) and E2(4, -4). Point E 1(3, -4) satisfies OE = AE, so it is a diamond; Point E2(4, -4) does not satisfy OE = AE, so it is not a diamond.

② When OA⊥EF and OA = EF, it is a square. At this time, the coordinates of point E can only be (3, -3).

The point with the coordinate (3, -3) is not on the parabola, so there is no such point E, so it is a square.

Special note: when several conditions need to be met at the same time, it is better to meet some of them first, and then see if other conditions are met.

Tracking Exercise 6 (Shenyang, Liaoning Province, 2007). It is known that the parabola Y = AX2+BX+C intersects with the X axis at points A and B, and intersects with the Y axis at point C, where point B is on the positive semi-axis of the X axis and point C is on the positive semi-axis of the Y axis, and the line segment lengths OB and OC (OB

(1) Find the coordinates of points A, B and C;

(2) Find the expression of this parabola;

(3) Connect AC and BC, if point E is a moving point on AB line.

(not coincident with point A and point B), point E is EF∑AC, and point BC is point F,

Connect CE, let the length of AE be m, and the area of △CEF be s, find the functional relationship between s and m, and write the value range of independent variable m;

(4) On the basis of (3), try to explain whether there is a maximum value of S, if so, request the maximum value of S, find the coordinates of point E at this time, and judge the shape of △BCE at this time; If it does not exist, please explain why.

Reference answer:

6.( 1) point B (2 2,0), point C (0 0,8), point a (-6,0), (2) the expression of parabola is y =-23x2-83x+8, (3) by EFAC=BEAB because AC = = 65438.

Let FG⊥AB and the vertical foot be g, then sin∠FEG=sin∠CAB=. So in Rt△EGF, fg = ef sin ∠ feg = 45.40-5m4 = 8-m, so s = =-65438.

(4) Existence, because S =- 12m2+4m and A =

△BCE is an isosceles triangle.