"Moving point problem" refers to an open problem that there are one or more moving points in a graph moving on a straight line, ray or arc. The key to solve this kind of problem is to seek quietness in movement.

The key: seek quietness while moving.

Mathematical thought: classification thought function thought equation thought combination of number and shape thought transformation thought

First, the establishment of parsing function

Function reveals the change law between quantity and quantity in the process of motion change, and the moving point problem embodies a function idea. Due to the change of the conditional motion of a point or figure, the relationship between the unknown quantity and the known quantity is caused.

Firstly, the Pythagorean theorem is applied to establish the resolution function.

Example 1 (Shanghai, 2000) As shown in figure 1, there is a moving point P,PH⊥OA, vertical foot H and center of gravity △ OPH on the arc AB of a fan-shaped OAB with a radius of 6 and a central angle of 90.

(1) When point P moves on arc AB, are there any line segments with the same length among line segments GO, GP and GH? If yes, please point out such a line segment and find out the corresponding length.

(2) Set the resolution function of PH, GP, and write the definition domain of the function (that is, the value range of the independent variable).

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Figure 1

(3) If △PGH is an isosceles triangle, try to find the length of line segment pH. 。

Solution: (1) When the point P moves on the arc AB, the OP remains unchanged, so there are segments with the same length in the segments GO, GP and GH, that is, GH= NH= OP=2.

(2) Rt△POH at ∴.

In units of Rt△MPH,

.

∴= gp = MP =(0 & lt； & lt6).

(3)△PGH is an isosceles triangle. There are three possible situations:

① When gp = pH, the solution is the root of the original equation, which accords with the meaning of the question.

② When gp = GH, the solution is obtained, which is the root of the original equation, but it does not meet the meaning of the question.

③ When pH = GH,.

To sum up, if △PGH is an isosceles triangle, then the length PH of the line segment is or 2.

Secondly, the resolution function is established by using the proportional formula.

Example 2 (Shandong, 2006) As shown in Figure 2, in △ABC, AB=AC= 1, and points D and E move on a straight line BC. Let BD= CE=

(1) If ∠ BAC = 30 and ∠ DAE = 105, try to determine the resolution function between and;

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E

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Figure 2

(2) If the degree of ∠BAC is 0, and the degree of ∠DAE is 0, is the resolution function between (1) still valid when what relation is satisfied? Try to explain why.

Solution: (1) In △ABC, AB = AC, ∠ BAC = 30,

∴∠abd=∠ace= 105 ∴∠abc=∠acb=75。

∠∠BAC = 30，∠DAE= 105，∴∠DAB+∠CAE=75，

∠ DAB +∠ ADB =∠ ABC = 75，

∴∠CAE=∠ADB，

∴△ADB∽△EAC，∴，

∴ , ∴ .

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3( 1)

(2) Because ∠DAB+∠CAE=, and ∠DAB+∠ADB=∠ABC=, and the functional relationship is established,

∴ =, organized

At that time, resolution function was established.

Example 3 (Shanghai, 2005) is shown in Figure 3( 1), where △ABC, ∠ ABC = 90, AB = 4 and BC = 3. Point O is a moving point on the AC side, with the center of point O as a semicircle, tangent to the AB side, and the intersection line OC at point E.

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(1) Verification: △ADE∽△AEP.

(2) Let OA=, AP=, find the resolution function about, and write its domain.

(3) When BF= 1, find the length of the line segment AP.

Solution: (1) linkage OD.

According to the meaning of the question, OD ⊥ AB, ∴∠ ODA = 90, ∠ ODA = ∠ DEP.

And from OD=OE, we get ∠ ode = ∠ OED. ∴∠ ade = ∠ AEP，∴△ ade ∽△ AEP。

(2)∫∠ABC = 90，AB=4，BC=3，∴AC=5.∠∠ABC =∠ado = 90，∴∴od∥bc，，

∴OD=, AD =. ∴AE= =。

∴∴△ade∽△AEP。 ∴ ( ).

③ When BF= 1,

① If the extension line of EP intersection CB is at point F, as shown in Figure 3( 1), then CF=4.

* ade = AEP，∴∠PDE=∠PEC.∠∠FBP =∠DEP = 90，∠FPB=∠DPE，

∴∠F=∠PDE、∴∠F=∠FEC、∴CF=CE.

The obtained ∴5- =4, that is, AP=2.

② If the EP intersection CB is at point F, as shown in Figure 3(2), then CF=2.

Similar to ①, CF=CE can be obtained.

5-= 2, yes.

You can get AP=6.

To sum up, when BF= 1, the length of line segment AP is 2 or 6.

Thirdly, the function relationship is established by finding the graphic area.

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Figure 8

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Example 4 (Shanghai, 2004) shows that in △ABC, ∠ BAC = 90, AB = AC =, and the radius of ⊙ A is 1. If point O moves on the edge of BC (not coincident with point B and point C), let BO= and area bo =, and △AOC is.

(1) and write the domain of the function.

(2) When the point O is taken as the center and the length of BO is taken as the radius to make a circle O, and the tangent of ⊙O and ⊙A is found,

△AOC area.

Solution: (1) the intersection point a is AH⊥BC, and the vertical foot is H.

∵∠BAC=90，AB=AC=，∴BC=4,AH= BC=2。 ∴OC=4-。

∵ , ∴ ( ).

(2)① When ⊙O and ⊙A are circumscribed,

At Rt△AOH, OA=, OH=, ∴.

At this time, the area of △AOC =.

② When ⊙O and ⊙A are inscribed,

At Rt△AOH, OA=, OH=, ∴.

At this time, the area of △AOC =.

To sum up, when ⊙O is tangent to ⊙A, the area of △AOC is or.

Two: dynamic geometry problems

The characteristics of dynamic geometry-the background of the problem is special graphics, (special angle, the nature of special graphics, the special position of graphics. The problem of moving points has always been a hot spot in the senior high school entrance examination. In recent years, we have investigated the particularity of motion: isosceles triangle, right triangle, similar triangles, parallelogram, trapezoid, special angle or the maximum of its trigonometric function, line segment or area.

First, the topic with dynamic geometry as the main line

(1) inching.

1. As shown in the figure, the points in the middle are on the edge, with the point as the vertex, the edge intersects with the point, and the rays intersect with the point respectively.

(1) When, find the length;

(2) When the center length of a point is tangent to the center length of a point,

Seek long;

(3) When the diameter of the edge is tangent to the line segment, find the length.

[Problem background and discrimination measurement point]

Solution: (1) Prove ∽∴, substitute data, ∴AF=2.

(2) Let BE=, and then use the method of (1).

Tangent line considers two cases: tangent line, tangent line and tangent line;

Cut.

When ∴ and ⊙ are tangent, the length of is or.

(3) When the diameter of the edge is tangent to the line segment.

(2) Linear motion problem

In right-angle ABCD, AB = 3, point O is on diagonal AC, and straight line L passes through point O and intersects AC vertically. AD at point E (1) If the straight line L passes through point B, fold △ABE along the straight line L, and point A coincides with the symmetrical center A' of rectangular ABCD, and find the length of BC;

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(2) If the straight line L and AB intersect at point F, AO= AC, let the length of AD be and the area of pentagonal BCDEF be S.① Find the functional relationship about S and point out the value range;

② Inquiry: Whether there is such a circle whose center is A and whose length and radius are tangent to the straight line L, and if so, find the value; If it does not exist, please explain why.

(1)∵A' is the symmetry center of the rectangle ABCD ∴ A' B = AA' = AC.

∵AB=A'B,AB=3∴AC=6

(2)① , , ,

∴ ,

( )

② If circle A is tangent to straight line L, there is no such situation, so circle A is tangent to straight line L. 。

(3) the problem of surface movement

As shown in the figure, in, and, there are two moving points (and, and don't coincide) on the edge and the top respectively, and they are kept, and the edge is a square on different sides of the point.

(1) experimental area;

(2) When the sides coincide, find the side length of the square;

(3) Let the area of the overlapping part with the square be, find the functional relation about, and write the definition domain;

(4) When it is an isosceles triangle, please write the length directly.

Solution: (1).

(2) Let the side length of the square be, then solve.

(3) When,

When.

(4) .

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Known: in △ABC, AB=AC, ∠B=30? , BC=6, point D is on the side BC, point E is on the line segment DC, DE=3, △DEF is an equilateral triangle, and sides DF and EF intersect with sides BA and CA at points M and N respectively.

(1) Verification: △ BDM ∽△ cen;

(2) Let BD =, the overlapping area of △ ABC and △DEF be the resolution function of, and write the definition domain.

(3) When points M and N are on sides BA and CA respectively, whether there is a point D, so that the circle with M as the center and BM as the radius is tangent to the straight line EF, and if there is, find the value of X; If it does not exist, please explain why.

Example 1: It is known that the length of the chord AB of ⊙O is equal to the radius of ⊙O, and the point C changes on ⊙O (not coincident with A and B), so find the size of ∠ACB.

Analysis: Does the change of point C affect the change of ∠ACB size? We might as well change to point C. How? It can change on the upper arc AB or the lower arc AB. Obviously, the results are different. Then, when the point C changes on the upper arc AB, the arc ∠ACB is the lower arc AB, and its size is half that of the lower arc AB, so its central angle will naturally be thought of to connect AO and BO, so because AB=OA=OB, that is, the triangle ABC is an equilateral triangle, then ∠AOB=600, then the central angle of the same arc is equal to.

When point C changes on the lower arc AB, the arc opposite to ∠ACB is the optimal arc AB, and its size is half that of the optimal arc AB, which is obtained by ∠AOB=600, and the degree of the optimal arc AB is 3600-600=3000, which is obtained from the relationship between the central angle and the circumferential angle opposite to the same arc: ∠ ACB = 6500.

So there are two answers to this question, which are 300 or 1500.

Topic 3: double moving point problem

The problem of inching, linear motion and shape motion is called dynamic geometry problem, which mainly takes geometric figures as the carrier and motion changes as the main line to integrate multiple knowledge points into one problem. This kind of problem is comprehensive and requires high ability, which can comprehensively examine students' practical operation ability, spatial imagination ability and problem-solving ability. Among them, the problem of double-moving points, which is famous for its flexibility, has become a hot topic in this year's senior high school entrance examination. Now select a few examples for classification.

1 discuss the problem of function image with double moving points as the carrier.

For example, in 1 (Hangzhou, 2007) right-angled trapezoidal ABCD, ∠ c = 90, and the height CD=6cm (as shown in figure 1). Moving points P and Q start from point B at the same time. Point P moves along BA, AD and DC to stop at point C, and point Q moves along BC to stop at point C, with the same speed. Q starts at point b at the same time. When the elapsed time is t(s), the area of △BPQ is y(cm)2 (as shown in Figure 2). Establish rectangular coordinate system with t and y as abscissa and ordinate respectively. It is known that when the point p moves from a to d on the edge of AD, the function image of y and t is the line segment MN in fig. 3.

(1) Find the lengths of BA and AD in the trapezoid respectively;

(2) Write the coordinates of two points, M and N, in Figure 3;

(3) Write the functional relationship between Y and T when the point P moves on the Pasteur and DC sides respectively (indicate the range of independent variables), and complete the approximate image of the functional relationship between Y and X in the whole movement in Figure 3.

2. Explore the open question of conclusion with double moving points as the carrier.

Example 2 (Taizhou, 2007) as shown in Figure 5, Rt△A→B→C, ∠ B = 90, ∠ Cab = 30. The coordinates of its vertex A are (10,0), the coordinates of its vertex B are (5,53), and AB= 10.

(1) Find the degree of ∠ guarantee.

(2) When point P moves on AB, the function image between the area s (square unit) of △OPQ and the time t (second) is a part of a parabola (as shown in Figure 6), and the moving speed of point P is found.

(3) Find the functional relationship between the area s and the time t in (2), and the coordinates of the point p when the area s takes the maximum value.

(4) If the velocity in (2) is kept constant at point P and point Q, the size of ∠OPQ increases with the increase of time t when point P moves along the AB side; When moving along the BC side, the size of ∠OPQ decreases with the increase of time T. When point P moves along these two sides, how many points P make ∠ OPQ = 90? Please explain the reason.

Solution (1) ∠ Bao = 60.

(2) The moving speed of point P is 2 units/second.

3. Explore the existing problems with double moving points as the carrier.

Example 3 (Yangzhou City, 2007) As shown in Figure 8, in the rectangular ABCD, AD = 3cm, AB = A cm(A >；; 3). Moving points M and N start from point B at the same time and move along B→A and B→C respectively at the speed of 1cm/sec. When crossing m, a straight line is perpendicular to AB, and points AN and CD intersect at P and Q respectively. When point N reaches the end point C, point M stops moving. Let the exercise time be t seconds.

(1) if a = 4cm and t = 1 sec, then PM = cm.

(2) If a = 5cm, find the time t for making △PNB∽△PAD, and find their similarity ratio;

(3) If there is a moment during the movement that makes the areas of trapezoidal PMBN and trapezoidal PQDA equal, find the range of A;

(4) Is there such a rectangle? Are there equal moments in the areas of trapezoidal PMBN, trapezoidal PQDA and trapezoidal PQCN? If it exists, find the value of a; If it does not exist, please explain why.

4. Taking double moving points as the carrier, the function extremum problem is explored.

Example 4 (Jilin Province in 2007) As shown in Figure 9, in a square ABCD with a side length of 82cm, E and F are two moving points on the diagonal AC, starting from point A and point C at the same time, respectively, and moving along the diagonal at the same speed of 1cm/s, and the right-angled side of the intersection point E of EH vertical AC and Rt△ACD is at H; Let f be the right angle of FG perpendicular to AC and Rt△ACD at G, connecting HG and EB. Let the graphic area surrounded by he, EF, FG and GH be s? 1, AE, EB, BA, what is the graphic area enclosed by S? 2 (here the area of the line segment is specified as 0). E stops when it reaches C, and F stops when it reaches A. If the movement time of E is x(s), answer the following questions:

(1) When 0

(2)① If y is S? 1 and s? 2. Find the functional relationship between Y and X; (Figure 10 is a standby figure)

② Find the maximum value of y 。

The quadrilateral of the solution (1) with vertices e, f, g and h is a rectangle. Because the side length of the square ABCD is 82, AC= 16. If b is BO⊥AC in O, OB=89. Because AE=x, s? 2=4x, because he =AE=x, EF= 16-2x, so s? 1=x( 16-2x), when s? 1=S？ 2,4x = x (16-2x), the solution is x 1=0 (truncation), x2=6, so when x=6, s? 1=S？ 2.

(2)① When 0 ≤ x

When 8≤x≤ 16, AE=x, CE=HE= 16-x, ef =16-2 (16-x) = 2x-16,

So s? 1= (16-x) (2x-16), so y = (16-x) (2x-16)+4x =-2x2+52x-256.

② When 0 ≤ x

When 8≤x≤ 16, y =-2x2+52x-256 =-2 (x-13) 2+82,

So when x= 13, the maximum value of y is 82.

To sum up, the maximum value of y is 82.

Comment on this problem is a function maximum problem created with double moving points as the carrier and squares as the background. Students are required to read the question carefully, understand the meaning of the question, draw graphs in different situations, establish the relationship between time variables and other related variables according to the graphs, and then construct the functional expression of area. This topic focuses on the maximum value of quadratic function in knowledge points, which requires students to have solid basic knowledge, flexible problem-solving methods and good thinking quality. Pay attention to the combination of numbers and shapes, classified discussion and flexible application of mathematical modeling in solving problems.

Fourth, the similar triangles problem caused by moving points in the function.

This example is shown in figure 1. It is known that the vertex of the parabola is A(2 1), passes through the origin O, and the other intersection with the X axis is b.

(1) Find the analytical formula of parabola; (The analytical formula of parabola obtained from vertex is)

(2) If point C is on the parabola symmetry axis, point D is on the parabola, and the quadrilateral of the four vertices of O, C, D and B is a parallelogram, find the coordinates of point D;

⑶ Connect OA and AB, as shown in Figure 2. Is there a point P on the parabola below the X axis that makes △OBP similar to △OAB? If it exists, find the coordinates of point P; If it does not exist, explain why.

Example 1 Theme Map

Figure 1

Figure 2

Analysis: 1. When two vertices of a quadrilateral are given, we should consider that the line connecting the two vertices is the side and diagonal of the quadrilateral. A quadrilateral with four vertices O, C, D and B is a parallelogram. There are two kinds of discussion: the side and diagonal of OB.

2. There are generally three solutions to the similar triangles problem caused by fixed points in functions.

① To find the third vertex of similar triangles, we must first analyze the characteristics of the sides and angles of the known triangle, and then get whether the known triangle is a special triangle. According to the known edges in the unknown triangle and the possible corresponding edges of the known triangle, the discussion is classified.

(2) Or using the corresponding angles in the known triangle, using the knowledge of Pythagorean theorem, trigonometric function, symmetry, rotation and so on, the size of the middle side of the unknown triangle can be deduced.

(3) If the sides of two triangles are not given, first set the coordinates of the points to be solved, then use the resolution function to represent the length of each side, and then use the similarity to solve the equations.

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