Several math problems (answers) in the third grade last semester are best Olympic math problems.

Seven, (this question out of 7 points)

23. It is known that the equation about X has two real roots, and the equation about Y has two real roots, and. When, find the value range of m.

Eight, (this question out of 8 points)

24. it is known that AB is the diameter of semicircle o, point c moves on the extension line of BA (point c does not coincide with point a), semicircle m with diameter OC intersects semicircle o at point d, and bisector of ∠DCB intersects semicircle m at point e. ..

(1) Verification: CD is the tangent of semicircle O (Figure1);

(2) Let EF⊥AB be at point F (Figure 2), guess that EF is equal to half of the existing line segment, and prove it;

(3) Under the above conditions, the parallel line passing through point E, CB, intersects with CD at point N. When NA is tangent to semicircle O (Figure 3), find the tangent value of ∠EOC.

Figure 1

Figure 2

Figure 3

23. Solution: The equation about x has two real roots, x 1 and x2.

Solve (1)

The equation about y has two real roots.

The solution is 0≤n≤4.

Relationship between dependent root and coefficient

Pack up, take it

It can be obtained from the image of quadratic function.

When 2

The range of m obtained from ① and ② is

Eight,

24.( 1) Prove: As shown in figure 1, connect OD, and OD is the radius of semicircle O.

Figure 1

∫OC is the diameter m of the semicircle.

∴∠CDO=90

∴CD is the tangent of semicircle O.

(2) guess:

Proof 3: As shown in the figure, connect OD and ME, and OD and ME intersect at H point.

Ce sharing ∠DCB

∴ ∴ME⊥OD,OH

∵EF⊥CO ∴∠MFE=∠MHO=90

∫∠ EMF =∠OMH, ME=MO

∴△MEF≌△MOH

∴EF=OH ∴

(3) Solution: As shown in Figure 3, extend the OE intersection CD to point K..

Figure 3

Let OF=x and EF=y, then OA=2y.

∵NE//CB, EF⊥CB, NA tangent semicircle O at point A.

∴ Quadrilateral Affin is a rectangle.

∴

Like the first proof of (2), e is the midpoint of OK.

∴N is the midpoint of CK

∴Rt△CEF∽Rt△EOF

∴

solve

∴tan∠EOC=3

25.( 1) solution: ∫ The parabola intersects the X axis at points A and B.

The equation about x has two unequal real roots.

solve

∵ point a is to the left of point b, m >；; 0,∴A(-m,0),B(2m,0)

Solution 2: As shown in Figure 2, the O-crossing point is OG//AC, and it is at G-point.

Figure 2

∴△CED∽△OGD ∴

Dc = duo ∴CE=OG

∫og//AC ∴△bog∽△bae∴

ob = 2m，AB=3m ∴

(3) Scheme 1: as shown in Figure 3.

Figure 3

∵ point c is on a parabola (not coincident with point a), and the distances between point c and point a and y axis are equal.

∴C(m,2m2)

Crossing point e is the high EP of DC edge, and crossing point a is the high AQ of OC edge.

∴EP//AQ

∴△CEP∽△CAQ

∴

∵

∴

The solution is m=2.

∴ The analytical formula of parabola is

The coordinates of point C are (2,8) and point B is (4,0).

Points D and C are perpendicular to the X-axis, and the X-axis intersects with points M and N respectively.

∴DM//CN

D is the midpoint of OC.

∴

∴ The coordinate of point D is (1, 4).

The analytical formula of a straight line is as follows

The analytical formula of ∴ straight line BE is

Solution 2: Connect OE as shown in Figure 4.

Figure 4

D is the midpoint of OC.

∴

The following is the same as the solution of (3) 1

23. As shown in Figure ①, OP is the bisector of ∠MON. Please draw a pair of congruent triangles with the line where OP is located as the symmetry axis. Please refer to congruent triangles's method and answer the following questions:

(1) As shown in Figure ②, in △ABC, ∠ACB is a right angle, ∠ B = 60, AD and CE are bisectors of ∠BAC and ∠BCA respectively, and AD and CE intersect at point F. Please judge and write the quantitative relationship between FE and FD.

(2) As shown in Figure ③, in △ABC, if ∠ACB is not right-angled, and other conditions in (1) remain unchanged, is your conclusion in (1) still valid? If yes, please prove it; If not, please explain why.

24. It is known that the parabola y=ax2+bx+c intersects the Y axis at point A (0 0,3) and the X axis at points B (1 0) and C (5 5,0) respectively.

(1) Find the analytical expression of this parabola;

(2) If point D is the bisector of line segment OA, find the analytical formula of straight line DC;

(3) If a moving point P starts from the midpoint m of OA, it first reaches a point on the X axis (set as point E), then reaches a point on the parabola symmetry axis (set as point F), and finally moves to point A. Find the coordinates of point E and point F which make the total path of point P shortest, and find the length of this shortest total path.

25. We give the following definition: If two diagonals of a quadrilateral are equal, it is called an equidiagonal quadrilateral. Please answer the following questions:

(1) Write the names of two kinds of figures of an equidiagonal quadrilateral in the special quadrilateral you have learned;

(2) Inquiry: When the acute angle of two diagonals in an equidiagonal quadrilateral is 60, prove the relationship between the sum of two sides facing the 60-degree angle and one of the diagonals, and prove your conclusion.

23. Solution: (1) The quantitative relationship between Fe and FD is Fe = FD.

(2) The conclusion Fe = FD in A: (1) still holds.

Proof 1: As shown below, intercept AG = AE on AC and connect FG.

Because ∠ 1 = ∠ 2, AF is the man.

Prove △ AEF △ AGF

So ∠ AFE =∠ AFG, Fe = FG.

When ∠ b = 60, AD and CE are bisectors of ∠BAC and ∠BCA, respectively.

Available < 2+< 3 = 60

So ∠ AFE =∠ CFD =∠ AFG = 60.

So ∠ CFG = 60.

If ∠ 3 = ∠ 4 and FC is a total of * * * sides, △ CFG △ CFD can be obtained.

So fg = FD

So Fe = FD

24. Solution: (1) According to the meaning of the question, C = 3.

therefore

solve

So the parabolic analytical formula is

(2) According to the meaning of the question, the bisectors of OA are (0, 1) and (0,2) respectively.

Let the analytical formula of straight CD be

When the coordinate of point D is (0, 1), the analytical formula of straight line CD is

When the coordinate of point D is (0,2), the analytical formula of straight line CD is

(3) As shown in the figure, from the meaning in the question, you can get

The symmetrical point of point m about x axis is

The symmetry point of point A about the parabola symmetry axis is a' (6,3).

Link A'M'

According to the axial symmetry and the shortest line segment between two points, the length of A'M' is the demand.

The length of the shortest total path for the movement of point P.

Therefore, the intersection of A'M' with the X axis is the point of E, and the intersection with the straight line X = 3 is the point of F. ..

The analytical formula of the straight line A'M' can be obtained by the following formula

The coordinates of point E and point F are (2,0) and (3,0) respectively.

From Pythagorean theorem, we can find that

Therefore, the length of the shortest total path (me+ef+fa) of point P is.

25. Solution: (1) omitted.

(2) Conclusion: When the acute angle between two diagonals in an equidiagonal quadrilateral is 60, the sum of the two sides facing the 60-degree angle is greater than or equal to the length of one diagonal.

It is known that in quadrilateral ABCD, diagonal AC and BD intersect at point O, and AC = BD.

And < aod = 60.

Verification: BC+AD ≥ AC

It is proved that the intersection d is DF‖AC, and the DE interception on DF makes DE = AC.

Add CE and BE

So ∠ edo = 60, quadrilateral ACED is a parallelogram.

So △BDE is an equilateral triangle, and CE = AD.

So de = be = AC

① When BC and CE are not in a straight line (as shown below)

△BCE, BC+CE > BE.

So BC+ad > AC

② When BC and CE are on the same straight line (as shown below)

Then BC+ce = be

Therefore BC+ad = AC.

Combining ① and ②, BC+AD ≥ AC is obtained.

That is, when the acute angle between two diagonals in an equidiagonal quadrilateral is 60, the sum of the two sides facing the 60-degree angle is greater than or equal to the length of one of the diagonals.

As shown in the figure, as we all know.

(1) Please sum the midpoint of two points (on the side respectively.

Outside), links, and writing make this graph have only two opposite sides.

Corresponding conditions of triangles with equal products, and show that the areas are equal.

Triangle;

(2) According to the corresponding conditions that make (1) hold,

certificate

As shown in the figure, as we all know.

(1) Please sum the midpoint of two points (on the side respectively.

Outside), links, and writing make this graph have only two opposite sides.

Corresponding conditions of triangles with equal products, and show that the areas are equal.

Triangle;

(2) According to the corresponding conditions that make (1) hold,

certificate

Solution:

The corresponding conditions of (1) are: BD = ce ≠ de;

Two pairs of triangles with equal area are △ABD and △ACE, △ABE and △ACD.

Proof 2: As shown in the figure, point A and point E are parallel lines of CB and CA, the two lines intersect at point F, EF and AB intersect at point G, and BF is connected. Then the quadrilateral FECA is a parallelogram, so FE = AC and AF = CE.

Because BD = CE

So BD = AF

So the quadrilateral FBDA is a parallelogram.

So FB = AD

In the delta era, AG+EG >AE.

At △BFG, BG+FG >FB.

It can be deduced that AG+EG+BG+FG >AE+FB.

So AB+AC >AD+AE

24. In a plane rectangular coordinate system, a parabola passes through two points.

(1) Find the analytical expression of this parabola;

(2) Let the vertex of the parabola be, and translate the straight line down by two units along the axial direction to obtain a straight line, and the straight line intersects with the parabola symmetry axis at this point, so as to find the analytical formula of the straight line;

(3) Under the condition of (2), find the coordinates of points with the same distance as the straight line.

Solution: (1) can be obtained from the meaning of the question.

Therefore, the analytical formula of parabola is:

(2) We know that the vertex coordinate of parabola is B (), so C (), the straight line passes through the origin. Let the analytical formula of straight line be, then there is. Therefore, the analytical formula of the straight line is.

(3) There are four points whose distances from straight lines OB, OC and BC are equal.

According to Pythagorean Theorem, OB=OC=BC=2, so △OBC is an equilateral triangle, quadrilateral ABCO is a diamond, and ∠ BCO = 60. Connecting AC and X axis at point M, it is easy to prove that the distances from point M to OB, OC and BC are equal. Since point A is on the bisector of ∠ BCO, it reaches BC and CO.

At the same time, it is not difficult to calculate the distance from point A to OB, so point A is one of them. Similarly, it is not difficult to imagine that the left and the bottom can be made into a rhombus congruent with ABCO (as shown in the figure, where △OBC is half of the new rhombus), and there must be two points at this time, so that the distances to the straight lines OB, OC and BC are equal.

The coordinates of these four points are: m (), a (0 0,2), (0,2), ().

We know that a triangle with two equilateral sides is called an isosceles triangle. Similarly, we define a quadrilateral with at least one set of equilateral sides as an equilateral quadrilateral.

(1) Please write the graphic name of equilateral quadrilateral in the special quadrilateral you have learned;

(2) As shown in the figure, in the middle, the point and the point are above and above respectively, and they are set and intersect with each other. If yes, please write an equilateral in the picture and guess which quadrilateral is equilateral.

(3) If it is not equal to 60 inches? Acute angle, point, and are on respectively. Explore whether there is an equilateral quadrilateral in the graph that meets the above conditions and prove your conclusion.

Solution:

(1) parallelogram, isosceles trapezoid, etc. Can meet the requirements.

(2) The angle equal to ∠A is ∠BOD (or ∠COE).

Quadrilateral DBCE is an equilateral quadrilateral.

③ At this time, DBCE has an equilateral quadrilateral.

Proof 1: As shown in the figure, CG⊥BE is at point G and BF⊥CD is at point F.

∠∠DCB =∠EBC =∠A, BC is the man.

∴△BGC≌△CFB

∴BF=CG

∠∠BDF =∠ABC+∠DCB =∠ Abbe +∠EBC+∠DCB =∠ Abbe+∠ A.

∠GEC =∠ Abe +∠ A.

∴△BDF≌△CEG