First attempt

1. Multiple-choice question: (The full score of this question is 42 points, and each small question is 7 points)

1. If,,, is known, the size relation is (c).

A.B. C. D。

2. The number of integer solutions of the equation is (b)

A.3. B.4. C.5. D.6

3. It is known that the side length of square ABCD is 1, e is a point on the extension line of BC, ce = 1, AE is connected, it intersects with CD at point F, BF is connected and extends to intersect with line segment de at point G, then the length of BG is (d).

A.B. C. D。

4. If the known real number is satisfied, the minimum value is (b)

A.。 b . 0 c . 1。 d。

5. If two unequal real roots of the equation are satisfied, the sum of all possible values of real numbers is (b).

A: 0.b. c。 d。

6. 1, 2, 3 and 4 constitute four digits (the digits can be reused), which are required to be met. This four-digit * * * has (c).

A.36, B.40, C.44 and D.48

Fill in the blanks: (The full score of this question is 28 points, and each small question is 7 points)

1. As we all know, mutually unequal real numbers are satisfied.

2. Let the number of integers of a complete square be 1.

3. In △AB=AC, it is known that AB=AC, ∠ A = 40, P is a point above AB, ∠ ACP = 20, then =.

4. It is known that real numbers satisfy,,, then =.

The second test (1)

1. As we all know, the sides of a right triangle are integers and the circumference is 30. Find the area of its circumscribed circle.

If the length of three sides of a right triangle is (), then.

Obviously, the diameter of the circumscribed circle of a triangle is the length of the hypotenuse.

Gradually, so.

Gradually, so.

Because it is an integer, so.

According to Pythagorean theorem, we can substitute and simplify, so

,

Because they are all integer sums, they can only be solved.

So the hypotenuse of a right triangle is long, and the area of the circumscribed circle of the triangle is.

(The full score of this question is 25 points) As shown in the figure, PA is the tangent of ⊙O, PBC is the secant of ⊙O, and AD⊥OP is at point D, which proves that:

Proof: connect OA, OB, OC

∵OA⊥AP, AD⊥OP, ∴ can be obtained by projective theorem.

It can also be concluded from the tangent theorem that ∴, ∴D, b, c and o are four-point * * * circles,

∴∠pdb=∠pco=∠obc=∠odc,∠pbd=∠cod,∴△pbd∽△cod，

∴ ,∴ .

3. (The full mark of this question is 25) It is known that the vertex of the parabola is p, which intersects with the positive semi-axis of the shaft at points A and B (), intersects with the shaft at point C, and PA is the tangent of the circumscribed circle of △ABC. Let m, if AM//BC, find the analytical formula of parabola.

The solution is easy to find point p and point C.

Let the center of the circumscribed circle of △ABC be d, then both point P and point D are on the perpendicular of line AB, and the coordinates of point D are.

Obviously, a variable has two quadratic equations, so the coordinate of the midpoint e of AB is, so AE =.

Because PA is tangent to ⊙D, PA⊥AD and AE⊥PD can be obtained by projective theorem, which is easy to know, so it can be obtained.

It is also derived from da = DC, that is, it can be solved after substitution (the other solution is discarded).

Because AM//BC, so that is.

Replace this solution (another solution is discarded).

Therefore, the analytical formula of parabola is.

The second test (b)

1. (The full mark of this question is 20) It is known that the side length of a right triangle is an integer and the circumference is 60. Find the area of its circumscribed circle.

If the length of three sides of a right triangle is (), then.

Obviously, the diameter of the circumscribed circle of a triangle is the length of the hypotenuse.

Gradually, so.

Gradually, so.

Because it is an integer, so.

According to Pythagorean theorem, we can substitute and simplify, so

,

Because they are all integers and, they can only be or.

Solve or

When, the area of the circumscribed circle of the triangle is;

When, the area of the circumscribed circle of a triangle is.

As shown in the figure, PA is the tangent of ⊙O, PBC is the secant of ⊙O, AD⊥OP is at point D, and the other intersection of the circumscribed circle of △ADC and BC is E. Proof: ∠ BAE = ∠ ACB.

Proof: connect OA, OB, OC, BD.

∵OA⊥AP, AD⊥OP and ∴ can all be obtained from the projective theorem.

, .

According to the tangent line theorem,

∴, ∴D, b, c, o four-point * * * cycle,

∴∠PDB=∠PCO=∠OBC=∠ODC，

∠PBD=∠COD,∴△PBD∽△COD，∴，

∴ ,∴ .

∠ BDA =∠ BDP+90 =∠ ODC+90 =∠ ADC，∴△BDA∽△ADC，

∴∠ Bad =∠ ACD, ∴AB is the tangent of the circumscribed circle of △ADC, ∴∠ BAE = ∠ ACB.

3. (The full mark of this question is 25) The question type and solution are the same as the third question in Volume (A).

The second test (c)

1. (The full mark of this question is 20) Questions and solutions are the same as the first question in Volume (B).

2. (The full score of this question is 25) The question type and solution are the same as the second question in Volume (B).

3. (Full score of this question is 25) It is known that the vertex of parabola is p, which intersects with the positive semi-axis of the shaft at point A and point B (), intersects with the shaft at point C, and PA is the tangent of the circumscribed circle of △ABC. Move the parabola to the left by one unit, and the new parabola will intersect the original parabola at point Q, and ∠ QBO = ∠ OBC. Analytical formula for finding parabola.

Solve the parabolic equation, that is, so point p and point C.

Let the center of the circumscribed circle of △ABC be d, then both point P and point D are on the perpendicular of line AB, and the coordinates of point D are.

Obviously, a variable has two quadratic equations, so the coordinate of the midpoint e of AB is, so AE =.

Because PA is tangent to ⊙D, PA⊥AD and AE⊥PD can be obtained by projective theorem, which is easy to know, so it can be obtained.

It is also derived from da = DC, that is, it can be solved after substitution (the other solution is discarded).

After translating the parabola to the left by units, the new parabola is

.

It is easy to find that the intersection of two parabolas is Q.

You can get ∠ qbo = ∠ obc from ∠ qbo = ∠ obc.

QN⊥AB, the vertical foot is n, then n, again, and so on.

∠QBO= =

.

∠ OBC = Again, so

.

Solve (another solution, give up).

Therefore, the analytical formula of parabola is.