1. Multiple-choice question: (The full score of this question is 42 points, and each small question is 7 points)
1. Let a2+ 1=3a, b2+ 1=3b, a≠b, then the value of algebraic+is ().
A.5 B.7 C.9 D. 1 1
2. As shown in the figure, let AD, BE and CF be the three heights of △ABC. If AB=6, BC=5, EF=3, the length of the line segment BE is ().
A. the fourth century BC.
3. Take two cards from five cards with the numbers 1, 2, 3, 4 and 5 written on them, take the number on the first card as ten digits, and the number on the second card as one digit to form a two-digit number, so the probability that this number is a multiple of 3 is ().
A.B. C. D。
4. In △ABC, ∠ ABC = 12, ∠ ACB = 132, BM and CN are bisectors of these two angles respectively, and points M and N are on straight lines AC and AB respectively, then ().
A.bm > CNb.bm = cnc.bm < cnd.bm and cn are uncertain.
From today, the prices of five different commodities with the same price will be reduced by 10% or 20% respectively. After a few days, the prices of these five commodities will be different. Let the ratio of the highest price to the lowest price be r, then the minimum value of r is ().
A.()3 B.()4 C.()5 D。
6. Given that real numbers x and y satisfy (x-)(y-)=2008, the value of 3x2-2y2+3x-3y-2007 is ().
A.-2008 B.2008 C.- 1 D. 1
Fill in the blanks: (The full score of this question is 28 points, and each small question is 7 points)
1. Let a =, then = _ _ _ _ _ _ _.
2. As shown in the figure, the side length of square ABCD is 1, m and n are two points on the straight line where BD is located, AM=, ∠ man = 135, then the area of quadrilateral AMCN is _ _ _ _ _ _ _ _.
3. It is known that the abscissas of the two intersections of the image of quadratic function y=x2+ax+b and the X axis are m and n respectively, and | m |+n | ≤ 1. Let the maximum and minimum values of b satisfying the above requirements be p and q, then | p |+| q | = _ _ _ _ _ _
4. Square the positive integer 1, 2,3, … in a string:1491625364964810012165438 …, ranking first.
answer
I. 1. B 2。 D 3。 C 4 explosive B 5。 B 6。 D
Two. 1.- 2 2.3.4. 1
Answer: 1. 1. According to the conditions, a2-3a+ 1=0, b2-3b+ 1=0, a≠b,
So A and B are the two roots of the unary quadratic equation x2-3x+ 1=0, so a+b=3 and ab= 1.
Therefore += = = 7.
2. Because AD, BE and CF are the three heights of △ABC, it is easy to know the four * * * cycles of B, C, E and F,
So △AEF∽△ABC, so = =, that is, cos∠BAC=, so sin∠BAC=.
In Rt△ABE, BE=ABsin∠BAC=6×=.
3. Two digits can be 12, 13, 14, 15, 2 1, 23, 24, 25, 3 1, 32, 34, 35. 15,21,24,42,45,51,54, * * 8, so the probability that this number is a multiple of 3 is =.
4. ∠ABC =12, BM is the bisector of ∠ ABC, ∠ MBC = (180-12) = 84.
∠BCM = 180-∠ACB = 180- 132 = 48，∴ BCM = 180-84-48。
∴BM=BC and ∠ ACN = (180-∠ ACB) = (180-132) = 24,
∴∠bnc= 180-∠ABC-∠BCN = 180- 12-(∠AC b+∠can)= 12 =∠ABC。
∴CN=CB. Therefore, BM=BC=CN.
It is easy to know that the prices of five kinds of goods can be different after 4 days.
Let the prices of five commodities before price reduction be a and n days later, the price of each commodity after n days can be expressed as a (1-10%) k (1-20%) n-k = a () k () n-k, where k is a natural number. a ()i+ 1 ()n-i- 1，a ()i+2 ()n-i-2，a ()i+3 ()n-i-3，a ()i+4 ()n-i-4。
Where I is a natural number not exceeding n, so the minimum value of R is =()4.
6.∫(x-)(y-)= 2008，
∴x-==y+,y-==x+.
X=y can be obtained from the above two formulas, so (x-)2=2008. The solution is x2=2008.
So 3x2-2Y2+3x-3y-2007 = 3x2-2x2+3x-3x-2007 = x2-2007 =1.
2. 1.∫a2 =()2 = = 1-a，∴a2+a= 1.
∴ Original formula =
= = =-=--( 1+a+a2)=-( 1+ 1)=-2。
2. Let the midpoint of BD be O and connected with AO, then AO⊥BD, AO=OB=, MO==,
∴MB=MO-OB=. and ∠ ABM = ∠ NDA = 135,
∠NAD =∠MAN-∠DAB-∠MAB = 135-90-∠MAB = 45-∠MAB =∠AMB，
So △ADN∽△MBA, so =, so DN = Ba =× 1 =. According to symmetry,
The area of the quadrangle AMCN is s = 2s△ man = 2××× Mn× ao = 2××× (++)× = )× =.
3. According to the meaning of the question, M and N are two roots of the unary quadratic equation x2+ax+b=0, so m+n=-a and Mn = B. 。
∵|m|+|n|≤ 1,∴|m+n|≤|m|+|n|≤ 1,|m-n|≤|m|+|n|≤ 1.
Equation x2+ax+b=0 △=a2-4b≥0 discriminant, ∴b≤=≤.
4b = 4mn = (m+n) 2-(m-n) 2 ≥1-(m-n) 2 ≥-1,so b≥- and the equal sign are obtained when m=-n=;
4b = 4mn = (m+n) 2-(m-n) 2 ≤1-(m-n) 2 ≤1,so b≤, when m=n=, an equal sign is obtained. So p=, q=-, and then | p || q.
4. 12 to 32, the result only accounts for 1 digit, and * * * accounts for 1×3=3 digits; 42 to 92, the results only account for 2 digits, and * * * accounts for 2×6= 12 digits; From 102 to 3 12, the result only takes 3 digits, and * * * takes 3×22=66 digits; From 322 to 992, the results only occupy 4 digits each, and * * * occupies 4×68=272 digits; From 1002 to 3 162, the result only takes 5 digits, and * * takes 5×2 17= 1085 digits; There is a gap of 2008-(3+12+66+272+1085) = 570 digits. 3 172 to 4 1 12, and the results only account for 6 digits, and * * accounts for 6× 95 =
Reference Answers and Grading Criteria of 2009 National Junior Middle School Mathematics Joint Competition
Note: Please follow this grading standard when grading papers. In the initial test, multiple-choice questions and fill-in-the-blank questions only have two grades of 7 points and 0 points; For the second test, please grade each question according to the grading scale specified in this grading standard. If the examinee's problem-solving method is different from this, as long as the thinking is reasonable and the steps are correct, please refer to the grading level specified in this grading standard and give corresponding scores when marking papers.
First attempt
First, multiple-choice questions (the full score of this question is 42 points, and each small question is 7 points)
1. Settings, and then ()
A.24。 c。 d。
Answer a.
By, by, therefore. So ...
.
2. In △ABC, the maximum angle ∠A is twice the minimum angle ∠C, and AB = 7, AC = 8, then BC = ().
A.。 B. c。 d。
Answer C.
Expand CA to d to make AD = AB, then, so △ CBD ∽△ AD=AB, so, so, so. Because so
3. When the maximum integer is not greater than, the number of solutions of the equation is ().
A. 1。 B. 2。 C. 3。 D. 4。
Answer C.
By the equation, and, so, that is, the solution, so you can only take the value.
When,, solution;
When there is no qualified solution;
When there is no qualified solution;
When,, solution;
When, when, solve.
Therefore, the original equation * * * has three solutions.
4. Let the center of the square ABCD be point O, and randomly select two from all triangles whose vertices are A, B, C, D and O.. The probability that their areas are equal is ().
A.。 B. c。 d。
Answer B.
Let's assume that the area of a square is 1. It is easy to know that a triangle with five vertices of A, B, C, D and O is an isosceles right triangle, which can be divided into two categories:
(1) The right vertex of an isosceles right triangle is one of the four vertices of a square ABCD, and there are four such triangles, all of which have an area of;
(2) The right vertex of an isosceles right-angled triangle is the center o of the square ABCD. There are four such triangles, all of which have the same area.
Therefore, with five points A, B, C, D and O as vertices, 4+4 = 8 triangles can be formed, and two of them can be selected at will, and there are 28 ways to take * * *.
To make the areas of two triangles equal, we can only take them from (1) or from (2), and there are two different ways to take1.
So, the probability of finding is.
5. As shown in the figure, in the rectangle ABCD, AB = 3, BC = 2, if BC is used as the diameter to make a semicircle in the rectangle, the tangent AE of the semicircle from point A, CBE= = ().
A.。 B. c。 d。
Answer D.
Let the midpoint of BC be O, connecting OE and CE.
Because AB⊥BC and AE⊥OE, A, B, O and E are * * * cycles, so ∠ BAE = ∠ COE.
AB = AE, OC=OE, so △ AB=AE ∽△ oce, therefore, that is.
And CE⊥BE, so, so CBE= =.
6. Let it be a positive integer greater than 1909, so that the number of complete squares is ()
A.3. BC5。 D. 6。
Answer B.
If, then, it is a complete square number, or it may be set to (where is a positive integer), then.
Verification is easy to know, the above formula is only available in. The corresponding values are 50,20,10,2.
So there are four * * * that are completely squared, namely 1959, 1989, 1999, 2007.
2. Fill in the blanks (the full score of this question is 28 points, and each small question is 7 points)
1. is known as a real number. If it is about two non-negative real roots of a quadratic equation, the minimum value of _ _ _ _ _ _ _ _.
Answer.
Because it is about two non-negative real roots of a quadratic equation, so
Solve.
,
When, get the minimum value.
2. Let D be a point on the AB side of △ABC, let DE//BC intersect with AC at point E, and DF//AC intersect with BC at point F. Given the sum of the areas of △ADE and △DBF, the area of quadrilateral DECF is _ _ _ _ _.
Answer.
Let the area of △ABC be, because △ADE∽△ABC, so.
And because of △BDF∽△BAC, so.
The sum of the two formulas is the solution.
So the area of DECF quadrilateral is.
3. If the real number meets the conditions, then _ _ _ _ _.
Answer.
Because, therefore, you can get it.
Therefore, the solution.
Thus, therefore, that is, sorting out and solving (the other is discarded).
Into the calculation, so.
4. If it is known to be a positive integer and it is an integer, then such an ordered number pair * * * has _ _ _ pairs.
Answer 7.
Let (be a positive integer), then, so it is a rational number.
Order, where all are positive integers, so, so, so, so.
In the same way (where is a positive integer), then.
Here we go again, so, so.
(1), yes, it is easy to find or (3,6) or (6,3).
(2), also can get.
(3), or (1, 2) can be obtained in the same way.
4. It can also be obtained.
So there are seven pairs of such ordered number pairs * * *, namely (240,240), (135,540), (540,135), (60,60), (60, 15.
The second test (1)
1. (The full mark of this question is 20) It is known that the intersection of the image of quadratic function with the axis is A and B, and the intersection with the axis is C. Let the center of the circumscribed circle of △ABC be point P. 。
(1) proves that the other intersection of ⊙P and the axis is a fixed point.
(2) If AB happens to be the diameter of ⊙P, sum.
Solution (1) It is easy to find the coordinates of the point as, let,, and then.
Let the other intersection point of ⊙P and the axis be d, because AB and CD are two intersecting chords of ⊙P, and their intersection point is point O, so OA× OB = OC× OD, then.
Because, so the point is on the negative semi-axis of the shaft, so the point D is on the positive semi-axis of the shaft, so the point D is a fixed point, and its coordinate is (0,1) .................................................................10 minute.
(2) Because AB⊥CD, if AB happens to be the diameter of ⊙P, then C and D are symmetrical about point O, then the coordinates of this point are
That is 15.
Again, so
,
............................... scored 20 points for the solution.
Let CD be the height on the hypotenuse AD of the right triangle ABC, and be the centers of △ADC and △BDC respectively, AC = 3, BC = 4, and find.
The answer is E⊥AB in e and F⊥AB in f.
In the right triangle ABC, AC = 3, BC = 4,.
And CD⊥AB can be obtained by projective theorem, therefore,
........................................, five points.
Because e is the radius of the inscribed circle of the right triangle ACD, =
........................... 10.
Connect D and D, then D and D are bisectors of ∠ADC and ∠BDC respectively, so ∠ DC = ∠ DA = ∠ DC = ∠ DB = 45, so ∠ D =90, so D∞.
............................. 15.
In the same way, you can get 20 points for ..........................................................................
So = ............................... 25 points.
3. (The full score of this question is 25 points) It is known as a positive number and meets the following two conditions:
①
②
It is proved that three sides can form a right triangle.
Prove that 1 is multiplied by ① ②,
That is 10.
That is to say,
That is to say, ..................................... 15.
That is to say,
That is to say,
That is, that is,
That is ... 20 points.
So or or, is or or.
Therefore, with three sides as the length, a right-angled triangle with .......................... of 25 points can be formed.
Proof 2 combined with formula 1 can be obtained from formula 2.
Deformation, ③ ........................10.
It is also obtained by the formula (1), that is,
Substituting into the formula (3),
That is 15.
, ... 20 points.
So either or.
Combined with ① formula, it can be obtained.
Therefore, with three sides as the length, a right-angled triangle with .......................... of 25 points can be formed.
The second test (b)
1. (The full mark of this question is 20) Questions and solutions are the same as the first question in Volume (a).
2. (The full mark of this question is 25) It is known that when △ABC, ∠ ACB = 90, the high line CH of AB and the bisectors AM and BN of △ABC intersect at P and Q, and the midpoints of PM and QN are E and F respectively. Verification: EF‖AB.
Solution because BN is the bisector of ∠ABC, so.
Because of CH⊥AB, so
,
Therefore, ............................. 10.
F is the midpoint of QN, so CF⊥QN, so c, f, h and b are * * * cycles.
........................... 15.
Similarly, so FC = FH, so point F is on the vertical line of CH. .........................................................................................................................................................
Similarly, point e is on the perpendicular of CH.
So EF⊥CH.
AB⊥CH again, so EF ‖ AB ............................................ scored 25 points.
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 (a).
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. (The full score of this question is 25 points) It is known as a positive number and meets the following two conditions:
①
②
Do you have a triangle with three sides? If it exists, find the maximum internal angle of the triangle.
Solution 1 multiplied by ① ②,
That is 10.
That is to say,
That is to say, ..................................... 15.
That is to say,
That is to say,
That is, that is,
That is ... 20 points.
So or or, is or or.
Therefore, three sides can form a right triangle with a maximum internal angle of 90.
25 points.
Solution 2 is combined with formula ①, which can be obtained from formula ②.
Deformation, ③ ........................10.
It is also obtained by the formula (1), that is,
Substituting into the formula (3),
That is 15.
, ... 20 points.
So either or.
Combined with ① formula, it can be obtained.
Therefore, three sides can form a right triangle with a maximum internal angle of 90.
25 points.
Kindergarten mathematics addition and subtraction open class teaching plan;
First,? Design inte