April 2008 13 8:30-9:30 am
1. Multiple-choice question: (The full score of this question is 42 points, and each small question is 7 points)
1, let a 2+1= 3a, b 2+1= 3b, a ≠ b, then the value of algebraic+is ().
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) (B)4 (C) (D)
3. If two cards are randomly selected from five cards with the numbers 1, 2, 3, 4 and 5 written on them, and the number on the first card is taken as ten digits, and the number on the second card is taken as single digits to form a two-digit number, the probability that the 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 & gt； CN(B)BM = CN(C)BM & lt； The relationship between CN (D)BM and CN is 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. It is known 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 with quadratic function y = x 2+ax+b and the X axis are m and n, respectively, | m |+| n | ≤ 1. Let the maximum and minimum values of b satisfying the above requirements be p and q respectively, then | p |+| q | =.
4. Square the positive integer 1, 2,3, … in a string:1491625364964810012165438 …, ranking first.
Answer: b, d, c, b, b, d; – 2、 、 、 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, a b = 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。 ∠ACN =( 180–∠ACB)=( 180– 132)= 24，∴∠BNC = 180–
It is easy to know that after four days, the prices of five commodities can be different from each other.
Let the prices of the first five commodities be A and N days, and the price of each commodity will be OK after N days.
Expressed as a? ( 1– 10%)k？ ( 1–20%)n–k = a？ ()k？ () n–k, where k is a natural number and 0 ≤ k ≤ n. To minimize the value of R, the prices of five commodities should be: a? () me? () n- me, 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+，
From the above two formulas, we can get X = y, so (X–) 2 = 2008, and we can solve x-)2 = 2008.
So 3x2–2y2+3x–3y–2007 = 3x2–2x2+3x–2007 = x2–2007 =1;
2. 1, ∫ a2 = () 2 =1–a, ∴a 2+a = 1, ∴ original formula =
= = =–=––( 1+a+a 2)= –( 1+ 1)=–2；
2. Let the midpoint of BD be O and connected with AO, then AO⊥BD, AO = OB =, MO = =,
∴MB = Mo-Aobu = .∞∠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 quadrilateral AMCN is s = 2s △ man = 2×× Mn× ao = 2×××× (++)× =;
3. according to the meaning of the question, m and n are the two roots of the unary quadratic equation x 2+a x+b = 0, so m+n =–a, m+n =-a b.
∵| m |+| n |≤ 1,∴| m+n |≤| m |+| n |≤ 1，| m–n |≤| m |+| n |≤ 1 .
The discriminant of the equation x 2+a x+b = 0 △ = a 2–4b ≥ 0, ∴b ≤ = ≤
4b = 4m n = (m+n) 2-(m–n) 2 ≥ (m+n) 2–1≥–1,so b ≥-–,when m =–n =, an equal sign is obtained; 4b = 4m n = (m+n) 2—(m–n) 2 ≤1—(m–n) 2 ≤1,so b ≤, and when m = n =, an equal sign is obtained. So p =, q =-–,so | p |+| q | =；;
4, 1 2 to 3 2, the result only accounts for1digit, * * * accounts for 1 × 3 = 3 digits; 4 2 to 9 2, the result only takes 2 digits, and * * * takes 2 × 6 = 12 digits; From 10 2 to 3 1 2, the result only takes 3 digits, and * * * takes 3 × 22 = 66 digits; From 32 2 to 99 2, the results only occupy 4 digits each, and * * * occupies 4 × 68 = 272 digits; From 100 2 to 3 16 2, the result only takes 5 digits, and * * takes 5 × 2 17 = 1085 digits; At this time, it is 2008-(3+12+66+272+1085) = 570 digits. From 3 17 2 to 4 1 1 2, the results are 6 bits each, and * * * takes 6 × 95 = 570 bits. Therefore, the number of the 2008th bit should be a single digit of 4112, that is,1;
2008 National Junior Middle School Mathematics League
April 2008, 65438+ 0: 00 a.m.-165438+0: 30 a.m.
The second test (1)
1. (The full mark of this question is 20) It is known that A 2+B 2 =1.For all real numbers X that satisfy the condition 0 ≤ x ≤ 1, the inequality A (1–x)–b x (b–x–b x).
Solution: sort out the inequality (1) and substitute it into a 2+b 2 = 1 to get (1+a+b) x2-(2a+1) x+a ≥ 0 (2).
In (2), let x = 0 and get a ≥ 0; ; Let x = 1 and get b ≥ 0. Yizhi 1+a+b > 0，0 & lt； & lt 1,
Therefore, the image (parabola) of the quadratic function y = (1+a+b) x2-(2a+1) x+a has an upward opening, and the abscissa of the vertex is between 0 and1. According to the problem, inequality (2) holds for all real numbers X that satisfy the condition 0 ≤ x ≤ 1, so its discriminant △ = (2a+1) 2–4a (1+a+b) ≤ 0, that is, a b ≥ is determined by equation group (3).
If b is excluded,16a4–16a2+1= 0, so a 2 = or a 2 = And because a ≥ 0,
So a 1 = or a 2 =, so b 1 = or b 2 =. So the minimum value of a b is 0, and the values of A and B are a =, b = and a =, b = respectively.
As shown in the figure, circle O and circle D intersect at points A and B, BC is the tangent of circle D, point C is on circle O, and AB = BC.
(1) Prove that point O is on the circumference of circle D;
(2) Let the area of △ABC be S, and find the minimum value of the radius r of circle D. ..
Solution: (1) even OA, OB, OC, AC, because O is the center of the circle and AB = BC, so △OBA∽△OBC, thus ∠OBA =∠OBC, because OD⊥AB, DB⊥BC, so.
(2) Let the radius of circle O be a, and the extension line of BO and AC intersect at point E, so it is easy to know that BE⊥AC. Let AC = 2 y (0
L2 = y2+(a+x)2 = y2+a2+2a x+x2 = 2a 2+2a x = 2a(a+x)= .
Because ∠ABC = 2∠OBA = 2∠OAB =∠BDO, AB = BC, DB = DO, so △BDO∽△ABC.
So =, that is =, so r =, so r 2 = =? = ? () 3 ≥, that is, r ≥, where the equal sign holds when a = y, and when AC is the diameter of the circle O. So the minimum value of the radius r of the circle D is.
3. (The full mark of this question is 25) Let A be a prime number, B be a positive integer, and 9 (2a+b) 2 = 509 (4a+511b) (1).
Find the values of a and b.
Solution: (1) formula is () 2 =, let m =, n =, then n = m 2,
B = = (2 a+b) 2 so 3n–5 11m+6a = 0, so 3m2–51m+6a = 0 (3), from the formula (1), (2
That is, the univariate quadratic equation (3) about m has integer roots, so its discriminant △ = 5112–72a is a complete square number.
Let △ = 5112–72a = t2 (natural number), then 72a = 5112–t2 = (511+t).
Because the parity of 5 1 1+t and 51–t is the same, 511+t ≥ 511,there are only the following situations.
①, ②, ③ and ④ can be obtained by adding two formulas respectively.
36 a+2 = 1022, 18 a+4 = 1022, 12 a+6 = 1022, 6 a+12 =/kloc.
⑤, ⑤, the two formulas are added to get 4 a+ 18 = 1022 respectively, and the solution is a = 251;
2 a+36 = 1022, get a = 493,493 =17× 29 is not a prime number, and discard it. A = 25 1。
At this time, the solution of equation (3) is m = 3 or m = (rounding).
Substitute a = 25 1 and m = 3 into equation (2) to get b = = 7.
The second test (b)
1. A 2+B 2 =1is known. For all real number pairs (x, y) that satisfy the condition of x+y = 1 and x y ≥ 0, the inequality ay2–xy+bx2 ≥ 0 (1) holds. When the product A+Y ≥ 0,
Solution: According to x+y = 1 and x y ≥ 0, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. In the formula (1), let x = 0, y = 1, and get a ≥ 0; ; Let x = 1 and y = 0, and b ≥ 0 is obtained. Substitute y = 1–x into the formula (1) to get a (1–x) 2–x (1–x)+bx2 ≥ 0.
That is, (1+a+b) x2-(2a+1) x+a ≥ 0 (2), Yizhi1+a+b >: 0,0 <; & lt 1,
Therefore, the image (parabola) of the quadratic function y = (1+a+b) x2-(2a+1) x+a has an upward opening, and the abscissa of the vertex is between 0 and1. According to the problem, inequality (2) holds for all real numbers x that satisfy the condition 0 ≤ x ≤ 1
So its discriminant △ = (2a+1) 2–4a (1+a+b) ≤ 0, that is, a b ≥ If B is eliminated from the equation set (3),16a4–16a2+/kloc-. And because a ≥ 0,
So a 1 = or a 2 =, so b 1 = or b 2 =. Therefore, the minimum value of a and b is 0, and the values of a and b are a =, b = and a =, b = respectively.
2. (The full score of this question is 25 points) The question type and solution are the same as the second question in Volume (A).
3. (The full score of this question is 25 points) 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 25 points) The question type and solution are the same as the first question in Volume (B).
2. (The full score of this question is 25 points) The question type and solution are the same as the second question in Volume (A).
3. (The full score of this question is 25 points) Let A be a prime number, B and C be positive integers, and satisfy, and find the value of A (b+c).
Solution: (1) formula is () 2 =, let m =, then 2b–c = (3), so 3n–511m+6a = 0, and n = m 2,
Therefore, 3m2–511m+6a = 0 (4). According to the formula (1), (2a+2b–c) 2 can be divisible by 509.
And 509 is a prime number, so 2a+2b–c can be divisible by 509, so m is an integer.
That is, the univariate quadratic equation (4) about m has integer roots, so its discriminant △ = 5112–72a is a complete square number.
Let △ = 5112–72a = t2 (natural number), then 72a = 5112–t2 = (511+t).
Because the parity of 5 1 1+t and 51–t is the same, 511+t ≥ 511,there are only the following situations.
①, ②, ③ and ④ can be obtained by adding two formulas respectively.
36 a+2 = 1022, 18 a+4 = 1022, 12 a+6 = 1022, 6 a+12 =/kloc.
⑤, ⑤, the two formulas are added to get 4 a+ 18 = 1022 respectively, and the solution is a = 251;
2 a+36 = 1022, get a = 493,493 =17× 29 is not a prime number, and discard it. A = 25 1。
At this time, the solution of equation (3) is m = 3 or m = (rounding).
Substitute a = 25 1 and m = 3 into formula (3) to get 2 b-c = = 7, that is, c = 2 b-7, and substitute into formula (2) to get b-(2 b-7) = 2, so b = 5 and c = 3, so a (b+c).