"Mathematics Weekly Cup" 20 10 National Junior High School Mathematics Competition Reference Answer
1. Multiple choice questions (***5 questions, 7 points for each question, ***35 points. Only one option is correct. Please put the code of the correct option in brackets after the question. If you don't fill in, fill in too much or fill in incorrectly, you will get 0 point).
1. If is, the value of is ().
(A) (B) (C) (D)
Solution: Set by the problem.
2. If real numbers A and B are satisfied, the range of A is ().
(A)a≤ (B)a≥4 (C)a≤ or a≥4 (D) ≤a≤4.
solution
Because b is a real number, the quadratic equation of one variable about b
The discriminant of is ≥0, and the solution is a≤ or a ≥ 4.
3. As shown in the figure, in quadrilateral ABCD, ∠ B = 135, ∠ C = 120, AB=, BC=, CD =, then the length of the AD side is ().
(A) (B)
(C) (D)
Solution: d
As shown in the figure, the intersection points A and D are AE, DF is perpendicular to the straight line BC, and the vertical feet are E and F respectively.
Known to be available
BE=AE=，CF=，DF=2，
So ef = 4+.
The intersection point a is AG⊥DF, and the vertical foot is g. In Rt△ADG, it is obtained according to Pythagorean theorem.
AD =。
4. In a column number, it is known that when k≥2,
(The integer symbol indicates the largest integer that does not exceed the real number, for example,) equals ().
1 (B) 2 (C) 3 (D) 4
Solution: b
Provide by sum
, , , ,
, , , ,
……
Because 20 10=4×502+2, it = 2.
5. As shown in the figure, in the plane rectangular coordinate system xOy, the vertex coordinates of the isosceles trapezoid ABCD are A( 1, 1), B(2,-1), C (-2,-1) and D (-/kloc-0) respectively. Point P 1 rotates around point b 180, point P2 rotates around point c 180, point P3 rotates around point d 180, …, and repeat the operation to get point P 1, P2, …, and then point P20655.
(A)(20 10，2) (B)(20 10，)
(C)(20 12)，(D)(0，2)
Solution: The coordinates of b and point sum are (2,0) and (2,0) respectively.
Remember, among them.
According to the symmetry relation, we can get:
, , , .
Order, you can also get that the coordinates of this point are (), that is, (),
Since 20 10=4 502+2, the coordinate of this point is (20 10,).
Second, fill in the blanks
6. Given a =- 1, the value of 2a3+7a2-2a- 12 is equal to.
Solution: 0
It is known that (A+ 1) 2 = 5, so A2+2A = 4, so
2 a3+7 a2-2a- 12 = 2 a3+4a 2+3 a2-2a- 12 = 3 a2+6a- 12 = 0。
7. A bus, a van and a car are driving in the same direction at a constant speed on a straight road. At a certain moment, the bus is in front, the car is behind, and the van is between the bus and the car. 10 minutes later, the car caught up with the van. After another 5 minutes, the car caught up with the bus; After another t minutes, the truck caught up with the bus, and then t =.
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Solution: 15
At a certain moment, the distance between the truck and the bus and the car is S kilometers, and the speed of the car, the truck and the bus is (km/min) respectively. It is set that the truck catches up with the bus within X minutes.
, ①
, ② .③
From ① ②, you get, so, x = 30. So (points).
8. As shown in the figure, in the plane rectangular coordinate system xOy, the vertex coordinates of polygon OABCDE are O (0 0,0), A (0 0,6), B (4 4,6), C (4 4,4), D (6 6,4) and E (6 6,0) respectively. If the straight line l passes through the point M (2, 3)
Solution:
As shown in the figure, extend the X axis of BC intersection point to point f; Connect OB, AF CE, DF and intersect at point n.
It is known that point M (2 2,3) is OB and the midpoint of AF, that is, point M is the center of rectangular ABFO, so the straight line divides rectangular ABFO into two parts with equal areas. Because point N (5 5,2) is the center of the rectangular CDEF,
A straight line passing through point n (5 5,2) divides the rectangular CDEF into two parts with equal areas.
Then, the straight line is the straight line sought.
Let the function expression of a straight line be, then
Solution, so the functional expression of the straight line is.
9. As shown in the figure, rays AM and BN are perpendicular to line AB, point E is a point above AM, perpendicular AC passing through point A intersects with BE respectively, BN is at points F and C, and perpendicular CD passing through point C is D. If CD = CF, then.
Solution:
See the picture, settings.
Because Rt△AFB∽Rt△ABC, so.
And because fc = DC = ab, that is,
Solve, or (give up).
Rt delta ∽ is rt delta again, so that's =
10. For i=2, 3, …, k, the remainder obtained by dividing the positive integer n by I is I- 1. If the minimum value of is satisfied, the minimum value of a positive integer is.
Solution: Because is a multiple of, the minimum value of is satisfied.
Where represents the least common multiple.
because
Therefore, the minimum value of the positive integer satisfied is.
III. Answer questions (***4 questions, 20 points for each question, 80 points for * * *)
1 1. As shown in the figure, △ABC is an isosceles triangle, AP is the height on the base BC, point D is the point on the line segment PC, and BE and CF are the circumscribed circle diameters of △ABD and △ACD respectively, connecting EF. Verification:.
Proof: Connect ED and FD as shown in the figure. Because Be and CF are both diameters, so
ED⊥BC，FD⊥BC，
Therefore, the three-point * * * line of D, E and F is .................. (5 points).
Connect Auto Exposure, Auto Focus and then
Therefore, △ ABC ∽△ AEF...........( 10)
Let AH⊥EF and vertical foot be h, then AH=PD. It can be obtained from △ABC∽△AEF.
Therefore,
So ........... (20 points)
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12. As shown in the figure, parabola (a 0) and hyperbola intersect at point A and point B. It is known that the coordinate of point A is (1 4), point B is in the third quadrant, and the area of △AOB is 3(O is the coordinate origin).
(1) The values of real numbers A, B and K;
(2) Take the point A on the parabola as the straight line AC‖x axis, intersect with the parabola at another point C, and find the coordinates of all points E satisfying △EOC∽△AOB.
Solution: (1) Because point A (1, 4) is on a hyperbola,
Therefore, k=4. So the functional expression of hyperbola is.
If the function expression of the line where point B(t,) and AB are located is, then there is
Solve,.
Therefore, the coordinate of the intersection of straight line AB and Y axis is 0, so
, organized,
Solution, or t = (rounding). So the coordinates of point B are (,).
Because both point A and point B are on the parabola (a 0), the solution is ............. (10 point).
(2) As shown in the figure, because AC‖x axis is C (,4), so Co = 4. BO=2, so.
Let the parabola (a 0) intersect with the negative semi-axis of X axis at point D, and the coordinate of point D is (0).
Because ∠ COD = ∠ BOD =, ∠COB=.
(i) Rotate △ clockwise around point O to get △. At this time, point (2) is the midpoint of CO, and the coordinates of this point are (4,).
Extend to point so that =, then point (8,) is the qualified point.
(ii) making a symmetrical figure δ about the X axis to obtain a point (1,); Extend to point so that =, then point E2(2,) is the qualified point.
So, the coordinates of this point are (8,) or (2,). ...................................................................................................................................................
13. Find all prime numbers p and positive integers m that are satisfied.
Solution: set by the problem,
Therefore, since P is a prime number, or ... (5 points)
(1) If, let, k is a positive integer, then,
Therefore, therefore.
So the solution is ............ (10)
(2) If, let and k are positive integers.
When, when,
Therefore, therefore, it is still 2.
Because it's strange, so, therefore.
therefore
It is out of the question.
When,,; If,, there is no positive integer solution; When there is no positive integer solution.
To sum up, the prime number p=5 and the positive integer M = 9....................(20 points).
What is the maximum number of 20 10 positive integers of 14. 1 2, …, 20 10 that can be taken out, so that the sum of any three numbers can be divisible by 33?
Solution: First of all, the following figures are 61:1,,,,(i.e. 199 1) ......................... (5 points).
On the other hand, suppose it is a number that satisfies the conditions listed in 1, 2, …, 20 10. For any four of these n numbers, because
, ,
So ...
So the difference between any two numbers is a multiple of 33 (10 point).
Let, i= 1, 2, 3, …, n.
By, by,
Therefore, it is ≥11............ (15 points).
≤ ,
So ≤60. Therefore, n≤6 1.
To sum up, the maximum value of n is 61....................... (20 points).