Who has information on the Mathematics Olympiad I want to participate in the Mathematics Olympiad

In 2004, the 9th China Cup Finals was the second test question of the first group.

1 Hospital A and Hospital B accept the same number of patients at the same time, and each patient suffers from one of the X or Y diseases, and after a few days**, Hospital A treats more patients than Hospital B treats patients. Q: After these days, is it possible that the **rate of disease X and the **rate of disease Y in Hospital A are lower than those in Hospital B? Illustrate. (** rate of disease X = 100%)

2 In a rectangular ABCD, BF=AE=3 cm, DE=6 cm, the area of a triangle GEC is 20 square centimeters, and the area of a triangle GFD is 16 square centimeters, so, what is the area of a rectangular ABCD in square centimeters?

3 A, B, and C respectively start from the apex a, b, and c of δabc, and choose a place to meet, and each car follows a straight road section to the meeting place (ab=c, ac=b, bc=a), and the fuel consumption per unit distance of the three cars is 1 3, 1 6, 1 8 respectively. To minimize the sum of the fuel used on the road for the three vehicles, where should they meet? What is the minimum fuel consumption (denoted by a, b, c)?

4 Some natural number expressed in decimal is equal to 16 times the sum of its digits, and the sum of all such natural numbers is found.

5 Find the natural numbers a,b:

1) a>b; (2);3) a+b is the square number.

6.As shown in the figure, 101 7 rectangular matrix, row spacing and column spacing are 1, on the 6th column (except for the intersection with column 0), each array has a target, and on the first 5 columns, all the array points are placed with obstacles. The sharpshooter stands in column 0 of row 0 and must first clear all obstacles in the trajectory (straight line) of the bullet in order to hit the target, if the sharpshooter can hit the target with each bullet, and each bullet can destroy but only one obstacle, then.

1) How many targets can be hit without clearing obstacles?

2) How many targets do you have to clear an obstacle to hit?

Line 7

Line 6

Line 5

Line 4

Line 3

Line 2

Line 1

Line 0

1.The volume of a cone is 45 cubic centimeters, and if its base radius is reduced to one-third and its height is expanded to twice its size, what is its volume?

2.If the two right-angled sides of a right-angled triangle are 5 cm and 12 cm long respectively, what is the volume of the figure obtained by rotating the right-angled side of the right-angled triangle with a length of 12 cm?

3.What is the surface area of a right-angled triangle with three sides 3 cm, 4 cm, and 5 cm long, and if you rotate the right triangle around the right-angled sides with a length of 3 cm?

4.After removing a quarter of a circle of iron sheet with a radius of 10cm, a conical chimney cap is made, and the bottom radius of the chimney cap is obtained.

5.What is the surface area of the largest cone excavated in a cylinder with a base radius of 3 cm and a height of 4 cm?

1. Analysis: The radius of the bottom surface is reduced to one-third of the original, the bottom area is reduced to one-ninth of the original, the height is expanded to 2 times of the original, and its volume is 1 9 2 times of the original.

Solution: 45 (1 3) 2=10 (cubic centimeters).

2. Analysis: Rotate around a right-angled edge with a length of 12cm to get a cone, with a height of 12cm and a bottom radius of 5cm.

Solution: 1 3 cubic centimeters).

3. Analysis: Rotate with a right-angled edge with a length of 3cm, and you will get a cone, with a height of 3cm and a bottom radius of 4cm.

Solution: base area = square centimeters).

side area = square centimeters).

surface area = square centimeters).

4. Analysis: The circular iron sheet is removed by a quarter of a circle, and the remaining arc length is the circumference of 3 4 circles, which is the circumference of the bottom surface of the cone, which can be found.

Solution: 2 cm).

Analysis: The radius is 3cm, the height is 4cm, and the hypotenuse is 5cm

Solution: The side area of the cone =

square centimeters) side area of the cylinder = 2

square centimeter) The base area of the cylinder = square centimeters).

The surface area of the remaining part = square centimeters).

1.The length of an arc is equal to 1 4 of the radius of the circle in which it is located, so what fraction of the circumference of the arc is?

2.If you put it on a straight line and flip it clockwise, then at least a few times will it fall on the straight line? And find the length of the route of motion. (A in the bottom left corner, B in the bottom right corner, C on top of the triangle).

Big brother, don't give 1 point.

The value range is that a belongs to (0,2] or a=2 plexus pure 2 (cannot be displayed.) is 2 times the root number 2).

Process: It is to ask the value range.

There are two cases: 1) A can take any value from 0 to 2, (excluding 0, but including 2), how to change you can see it in a diagram, you can know that when 090 degrees), when a takes the value of 2, bc=ac, c is 90 degrees. These are all triangles with and without one triangle.

2) a = 2 2 (this is 2 times the root number 2...) There is a small tick in front of the second 2. In this case, b is at a right angle.

Analysis: If at 22 2, it is not possible to make the side length of the AB line segment with a length of 2, because the perpendicular lines are larger than 2

Also, when it's a right triangle, a = 2 times the first knock root, and the bridge carries the number 2

Teach you how to solve it: 1: make ray ab, 2: take c as the center of the circle, and 2 as the radius to make a circle.

3: Find the range of a with the intersection of a circle and rays.

I'm old, and the children are going to calculate it, but I know that A can't be equal to 2, if A is equal to 2, ABC is not a triangle, but a line split, A must be a small cover in 2!

Let the sum of k integers be a, then.

a1+a2+……ak）*（k-1）- 2014*k = a1+a2+……ak）

i.e. a (k-1)- 2014 k = ak= 2a a - 2014) =2 + 4028 (a-2014).

So when a is 2015, k is the largest, at 4030.

At this time, the number of 4030 can be: -2014, -2013, -2012, ......Until 2015.

The number of male students in the third grade of junior high school accounts for 1 4 of the number of male students in the three grades, that is: the number of male students in the third grade accounts for 1 (4-1) of the number of male students in the first, second and second grades of junior high school, and the number of students in the first and second grades of junior high school is the same, and it is known that the number of male students in the first and second grades of junior high school is the same, that is, the total number of male and female students in the first and second grades of junior high school is the same, therefore: the number of male students in the third grade accounts for 1 [(4-1)*2] of the number of male students in the first and second grades of junior high school [(4-1)*2], obtained; The number of boys in the three grades accounts for 1 of the number of students in the first and second grades of junior high school [(4-1)*2] +1 2, and the number of boys in the first and second grades of junior high school accounts for (1+1) (1+1 + 4 5) of the number of students in the three grades

Therefore, the number of boys in the three grades accounts for *(1+1) (1+1+ 4 5) of the number of students in the three grades

Then: the number of girls in the three grades accounts for 1-*(1+1) (1+1+ 4 5)=11 21

I wish you progress in your studies and go to the next level! (

For the convenience of calculation, the number of junior high school students in the first and second years of junior high school is 5, then the number of junior high school students is 4 junior high school boys = junior high school girls, then there are junior high school girls = junior high school boys, junior high school boys + junior high school boys = junior high school boys + junior high school girls = 5

If the boy in the third year of junior high school is x, then there is x (5+x)=1 4, and the solution is x=5 3, then the girl in the third year of junior high school is 4-5 3=7 3

Then the three grades of girls are 5 (first year girl + second year girl) + 7 3 = 22 3, the proportion is 22 3 14 = 11 21

I don't know how to continue to ask.

You'd better draw a diagram to solve this problem.

We set up x people in the first year of junior high school, then: x people in the second year of junior high school, 4 5 x people in the third year of junior high school.

First of all, the number of people in the first year of junior high school and the second year of junior high school is the same.

Then the total number of people is 2 8x

Now ask for girls:

The third day of junior high is 4 5 * 1 - 1 4).

If there are x people in the first year, then there will be x people in the second year of the junior high school, and 4 5 x people in the third year of the junior high school.

Total number of people 2 8x

Girls: 3rd year of junior high school 4 5 * 1- 1 4) 1st year + 2nd year = total number of students in the first year of junior high school.

According to the conditions, the number of students in three grades is x, x, and the number of boys and girls in the first year of junior high school is a and b

Then the second day of junior high is b, a

The third year of junior high school is C and D for men and women

then c (a+b+c) = 1 4 = > x = 3c solvable (a+b+d) (

m=2005*2005*2005-2005=2005*(2005*2005-1) is divisible by 2005.

2006*(2005*2005-2005) is divisible by 2006.

m=(2004+1)*2005*2005-2005 can be divisible by 2004 by the above method, so d is selected

The real number m = the cubic of 2005 - 2005, which is not divisible by m.

m = the third power of 2005 - 2005 = 2005 (2005 2-1) = 2005 * 2006 * 2004

Not divisible by m.