A few math Olympiad questions in the second year of junior high school ask for help from the master

Well, this question is really simple enough.

Intercept dg=be on the edge DC

Triangle ABE and triangle ADG congruence.

So angular gad=angular bae=30

And because the angle daf=15

So angular gaf=15

So the angular triangle daf and the triangle haf are congruent.

So the angle AFD = angle AFH = 75

So the angular EFC 30

Because ab root number 3

So be 1

EC (root number 3) 1

fc 3 root number 3

df (2 times root number 3) 3

So the area of the triangle abe (root number 3) 2

The area of the triangular ECF (2 times the root number 3) 3

The area of the triangle FAD (6 3 times the root number 3) 2 i.e.: the area of the triangle AEF The area of the square The area of the triangle ABE The area of the triangle ECF The area of the triangle FAD.

3 (root number 3) 2 (2 times root number 3) 3 (6 times 2 times root number 3) 2 3 root number 3

Question 2, steps.

From the inequality of eleven sixths z x+y 2z z z is not the smallest, if z is smallest then x+y>2z

In the same way, the inequality is three-two, x, y+z, five-thirds, x, x, x, x,

Then y is the smallest.

Suppose z>x then we get 11/6 x x+y 2z 3/2 x y + z 5/3 x

x-z> 11/6 x minus 5/3 x = 1/6 x>0 to get x>z

So z>x is not true.

Get x z y

Question 5, yes.

1.I'm still thinking that it should be squared.

2 by 11 6zz by y+z<2 x>y+z of 3x, x>z of 2x+y>y+2z

Same reason as 11 6zz and x+z eleven quarters of y's x+z< 3y's y>z

y+z five-thirds x x y z is a positive energy number, so a range of y=0 Let the solution be x1 and x2 There are three formulas: x1+x2=-1 x1 x2=(a+2010) a and |x1-x2|>=4k-1 can get a range and solve it yourself.

5 Yes.

1.Use the root axis method (students who take the high school entrance examination do not need to know).

x 0 or x -1;x 1 or x 0;x 1 or -2 x 0;x 0 or -2 x -1

2.In the ordinary way of solving the system of equations: x = the square of a, y = thirds (a-1).

Solution a 13A rational number with an absolute value less than 12? There are an infinite number of them, -12 x 12 rational numbers, such as -11, -10,,, these can be, you can't find the product. Less than 5 is also the same. (You're asking about integers, right?).

4.Are these two questions? I didn't quite understand. Also, this problem should also be an integer, otherwise it can't be solved, you have to explain ...... clearly

1:5s dmn s mbc = 1:16 (by exploiting the midpoint) s dmn s mec = 1:

3 (equal bottom, the ratio of height 1:3) s ade s quadrilateral anme=1:3 (easy to obtain) let s dmn=k, s quadrilateral anme=x

Then (k+x) (15k+3k)=1:3

The solution is dmn s quadrilateral anme=1:5