Urgent demand: 1999 national junior high school mathematics competition, national junior high school mathematics joint competition and other papers.

Reference answers to the questions in the 2006 national junior high school mathematics competition

1. Multiple choice questions (***5 questions, 6 points for each question, out of 30 points. Each of the following questions has four options, code-named A, B, C and D, only one of which is correct. Please fill in the code of the correct option in the brackets after the question. Don't fill in, fill in too much or fill in the wrong, get zero)

1. On the expressway, start at 3 kilometers and pass a speed limit sign every 4 kilometers; And from 10 km, it passes through the speed monitor every 9 km. Just at 19 km, passing through these two facilities for the first time, then the kilometers passing through these two facilities for the second time are ().

36 (B)37 (C)55 (D)90

A: C.

Solution: Because the least common multiple of 4 and 9 is 36, 19+36 = 55, it is 55 kilometers to pass through these two facilities for the second time.

So choose C.

2. It is known that,, and then the value is equal to ().

(A)-5(B)-5(C)-9(D)9

A: C.

Solution: From what is known, it is available again.

So,

Solve.

So choose C.

3. The three vertices of 3.RT △ ABC are all on parabola, and the hypotenuse AB is parallel to the X axis. If the height on the hypotenuse is, then ().

(A) (B) (C) (D)

A: B.

Solution: Let the coordinates of point A and point C be (), then the coordinates of point B are, which is obtained by Pythagorean theorem.

So ...

Because of this, the hypotenuse AB is high.

So choose B.

4. a square piece of paper, cut it into two parts along a straight line without any vertices with scissors; Take out a part of it, and then cut it into two parts along a straight line that does not exceed any vertex; And take out one of the three parts, or cut it into two parts along a straight line that doesn't pass through any vertex ... So, you finally get 34 62-sided papers and some polygonal papers, and at least the number of knives to be cut is ().

A: B.

Solution: according to the meaning of the question, when scissors are cut into two parts along a straight line that does not pass through the vertex, the sum of the internal angles of each part increases by 360. So after several cuts, (+1) polygons can be obtained, and the sum of the internal angles of these polygons is (+1) × 360.

Because there are 34 polygons with 62 sides in this (+1) polygon, the sum of their internal angles is

34×(62-2)× 180 =34×60× 180 ,

The remaining polygons are (+1)-34 =-33 (pieces), and the sum of the internal angles of these polygons is not less than (-33) × 180. Therefore,

( + 1)×360 ≥34×60× 180 +( -33)× 180 ,

Solution ≥ 2005.

When we use the following method to cut 2005 knives, we can get a qualified conclusion. First, we cut out 1 triangles from the square and get 1 triangles and 1 pentagons. Then cut 1 triangle from the Pentagon to get 2 triangles and 1 hexagon ... and so on, cut 58 knives to get 58 triangles and 1 62 hexagons. Then take out 33 triangles and cut a knife on each triangle to get 33 triangles and 33 quadrangles. These 33 quadrangles, press the top.

58+33+33× 58 = 2005 (knife).

So choose B.

5. As shown in the figure, the square is inscribed in ⊙, the point is on the bad arc, and it is connected and intersects with the point. If is, the value of is ().

(A) (B)

A: D.

Solution: As shown in the figure, let the radius of ⊙ be,,,.

In ⊙, according to the chord intersection theorem, we get.

That is to say,

So ...

Add DO, and from Pythagorean theorem, we get

That is to say,

Solve.

So ...

So choose D.

Fill in the blanks (***5 small questions, 6 points for each small question, out of 30 points)