Junior high school mathematics competition test paper

1. Suppose a is in the first quadrant.

By solving the equations y=kx and y= 1/x, we can get x (a) = under the radical sign (1/k) and y (a) = under the radical sign, etc. X(c)=- under the radical sign (1/k), Y (c) = under the radical sign.

Triangle ABC can be divided into two triangles, OAB and OBC. The area of both triangles is 1/2, which is easy to see. Of course, we can also obtain by calculation:

For example, under the root number, the triangle OAB =1/2 * ob * ab =1/2 * x (a) * y (a) =1/2 * under the root number, K = 65438+.

The triangle OBC is the same, the length of the base OBC is equal to the abscissa of A, and the height is the absolute value of the ordinate of C, so the area can also be 1/2.

Therefore, the area of the triangle ABC =1/2+1/2 =1.

2. Set a>b, then a-b= 120,

Let the greatest common divisor be k, a = MK and b = NK.

Then (m-n)k = 120- formula 1.

In addition, the minimum common multiple [a, b] of ab = mnk, and the maximum common divisor (a, b) = k..

Therefore, mnk/k = 105, that is

mn= 105=3*5*7

Substitute different values of m and n into the equation 1 to see if it holds.

When n = 1, m=3*5*7, but 3*5*7- 1 cannot be divisible by 120, so the formula 1 is not valid.

When n = 7, m=3*5= 15, and 3*5-7=8 is divisible by 120.

When n=5, 3*7-5= 16 cannot be divisible by 120.

When n = 3, 5*7-3=32 cannot be divisible by 120.