For example, given a convex quadrilateral, it is a moving point on the plane, so.
(1) Verification: When the minimum value is reached, the four points are * * * circles;
(Ⅱ) Let it be the upper point of the circumscribed circle and satisfy the minimum tangent of:,, and.
[Solution 1] (i) If you answer graph1,from Ptolemy inequality, for any point on the plane, there is
.
therefore
.
Because the above inequality is equal if and only if the circle in the sequence is * * *, if and only if the circumscribed circle is above,
... 10 point
In addition, this inequality has an equal sign if and only if the * * * line is above. Therefore, it takes the minimum value if and only if it is the intersection of the circumscribed circles of and.
Therefore, when the minimum value is reached, the four points are * * * circles ... 20 points.
(2) Remember, then, by sine theorem, so, that's it.
,
Finishing, ... 30 points
Solve or (give up),
Therefore,.
From the known =, yes, it is sorted out, so you can get …40 points.
Therefore, it is an isosceles right triangle.
It is also an isosceles right triangle, so …
So ... 50 points
[Solution 2] (1) As shown in Figure 2, the circumscribed circle connecting the intersection points is on the point (because it is outside, it is above).
If the vertical lines made separately are crossed, it is easy to know, so the inner corner of the third part is, then, because, by the same token,
So ?... 10/0 point
Set,,,
For any point on the plane, there is
,
Therefore.
From the point of arbitrariness, it is known that the point is the point that makes it reach the minimum value.
From point to top, it is * * * at four o'clock ... 20 minutes.
(2) The minimum value of item (1)
,
Remember, according to sine theorem, therefore, that's it.
,
Finishing, ... 30 points
Solve or (give up),
Therefore,.
From the known =, yes, it is sorted out, so you can get …40 points.
So, it is an isosceles right triangle, because the point is on it, so it is a rectangle, so, so ... 50 minutes.
[Solution 3] (1) A complex plane is introduced, and the corresponding complex number is still expressed by an equation.
According to the triangle inequality, for complex numbers, there are
,
Take an equal sign if and only if it is in the same direction as (a complex vector).
Yes,
therefore
( 1)
,
therefore
. (2) ... 10 point
(1) is equal to the following conditions
Complex sum
The same direction, so there are real numbers, which makes
,
,
So,
The angle to which a vector rotates is equal to the angle to which it rotates,
So that all four points are * * * round.
(2) The conditions for the formula to take the equal sign are obviously * * * lines and above.
So when the minimum value is reached, the point is on the circumscribed circle, and the circle at four o'clock is * * *...20 points.
(ii) from (i).
The following is the same solution.
Second, (the full score of this question is 50 points)
Let it be a periodic function and 1 be a periodic function. Prove:
(i) If it is a rational number, it has a prime number, making it a period;
(ii) If it is an irrational number, then there is a series of irrational numbers with one term, and each term is a period.
[Proof] (i) If it is a rational number, it has a positive integer, so it has an integer.
.
therefore
Yes, that period ... 10 points.
Let the prime factor be, then, like this.
Yes, that period ... 20 minutes.
(ii) if it is unreasonable, make
,
Then, it is irrational to make
,
……
,
30 points for ........................
From mathematical induction, it is easy to know that they are irrational sums. So, that is to say, they are decreasing sequence ... 40 points.
The last proof: each one is a period. In fact, because 1 and are periods, they are also periods. If it's a period, it's also a period. Through mathematical induction, it is proved that everything is a cycle ... 50 minutes.
Iii. (Full score for this question is 50 points)
Assumption, proof: If and only if there is a sequence that satisfies the following conditions:
(ⅰ) , ;
(ii) Existence;
(ⅲ) , .
[Proof] Necessity: Assuming that there are conditions (1), (2) and (3). Please note that the formula in (iii) can be changed to
, ,
One of them is.
Add the above formula in the project 1 to the project, and note that
... 10 point
Can be set by (II), limited by the above formula.
,
So ... 20 points
Adequacy: Hypothesis. Polynomial functions are defined as follows:
, ,
It is a increasing function on [0, 1], and
, .
Therefore, the equation has a unique root in [0, 1], that is, ... 30 points.
If the sequence is 0, the condition (i) is obviously satisfied, and
.
Because, therefore, that is, the limit exists, satisfying (Ⅱ) ... 40 points.
Finally, it is verified that (III) is satisfied because, that is, therefore.
.
To sum up, there are sequences satisfying (i), (ii) and (iii) ... 50 points.