First attempt
First, multiple-choice questions (the full score of this question is 42 points, and each small question is 7 points)
1. If, then (1)
A.24。 c。 d。
2. In △ABC, if the maximum angle ∠A is twice the minimum angle ∠C, and AB = 7 and AC = 8, then BC = (c).
A.。 B. c。 d。
3. The largest integer is not greater than, and the number of solutions of the equation is (c).
A. 1。 B. 2。 C. 3。 D. 4。
4. Let the center of the square ABCD be point O, and randomly select two from all triangles whose vertices are A, B, C, D and O.. The probability that their areas are equal is (b).
A.。 B. c。 d。
5. As shown in the figure, in the rectangular ABCD, AB = 3, BC = 2. If a semicircle is made in the rectangle with BC as the diameter, the tangent AE of the semicircle from point A, CBE = (d).
A.。 B. c。 d。
6. Let it be a positive integer greater than 1909, so that the number of complete squares is (b).
A.3. BC5。 D. 6。
2. Fill in the blanks (the full score of this question is 28 points, and each small question is 7 points)
1. is known as a real number. If it is about two non-negative real roots of a quadratic equation, the minimum value of _ _ _ _ _ _ _ _.
2. Let D be a point on the AB side of △ABC, let DE//BC intersect with AC at point E, and DF//AC intersect with BC at point F. Given the sum of the areas of △ADE and △DBF, the area of quadrilateral DECF is _ _ _ _ _.
3. If the real number meets the conditions, then _ _ _ _ _.
4. If it is known to be a positive integer and it is an integer, then such an ordered number pair * * * has _ _ 7 _ _ pairs.
The second test (1)
1. (The full mark of this question is 20) It is known that the intersection of the image of quadratic function with the axis is A and B, and the intersection with the axis is C. Let the center of the circumscribed circle of △ABC be point P. 。
(1) proves that the other intersection of ⊙P and the axis is a fixed point.
(2) If AB happens to be the diameter of ⊙P, sum.
Solution (1) It is easy to find the coordinates of the point as, let,, and then.
Let the other intersection point of ⊙P and the axis be d, because AB and CD are two intersecting chords of ⊙P, and their intersection point is point O, so OA× OB = OC× OD, then.
Because, so the point is on the negative semi-axis of the shaft, so the point D is on the positive semi-axis of the shaft, so the point D is a fixed point, and its coordinate is (0, 1).
(2) Because AB⊥CD, if AB happens to be the diameter of ⊙P, then C and D are symmetrical about point O, then the coordinates of this point are
Namely.
Again, so
, the solution.
Let CD be the height on the hypotenuse AD of the right triangle ABC, and be the centers of △ADC and △BDC respectively, AC = 3, BC = 4, and find.
The answer is E⊥AB in e and F⊥AB in f.
In the right triangle ABC, AC = 3, BC = 4,.
And CD⊥AB can be obtained by projective theorem, therefore,
.
Because e is the radius of the inscribed circle of the right triangle ACD, =
Connect D and D, then D and D are bisectors of ∠ADC and ∠BDC respectively, so ∠ DC = ∠ DA = ∠ DC = ∠ DB = 45, so ∠ D =90, so D∞.
The same is true. So =
3. (The full score of this question is 25 points) It is known as a positive number and meets the following two conditions:
①
②
It is proved that three sides can form a right triangle.
Prove that 1 is multiplied by ① ②,
That is to say,
That is to say,
That is to say,
That is to say,
That is to say,
That is, that is,
That is to say,
So or or, is or or.
So with three sides as long, you can form a right triangle.
Proof 2 combined with formula 1 can be obtained from formula 2.
Deformation, get ③
It is also obtained by the formula (1), that is,
Substitute in equation 3, and you get, that is.
,
So either or.
Combined with ① formula, it can be obtained.
So with three sides as long, you can form a right triangle.
The second test (b)
1. (The full mark of this question is 20) Questions and solutions are the same as the first question in Volume (a).
2. (The full mark of this question is 25) It is known that when △ABC, ∠ ACB = 90, the high line CH of AB and the bisectors AM and BN of △ABC intersect at P and Q, and the midpoints of PM and QN are E and F respectively. Verification: EF‖AB.
Solution because BN is the bisector of ∠ABC, so.
Because of CH⊥AB, so
,
Therefore.
F is the midpoint of QN, so CF⊥QN, so c, f, h and b are * * * cycles.
Again, so FC = FH, so point F is on the middle vertical line of CH.
Similarly, point e is on the perpendicular of CH.
So EF⊥CH. and AB⊥CH, so ef ab.
3. (The full mark of this question is 25) The question type and solution are the same as the third question in Volume (A).
The second test (c)
1. (The full mark of this question is 20) Questions and solutions are the same as the first question in Volume (a).
2. (The full score of this question is 25) The question type and solution are the same as the second question in Volume (B).
3. (The full score of this question is 25 points) It is known as a positive number and meets the following two conditions:
①
②
Do you have a triangle with three sides? If it exists, find the maximum internal angle of the triangle.
Solution 1 multiplied by ① ②,
That is to say,
That is to say,
That is to say,
That is to say,
That is to say,
That is, that is,
That is to say,
So or or, is or or.
Therefore, three sides can form a right triangle with a maximum internal angle of 90.
Solution 2 is combined with formula ①, which can be obtained from formula ②.
Deformation, get ③
It is also obtained by the formula (1), that is,
Substitute in equation 3, and you get, that is.
,
So either or.
Combined with ① formula, it can be obtained.
Therefore, three sides can form a right triangle with a maximum internal angle of 90.