Just try it.
The outline of the preliminary test competition of the national senior high school mathematics league matches the teaching requirements and contents stipulated in the full-time middle school mathematics syllabus, that is, the knowledge scope and methods stipulated in the college entrance examination are slightly improved, and the preliminary test of probability and calculus is not taken.
Second division
1, plane geometry
Basic requirements: master all the contents determined by the outline of junior high school mathematics competition.
Supplementary requirements: area and area method.
Several important theorems: Menelius Theorem, Seva Theorem, Ptolemy Theorem and siemsen Theorem.
Several important extreme values: the point with the smallest sum of the distances to the three vertices of a triangle-fermat point. The center of gravity is the point where the sum of squares of the distances to the three vertices of a triangle is the smallest. The center of gravity is the point in the triangle where the distance product of three sides is the largest.
Geometric inequality.
Simple isoperimetric problem. Understand the following theorem:
In the set of N-polygons with a certain circumference, the area of the regular N-polygon is the largest.
In a set of simple closed curves with a certain perimeter, the area of the circle is the largest.
In a group of N-sided polygons with a certain area, the perimeter of the regular N-sided polygon is the smallest.
In a set of simple closed curves with a certain area, the circumference of a circle is the smallest.
Motion in geometry: reflection, translation and rotation.
Complex number method and vector method.
Planar convex set, convex hull and their applications.
2. Algebra
Other requirements based on the first test outline:
Image of periodic function and periodic and absolute value function.
Triple angle formula, some simple identities of triangle, triangle inequality.
The second mathematical induction.
Recursion, first and second order recursion, characteristic equation method.
Function iteration, finding n iterations, simple function equation.
N-element mean inequality, Cauchy inequality, rank inequality and their applications.
Exponential form of complex number, Euler formula, Dimov theorem, unit root, application of unit root.
Cyclic permutation, repeated permutation and combination, simple combinatorial identity.
The number of roots of an unary n-degree equation (polynomial), the relationship between roots and coefficients, and the pairing theorem of imaginary roots of real coefficient equations.
Simple elementary number theory problems should include infinite descent method, congruence, Euclid division, nonnegative minimum complete residue class, Gaussian function, Fermat's last theorem, Euler function, Sun Tzu's theorem, lattice points and their properties.
3. Solid geometry
Polyhedral angle, properties of polyhedral angle. Basic properties of trihedral angle and straight trihedral angle.
Regular polyhedron, euler theorem.
Proof method of volume.
Sections, sections, and surface flat patterns will be made.
4. Plane analytic geometry
Normal formula of straight line, polar coordinate equation of straight line, straight line bundle and its application.
The region represented by binary linear inequality.
The area formula of triangle.
Tangents and normals of conic curves.
Power and root axis of a circle.
5. Others
Dove cage principle
The principle of gold tolerance.
Extreme principle.
Division of sets.
Cover.
Spartan king
Menelaus, an ancient Greek mathematician, first proved Menelaus theorem. It points out that if a straight line intersects with three sides AB, BC, CA of △ABC or its extension lines at points F, D and E, then (AF/FB )× (BD/DC )× (CE/EA) =1.
Prove:
Passing point a is the extension line of AG∨BC passing through DF at g,
Then af/FB = ag/BD, BD/DC = BD/DC, ce/ea = DC/ag.
Three formulas are multiplied: AF/FB× BD/DC× CE/EA = Ag/BD× BD/DC× DC/Ag =1.
The inverse theorem also holds: if there are three points F, D and E on the edges of AB, BC and CA or their extension lines, and they satisfy (AF/FB )× (BD/DC )× (CE/EA) =1,then these three points F, D and E are * * lines. Using this inverse theorem, we can judge the trisection line.
In addition, many people will think that writing this formula is too complicated to remember without reading it. Here are some methods provided by others to help them write.
In order to illustrate the problem and impress everyone, we assume that A, B, C, D, E and F in the picture are six tourist attractions, which are connected by roads. We flew over these scenic spots by helicopter and then chose any one to land. We transferred to the bus, went to every scenic spot along the expressway, and finally returned to the starting point, where the helicopter stopped waiting for us to go back.
We don't have to consider how to take the shortest route, we just need to "visit" all the scenic spots. A scenic spot that only "passes by" without stopping to watch is not a "tour".
For example, when the helicopter landed at point A, we set off from point A, "visited" the scenic spots represented by the other five letters, and finally returned to the starting point A. ..
Another requirement is that three scenic spots on the same straight line must swim continuously before they can be changed to other scenic spots on the straight line.
There are four kinds of travel plans from point A, which are explained one by one below:
Scheme ①-from A to F (non-stop), then back to B (non-stop), then to D (non-stop), then from B (non-stop) to C (non-stop), then to E (non-stop), and finally from E to C (non-stop) to the starting point A. ..
According to this scheme, you can write the following relationship:
(AF:FB)*(BD:DC)*(CE:EA)= 1 .
Now, you know how to write the formula of Menelaus theorem.
The travel plan from point A also includes:
Scheme ② can be abbreviated as: A→B→F→D→E→C→A, from which the following formula can be written:
(AB: BF) * (FD: DE) * (EC: CA) =1.Starting from A, you can also go in the direction of "C", so there are:
Schema ③-A → C → E → D → F → B → A, from which the formula can be written:
(AC: CE) * (ED: DF) * (FB: BA) =1.Starting from A, there is one last plan:
Schema ④-A → E → C → D → B → F → A, from which the formula is written:
(AE:EC)*(CD:DB)*(BF:FA)= 1 .
Our helicopter can also choose to land at any point of B, C, D, E and F, so there are some other formulas in the figure.
It is worth noting that some formulas contain four factors instead of three in Menelaus theorem. When the helicopter lands at point B, there are four factors. At points C and F, there are three formulas and four formulas. In Formula 4, some scenic spots will be visited twice.
I don't know if Menelaus thinks so, but I just listed one or two typical formulas for everyone to see.
It can also be seen from the counterclockwise direction that the distance from the first vertex to the first intersection is greater than the distance from the next vertex, and so on, three proportions can be obtained, and their product is 1.
Can we say that we have a deeper understanding of Menelaus theorem now? Those complicated multiplication and division relationships are written incorrectly and can't be remembered.
Ptolemy theorem in general geometry textbooks was actually written by Hipparchus, and Ptolemy only extracted it from his book.
Ptolemy theorem points out that the sum of the products of two opposite sides of a convex quadrilateral inscribed in a circle is equal to the product of two diagonal lines.
In a quadrilateral inscribed in a circle, the area of a rectangle surrounded by two diagonal lines is equal to the sum of the area of a rectangle surrounded by one set of opposite sides and the area of a rectangle surrounded by another set of opposite sides.
From this theorem, we can derive the sum and difference formula of sine and cosine and a series of trigonometric identities. Ptolemy theorem is essentially about the basic properties of circles. [Edit this paragraph] Proof (The following is the proof of inference, Ptolemy theorem can be regarded as a special case. )
Let △ABE make ∠BAE=∠CAD ∠ABE=∠ ACD in any quadrilateral ABCD.
Because △ABE∽△ACD
So BE/CD=AB/AC, which means be AC = AB CD (1).
There is also a proportional formula AB/AC=AE/AD.
And ∠BAC=∠DAE.
So △ABC∽△AED is similar.
BC/ED=AC/AD means ed AC = BC ad (2).
The same is true of (1)+(2)
AC(BE+ED)=AB CD+AD BC
And because BE+ED≥BD.
(Only when the quadrilateral ABCD is inscribed with a circle, the equal sign holds, that is, "Ptolemy Theorem")
So the proposition can be proved that the inference of [edit this paragraph] is 1. Any convex quadrilateral ABCD must have AC BD ≤ ABCD+AD BC. If and only if ABCD is a four-point * * * circle, take the equal sign.
2. The inverse theorem of Ptolemy's theorem also holds: the sum of the products of two opposite sides of a convex quadrilateral is equal to the product of two diagonal lines, then this convex quadrilateral is inscribed with a circle. [Edit this paragraph] The Ptolemy inequality is generalized as: the product of any two opposite sides of a quadrilateral is not less than the product of other opposite sides, if and only if it is a * * * circle or a * * * line.
Simple proof: complex number identity: (a-b)(c-d)+(a-d)(b-c)=(a-c)(b-d), modulus on both sides,
The inequality acbd ≤| (a-b) (c-d) | +| (b-c) (a-d) | = abcd+bcad.
note:
The equal sign 1. holds if the radial angles of (a-b)(c-d) and (a-d)(b-c) are equal, which is equivalent to the four-point circle of A, B, C and D. ..
2. Four points are not limited to the same plane.
Euler theorem: On the AD of a line segment, mark two points, B and C, and then AD BC+AB CD = AC BD.
There is a triangle ABC with a small P on the plane. The projection of P on three sides of a triangle (that is, the vertical foot from P to the side) * * line (this line is called Simson line) if and only if P is on the circumscribed circle of the triangle.
The relevant results are as follows:
The vertical center of the triangle is called H, and the intersection of the Simpson line and PH is the midpoint of the line segment PH, which is on the nine-point circle.
The intersection angle of two Simpson lines is equal to the circumferential angle of two points.
If the circumscribed circles of two triangles are the same, the point P on the circumscribed circle corresponds to the intersection angle of the two triangles' Simpson lines, regardless of the position of P. ..
The necessary and sufficient condition for drawing a perpendicular from a point to three sides of a triangle is that the point falls on the circumscribed circle of the triangle. [Edit this paragraph] Prove that 1: there is a little p on the circumscribed circle of △ABC, and PE⊥AC is in E, PF⊥AB is in F, PD⊥BC is in D, which are connected with DE and DF respectively.
It is easy to prove that P, B, F, D, P, D, C, E and A, B, P and C are * * cycles respectively, so ∠FDP=∠ACP ①, ∫ is ∠ABP's complement) and ∠PDE =∞.
② and ∠ ACP+∠ PCE = 180.
③ ∴∠FDP+∠PDE= 180
④ lines are f, d and E***. On the other hand, when the straight lines are F, D and E***, the positions of circles A, B, P and C*** can be seen from ④→→→→→→→→→→③→①.
Proof 2: As shown in the figure, if the line of L, M and N connects BP and CP, then because PL is perpendicular to BC, PM is perpendicular to AC and PN is perpendicular to AB, there are B, P, L, N and.
M, P, L and C are four * * * cycles respectively, and there are
Angle PBN = 180- angle PLN = angle PLM = angle PCM.
So A, B, P and C are four * * * circles.
If a, b, p and c are four * * * circles, then the angle PBN = the angle PCM. Because PL is perpendicular to BC, PM is perpendicular to AC and PN is perpendicular to AB, there are four * * * cycles of B, P, L, N and M, P, L and C, and there are
Angle PBN = 180- angle PLN = angle PLM = angle PCM.
So, l, m, n three-point * * * line.
For any point o on the ABC plane of triangle, COnnect AO, BO and co and extend. If the other side of the triangle passes through p, q and r respectively, BP/PC CQ/QA AR/Rb = 1.
The inverse proposition of the above theorem also holds.
Seva (1648- 1734) theorem and its inverse theorem can be used to prove the * * * point problem about three straight lines. Hehe, there are still a few months. Come on, I wish you all the best and win the grand prize!