It is proved that the sum of the vectors of the center of gravity of a triangle and the lines of its three vertices is a zero vector

Let a (x1, Y1), B (X2, Y2), C (X3, Y3) be three vertices, then the center of gravity o [(x1 + x2 + x3) / 3, (Y1 + Y2 + Y3) / 3] vector OA = [X1 - (x1 + x2 + x3) / 3, Y1 - (Y1 + Y2 + Y3) / 3] vector ob = [X2 - (x1 + x2 + x3) / 3, Y2 - (Y1 + Y2 + Y3) / 3] vector OC = [X3 - (x1 + x2 + x3) / 3, Y3 - (Y1 + Y2 + Y3) / 3], then the vector OA +

Please prove by vector that the center of gravity of the triangle connected by the same divisor coincides with the center of gravity of the original triangle

Let BD: DC = Ce: EA = AF: FB = γ
According to vector addition, vector BD + vector CE + vector AF = (γ / (1 + γ)) (vector BC + vector Ca + vector AB) = (γ / (1 + γ)) * 0 = 0
Let o be the center of gravity of △ ABC, with vector OA + vector ob + vector OC = 0
And the vector od = vector ob + vector BD
Vector OE = vector OC + vector CE
Vector of = vector OA + vector AF
therefore
Vector od + vector OE + vector of
=(vector ob + vector BD) + (vector OC + vector CE) + (vector OA + vector AF)
=(vector OA + vector ob + vector OC) + (vector BD + vector CE + vector AF)
= 0 + 0 = 0
So o is also the center of gravity of △ def
Note: a theorem is used here
O is the center of gravity of a triangle if and only if the vector OA + ob + OC = 0

What are the inner center, outer center, center of gravity and perpendicular center of triangle ABC? How to prove them?

1. The intersection of bisectors of three internal angles of a triangle
2. The intersection of the vertical bisectors of the three sides of a triangle
3. [center of gravity] the intersection of the three midlines of a triangle;
4. The intersection of the three heights of a triangle

As shown in the figure, if △ ABC rotates 90 ° around point O to get △ def, and D and a are corresponding points, ad = 4cm, then s △ AOD=______ ．

After ∵ ABC rotates 90 ° around point O, we can get ∵ def, D and a are corresponding points, ∵ Ao = do, ∵ AOD = 90 °, ∵ AOD is isosceles right triangle, ∵ ad = 4cm, ∵ ad side high line = 12ad = 12 × 4 = 2cm, ∵ s △ AOD = 12 × 4 × 2 = 4cm2

(1 / 2) in the plane rectangular coordinate system, rotate the triangle ABC 180 degrees around the point C (0, - 1) to get the triangle a'b'c ', and let the coordinates of a' be (a, b)
(1 / 2) in the plane rectangular coordinate system, rotate the triangle ABC 180 degrees around the point C (0, - 1) to get the triangle a'b'c '. Suppose that the coordinates of a' are (a, b), what are the coordinates of point a?

The coordinate of point A1 corresponding to a is (a, B + 1) by translating AA 'upward one unit. Because A1 and A2 are symmetrical about the origin, a' corresponds to point A2 (- A, - B-1).. a '(- A, - b-2)

If the function f (x) = x + 3x increases monotonically on [P, + ∞), then the minimum value of real number P is___ ．

In order to study the monotonicity on [P, + ∞), then x ＞ 0f (x) = x + 3x ≥ 23, if and only if x = 3, the function with equal sign  decreases monotonically on (0,3), and increases monotonically on [3, + ∞). The minimum value of real number P is 3, so the answer is: 3

The coordinates of the three vertices of triangle ABC are a (- 1,0) B (- 2,0) C (- 2,2) rotated 180 degrees around a point P to get the def, so that DF can find the DF coordinates on the hyperbola y = 12  X

Let P (P, q) d (a, b), f (C, d) be the midpoint of AD and CF, so (- 1 + a) / 2 = P (0 + b) / 2 = q (- 2 + C) / 2 = P (2 + D) / 2 = q, so - 1 + a = - 2 + C0 + B = 2 + DDF in y = 12 / XAB = 12, CD = 12, substitute C = a + 1, d = B-2 into ab-2a + B-2 = 12ab = 12, so substitute B = 2A + 2 into AB = 12a & sup2; + a-6 = 0A = - 3, a = 2A = - 3

Help summarize the first chapter of high school mathematics (function parity and increase and decrease and set) and the correct writing format

In fact, according to the book, try to draw the figure below! No problem!

The coordinates of the three vertices of the triangle ABC are a (x1, Y1), B (X2, Y2), C (X3, Y3), and the coordinates of the center of gravity g of the triangle ABC are obtained

There's a formula for this
G( (x1+x2+x3)/3,(y1+y2+y3)/3 )

New one mathematics, is about the parity of the function and prove some problems of increasing and decreasing function~~
1. Given that f (x) is an even function, G (x) is an odd function, and f (x) + G (x) = x & # 178; + X-2, find the expression of F (x), G (x)
2. If f (x) = (x + 1) (x + a) is even, then a= .
3. Let even function f (x) satisfy f (x) = x cubic - 8 (x ≥ 0), then {x | f (X-2) > 0} = ()
A. {x | x ＜ - 2 or X ＞ 4} B. {x | x ＜ 0 or X ＞ 4} C. {x | x ＜ 0 or X ＞ 6} D. {x | x ＜ - 2 or X ＞ 2}
4. The function f (x) defined on R satisfies: for any α, β∈ R, there is always f (α + β) - [f (α) + F (β)] = 2010, then the following statement is correct: a.f (x) - 1 is odd function, B.F (x) + 1 is odd function, C.F (x) - 2010 is odd function
D. F (x) + 2011 is an odd function
5. If the image of the even function y = f (x) has three intersections with the X axis, then the sum of all the following of the equation f (x) = O is .
6. Let f (x) be an odd function defined on R, and if x > 0, f (x) = (x power of 2) - 3, then the value of F (- 2) is equal to .

(1)f(x)=x^2-2,g(x)=x
(2) Because f (x) = (x + 1) (x + a) is even function
So f (x) = f (- x)
A = - 1
(3) F (x) MS is not even function!
It can be done without this condition
f(x-2)=(x-4)(x^2-2x+4)>0
Then x > 4 (none of the answers are right, please confirm the question)
(4) Let a = b = 0, then f (0) = - 2010
Because if an odd function is meaningful at x = 0, it must pass through the origin
So f (x) + 2010 is an odd function
(5) Because f (x) is an even function, x = 0 is one solution, and the other two solutions are symmetric about the Y axis, so the sum of all the following solutions of the equation f (x) = O is 0
(6) Because f (x) is an odd function, f (- x) = - f (x), so f (- 2) = - f (2) = - (2 ^ 2-3) = - 1

It is proved that the sum of the vectors of the center of gravity of a triangle and the lines of its three vertices is a zero vector