The problem of mathematical vectors!

Some conclusions: The following are all vectors.

1 if p is the center of gravity of △ABC, PA+PB+PC=0.

2 if p is PA of △ABC? PB=PB？ PC=PA？ PC (internal product)

3 If p is the inner APA+BP B+CPC of △ABC = 0 (ABC is trilateral).

4 If P is the outer center of △ABC |PA|? =|PB|？ =|PC|？

(AP means that the AP vector |AP| is its module)

5 AP=λ(AB/|AB|+AC/|AC|), λ∈[0, +∞) then the straight line AP passes through the heart of △ABC.

6 AP = λ (AB/| AB | COSB+AC/| AC | COSC), λ ∈ [0, +∞) passes through the center of gravity.

7 AP=λ(AB/|AB|sinB+AC/|AC|sinC)，λ∈[0，+∞)

Or AP=λ(AB+AC), λ∈[0, +∞) passes through the center of gravity.

8. If aOA=bOB+cOC, then 0 is the intersection of the centroid of ∠A and the bisector of ∠ B and C.

The following is the relevant proof of some conclusions.

O is the interior of a triangle if and only if aOA vector +bOB vector +cOC vector =0 vector.

Adequacy:

Known aOA vector +bOB vector +cOC vector =0 vector,

AcCOrding to vector addition, the co intersection AB is extended to d:

OA=OD+DA, OB=OD+DB, substituting for known:

a(OD+DA)+b(OD+DB) +cOC=0，

Because of the connection between OD and OC***, OD=kOC can be set.

The above formula can be replaced by a vector of (ka+kb+c) OC+( aDA+bDB)=0.

Vector DA and DB*** lines, vector OC and vectors DA and DB are not * * * lines,

So there can only be: ka+kb+c=0, aDA+bDB=0 vectors,

According to the vector of aDA+bDB=0, the ratio of the length of DA to DB is b/a,

So CD is the bisector of ∠ACB, which proves that the other two are also angular bisectors.

Necessity:

As we all know, O is the center of a triangle.

Let BO and AC intersect at e, and CO and AB intersect at f,

O is the heart

∴b/a=AF/BF,c/a=AE/CE

The parallel lines passing through a as CO intersect with the extension line of B0 at n, and the parallel lines passing through a as BO intersect with the extension line of c0 at m,

So the quadrilateral Oman is a parallelogram.

According to the parallelogram law, it is concluded that

Vector OA

= vector OM+ vector ON

=(OM/CO)* vector CO+(ON/BO)* vector BO

=(AE/CE)* Vector CO+(AF/BF)* Vector BO

=(c/a)* vector CO+(b/a)* vector BO∴a* vector OA=b* vector BO+c* vector CO.

∴a* vector OA+b* vector OB+c* vector OC= vector 0.

It is known that △ABC is an oblique triangle, O is a fixed point on the plane where △ABC lies, and the moving point P satisfies the vector OP = OA+ENTER {(AB/| AB | 2 * SIN2B)+AC/(| AC | 2 * SIN2C)}.

Find the vertical center of the locus of point P passing through the triangle.

OP = OA+enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

OP-OA = enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

AP = enter {(ab/| ab | 2 * sin2b)+AC/(| AC | 2 * sin2c)}，

AP？ BC= enter {(AB？ BC/| AB | 2 * SIN2B)+AC? BC/(| AC | 2 * sin2c)},

AP？ BC= enter {|AB|？ | BC | cos( 180-b)/(|ab|^2*sin2b)+| AC |？ | BCE | COSC/(| AC | 2 * SIN2C)},

AP？ BC= enter {-|AB|？ | BC | cos b/(|ab|^2*2sinb cos b)+| AC |？ |BC| cosC/(|AC|^2*2sinC cosC)}，

AP？ BC = enter {-| BC |/(| AB | * 2 sinb)+| BC |/(| AC | * 2 sinc)}，

According to sine theorem: |AB|/sinC=|AC|/ sinB, so |AB|*sinB=|AC|*sinC.

∴-|bc|/(| ab | * 2 sinb)+| BC |/(| AC | * 2 sinc)= 0，

Is that AP? BC=0，

The locus of point P passes through the vertical center of the triangle.

OP = OA+λ(AB/(| AB | sinB)+AC/(| AC | sinC))

OP-OA =λ(AB/(| AB | sinB)+AC/(| AC | sinC))

AP =λ(AB/(| AB | sinB)+AC/(| AC | sinC))

AP and AB/|AB|sinB+AC/|AC|sinC***

According to sine theorem: |AB|/sinC=|AC|/sinB,

So |AB|sinB=|AC|sinC,

So AP and AB+AC***

AB+AC passes through the midpoint D of BC, so the trajectory of point P also passes through the midpoint D,

Point p passes through the center of gravity of the triangle.

OP = OA+λ(ABC OSC/| AB |+ACcosB/| AC |)

AP=λ(ABcosC/|AB|+ACcosB/|AC|)

AP？ BC=λ(AB？ BC cosC/|AB|+AC？ BC cosB/|AC|)

=λ([|AB|？ | BC | cos( 180-B)cosC/| AB |+| AC |？ |BC| cosC cosB/|AC|]

=λ[-|BC|cosBcosC+|BC| cosC cosB]

=0,

The vector AP is perpendicular to the vector BC,

The trajectory of point p is too vertical.

OP=OA+λ(AB/|AB|+AC/|AC|)

OP-OA =λ(AB/|AB|+AC/|AC|)

AP=λ(AB/|AB|+AC/|AC|)

AB/|AB| and AC/|AC| are unit length vectors in AB and AC directions,

The sum vector of unit vectors of vectors AB and AC,

Because it is a unit vector, the module lengths are all equal, forming a diamond.

The sum vector of the unit vectors of vectors AB and AC is a rhombic diagonal,

It is easy to know that it is an angular bisector, so the trajectory of point P passes through the heart.

Triangle barycenter expression: vector OA+ vector OB+ vector OC= zero vector.

Prove: Let AD be the center line of BC side in triangle ABC, and O be the center of gravity of triangle.

Extend OD to e, make OD=DE, and connect BE and CE.

And BD=DC, so the quadrilateral BOCE is a parallelogram.

So vector OB+ vector OC= vector OE

O is the center of gravity, and AD is divided into two parts: 2: 1, that is, AO = 2OD = OE.

To sum up, vector OA=- vector OE=- (vector OB+ vector OC)

Namely: vector OA+ vector OB+ vector OC=0.

So o is the center of gravity of the triangle

O is the vertical center of the triangle: the square of vector OA+the square of vector BC = the square of vector OB+the square of vector CA = the square of vector OC+the square of vector AB.

Prove: vector OA square+vector BC square = vector OB square+vector CA square.

That is, the square of vector OA-the square of vector OB = the square of vector CA-the square of vector BC.

That is, (vector OA- vector OB) (vector OA+ vector OB)= (vector CA- vector BC) (vector CA+ vector BC)

Vector BA? (vector OA+ vector OB)= (vector CA- vector BC)? Vector BA

Vector BA? (vector OA- vector CA+ vector OB+ vector BC)=0

That is, 2 vector BA? Vector OC=0

∴OC⊥AB

Similarly, OA⊥CB and OB⊥AC can prove it.

So o is the center of the triangle.

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