Barycenter of points : corrected 1st grade math exercises.

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The answer key to the math exercises in 1st grade on the barycenter n weighted points. Use the stability and associativity properties of the barycenter in first grade.

Exercise 1:

It is up to you to make these constructions knowing that the barycenter is necessarily aligned with points A and B.

Exercise 2:

1. Describe the set of points M in the plane such that $\,\%7C5\vec{MA}+6\vec{MB}\%7C=22\,.$
Consider I barycenter of points (A,5) and (B,6)
$\,5\vec{MA}+6\vec{MB}=11\vec{MI}+5\vec{IA}+6\vec{IB}=11\vec{MI}+\vec{0}=11\vec{MI}\,.$
so the set corresponds to the circle of center I and radius 2 .

2. Describe the set of points M in the plane such that $\,\%7C-5\vec{MA}+8\vec{MB}\%7C=12\,.$
Consider I barycenter of points (A,-5) and (B,8)
$\,-5\vec{MA}+8\vec{MB}=3\vec{MI}-5\vec{IA}+8\vec{IB}=3\vec{MI}+\vec{0}=3\vec{MI}\,.$
so the set corresponds to the circle of center I and radius 4 .

3. Describe the set of points M in the plane such that $\,\%7C5\vec{MA}-6\vec{MB}\%7C=\%7C7\vec{MA}-6\vec{MB}\%7C\,.$
Consider I barycenter of points (A,5) and (B,-6) and J barycenter of points (A,7) and (B;-6)
$\,5\vec{MA}-6\vec{MB}=-\vec{MI}+5\vec{IA}-6\vec{IB}=-\vec{MI}+\vec{0}=-\vec{MI}\,.$
Also: $\,7\vec{MA}-6\vec{MB}=\vec{MJ}+7\vec{JA}-6\vec{JB}=-\vec{MJ}+\vec{0}=\vec{MJ}\,.$
so we are looking for the points M such that MI=MJ, the set is therefore the bisector of the segment [IJ].

Exercise 3:
Let R be an orthonormal reference frame of the plane .

1. Carry out the construction

2. We note $\,C_1$ the set of points M of the plane such that $\,\%7C4\vec{MA}+5\vec{MB}\%7C=45\,.$ .
$4\vec{MA}+5\vec{MB}=(4(3-x)+5(-1-x);4(4-y)+5(2-y))\,\\=(-9x+7;-9y+26).$ .
$\,\%7C4\vec{MA}+5\vec{MB}\%7C=\sqrt{(-9x+7)^2+(-9y+26)^2}\\=\sqrt{81%5B(x-\frac{7}{9})^2+(y-\frac{26}{9})^2%5D}.$

Determine the equation of the set $\,C_1$ .
The equation of this set is therefore :
$(x-\frac{7}{9})^2+(y-\frac{26}{9})^2=(\frac{45}{81})^2.$

2. Show that it is a mediator.

Exercise 4:

Consider the point G, barycenter of (A,1); (B,1) and (C,2)

$\vec{MA}+\vec{MB}+2\vec{MC}=\vec{MG}+\vec{GA}+\vec{MG}+\vec{GB}+2\vec{MG}+2\vec{GC}=4\vec{MG}$

Consider the point K, barycenter of (B,1) and (C,3)

$\vec{MB}+3\vec{MC}=\vec{MK}+\vec{KB}+3\vec{MK}+3\vec{KC}=4\vec{MK}$

so it is the same as looking for the set of points in the plane such that :

$,\%7C,4\vec{MG},,\%7C=,\%7C4\vec{MK},,\%7C$

$,\%7C,\vec{MG},,\%7C=,\%7C\vec{MK},,\%7C$

Conclusion: the set of points M is the bisector of the segment [MK]

Exercise 7:

Let’s use the associativity of the barycenter

Let K be the barycenter of (A,1) (B,1) and L the barycenter of (C,3) (D,3)

The points K and L are isobarycenters so they are the midpoints of segments.

Moreover, these three barycenters exist because the sum of their masses is non-zero.

By associativity of the barycenter, G is the barycenter of (K,2) (L,6)

To construct G, it is sufficient to place the point K in the middle of [AB] and L in the middle of [CD].

and

$2\vec{GK}+6\vec{GL}=\vec{0}$

$2\vec{GL}+2\vec{LK}+6\vec{GL}=\vec{0}$

$8\vec{GL}=-2\vec{LK}$

$8\vec{GL}=2\vec{KL}$

$\vec{LG}=\frac{2}{8}\vec{LK}$

$\vec{LG}=\frac{1}{4}\vec{LK}$

Then it is enough to place the point K at the quarter of the segment [LK] starting from the point L.

Exercise 8:

Hint: use the associativity of the barycenter.
ABCD is a quadrilateral.
We note G its isobarycenter.
The purpose of this exercise is to clarify the position of G.

1) Let I be the midpoint of [AB] and J the midpoint of [CD].

Show that G is the barycenter of I and J with coefficients to be specified.

2) Conclude and make a figure.

Exercise 9:
1. For each of the following cases, where should the hook G be attached to the segment [AB] to achieve balance?
(M = 2 kg)

CASE 1:

Let G be the barycenter of (A,2) and (B,3)
G exists because $2+3=5\neq\,0$
$2\vec{GA}+3\vec{GB}=\vec{0}$

$2\vec{GA}+3\vec{GA}+3\vec{AB}=\vec{0}$

$5\vec{GA}=-3\vec{AB}$

$\vec{GA}=-\frac{3}{5}\vec{AB}$

${\color{DarkRed}\,\vec{AG}=\,\frac{3}{5}\vec{AB}}$

Conclusion: the hook must be placed at $\frac{3}{5}$ of [AB] starting from point A.

Case 2:

Let G be the barycenter of (A,2) and (B,5)
G exists because $2+5=7\neq\,0$
$2\vec{GA}+5\vec{GB}=\vec{0}$

$2\vec{GA}+5\vec{GA}+5\vec{AB}=\vec{0}$

$7\vec{GA}=-5\vec{AB}$

$\vec{GA}=-\frac{5}{7}\vec{AB}$

${\color{DarkRed}\,\vec{AG}=\,\frac{5}{7}\vec{AB}}$

Conclusion: the hook must be placed at $\frac{5}{7}$ of [AB] starting from point A.

These diagrams can be reproduced at any scale.

Exercise 10:
1. Let I be the middle of [BC].

Show that :

$\vec{GB}+\vec{GC}=2\vec{GI}$

Using the associativity of the barycenter, we can state that

So we have :
$2\vec{GA}+2\vec{GI}=\vec{0}$
or
$2\vec{GA}+\vec{GB}+\vec{GC}=\vec{0}$ (definition of the barycentre G)

${\color{DarkRed}\,\vec{GB}+\vec{GC}=2\vec{GI}}$

2. Deduce that G is the barycenter of A and I with coefficients to be specified.

3. Conclude.

$2\vec{GA}+2\vec{GI}=\vec{0}$

$\Leftrightarrow\,\vec{GA}+\vec{GI}=\vec{0}$

Conclusion : G is the middle of the segment [AI].

Exercise 11:

Hint: G is the middle of [AI] and use the associativity of the barycenter.
Consider a triangle ABC and designate by G the barycenter of (A; 1), (B; 4) and (C; – 3).

1. Construct the barycenter I of (B; 4) and (C; – 3).

2. Show that $\vec{GA}+\vec{GI}=\vec{0}$ .

3. Deduce the position of G on (AI).

Exercise 12:
In triangle ABC, E is the middle of [AB] and G is the barycenter of (A; – 2), (B; – 2) and (C; 15).

Show that G, C and E are aligned.

First, the barycenter G exists because $-2-2+15=11\neq0$

We have:

$-2\vec{GA}-2\vec{GB}+15\vec{GC}=\vec{0}$

and $\vec{AE}=\frac{1}{2}\vec{AB}$

Let’s use the Chasles relation:

$-2\vec{GE}-2\vec{EA}-2\vec{GE}-2\vec{EB}+15\vec{GC}=\vec{0}$

$-2\,(\vec{EA}+\vec{EB}\,\,)-4\vec{GE}+15\vec{GC}=\vec{0}$

or $\vec{EA}+\vec{EB}=\vec{0}$ because E is the middle of [AB].

so we have the equality :

$-4\vec{GE}+15\vec{GC}=\vec{0}$

$\vec{GC}=\frac{4}{15}\vec{GE}$

We conclude that the vectors $\vec{GC}$ and $\vec{GE}$ are collinear.

Moreover the point G belongs to these two vectors

therefore the points G, C and E are aligned.

Exercise 13:
B is the middle of [AC].

Show that the barycenter of (A; 1) and (C; 3) is coincident with that of (B; 2) and (C; 2).

Let $H=bary\,\{\,(A;1),(C;3)\,\,\}$ and $G=bary\,\{\,(B;2),(C;2)\,\,\}$

We have:

$\vec{HA}+3\vec{HC}=\vec{0}$

$2\vec{GB}+2\vec{GC}=\vec{0}$

or
$\vec{GB}+\vec{GC}=\vec{0}$ and $\vec{HA}+3\vec{HA}+3\vec{AC}=\vec{0}$

$4\vec{HA}+3\vec{AC}=\vec{0}$

$\vec{GA}+\vec{AB}+\vec{GA}+\vec{AC}=\vec{0}$

IN SUMMARY:

$2\vec{GA}+\vec{AB}+\vec{AC}=\vec{0}$ and $\vec{AH}=\frac{3}{4}\vec{AC}$

$2\vec{GA}+\frac{3}{2}\vec{AC}=\vec{0}$ (because B is the middle of [AC] ) and $2\vec{AH}=\frac{3}{2}\vec{AC}$

thus:

$2\vec{GA}+2\vec{AH}=\vec{0}$

$2\,(\vec{GA}+\vec{AH}\,\,)=\vec{0}$

$\vec{GA}+\vec{AH}\,=\vec{0}$

$\vec{GH}\,=\vec{0}$

Conclusion: the points G and H are merged.

Exercise 14:
ABC is a triangle.

1. G is the barycenter of (A; 1), (B; 2) and (C; 3). Construct the point G. Explain.

Let’s use the associativity of the barycenter.
G is the barycenter of (A,1) (I,5) with I the barycenter of (B,2) and (C,3)

2. G ‘ is the barycenter of (A ; 1), (B ; 3) and (C ; – 3). Construct the point G ‘ . Explain.
Let’s use the associativity of the barycenter,
G’ is the barycenter of (K,4) and (C,-3) with K being the barycenter of (A,1) and (B,3)

3. Show that (AG’) is parallel to (BC).

Hint: show that the vectors $\vec{AG'}$ and $\vec{BC}$ are collinear.

Exercise 15:

First of all G exists because $1+1+3+3=8\neq0$

Note $I=bary\,\,{\(A;1),(B;1)\,\,\}$ and $J=bary\,\,{\(C;3),(D;3)\,\,\}$

Let’s use the associativity of the barycenter.

then
$G=bary\,\,{\(I;2),(J;6)\,\,\}$

We place I which is the middle of [AB].
We place J in the middle of [CD]

then :

$2\vec{GI}+6\vec{GJ}=\vec{0}$

$2\vec{GI}+6\vec{GI}+6\vec{IJ}=\vec{0}$

$8\vec{GI}=-6\vec{IJ}$

$\vec{IG}=\frac{6}{8}\vec{IJ}$

$\vec{IG}=\frac{3}{4}\vec{IJ}$

The segment [IJ] is divided into four equal parts
and we place the point G at three quarters of [IJ] starting from I.

Exercise 16:

Exercise 17:

Exercise 18:

Let ABC be an equilateral triangle of side 3 cm.

1) Locate, with justification, the barycenter Z of (A; 1), (B; 3) and (C; – 3).

z exists because $1+3-3=1\neq\,0$

$\vec{ZA}+3\vec{ZB}-3\vec{ZC}=\vec{0}$

$\vec{ZA}+\,3\vec{ZA}+\,3\vec{AB}\,-3\vec{ZA}\,-3\vec{AC}=\vec{0}$

$\vec{ZA}\,+\,3\vec{AB}\,-3\vec{AC}=\vec{0}$

$\vec{ZA}\,+\,3\vec{AB}\,+3\vec{CA}=\vec{0}$

$\vec{ZA}\,+\,3\vec{CB}\,=\vec{0}$
$\vec{ZA}=-\,3\vec{CB}$

$\vec{ZA}=\,3\vec{BC}$

2) Show that the lines (AZ) and (BC) are parallel.

According to the above, $\vec{ZA}=\,3\vec{BC}$ .

These vectors are collinear, so we have (AZ) parallel to (BC).

Exercise 19:

Hint: use the fact that G is the isobarycenter of the system {(A,1);(B,1);(C,1)}

and use the associativity of the barycenter.

ABC is a triangle with center of gravity G.

I, J, M, N, R and S are the points defined by :

$\vec{AI}=\frac{1}{3}\vec{AB};\vec{AJ}=\frac{2}{3}\vec{AB};\vec{AM}=\frac{1}{3}\vec{AC}\\\vec{AN}=\frac{2}{3}\vec{AC};\vec{BR}=\frac{1}{3}\vec{BC};\vec{BS}=\frac{2}{3}\vec{BC}$

Show that the lines (IS), (MR) and (NJ) are concurrent at G.

Exercise 20:

Hint: we can consider the barycenter G of (A; 5), (B; 2) and (C; – 3).
ABC is a triangle.
Consider the barycenter A’ of (B; 2) and (C; – 3), the barycenter B ‘ of (A; 5) and (C; – 3)
and the barycenter C ‘ of (A; 5) and (B; 2).

Show that the lines (AA ‘), (BB ‘) and (CC ‘) are concurrent.

Exercise 21:
ABC is a triangle. Let G be the barycenter of (A; 1), (B; 3) and (C; – 3).

Show that the lines (AG) and (BC) are parallel.

First, the barycenter G exists because $1+3-3=1\neq0.$

We have:

$\vec{GA}+3\vec{GB}-3\vec{GC}=\vec{0}$

Let’s use the Chasles relation:

$\vec{GA}+3\vec{GB}-3\vec{GB}-3\vec{BC}=\vec{0}$

$\vec{GA}-3\vec{BC}=\vec{0}$

$\vec{GA}=3\vec{BC}$

The vectors $\vec{GA}$ and $\vec{BC}$ are thus collinear

the lines (AG) and (BC) are parallel.

Exercise 22:

1. Place the points A(1 ; 2), B( – 3 ; 4) and C( – 2 ; 5) in a reference frame.
Let G be the barycenter of the weighted points (A; 3), (B; 2) and (C; – 4).

2. What are the coordinates of G? Place G.

3. Does the line (BG) pass through the origin of the reference frame? Justify.

Exercise 23:

[AB] is a segment of length 10 cm and G bar {(A ; 2) , (B ; 3)}

1. Expand and reduce $2(\vec{MG}+\vec{GA})^2+3(\vec{MG}+\vec{GB})^2$

$=5MG^2+2GA^2+3GB^2+4\vec{MG}.\vec{GA}+6\vec{MG}.\vec{GB}$

2. Then prove that for any point M in the plane we have 2MA² + 3MB² = 5MG² + 120

3. Determine then and represent the set of points M of the plane such that 2MA² + 3MB² = 245

Exercise 24:

Hint: use the definition and properties of the barycenter.

A, B and C are 3 points of the plane not aligned and k is any real number.

I bar { (B ;1), (C ;2)} and G the barycenter of(A, k),(B, 1- k) and(C, 2)

1. Express $\vec{IG}$ as a function of $\vec{IA}$ , $\vec{IB}$ and $\vec{IC}$ .

2. Simplify the expression obtained in 1. and deduce the set (E) of points G when k describes $\mathbb{R}$ .

3. Graphically represent (E) in the case AB = 5 cm, BC = 6 cm, AC = 5.5 cm

Exercise 25:

Tip: Don’t complicate your life with hellish calculations, use cleverly

barycentric associativity.

A, B, C and D are four distinct points.

Let K be the barycenter of (A, 3) (B, 1), J the midpoint of [DC], G the center of gravity of BCD and I the midpoint of [AG].

Show that the points I, J and K are aligned.

Exercise 26:

Hints: use the properties of the barycenter and associativity.

ABC a triangle; to any real m, we associate the point G
_m
the barycenter of (A; 2); (B; m) and (C; – m).

We note O the middle of [BC].
Explain why _Gm always exists and show that, when m describes $\mathbb{R}$ , G
_m
describes a line D which you will specify. >
2. a) Construct _G2 and _G-2. With AB= 4cm , AC = 3cm and BC = 6cm >
b) Assume m is different from 2 and -2. >
Let G
_m
be a point of D distinct from A, _G2 _andG-2. >
Prove that (BG
_m
) intersects (AC) at a point I and that (CG
_m
) intersects (AB) at a point noted J. >
3. in the frame $(A,\vec{AB},\vec{AC})$ , >
calculate the coordinates of I and J as a function of m. >
Deduce that the points O, I and J are aligned. >
(We can use the analytical condition of collinearity of 2 vectors) >

Exercise 27:

Hint: use the fact that the center of gravity is an isobarycenter and then the associativity of the barycenter.

Let ABC be a triangle, A’ , B’ , and C’ the midpoints of the sides opposite A, B and C respectively, M a given point.

We note_A1,_B1and_C1the symmetrical of point M with respect to A’ , B’ , and C’ .

M’ is the barycenter of the points (A,1) (B,1) (C,1) and (M,-1)
1. Show that the lines (_AA1), (_BB1) and (_CC1) are concurrent at M ‘. >
2. Let G be the center of gravity of ABC. Show that M ‘ , M and G are aligned and specify the position of M ‘ on the line (MG). >

Exercise 28:
ABCD is a square.

1. What is the set E of points M of the plane such that :

$\,\%7C\,2\vec{MA}-\vec{MB}+\vec{MC}\,\,\%7C=AB$

Note $G=bary\,\{(A,2),(B,-1),(C,1)\,\,\}$

G exists because $2-1+1=2\neq0$

$\,\%7C\,2\vec{MA}-\vec{MB}+\vec{MC}\,\,\%7C=AB$

$\,\%7C\,2\vec{MG}+2\vec{GA}-\vec{MG}-\vec{GB}+\vec{MG}+\vec{GC}\,\,\%7C=AB$

$\,\%7C\,2\vec{MG}-\vec{MG}+\vec{MG}+2\vec{GA}-\vec{GB}+\vec{GC}\,\,\%7C=AB$

$\,\%7C\,2\vec{MG}+\vec{0}\,\,\%7C=AB$ because $G=bary\,\{(A,2),(B,-1),(C,1)\,\,\}$

$2\,\%7C\,\vec{MG}\,\,\%7C=AB$

$\,\%7C\,\vec{MG}\,\,\%7C=\frac{AB}{2}$

Conclusion: the set of points we are looking for is the circle with center G and radius $\frac{AB}{2}$ .

2. Represent this set E.

Exercise 29:
Let ABCD be a square and K the barycenter of the weighted points (A; 2), (B; – 1), (C; 2) and (D; 1).

Let I be the barycenter of the weighted points (A; 2) and (B; – 1), and J the barycenter of (C; 2) and (D; 1).

1. Place I and J and justify.
First of all the barycenters I and J exist because $2-1=1\neq0$ and $2+1=3\neq0$ .

$2\vec{IA}-\vec{IB}=\vec{0}$

$2\vec{IA}-\vec{IA}-\vec{AB}=\vec{0}$

$\vec{IA}=\vec{AB}$

Conclusion : I is the symmetrical of point A with respect to B or A is the middle of [IB].

$2\vec{JC}+\vec{JD}=\vec{0}$

$2\vec{JC}+\vec{JC}+\vec{CD}=\vec{0}$

$3\vec{JC}=-\vec{CD}$

$\vec{CJ}=\,\frac{1}{3}\vec{CD}$

Conclusion: J is located at one third of [CD] starting from C.

2. Reduce the writing of the following vectors:

$2\vec{KA}-\vec{KB}\,et\,2\vec{KC}+\vec{KD}.$

$2\,\vec{KA}-\,\vec{KB}=2\,\vec{KI}+2\,\vec{IA}-\,\vec{KI}-\,\vec{IB}=\,\vec{KI}$

and

$2\,\vec{KC}+\,\vec{KD}=2\,\vec{KJ}+2\,\vec{JC}+\vec{KJ}+\vec{JD}=\,3\vec{KJ}$

Deduce that K is the barycenter of (I; 1) and (J; 3).

let’s take one of the two equalities:

$2\,\vec{KA}-\,\vec{KB}=\,\vec{KI}$

$2\,\vec{KA}-\,\vec{KB}-\vec{KI}=\vec{0}$

so K is the barycenter of (I; 1) and (J; 3).

Exercise 30:
Let ABC be a triangle and G a point verifying :

$\vec{AB}-4\vec{GA}-2\vec{GB}-3\vec{GC}=\vec{0}$

Is point G the barycenter of the weighted points (A; 5), (B; 1) and (C; 3)? Justify.

$\vec{AB}-4\vec{GA}-2\vec{GB}-3\vec{GC}=\vec{0}$

$\vec{AG}+\vec{GB}-4\vec{GA}-2\vec{GB}-3\vec{GC}=\vec{0}$

$-\vec{GA}+\vec{GB}-4\vec{GA}-2\vec{GB}-3\vec{GC}=\vec{0}$

$-5\vec{GA}-\vec{GB}-3\vec{GC}=\vec{0}$

Let’s multiply this vector equality by – 1 .

$5\vec{GA}+\vec{GB}+3\vec{GC}=\vec{0}$

$G=bary\,\{(A;5),(B;1),(C;3)\,\,\}$

Exercise 31:

In a benchmark $(O;\vec{i},\vec{j})$ ,
1.Place the points A(2; 1), B( – 1; 5), C(5; 7) and G(1; $\frac{5}{2}$ ).

2. Determine the coordinates of the isobarycenter I of points B and C.

3. Determine the coordinates of the center of gravity H of triangle ABC.

hint: use the fact that H is the barycenter of (A,1) (B,1) (C,1)

4. Is there a real k such that G is barycenter of (A; 1) and (B; k)? Justify.

Exercise 32:

Let ABC be an isosceles triangle at A such that BC = 8 cm and BA = 5 cm. Let I be the middle of [BC].

1. Place the point F such that $\vec{BF}=-\frac{1}{3}\vec{BA}$ .

and show that F is the barycenter of the points A and B weighted by real numbers that we will determine.

Hint: insert Point F into the vector $\vec{BA}$ and then transpose it into the first member.
2. P being a point of the plane, reduce each of the following sums:

$\frac{1}{2}\vec{PB}+\frac{1}{2}\vec{PC}\\-\vec{PA}+2\vec{PB}\\2\vec{PB}-2\vec{PA}$

3. Determine and represent the set of points M of the plane verifying :

Hint: consider the barycenter of (B,1) (C,1) and the barycenter of (A,-1) (B,2)

$\%7C\frac{1}{2}\vec{MB}+\frac{1}{2}\vec{MC}\,\,\%7C=\,\%7C\,-\vec{MA}+2\vec{MB}\,\,\%7C$

4. Determine and represent the set of points M of the plane verifying :

$\%7C\vec{NB}+\vec{NC}\,\,\%7C=\,\%7C\,2\vec{NB}-2\vec{NA}\,\,\%7C$

The answers to the exercises on the barycenter in 1st grade.

After consulting the answers to these exercises on the barycenter of n weighted points in 1st grade, you can return to the exercises in 1st grade.

The exercises in the first year.

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