Barycenter: answer key to 1st grade math exercises in PDF.

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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.

Isobarycentre.

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 G=bary{(A,2);(I,2))}

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)

so

{\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.

G=bary{(A,2);(I,2))}

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].

Barycentre

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).

Barycentre

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.

Barycenter and triangle

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.

Barycenter and triangle

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.

Quadrilateral
Exercise 16:

Answer key to the exercise

Exercise 17:

Answer key to the exercise

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).

triangle

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.

Concurrent lines

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.

Triangle and barycenter

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.

Triangle

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).
Benchmark and barycenter

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

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

benchmark and barycenter

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 noteA1 ,B1 andC1 the 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.

Square

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).

Square

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)\,\,\}
Triangle

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.

Triangle

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

set of points

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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