Proof that the triangle formed by mass centers is equilateral. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Centroids of triangleCompetition style problem circa 1992there concurrent lines, perpendicular to the sides of a triangleProve that $S = 8S_1-4S_2$ in the geometrical figureCircle inscribed in Equilateral TrianglesProve triangle similiarity by given expressionAAA similarity Theorem.Similarity of nested trianglesProving similarity for a triangle inside another triangle given ratio between sides.Triangle equilateral proof
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Proof that the triangle formed by mass centers is equilateral.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Centroids of triangleCompetition style problem circa 1992there concurrent lines, perpendicular to the sides of a triangleProve that $S = 8S_1-4S_2$ in the geometrical figureCircle inscribed in Equilateral TrianglesProve triangle similiarity by given expressionAAA similarity Theorem.Similarity of nested trianglesProving similarity for a triangle inside another triangle given ratio between sides.Triangle equilateral proof
$begingroup$
Let $ABC$ be a triangle and consider $A_1$, $B_1$, $C_1$ outside the triangle such that triangles $ABC_1$, $BCA_1$ and $ACB_1$ are equilaterals.
Consider now $A_2$, $B_2$ and $C_2$ mass centres of $BCA_1$, $ACB_1$ and $ABC1$, respectively. Prove that triangle $A_2B_2C_2$ is a equilateral triangle.
geometry euclidean-geometry centroid geometric-transformation
edited Apr 2 at 19:02
SMM
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle and consider $A_1$, $B_1$, $C_1$ outside the triangle such that triangles $ABC_1$, $BCA_1$ and $ACB_1$ are equilaterals.
Consider now $A_2$, $B_2$ and $C_2$ mass centres of $BCA_1$, $ACB_1$ and $ABC1$, respectively. Prove that triangle $A_2B_2C_2$ is a equilateral triangle.
geometry euclidean-geometry centroid geometric-transformation
edited Apr 2 at 19:02
SMM
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle and consider $A_1$, $B_1$, $C_1$ outside the triangle such that triangles $ABC_1$, $BCA_1$ and $ACB_1$ are equilaterals.
Consider now $A_2$, $B_2$ and $C_2$ mass centres of $BCA_1$, $ACB_1$ and $ABC1$, respectively. Prove that triangle $A_2B_2C_2$ is a equilateral triangle.
geometry euclidean-geometry centroid geometric-transformation
edited Apr 2 at 19:02
SMM
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
$endgroup$
Let $ABC$ be a triangle and consider $A_1$, $B_1$, $C_1$ outside the triangle such that triangles $ABC_1$, $BCA_1$ and $ACB_1$ are equilaterals.
Consider now $A_2$, $B_2$ and $C_2$ mass centres of $BCA_1$, $ACB_1$ and $ABC1$, respectively. Prove that triangle $A_2B_2C_2$ is a equilateral triangle.
geometry euclidean-geometry centroid geometric-transformation
geometry euclidean-geometry centroid geometric-transformation
edited Apr 2 at 19:02
SMM
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
edited Apr 2 at 19:02
SMM
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
edited Apr 2 at 19:02
SMM
3,268512
edited Apr 2 at 19:02
SMM
3,268512
edited Apr 2 at 19:02
SMM
3,268512
3,268512
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
asked Apr 2 at 18:03
Andrei AlexandruAndrei Alexandru
134
134
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the compositions of rotations $I= R_A_2,120^circcirc R_B_2,120^circ$; note $I(A)=B$. Let $S$ be the point such that oriented angle $angle(SB_2,B_2A_2)=60^circ$ and oriented angle $angle (B_2A_2,SA_2)=60^circ$. We have:
$$I=R_A_2,120^circcirc R_B_2,120^circ= S_SA_2circ S_A_2B_2circ S_A_2B_2circ S_SB_2= S_SA_2circ S_SB_2= R_S,240^circ.$$
Therefore, $R_S,240^circ(A)=B$, so $SA=SB$ and oriented angle $angle(SA,SB)=240^circ$. Now we conclude that $S=C_2$. So $angle(C_2B_2,B_2A_2)=60^circ$ and $angle (B_2A_2,C_2A_2)=60^circ$, hence $triangle A_2B_2C_2$ is equilateral.
answered Apr 2 at 18:59
SMMSMM
3,268512
$endgroup$
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
add a comment |
$begingroup$
I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $overlineB_2C_2$ an angle of 60 anti-clockwise we get $overlineB_2A_2$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute beginalignR(C_2-B_2)&=R(fracC+B_1-B-C_13)=R(fracC-B+B_1-A+A-C_13)\
&=fracR(C-B)3+fracR(B_1-A)3+fracR(A-C_1)3
endalign
Now, note that $R(C-B)=A_1-B$, $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get
beginalignR(C_2-B_2)=fracA_1-B3+fracC-A3+fracB-C_13=fracB+C+A_13-fracA+B+C_13=A_2-B_2.
endalign
This is what we wanted to prove.
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
$endgroup$
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the compositions of rotations $I= R_A_2,120^circcirc R_B_2,120^circ$; note $I(A)=B$. Let $S$ be the point such that oriented angle $angle(SB_2,B_2A_2)=60^circ$ and oriented angle $angle (B_2A_2,SA_2)=60^circ$. We have:
$$I=R_A_2,120^circcirc R_B_2,120^circ= S_SA_2circ S_A_2B_2circ S_A_2B_2circ S_SB_2= S_SA_2circ S_SB_2= R_S,240^circ.$$
Therefore, $R_S,240^circ(A)=B$, so $SA=SB$ and oriented angle $angle(SA,SB)=240^circ$. Now we conclude that $S=C_2$. So $angle(C_2B_2,B_2A_2)=60^circ$ and $angle (B_2A_2,C_2A_2)=60^circ$, hence $triangle A_2B_2C_2$ is equilateral.
answered Apr 2 at 18:59
SMMSMM
3,268512
$endgroup$
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
add a comment |
$begingroup$
Consider the compositions of rotations $I= R_A_2,120^circcirc R_B_2,120^circ$; note $I(A)=B$. Let $S$ be the point such that oriented angle $angle(SB_2,B_2A_2)=60^circ$ and oriented angle $angle (B_2A_2,SA_2)=60^circ$. We have:
$$I=R_A_2,120^circcirc R_B_2,120^circ= S_SA_2circ S_A_2B_2circ S_A_2B_2circ S_SB_2= S_SA_2circ S_SB_2= R_S,240^circ.$$
Therefore, $R_S,240^circ(A)=B$, so $SA=SB$ and oriented angle $angle(SA,SB)=240^circ$. Now we conclude that $S=C_2$. So $angle(C_2B_2,B_2A_2)=60^circ$ and $angle (B_2A_2,C_2A_2)=60^circ$, hence $triangle A_2B_2C_2$ is equilateral.
answered Apr 2 at 18:59
SMMSMM
3,268512
$endgroup$
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
add a comment |
$begingroup$
Consider the compositions of rotations $I= R_A_2,120^circcirc R_B_2,120^circ$; note $I(A)=B$. Let $S$ be the point such that oriented angle $angle(SB_2,B_2A_2)=60^circ$ and oriented angle $angle (B_2A_2,SA_2)=60^circ$. We have:
$$I=R_A_2,120^circcirc R_B_2,120^circ= S_SA_2circ S_A_2B_2circ S_A_2B_2circ S_SB_2= S_SA_2circ S_SB_2= R_S,240^circ.$$
Therefore, $R_S,240^circ(A)=B$, so $SA=SB$ and oriented angle $angle(SA,SB)=240^circ$. Now we conclude that $S=C_2$. So $angle(C_2B_2,B_2A_2)=60^circ$ and $angle (B_2A_2,C_2A_2)=60^circ$, hence $triangle A_2B_2C_2$ is equilateral.
answered Apr 2 at 18:59
SMMSMM
3,268512
$endgroup$
Consider the compositions of rotations $I= R_A_2,120^circcirc R_B_2,120^circ$; note $I(A)=B$. Let $S$ be the point such that oriented angle $angle(SB_2,B_2A_2)=60^circ$ and oriented angle $angle (B_2A_2,SA_2)=60^circ$. We have:
$$I=R_A_2,120^circcirc R_B_2,120^circ= S_SA_2circ S_A_2B_2circ S_A_2B_2circ S_SB_2= S_SA_2circ S_SB_2= R_S,240^circ.$$
Therefore, $R_S,240^circ(A)=B$, so $SA=SB$ and oriented angle $angle(SA,SB)=240^circ$. Now we conclude that $S=C_2$. So $angle(C_2B_2,B_2A_2)=60^circ$ and $angle (B_2A_2,C_2A_2)=60^circ$, hence $triangle A_2B_2C_2$ is equilateral.
answered Apr 2 at 18:59
SMMSMM
3,268512
answered Apr 2 at 18:59
SMMSMM
3,268512
answered Apr 2 at 18:59
SMMSMM
3,268512
answered Apr 2 at 18:59
SMMSMM
3,268512
3,268512
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
add a comment |
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
+1...............
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
$begingroup$
I need a proof for 5th grade. You can't use rotations or this kind of abstract tools
$endgroup$
– Andrei Alexandru
Apr 2 at 19:29
add a comment |
$begingroup$
I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $overlineB_2C_2$ an angle of 60 anti-clockwise we get $overlineB_2A_2$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute beginalignR(C_2-B_2)&=R(fracC+B_1-B-C_13)=R(fracC-B+B_1-A+A-C_13)\
&=fracR(C-B)3+fracR(B_1-A)3+fracR(A-C_1)3
endalign
Now, note that $R(C-B)=A_1-B$, $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get
beginalignR(C_2-B_2)=fracA_1-B3+fracC-A3+fracB-C_13=fracB+C+A_13-fracA+B+C_13=A_2-B_2.
endalign
This is what we wanted to prove.
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
$endgroup$
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
add a comment |
$begingroup$
I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $overlineB_2C_2$ an angle of 60 anti-clockwise we get $overlineB_2A_2$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute beginalignR(C_2-B_2)&=R(fracC+B_1-B-C_13)=R(fracC-B+B_1-A+A-C_13)\
&=fracR(C-B)3+fracR(B_1-A)3+fracR(A-C_1)3
endalign
Now, note that $R(C-B)=A_1-B$, $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get
beginalignR(C_2-B_2)=fracA_1-B3+fracC-A3+fracB-C_13=fracB+C+A_13-fracA+B+C_13=A_2-B_2.
endalign
This is what we wanted to prove.
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
$endgroup$
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
add a comment |
$begingroup$
I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $overlineB_2C_2$ an angle of 60 anti-clockwise we get $overlineB_2A_2$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute beginalignR(C_2-B_2)&=R(fracC+B_1-B-C_13)=R(fracC-B+B_1-A+A-C_13)\
&=fracR(C-B)3+fracR(B_1-A)3+fracR(A-C_1)3
endalign
Now, note that $R(C-B)=A_1-B$, $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get
beginalignR(C_2-B_2)=fracA_1-B3+fracC-A3+fracB-C_13=fracB+C+A_13-fracA+B+C_13=A_2-B_2.
endalign
This is what we wanted to prove.
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
$endgroup$
I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $overlineB_2C_2$ an angle of 60 anti-clockwise we get $overlineB_2A_2$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute beginalignR(C_2-B_2)&=R(fracC+B_1-B-C_13)=R(fracC-B+B_1-A+A-C_13)\
&=fracR(C-B)3+fracR(B_1-A)3+fracR(A-C_1)3
endalign
Now, note that $R(C-B)=A_1-B$, $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get
beginalignR(C_2-B_2)=fracA_1-B3+fracC-A3+fracB-C_13=fracB+C+A_13-fracA+B+C_13=A_2-B_2.
endalign
This is what we wanted to prove.
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
answered Apr 2 at 18:49
Julian MejiaJulian Mejia
76729
76729
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
add a comment |
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
+1..................
$endgroup$
– Maria Mazur
Apr 2 at 19:02
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
I need a 5th grade proof using only elementary geometry
$endgroup$
– Andrei Alexandru
Apr 2 at 19:37
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
$begingroup$
This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening.
$endgroup$
– Julian Mejia
Apr 2 at 19:44
add a comment |
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