Mixing
$z_1^2+z_2^2+z_3^2=3z_0^2$ if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the...
$begingroup$
Prove that, if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre, then $z_1^2+z_2^2+z_3^2=3z_0^2$
My Attempt
$$
z_0=frac{z_1+z_2+z_3}{3}implies 3z_0=z_1+z_2+z_3\
9z_0^2=z_1^2+z_2^2+z_3^2+2z_1z_2+2z_2z_3+2z_3z_1
$$
How do I proceed further and complete the proof ?
complex-numbers triangle centroid
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
$endgroup$
add a comment |
$begingroup$
Prove that, if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre, then $z_1^2+z_2^2+z_3^2=3z_0^2$
My Attempt
$$
z_0=frac{z_1+z_2+z_3}{3}implies 3z_0=z_1+z_2+z_3\
9z_0^2=z_1^2+z_2^2+z_3^2+2z_1z_2+2z_2z_3+2z_3z_1
$$
How do I proceed further and complete the proof ?
complex-numbers triangle centroid
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
$endgroup$
$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48
add a comment |
$begingroup$
Prove that, if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre, then $z_1^2+z_2^2+z_3^2=3z_0^2$
My Attempt
$$
z_0=frac{z_1+z_2+z_3}{3}implies 3z_0=z_1+z_2+z_3\
9z_0^2=z_1^2+z_2^2+z_3^2+2z_1z_2+2z_2z_3+2z_3z_1
$$
How do I proceed further and complete the proof ?
complex-numbers triangle centroid
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
$endgroup$
Prove that, if $z_1,z_2,z_3$ be the vertices of an equilateral triangle and $z_0$ be the circumcentre, then $z_1^2+z_2^2+z_3^2=3z_0^2$
My Attempt
$$
z_0=frac{z_1+z_2+z_3}{3}implies 3z_0=z_1+z_2+z_3\
9z_0^2=z_1^2+z_2^2+z_3^2+2z_1z_2+2z_2z_3+2z_3z_1
$$
How do I proceed further and complete the proof ?
complex-numbers triangle centroid
complex-numbers triangle centroid
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
edited Jan 16 at 13:16
ss1729
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
asked Jan 16 at 9:33
ss1729ss1729
1,9721923
1,9721923
$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48
add a comment |
$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48
$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48
$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here, we clearly know (by rotation theorem):
$$frac{z_2-z_0}{z_1-z_0} = frac{z_3-z_0}{z_2-z_0} = frac{z_1-z_0}{z_3-z_0}$$
On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$
Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$
Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$
Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
$endgroup$
add a comment |
$begingroup$
Let $$z_1=z_0+r(costheta+isintheta),$$ $$z_2=z_0+r(cos(120^{circ}+theta)+isin(120^{circ}+theta))$$ and $$z_3=z_0+r(cos(240^{circ}+theta)+isin(240^{circ}+theta)).$$
Id est, it's enough to prove that
$$cos2theta+cos(240^{circ}+2theta)+cos(480^{circ}+2theta)=0,$$
$$sin2theta+sin(240^{circ}+2theta)+sin(480^{circ}+2theta)=0,$$
$$costheta+cos(120^{circ}+theta)+cos(240^{circ}+theta)=0$$ and
$$sintheta+sin(120^{circ}+theta)+sin(240^{circ}+theta)=0.$$
Can you end it now?
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
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$begingroup$
Here, we clearly know (by rotation theorem):
$$frac{z_2-z_0}{z_1-z_0} = frac{z_3-z_0}{z_2-z_0} = frac{z_1-z_0}{z_3-z_0}$$
On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$
Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$
Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$
Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
$endgroup$
add a comment |
$begingroup$
Here, we clearly know (by rotation theorem):
$$frac{z_2-z_0}{z_1-z_0} = frac{z_3-z_0}{z_2-z_0} = frac{z_1-z_0}{z_3-z_0}$$
On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$
Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$
Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$
Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
$endgroup$
add a comment |
$begingroup$
Here, we clearly know (by rotation theorem):
$$frac{z_2-z_0}{z_1-z_0} = frac{z_3-z_0}{z_2-z_0} = frac{z_1-z_0}{z_3-z_0}$$
On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$
Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$
Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$
Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
$endgroup$
Here, we clearly know (by rotation theorem):
$$frac{z_2-z_0}{z_1-z_0} = frac{z_3-z_0}{z_2-z_0} = frac{z_1-z_0}{z_3-z_0}$$
On Solving you'll get:
$$z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1$$
Further, we know:
$$(z_1-z_0)+(z_2-z_0)+(z_3-z_0) = 0$$
Transferring terms and squaring both sides we get (after using the prev. result):
$$3(z_1^2+z_2^2+z_3^2) = 9z_0^2$$
Or, we get:
$$z_1^2+z_2^2+z_3^2= 3z_0^2$$
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
answered Jan 16 at 13:51
Mayank M.Mayank M.
493413
493413
add a comment |
add a comment |
$begingroup$
Let $$z_1=z_0+r(costheta+isintheta),$$ $$z_2=z_0+r(cos(120^{circ}+theta)+isin(120^{circ}+theta))$$ and $$z_3=z_0+r(cos(240^{circ}+theta)+isin(240^{circ}+theta)).$$
Id est, it's enough to prove that
$$cos2theta+cos(240^{circ}+2theta)+cos(480^{circ}+2theta)=0,$$
$$sin2theta+sin(240^{circ}+2theta)+sin(480^{circ}+2theta)=0,$$
$$costheta+cos(120^{circ}+theta)+cos(240^{circ}+theta)=0$$ and
$$sintheta+sin(120^{circ}+theta)+sin(240^{circ}+theta)=0.$$
Can you end it now?
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
add a comment |
$begingroup$
Let $$z_1=z_0+r(costheta+isintheta),$$ $$z_2=z_0+r(cos(120^{circ}+theta)+isin(120^{circ}+theta))$$ and $$z_3=z_0+r(cos(240^{circ}+theta)+isin(240^{circ}+theta)).$$
Id est, it's enough to prove that
$$cos2theta+cos(240^{circ}+2theta)+cos(480^{circ}+2theta)=0,$$
$$sin2theta+sin(240^{circ}+2theta)+sin(480^{circ}+2theta)=0,$$
$$costheta+cos(120^{circ}+theta)+cos(240^{circ}+theta)=0$$ and
$$sintheta+sin(120^{circ}+theta)+sin(240^{circ}+theta)=0.$$
Can you end it now?
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
add a comment |
$begingroup$
Let $$z_1=z_0+r(costheta+isintheta),$$ $$z_2=z_0+r(cos(120^{circ}+theta)+isin(120^{circ}+theta))$$ and $$z_3=z_0+r(cos(240^{circ}+theta)+isin(240^{circ}+theta)).$$
Id est, it's enough to prove that
$$cos2theta+cos(240^{circ}+2theta)+cos(480^{circ}+2theta)=0,$$
$$sin2theta+sin(240^{circ}+2theta)+sin(480^{circ}+2theta)=0,$$
$$costheta+cos(120^{circ}+theta)+cos(240^{circ}+theta)=0$$ and
$$sintheta+sin(120^{circ}+theta)+sin(240^{circ}+theta)=0.$$
Can you end it now?
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
Let $$z_1=z_0+r(costheta+isintheta),$$ $$z_2=z_0+r(cos(120^{circ}+theta)+isin(120^{circ}+theta))$$ and $$z_3=z_0+r(cos(240^{circ}+theta)+isin(240^{circ}+theta)).$$
Id est, it's enough to prove that
$$cos2theta+cos(240^{circ}+2theta)+cos(480^{circ}+2theta)=0,$$
$$sin2theta+sin(240^{circ}+2theta)+sin(480^{circ}+2theta)=0,$$
$$costheta+cos(120^{circ}+theta)+cos(240^{circ}+theta)=0$$ and
$$sintheta+sin(120^{circ}+theta)+sin(240^{circ}+theta)=0.$$
Can you end it now?
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
edited Jan 16 at 10:16
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 9:50
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
add a comment |
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
"Let $z_0=0$" it deserves an explanation why you can assume that. The equation is not invariant under translations.
$endgroup$
– Wojowu
Jan 16 at 10:06
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
$begingroup$
@Wojowu We can write $z_1=z_0+r(costheta+isintheta)$ and similar and we'll get the same computations. If you wish a can fix my post, but I think it's the same.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:11
1
1
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
I disagree that they are the same computations. Similar - perhaps, but not identical.
$endgroup$
– Wojowu
Jan 16 at 10:14
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
$begingroup$
@Wojowu I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 10:19
add a comment |
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$begingroup$
$$sqrt3=dfrac{z_1-z_0}{z_0-z_2}$$ Squaring we get $$3(z_0-z_2)^2=(z_1-z_0)^2iff?$$
$endgroup$
– lab bhattacharjee
Jan 16 at 9:48