Applying Pick's TheoremPick's Theorem on a triangular (or hex) gridFind the 4th vertex of the parallelogramA clarification on a problem on proving existence of regular polygonsProving Pick's theoremRemove points in square matrix in order to get a circleTriangle with same black and white areasNapoleon's theorem with quadrilateralsThree vertices of a ParalellagramParallelogram gridGeneral Form of Pick's Theorem
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Applying Pick's Theorem
Pick's Theorem on a triangular (or hex) gridFind the 4th vertex of the parallelogramA clarification on a problem on proving existence of regular polygonsProving Pick's theoremRemove points in square matrix in order to get a circleTriangle with same black and white areasNapoleon's theorem with quadrilateralsThree vertices of a ParalellagramParallelogram gridGeneral Form of Pick's Theorem
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Let's say that $X$ is a parallelogram with vertices that have integer coordinates, how could I prove that $X$'s area is an integer?
The vertices are
$0, A, B$ and $A + B$. How would I do this?
geometry
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add a comment |
$begingroup$
Let's say that $X$ is a parallelogram with vertices that have integer coordinates, how could I prove that $X$'s area is an integer?
The vertices are
$0, A, B$ and $A + B$. How would I do this?
geometry
$endgroup$
$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
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– Eevee Trainer
Mar 22 at 4:06
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I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
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– Eevee Trainer
Mar 22 at 4:07
add a comment |
$begingroup$
Let's say that $X$ is a parallelogram with vertices that have integer coordinates, how could I prove that $X$'s area is an integer?
The vertices are
$0, A, B$ and $A + B$. How would I do this?
geometry
$endgroup$
Let's say that $X$ is a parallelogram with vertices that have integer coordinates, how could I prove that $X$'s area is an integer?
The vertices are
$0, A, B$ and $A + B$. How would I do this?
geometry
geometry
edited Mar 22 at 12:01
asked Mar 13 at 15:44
user645044
$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
$endgroup$
– Eevee Trainer
Mar 22 at 4:06
$begingroup$
I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
$endgroup$
– Eevee Trainer
Mar 22 at 4:07
add a comment |
$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
$endgroup$
– Eevee Trainer
Mar 22 at 4:06
$begingroup$
I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
$endgroup$
– Eevee Trainer
Mar 22 at 4:07
$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
$endgroup$
– Eevee Trainer
Mar 22 at 4:06
$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
$endgroup$
– Eevee Trainer
Mar 22 at 4:06
$begingroup$
I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
$endgroup$
– Eevee Trainer
Mar 22 at 4:07
$begingroup$
I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
$endgroup$
– Eevee Trainer
Mar 22 at 4:07
add a comment |
2 Answers
2
active
oldest
votes
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The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors:
$$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$
Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
$begingroup$
If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then
$Abar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$
so $|Im(Abar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors:
$$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$
Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
$begingroup$
The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors:
$$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$
Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
add a comment |
$begingroup$
The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors:
$$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$
Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
$endgroup$
The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors:
$$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$
Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
answered Mar 13 at 15:49
Parcly TaxelParcly Taxel
44.7k1376109
44.7k1376109
add a comment |
add a comment |
$begingroup$
If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then
$Abar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$
so $|Im(Abar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
$endgroup$
add a comment |
$begingroup$
If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then
$Abar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$
so $|Im(Abar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
$endgroup$
add a comment |
$begingroup$
If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then
$Abar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$
so $|Im(Abar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
$endgroup$
If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then
$Abar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$
so $|Im(Abar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
answered Mar 13 at 16:05
gandalf61gandalf61
9,199825
9,199825
add a comment |
add a comment |
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$begingroup$
In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.]
$endgroup$
– Eevee Trainer
Mar 22 at 4:06
$begingroup$
I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- math.stackexchange.com/posts/3146771/revisions
$endgroup$
– Eevee Trainer
Mar 22 at 4:07