Proving the Shoelace Method at the Precalculus Level
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Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $frac{1}{2}cdotleft|x_1cdot y_2 - x_2cdot y_1right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?
geometry algebra-precalculus contest-math analytic-geometry
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
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add a comment |
$begingroup$
Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $frac{1}{2}cdotleft|x_1cdot y_2 - x_2cdot y_1right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?
geometry algebra-precalculus contest-math analytic-geometry
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
$endgroup$
add a comment |
$begingroup$
Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $frac{1}{2}cdotleft|x_1cdot y_2 - x_2cdot y_1right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?
geometry algebra-precalculus contest-math analytic-geometry
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
$endgroup$
Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $frac{1}{2}cdotleft|x_1cdot y_2 - x_2cdot y_1right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?
geometry algebra-precalculus contest-math analytic-geometry
geometry algebra-precalculus contest-math analytic-geometry
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
edited Mar 22 '13 at 14:50
Isaac
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
asked Jul 28 '10 at 18:26
IsaacIsaac
29.9k1285128
29.9k1285128
add a comment |
add a comment |
1 Answer
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One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $frac12 oint_{polygon} x,dy - y,dx$ here.) By induction, you then only have to verify the formula for a triangle.
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $frac12 oint_{polygon} x,dy - y,dx$ here.) By induction, you then only have to verify the formula for a triangle.
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
$endgroup$
add a comment |
$begingroup$
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $frac12 oint_{polygon} x,dy - y,dx$ here.) By induction, you then only have to verify the formula for a triangle.
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
$endgroup$
add a comment |
$begingroup$
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $frac12 oint_{polygon} x,dy - y,dx$ here.) By induction, you then only have to verify the formula for a triangle.
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
$endgroup$
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $frac12 oint_{polygon} x,dy - y,dx$ here.) By induction, you then only have to verify the formula for a triangle.
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
edited Dec 11 '18 at 11:02
Glorfindel
3,41981830
3,41981830
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
answered Jul 30 '10 at 20:39
Greg GravitonGreg Graviton
3,66121733
3,66121733
add a comment |
add a comment |
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