Question on convex pentagons with lattice points as its vertices
I have the following question which I found as an example in a book with me:
"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even"
The book starts the solution like this:
"Colour the lattices as in a chess board and erect right triangles on the sides of the pentagon with the sides of the pentagon as the longest side. With the other two sides along the sides of the square trace the ten shorter sides. Since at the end we return to the point we started we must have traced an even number of lattice points."
How is the last statement concluded?
graph-theory
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
add a comment |
I have the following question which I found as an example in a book with me:
"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even"
The book starts the solution like this:
"Colour the lattices as in a chess board and erect right triangles on the sides of the pentagon with the sides of the pentagon as the longest side. With the other two sides along the sides of the square trace the ten shorter sides. Since at the end we return to the point we started we must have traced an even number of lattice points."
How is the last statement concluded?
graph-theory
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
add a comment |
I have the following question which I found as an example in a book with me:
"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even"
The book starts the solution like this:
"Colour the lattices as in a chess board and erect right triangles on the sides of the pentagon with the sides of the pentagon as the longest side. With the other two sides along the sides of the square trace the ten shorter sides. Since at the end we return to the point we started we must have traced an even number of lattice points."
How is the last statement concluded?
graph-theory
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
I have the following question which I found as an example in a book with me:
"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even"
The book starts the solution like this:
"Colour the lattices as in a chess board and erect right triangles on the sides of the pentagon with the sides of the pentagon as the longest side. With the other two sides along the sides of the square trace the ten shorter sides. Since at the end we return to the point we started we must have traced an even number of lattice points."
How is the last statement concluded?
graph-theory
graph-theory
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
asked Dec 1 '18 at 7:21
saisanjeevsaisanjeev
947212
947212
add a comment |
add a comment |
1 Answer
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There are two parts to that reasoning:
- Replacing a slanted edge with two edges in the axial directions doesn't change whether the perimeter is odd. This follows from the fact that every primitive Pythagorean triple has exactly one odd leg and an odd hypotenuse; we either replace odd with odd+even or even with even+even.
- A closed lattice polygon with all sides in the axial directions must have even perimeter. This is what the chessboard coloring (best done with the squares centered at lattice points) is for; we alternate colors with every 1 unit we travel, and eventually return to the start. That's an even number of changes, for an even total distance.
Alternately, we can show that the sum of the edges in the $x$-direction and the sum of the edges in the $y$-direction are both even, by similar reasoning in one dimension.
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
add a comment |
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There are two parts to that reasoning:
- Replacing a slanted edge with two edges in the axial directions doesn't change whether the perimeter is odd. This follows from the fact that every primitive Pythagorean triple has exactly one odd leg and an odd hypotenuse; we either replace odd with odd+even or even with even+even.
- A closed lattice polygon with all sides in the axial directions must have even perimeter. This is what the chessboard coloring (best done with the squares centered at lattice points) is for; we alternate colors with every 1 unit we travel, and eventually return to the start. That's an even number of changes, for an even total distance.
Alternately, we can show that the sum of the edges in the $x$-direction and the sum of the edges in the $y$-direction are both even, by similar reasoning in one dimension.
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
add a comment |
There are two parts to that reasoning:
- Replacing a slanted edge with two edges in the axial directions doesn't change whether the perimeter is odd. This follows from the fact that every primitive Pythagorean triple has exactly one odd leg and an odd hypotenuse; we either replace odd with odd+even or even with even+even.
- A closed lattice polygon with all sides in the axial directions must have even perimeter. This is what the chessboard coloring (best done with the squares centered at lattice points) is for; we alternate colors with every 1 unit we travel, and eventually return to the start. That's an even number of changes, for an even total distance.
Alternately, we can show that the sum of the edges in the $x$-direction and the sum of the edges in the $y$-direction are both even, by similar reasoning in one dimension.
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
add a comment |
There are two parts to that reasoning:
- Replacing a slanted edge with two edges in the axial directions doesn't change whether the perimeter is odd. This follows from the fact that every primitive Pythagorean triple has exactly one odd leg and an odd hypotenuse; we either replace odd with odd+even or even with even+even.
- A closed lattice polygon with all sides in the axial directions must have even perimeter. This is what the chessboard coloring (best done with the squares centered at lattice points) is for; we alternate colors with every 1 unit we travel, and eventually return to the start. That's an even number of changes, for an even total distance.
Alternately, we can show that the sum of the edges in the $x$-direction and the sum of the edges in the $y$-direction are both even, by similar reasoning in one dimension.
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
There are two parts to that reasoning:
- Replacing a slanted edge with two edges in the axial directions doesn't change whether the perimeter is odd. This follows from the fact that every primitive Pythagorean triple has exactly one odd leg and an odd hypotenuse; we either replace odd with odd+even or even with even+even.
- A closed lattice polygon with all sides in the axial directions must have even perimeter. This is what the chessboard coloring (best done with the squares centered at lattice points) is for; we alternate colors with every 1 unit we travel, and eventually return to the start. That's an even number of changes, for an even total distance.
Alternately, we can show that the sum of the edges in the $x$-direction and the sum of the edges in the $y$-direction are both even, by similar reasoning in one dimension.
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
answered Dec 1 '18 at 7:56
jmerryjmerry
2,894312
2,894312
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
add a comment |
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
What if one of the sides of the pentagon is along one of the axial directions?
– saisanjeev
Dec 2 '18 at 7:09
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
If one of the sides is already along an axial direction, we don't bother replacing it. Note that the number of edges never actually matters. It might say "the ten shorter sides", but we never use that there are ten of them.
– jmerry
Dec 2 '18 at 7:28
add a comment |
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