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How many times can I cut a shape in half before getting a new shape?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Forming a ring with equilateral triangle of the same sizeTrapezoid area proof by dividing it into two triangles?Is there a 4-dimensional picture making a geometrical proof of Heron's formulaLinear Algebra Tetrahedron Centroid 3/4 Down Segment ProofThe reverse pizza problem .finding the top of a triangular pyramidEqual Equilateralizer Point in a tetrahedronIrrationality of square root of 2, as explained in the book, “How not to be wrong, the power of mathematical thinking”Tiling a floor with squaresArea of a regular hexagon via area of triangles
$begingroup$
If you take triangle and slice it from one corner to the opposite base, you get another triangle. That leaves you with two new triangles. If I cut those triangles in half, I'll have four.
On the other hand, if I cut a circle in half, I immediately have a different shape.
Is there a formula/ name of concept that would tell me how many times I could cut a shape in half before I no longer receive that shape?
Also, for example, would the formula take cutting an equilateral shape in half and getting another equilateral triangle (which I don't think is possible)? Or does it only take into account how many times you can cut the equilateral shape in half and get another triangle (e.g. acute or right)?
geometry recreational-mathematics
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
$endgroup$
|
show 1 more comment
$begingroup$
If you take triangle and slice it from one corner to the opposite base, you get another triangle. That leaves you with two new triangles. If I cut those triangles in half, I'll have four.
On the other hand, if I cut a circle in half, I immediately have a different shape.
Is there a formula/ name of concept that would tell me how many times I could cut a shape in half before I no longer receive that shape?
Also, for example, would the formula take cutting an equilateral shape in half and getting another equilateral triangle (which I don't think is possible)? Or does it only take into account how many times you can cut the equilateral shape in half and get another triangle (e.g. acute or right)?
geometry recreational-mathematics
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
$endgroup$
3
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
1
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
1
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57
|
show 1 more comment
$begingroup$
If you take triangle and slice it from one corner to the opposite base, you get another triangle. That leaves you with two new triangles. If I cut those triangles in half, I'll have four.
On the other hand, if I cut a circle in half, I immediately have a different shape.
Is there a formula/ name of concept that would tell me how many times I could cut a shape in half before I no longer receive that shape?
Also, for example, would the formula take cutting an equilateral shape in half and getting another equilateral triangle (which I don't think is possible)? Or does it only take into account how many times you can cut the equilateral shape in half and get another triangle (e.g. acute or right)?
geometry recreational-mathematics
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
$endgroup$
If you take triangle and slice it from one corner to the opposite base, you get another triangle. That leaves you with two new triangles. If I cut those triangles in half, I'll have four.
On the other hand, if I cut a circle in half, I immediately have a different shape.
Is there a formula/ name of concept that would tell me how many times I could cut a shape in half before I no longer receive that shape?
Also, for example, would the formula take cutting an equilateral shape in half and getting another equilateral triangle (which I don't think is possible)? Or does it only take into account how many times you can cut the equilateral shape in half and get another triangle (e.g. acute or right)?
geometry recreational-mathematics
geometry recreational-mathematics
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
asked Mar 23 at 15:32
JustBlossomJustBlossom
1104
1104
3
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
1
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
1
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57
|
show 1 more comment
3
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
1
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
1
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57
3
3
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
1
1
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
1
1
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Assuming that you mean "shape" as a polygon. Also assuming that when you say "cut in half", you mean dividing into two pieces, not necessarily with the same area, with a single straight line. Also assuming that the problem involves dividing into exactly two pieces and that the pieces must both be similar to the original shape, (similar meaning that one shape can be converted into another via reflection, rotation, and/or uniform stretching).
If a cut does not pass through a vertex with an internal angle of greater than 180 degrees, then, based on whether its points of intersection are vertices or intersect a side, will increase the number of angles by anywhere from 2 (if both endpoints are vertices) to 4 (neither endpoint is a vertex). In order to create two similar shapes to the original, the new number of angles bust be twice the original. This allows for a triangle or a quadrilateral only.
For a triangle, the cut must pass through a vertex and a side (geometrically, this must be the opposite side). The cut must intersect the side at either a right angle or form an obtuse and an acute angle. If the cut forms an obtuse and an acute angle, then the original triangle must be an obtuse triangle. If the cut passes through an acute vertex, then the obtuse angle in the original triangle and the obtuse angle formed by the cut must be different (if the original triangle ABC had its obtuse angle at A and the cut passes through vertex B and line AC at point D, then the two obtuse angles are BAC and BDC and must be different as D is on AC). If the cut passes through the obtuse vertex, the resulting triangles also will not be similar (the cut passes through AE, where E is on BC in the above example. The obtuse angles BAC and BAE are not the same angle). However, for a right triangle, the altitude from the right angle to the hypotenuse forms two similar right triangles.
For a quadrilateral, the cut must pass through two sides (geometrically, they must be opposite sides in order for each piece to have the same number of sides). If the quadrilateral was a trapezoid (exactly one pair of parallel sides), then we need to consider two possibilities: either the cut passes through the two parallel sides or it doesn't. If it passes through the two parallel sides, then because the cut intersects two parallel lines, opposite angles formed by the cut are the same. In order to have similar shapes, the sequence of angles must be the same for each pieces (although in the reverse order). This results in the original shape being a parallelogram, which is not a trapezoid. If the cut passed through the non-parallel sides, in order for both new shapes to be trapezoids, the cut must be parallel to the pair of parallel sides. The parallel side of the original trapezoid which is longer (the two parallel sides must be different lengths in a trapezoid) remains the longer of the two parallel sides in one of the pieces, although the length of the shorter side is different. This means that the size ratio is different, so the shapes are not similar. If the original shape is a trapezium (a quadrilateral with no parallel sides), then the cut cannot be parallel to any of the original sides (otherwise it would form a trapezoid). This creates an impossibility as the sequence of angles between the original shape and the pieces must remain the same (except for a possible reversal of the order); each piece shares two consecutive angles with the original trapezium but has angles of a different size for each of the other two angles. This means that for quadrilaterals, the only possible similar cutting shape is a parallelogram (of any angle ratio) and a side ratio of the golden ratio; the dividing cut is parallel to each of the smaller sides and divides each of the larger sides into pieces with a size ratio of the golden ratio.
For polygons with a vertex with an internal angle of at least 180 degrees (and the cut passing through such a vertex), the cut cannot create any new such angles. As a result, the number of such angles cannot double and the resulting pieces cannot both be similar to the original shape.
answered Apr 1 at 13:52
Moko19Moko19
3688
$endgroup$
add a comment |
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$begingroup$
Assuming that you mean "shape" as a polygon. Also assuming that when you say "cut in half", you mean dividing into two pieces, not necessarily with the same area, with a single straight line. Also assuming that the problem involves dividing into exactly two pieces and that the pieces must both be similar to the original shape, (similar meaning that one shape can be converted into another via reflection, rotation, and/or uniform stretching).
If a cut does not pass through a vertex with an internal angle of greater than 180 degrees, then, based on whether its points of intersection are vertices or intersect a side, will increase the number of angles by anywhere from 2 (if both endpoints are vertices) to 4 (neither endpoint is a vertex). In order to create two similar shapes to the original, the new number of angles bust be twice the original. This allows for a triangle or a quadrilateral only.
For a triangle, the cut must pass through a vertex and a side (geometrically, this must be the opposite side). The cut must intersect the side at either a right angle or form an obtuse and an acute angle. If the cut forms an obtuse and an acute angle, then the original triangle must be an obtuse triangle. If the cut passes through an acute vertex, then the obtuse angle in the original triangle and the obtuse angle formed by the cut must be different (if the original triangle ABC had its obtuse angle at A and the cut passes through vertex B and line AC at point D, then the two obtuse angles are BAC and BDC and must be different as D is on AC). If the cut passes through the obtuse vertex, the resulting triangles also will not be similar (the cut passes through AE, where E is on BC in the above example. The obtuse angles BAC and BAE are not the same angle). However, for a right triangle, the altitude from the right angle to the hypotenuse forms two similar right triangles.
For a quadrilateral, the cut must pass through two sides (geometrically, they must be opposite sides in order for each piece to have the same number of sides). If the quadrilateral was a trapezoid (exactly one pair of parallel sides), then we need to consider two possibilities: either the cut passes through the two parallel sides or it doesn't. If it passes through the two parallel sides, then because the cut intersects two parallel lines, opposite angles formed by the cut are the same. In order to have similar shapes, the sequence of angles must be the same for each pieces (although in the reverse order). This results in the original shape being a parallelogram, which is not a trapezoid. If the cut passed through the non-parallel sides, in order for both new shapes to be trapezoids, the cut must be parallel to the pair of parallel sides. The parallel side of the original trapezoid which is longer (the two parallel sides must be different lengths in a trapezoid) remains the longer of the two parallel sides in one of the pieces, although the length of the shorter side is different. This means that the size ratio is different, so the shapes are not similar. If the original shape is a trapezium (a quadrilateral with no parallel sides), then the cut cannot be parallel to any of the original sides (otherwise it would form a trapezoid). This creates an impossibility as the sequence of angles between the original shape and the pieces must remain the same (except for a possible reversal of the order); each piece shares two consecutive angles with the original trapezium but has angles of a different size for each of the other two angles. This means that for quadrilaterals, the only possible similar cutting shape is a parallelogram (of any angle ratio) and a side ratio of the golden ratio; the dividing cut is parallel to each of the smaller sides and divides each of the larger sides into pieces with a size ratio of the golden ratio.
For polygons with a vertex with an internal angle of at least 180 degrees (and the cut passing through such a vertex), the cut cannot create any new such angles. As a result, the number of such angles cannot double and the resulting pieces cannot both be similar to the original shape.
answered Apr 1 at 13:52
Moko19Moko19
3688
$endgroup$
add a comment |
$begingroup$
Assuming that you mean "shape" as a polygon. Also assuming that when you say "cut in half", you mean dividing into two pieces, not necessarily with the same area, with a single straight line. Also assuming that the problem involves dividing into exactly two pieces and that the pieces must both be similar to the original shape, (similar meaning that one shape can be converted into another via reflection, rotation, and/or uniform stretching).
If a cut does not pass through a vertex with an internal angle of greater than 180 degrees, then, based on whether its points of intersection are vertices or intersect a side, will increase the number of angles by anywhere from 2 (if both endpoints are vertices) to 4 (neither endpoint is a vertex). In order to create two similar shapes to the original, the new number of angles bust be twice the original. This allows for a triangle or a quadrilateral only.
For a triangle, the cut must pass through a vertex and a side (geometrically, this must be the opposite side). The cut must intersect the side at either a right angle or form an obtuse and an acute angle. If the cut forms an obtuse and an acute angle, then the original triangle must be an obtuse triangle. If the cut passes through an acute vertex, then the obtuse angle in the original triangle and the obtuse angle formed by the cut must be different (if the original triangle ABC had its obtuse angle at A and the cut passes through vertex B and line AC at point D, then the two obtuse angles are BAC and BDC and must be different as D is on AC). If the cut passes through the obtuse vertex, the resulting triangles also will not be similar (the cut passes through AE, where E is on BC in the above example. The obtuse angles BAC and BAE are not the same angle). However, for a right triangle, the altitude from the right angle to the hypotenuse forms two similar right triangles.
For a quadrilateral, the cut must pass through two sides (geometrically, they must be opposite sides in order for each piece to have the same number of sides). If the quadrilateral was a trapezoid (exactly one pair of parallel sides), then we need to consider two possibilities: either the cut passes through the two parallel sides or it doesn't. If it passes through the two parallel sides, then because the cut intersects two parallel lines, opposite angles formed by the cut are the same. In order to have similar shapes, the sequence of angles must be the same for each pieces (although in the reverse order). This results in the original shape being a parallelogram, which is not a trapezoid. If the cut passed through the non-parallel sides, in order for both new shapes to be trapezoids, the cut must be parallel to the pair of parallel sides. The parallel side of the original trapezoid which is longer (the two parallel sides must be different lengths in a trapezoid) remains the longer of the two parallel sides in one of the pieces, although the length of the shorter side is different. This means that the size ratio is different, so the shapes are not similar. If the original shape is a trapezium (a quadrilateral with no parallel sides), then the cut cannot be parallel to any of the original sides (otherwise it would form a trapezoid). This creates an impossibility as the sequence of angles between the original shape and the pieces must remain the same (except for a possible reversal of the order); each piece shares two consecutive angles with the original trapezium but has angles of a different size for each of the other two angles. This means that for quadrilaterals, the only possible similar cutting shape is a parallelogram (of any angle ratio) and a side ratio of the golden ratio; the dividing cut is parallel to each of the smaller sides and divides each of the larger sides into pieces with a size ratio of the golden ratio.
For polygons with a vertex with an internal angle of at least 180 degrees (and the cut passing through such a vertex), the cut cannot create any new such angles. As a result, the number of such angles cannot double and the resulting pieces cannot both be similar to the original shape.
answered Apr 1 at 13:52
Moko19Moko19
3688
$endgroup$
add a comment |
$begingroup$
Assuming that you mean "shape" as a polygon. Also assuming that when you say "cut in half", you mean dividing into two pieces, not necessarily with the same area, with a single straight line. Also assuming that the problem involves dividing into exactly two pieces and that the pieces must both be similar to the original shape, (similar meaning that one shape can be converted into another via reflection, rotation, and/or uniform stretching).
If a cut does not pass through a vertex with an internal angle of greater than 180 degrees, then, based on whether its points of intersection are vertices or intersect a side, will increase the number of angles by anywhere from 2 (if both endpoints are vertices) to 4 (neither endpoint is a vertex). In order to create two similar shapes to the original, the new number of angles bust be twice the original. This allows for a triangle or a quadrilateral only.
For a triangle, the cut must pass through a vertex and a side (geometrically, this must be the opposite side). The cut must intersect the side at either a right angle or form an obtuse and an acute angle. If the cut forms an obtuse and an acute angle, then the original triangle must be an obtuse triangle. If the cut passes through an acute vertex, then the obtuse angle in the original triangle and the obtuse angle formed by the cut must be different (if the original triangle ABC had its obtuse angle at A and the cut passes through vertex B and line AC at point D, then the two obtuse angles are BAC and BDC and must be different as D is on AC). If the cut passes through the obtuse vertex, the resulting triangles also will not be similar (the cut passes through AE, where E is on BC in the above example. The obtuse angles BAC and BAE are not the same angle). However, for a right triangle, the altitude from the right angle to the hypotenuse forms two similar right triangles.
For a quadrilateral, the cut must pass through two sides (geometrically, they must be opposite sides in order for each piece to have the same number of sides). If the quadrilateral was a trapezoid (exactly one pair of parallel sides), then we need to consider two possibilities: either the cut passes through the two parallel sides or it doesn't. If it passes through the two parallel sides, then because the cut intersects two parallel lines, opposite angles formed by the cut are the same. In order to have similar shapes, the sequence of angles must be the same for each pieces (although in the reverse order). This results in the original shape being a parallelogram, which is not a trapezoid. If the cut passed through the non-parallel sides, in order for both new shapes to be trapezoids, the cut must be parallel to the pair of parallel sides. The parallel side of the original trapezoid which is longer (the two parallel sides must be different lengths in a trapezoid) remains the longer of the two parallel sides in one of the pieces, although the length of the shorter side is different. This means that the size ratio is different, so the shapes are not similar. If the original shape is a trapezium (a quadrilateral with no parallel sides), then the cut cannot be parallel to any of the original sides (otherwise it would form a trapezoid). This creates an impossibility as the sequence of angles between the original shape and the pieces must remain the same (except for a possible reversal of the order); each piece shares two consecutive angles with the original trapezium but has angles of a different size for each of the other two angles. This means that for quadrilaterals, the only possible similar cutting shape is a parallelogram (of any angle ratio) and a side ratio of the golden ratio; the dividing cut is parallel to each of the smaller sides and divides each of the larger sides into pieces with a size ratio of the golden ratio.
For polygons with a vertex with an internal angle of at least 180 degrees (and the cut passing through such a vertex), the cut cannot create any new such angles. As a result, the number of such angles cannot double and the resulting pieces cannot both be similar to the original shape.
answered Apr 1 at 13:52
Moko19Moko19
3688
$endgroup$
Assuming that you mean "shape" as a polygon. Also assuming that when you say "cut in half", you mean dividing into two pieces, not necessarily with the same area, with a single straight line. Also assuming that the problem involves dividing into exactly two pieces and that the pieces must both be similar to the original shape, (similar meaning that one shape can be converted into another via reflection, rotation, and/or uniform stretching).
If a cut does not pass through a vertex with an internal angle of greater than 180 degrees, then, based on whether its points of intersection are vertices or intersect a side, will increase the number of angles by anywhere from 2 (if both endpoints are vertices) to 4 (neither endpoint is a vertex). In order to create two similar shapes to the original, the new number of angles bust be twice the original. This allows for a triangle or a quadrilateral only.
For a triangle, the cut must pass through a vertex and a side (geometrically, this must be the opposite side). The cut must intersect the side at either a right angle or form an obtuse and an acute angle. If the cut forms an obtuse and an acute angle, then the original triangle must be an obtuse triangle. If the cut passes through an acute vertex, then the obtuse angle in the original triangle and the obtuse angle formed by the cut must be different (if the original triangle ABC had its obtuse angle at A and the cut passes through vertex B and line AC at point D, then the two obtuse angles are BAC and BDC and must be different as D is on AC). If the cut passes through the obtuse vertex, the resulting triangles also will not be similar (the cut passes through AE, where E is on BC in the above example. The obtuse angles BAC and BAE are not the same angle). However, for a right triangle, the altitude from the right angle to the hypotenuse forms two similar right triangles.
For a quadrilateral, the cut must pass through two sides (geometrically, they must be opposite sides in order for each piece to have the same number of sides). If the quadrilateral was a trapezoid (exactly one pair of parallel sides), then we need to consider two possibilities: either the cut passes through the two parallel sides or it doesn't. If it passes through the two parallel sides, then because the cut intersects two parallel lines, opposite angles formed by the cut are the same. In order to have similar shapes, the sequence of angles must be the same for each pieces (although in the reverse order). This results in the original shape being a parallelogram, which is not a trapezoid. If the cut passed through the non-parallel sides, in order for both new shapes to be trapezoids, the cut must be parallel to the pair of parallel sides. The parallel side of the original trapezoid which is longer (the two parallel sides must be different lengths in a trapezoid) remains the longer of the two parallel sides in one of the pieces, although the length of the shorter side is different. This means that the size ratio is different, so the shapes are not similar. If the original shape is a trapezium (a quadrilateral with no parallel sides), then the cut cannot be parallel to any of the original sides (otherwise it would form a trapezoid). This creates an impossibility as the sequence of angles between the original shape and the pieces must remain the same (except for a possible reversal of the order); each piece shares two consecutive angles with the original trapezium but has angles of a different size for each of the other two angles. This means that for quadrilaterals, the only possible similar cutting shape is a parallelogram (of any angle ratio) and a side ratio of the golden ratio; the dividing cut is parallel to each of the smaller sides and divides each of the larger sides into pieces with a size ratio of the golden ratio.
For polygons with a vertex with an internal angle of at least 180 degrees (and the cut passing through such a vertex), the cut cannot create any new such angles. As a result, the number of such angles cannot double and the resulting pieces cannot both be similar to the original shape.
answered Apr 1 at 13:52
Moko19Moko19
3688
answered Apr 1 at 13:52
Moko19Moko19
3688
answered Apr 1 at 13:52
Moko19Moko19
3688
answered Apr 1 at 13:52
Moko19Moko19
3688
3688
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3
$begingroup$
If you can do it once, you can do it infinitely many times. One example would be the triangle formed by cutting a square along a diagonal.
$endgroup$
– Jens
Mar 23 at 15:54
$begingroup$
You would need to rigously define "shape" and "cut in half" first, otherwise the question can not be answered.
$endgroup$
– Somos
Mar 23 at 17:36
$begingroup$
@Somos I can see how to define "cut in half" in more detail (e.g. cut in half from one corner or edge to another vs. just cut the area in half like when you cut a square in half using diamond). But what do you mean by defining "shape" more rigorously?
$endgroup$
– JustBlossom
Mar 23 at 17:44
1
$begingroup$
I mean exactly what I wrote. what do you mean by "shape"? Does it mean exactly the same set or what? Are two circles in different places and sizes the same "shape"? Are two triangles in different places and rotated the same "shape"? Do you know Banch-Tarski paradox?
$endgroup$
– Somos
Mar 23 at 17:45
1
$begingroup$
What Somos is trying to get at is what do you mean by saying that two shapes are the same? Yes, you can cut a triangle as described, but the new triangles are not the same size, nor are they even similar to the old. So by saying you "get the same shape", you have chosen to ignore size and dismilarity, and focus only on "closed curve consisting of 3 joined line segments" as being the definition of "shape". Cut a circle in two, and you get two closed convex curves, just like the original. Maybe that could count as meaning of "same shape"? It is a matter of what you decide "same shape" means.
$endgroup$
– Paul Sinclair
Mar 24 at 1:57