Mixing
Special triangles in convex polygons
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Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but I’m totally stuck right now. The only thing I’m pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
asked 5 hours ago
Francesco Arnaudo
212
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Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
4
down vote
favorite
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but I’m totally stuck right now. The only thing I’m pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
asked 5 hours ago
Francesco Arnaudo
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but I’m totally stuck right now. The only thing I’m pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
asked 5 hours ago
Francesco Arnaudo
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given equal-sided 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?
This seems a very easy task by first look, but I’m totally stuck right now. The only thing I’m pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a decagon can be filled with 30-60-90 triangles.
Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.
mathematics geometry triangle
mathematics geometry triangle
asked 5 hours ago
Francesco Arnaudo
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Francesco Arnaudo
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
asked 5 hours ago
Francesco Arnaudo
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Francesco Arnaudo
212
asked 5 hours ago
Francesco Arnaudo
212
212
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Francesco Arnaudo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago
add a comment |Â
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
1
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
answered 1 hour ago
Penguino
6,8121866
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
Feel free to edit Hugh.
– Penguino
11 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
answered 1 hour ago
Penguino
6,8121866
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
Feel free to edit Hugh.
– Penguino
11 mins ago
add a comment |Â
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
answered 1 hour ago
Penguino
6,8121866
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
Feel free to edit Hugh.
– Penguino
11 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
answered 1 hour ago
Penguino
6,8121866
It is possible to do better than a hexagon, if an irregular polygon is acceptable.
One can construct an irregular dodecagon with $180$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $4:sqrt12$; as a specific size for the 30-60-90 triangle is not given.
In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.
It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. See below for a rectangle with side length ratio $~1.0785:1$. By taking $sqrt3$ by $1$ squares, each made from two triangles, you could stack them $n$ by $fracnsqrt3$ which, for sufficiently large $n$, would approach $1:1$.
answered 1 hour ago
Penguino
6,8121866
edited 6 mins ago
answered 1 hour ago
Penguino
6,8121866
answered 1 hour ago
Penguino
6,8121866
answered 1 hour ago
Penguino
6,8121866
6,8121866
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
Feel free to edit Hugh.
– Penguino
11 mins ago
add a comment |Â
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
Feel free to edit Hugh.
– Penguino
11 mins ago
1
1
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + nsqrt3 neq 2n + nsqrt3 + nsqrt3$.
– Hugh
46 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
Yep - you are right - I sketched to soon :)
– Penguino
42 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
I've just suggested an edit, it corrects some grammar and adds math formatting. :)
– Hugh
16 mins ago
1
1
Feel free to edit Hugh.
– Penguino
11 mins ago
Feel free to edit Hugh.
– Penguino
11 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
The answer is that
a regular dodecagon can indeed be made out of 30-60-90 tiles; you can see one rather complicated construction here.
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
answered 2 hours ago
Gareth McCaughan♦
57.8k3144221
57.8k3144221
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
add a comment |Â
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4sqrt3approx6.928$.
– Jaap Scherphuis
34 mins ago
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
add a comment |Â
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
Here is a convex dodecahedron made of $76$ of those triangles.
Can it be done with fewer?
answered 8 mins ago
Jaap Scherphuis
13.1k12259
answered 8 mins ago
Jaap Scherphuis
13.1k12259
answered 8 mins ago
Jaap Scherphuis
13.1k12259
answered 8 mins ago
Jaap Scherphuis
13.1k12259
13.1k12259
add a comment |Â
add a comment |Â
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
Francesco Arnaudo is a new contributor. Be nice, and check out our Code of Conduct.
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Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon
– Hugh
5 hours ago
1
The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon.
– Francesco Arnaudo
5 hours ago
If say six — I can't think of anything better.
– Hugh
4 hours ago