Mixing
Five Similar Triangles from 4-5-6
$begingroup$
A 4-5-6 triangle can be divided into 5 similar scalene triangles. I have the solution, but I figured out it was a 4-5-6 after putting together the component triangles. Can anyone else find it?
The Zak triangle can be divided into 6 similar triangles in 2 ways. It leads to an interesting fractal.
I have a collection of unusual similar triangle polygons. I believe only three quadrilaterals can be divided into 4 similar obtuse triangles with a common vertex. One is used in the 4-5-6, the Zak triangle uses $rho$, the plastic constant, the third uses $psi$, the supergolden ratio. Below the numbers are powers of $sqrtpsi$. This quadrilateral leads to interesting fractals.
What other triangles can be divided into similar obtuse triangles in non-trivial ways?
This is related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.
geometry triangle recreational-mathematics puzzle
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
$endgroup$
|
show 1 more comment
$begingroup$
A 4-5-6 triangle can be divided into 5 similar scalene triangles. I have the solution, but I figured out it was a 4-5-6 after putting together the component triangles. Can anyone else find it?
The Zak triangle can be divided into 6 similar triangles in 2 ways. It leads to an interesting fractal.
I have a collection of unusual similar triangle polygons. I believe only three quadrilaterals can be divided into 4 similar obtuse triangles with a common vertex. One is used in the 4-5-6, the Zak triangle uses $rho$, the plastic constant, the third uses $psi$, the supergolden ratio. Below the numbers are powers of $sqrtpsi$. This quadrilateral leads to interesting fractals.
What other triangles can be divided into similar obtuse triangles in non-trivial ways?
This is related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.
geometry triangle recreational-mathematics puzzle
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
$endgroup$
2
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
1
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
1
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47
|
show 1 more comment
$begingroup$
A 4-5-6 triangle can be divided into 5 similar scalene triangles. I have the solution, but I figured out it was a 4-5-6 after putting together the component triangles. Can anyone else find it?
The Zak triangle can be divided into 6 similar triangles in 2 ways. It leads to an interesting fractal.
I have a collection of unusual similar triangle polygons. I believe only three quadrilaterals can be divided into 4 similar obtuse triangles with a common vertex. One is used in the 4-5-6, the Zak triangle uses $rho$, the plastic constant, the third uses $psi$, the supergolden ratio. Below the numbers are powers of $sqrtpsi$. This quadrilateral leads to interesting fractals.
What other triangles can be divided into similar obtuse triangles in non-trivial ways?
This is related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.
geometry triangle recreational-mathematics puzzle
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
$endgroup$
A 4-5-6 triangle can be divided into 5 similar scalene triangles. I have the solution, but I figured out it was a 4-5-6 after putting together the component triangles. Can anyone else find it?
The Zak triangle can be divided into 6 similar triangles in 2 ways. It leads to an interesting fractal.
I have a collection of unusual similar triangle polygons. I believe only three quadrilaterals can be divided into 4 similar obtuse triangles with a common vertex. One is used in the 4-5-6, the Zak triangle uses $rho$, the plastic constant, the third uses $psi$, the supergolden ratio. Below the numbers are powers of $sqrtpsi$. This quadrilateral leads to interesting fractals.
What other triangles can be divided into similar obtuse triangles in non-trivial ways?
This is related to New Substitution Tilings Using 2, φ, ψ, χ, ρ.
geometry triangle recreational-mathematics puzzle
geometry triangle recreational-mathematics puzzle
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
edited Mar 7 at 19:59
Ed Pegg
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
asked Jan 23 at 14:38
Ed PeggEd Pegg
10k32593
10k32593
2
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
1
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
1
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47
|
show 1 more comment
2
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
1
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
1
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47
2
2
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
1
1
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
1
1
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
For the secondary question of the 4-5-6 triangle:
Start with $triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $angle A$ and $angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.
I have not formally proved it, but it seems to work.
The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
$endgroup$
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the secondary question of the 4-5-6 triangle:
Start with $triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $angle A$ and $angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.
I have not formally proved it, but it seems to work.
The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
$endgroup$
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
add a comment |
$begingroup$
For the secondary question of the 4-5-6 triangle:
Start with $triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $angle A$ and $angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.
I have not formally proved it, but it seems to work.
The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
$endgroup$
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
add a comment |
$begingroup$
For the secondary question of the 4-5-6 triangle:
Start with $triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $angle A$ and $angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.
I have not formally proved it, but it seems to work.
The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
$endgroup$
For the secondary question of the 4-5-6 triangle:
Start with $triangle ABC$ with $|AB|=4, |AC|=5, |BC|=6$. Bisect $angle A$ and $angle B$, drawing their rays to their point if intersection at the in center $I$. From $I$ construct ray $ID$ parallel to $AC$ where $D$ lies on $BC$. Construct point $E$ on $AC$ such that $|AE|=4$ or equivalently, $IE$ is congruent to $IB$. Alternatively, place $E$ so that $|AE|=1$ or $DE$ is congruent to $IB$ (this alternative is shown in the diagram below). Draw segments $IE$ and $DE$ to complete the dissection.
I have not formally proved it, but it seems to work.
The existence of two solutions for the 4-5-6 triangle, with either $|AE|=1$ or $|AE|=4$, implies a fourth quadrilateral divisible into four similar obtuse triangles with a common vertex. Two of the quadrilaterals are the possible $ABDE$ quadrilaterals from the 4-5-6 triangle solutions, the other two are the quadrilaterals based on the cubic roots as given by the OP.
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
edited Jan 23 at 23:58
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
answered Jan 23 at 21:43
Oscar LanziOscar Lanzi
13.1k12136
13.1k12136
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
add a comment |
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
1
1
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
Yep. I added a diagram.
$endgroup$
– Ed Pegg
Jan 23 at 22:19
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
$begingroup$
$((0,0), (1,3sqrt7)/2, (5,0), (3,sqrt7)/2, (7,sqrt7)/2,(1,0))$
$endgroup$
– Ed Pegg
Jan 23 at 22:32
1
1
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
$begingroup$
The diagram does not match my description. I am thinking now the solution is nonunique (you can reflect $triangle IDE$)!
$endgroup$
– Oscar Lanzi
Jan 23 at 22:35
1
1
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
$begingroup$
My code was looking for algebraic roots. I remember errors caused by $sqrt2$, and this was why, I thought those two quadrilaterals were the same.
$endgroup$
– Ed Pegg
Jan 24 at 15:46
add a comment |
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2
$begingroup$
Looks like multiple questions, and isn't there an SE for puzzles?
$endgroup$
– Oscar Lanzi
Jan 23 at 14:44
$begingroup$
My main question is at the end.
$endgroup$
– Ed Pegg
Jan 23 at 15:00
1
$begingroup$
I'll edited the CA to remive a downvote, but somehow "sqrt(plastic)" came up. Please correct, thanks.
$endgroup$
– Oscar Lanzi
Jan 23 at 21:50
1
$begingroup$
There are (at least) four of the quadrilaterals you describe, because of two solutions for the 4-5-6 triangle. See my edited answer.
$endgroup$
– Oscar Lanzi
Jan 24 at 0:13
$begingroup$
Why the restriction to obtuse triangles?
$endgroup$
– WRSomsky
Jan 28 at 17:47