Mixing
Find the atractor of the triangles formed by joining the feet of altitudes of the previous triangle?
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Triangle 1 (see the picture) is given. Find the point toward which the vertices of triangle n -> infinity converge, assuming that triangle n is constructed by uniting the feet of the altitudes of triangle n-1.
Sequence of triangles formed by the above mentioned rule.
For the definition of "foot of an altitude" please see: Perpendicular Foot
geometry triangle
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
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add a comment |
$begingroup$
Triangle 1 (see the picture) is given. Find the point toward which the vertices of triangle n -> infinity converge, assuming that triangle n is constructed by uniting the feet of the altitudes of triangle n-1.
Sequence of triangles formed by the above mentioned rule.
For the definition of "foot of an altitude" please see: Perpendicular Foot
geometry triangle
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
$endgroup$
$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
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– ShapeOfMatter
Jan 18 at 21:32
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@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37
add a comment |
$begingroup$
Triangle 1 (see the picture) is given. Find the point toward which the vertices of triangle n -> infinity converge, assuming that triangle n is constructed by uniting the feet of the altitudes of triangle n-1.
Sequence of triangles formed by the above mentioned rule.
For the definition of "foot of an altitude" please see: Perpendicular Foot
geometry triangle
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
$endgroup$
Triangle 1 (see the picture) is given. Find the point toward which the vertices of triangle n -> infinity converge, assuming that triangle n is constructed by uniting the feet of the altitudes of triangle n-1.
Sequence of triangles formed by the above mentioned rule.
For the definition of "foot of an altitude" please see: Perpendicular Foot
geometry triangle
geometry triangle
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
edited Jan 18 at 21:37
Robert Werner
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
asked Jan 18 at 21:26
Robert WernerRobert Werner
1185
1185
$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
$endgroup$
– ShapeOfMatter
Jan 18 at 21:32
$begingroup$
@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37
add a comment |
$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
$endgroup$
– ShapeOfMatter
Jan 18 at 21:32
$begingroup$
@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37
$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
$endgroup$
– ShapeOfMatter
Jan 18 at 21:32
$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
$endgroup$
– ShapeOfMatter
Jan 18 at 21:32
$begingroup$
@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37
$begingroup$
@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37
add a comment |
1 Answer
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By denoting as $T_n$ the triangle at the $n$-th iteration we can easily describe the angles of $T_{n+1}$, orthic triangle of $T_n$, in terms of the angles of $T_n$. We may check that the area and the perimeter of $T_n$ converge to zero, but the "shape" of $T_n$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $P$, from some $n$ onward the orthocenter of $T_n$ has to lie in the interior of $T_n$, meaning that $T_n$ is acute-angled for any $n$ sufficiently large. On the other hand the shape of $T_n$ changes according to
$$(A,B,C)to (pi-2A,pi-2B,pi-2C) $$
and almost surely the map sending $x$ into $-2xpmod{pi}$ is not convergent.
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
By denoting as $T_n$ the triangle at the $n$-th iteration we can easily describe the angles of $T_{n+1}$, orthic triangle of $T_n$, in terms of the angles of $T_n$. We may check that the area and the perimeter of $T_n$ converge to zero, but the "shape" of $T_n$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $P$, from some $n$ onward the orthocenter of $T_n$ has to lie in the interior of $T_n$, meaning that $T_n$ is acute-angled for any $n$ sufficiently large. On the other hand the shape of $T_n$ changes according to
$$(A,B,C)to (pi-2A,pi-2B,pi-2C) $$
and almost surely the map sending $x$ into $-2xpmod{pi}$ is not convergent.
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
By denoting as $T_n$ the triangle at the $n$-th iteration we can easily describe the angles of $T_{n+1}$, orthic triangle of $T_n$, in terms of the angles of $T_n$. We may check that the area and the perimeter of $T_n$ converge to zero, but the "shape" of $T_n$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $P$, from some $n$ onward the orthocenter of $T_n$ has to lie in the interior of $T_n$, meaning that $T_n$ is acute-angled for any $n$ sufficiently large. On the other hand the shape of $T_n$ changes according to
$$(A,B,C)to (pi-2A,pi-2B,pi-2C) $$
and almost surely the map sending $x$ into $-2xpmod{pi}$ is not convergent.
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
By denoting as $T_n$ the triangle at the $n$-th iteration we can easily describe the angles of $T_{n+1}$, orthic triangle of $T_n$, in terms of the angles of $T_n$. We may check that the area and the perimeter of $T_n$ converge to zero, but the "shape" of $T_n$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $P$, from some $n$ onward the orthocenter of $T_n$ has to lie in the interior of $T_n$, meaning that $T_n$ is acute-angled for any $n$ sufficiently large. On the other hand the shape of $T_n$ changes according to
$$(A,B,C)to (pi-2A,pi-2B,pi-2C) $$
and almost surely the map sending $x$ into $-2xpmod{pi}$ is not convergent.
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
By denoting as $T_n$ the triangle at the $n$-th iteration we can easily describe the angles of $T_{n+1}$, orthic triangle of $T_n$, in terms of the angles of $T_n$. We may check that the area and the perimeter of $T_n$ converge to zero, but the "shape" of $T_n$ (i.e. the triple of the angles) does not converge, in the general case.
Actually it is known that such iteration is usually chaotic, and not difficult to prove: assuming that our sequence is convergent to a point $P$, from some $n$ onward the orthocenter of $T_n$ has to lie in the interior of $T_n$, meaning that $T_n$ is acute-angled for any $n$ sufficiently large. On the other hand the shape of $T_n$ changes according to
$$(A,B,C)to (pi-2A,pi-2B,pi-2C) $$
and almost surely the map sending $x$ into $-2xpmod{pi}$ is not convergent.
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
edited Jan 18 at 22:16
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 18 at 22:02
Jack D'AurizioJack D'Aurizio
290k33282664
290k33282664
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$begingroup$
Can you explain, perhaps in a separate paragraph from the question, the rule for constructing additional triangles? An algorithm for triangle $n$, in terms on $n-1$, would fine. Basically I have no idea what you mean by "uniting the feet of the altitudes".
$endgroup$
– ShapeOfMatter
Jan 18 at 21:32
$begingroup$
@ShapeOfMatter: I believe it is quite clear: if $T_n$ is the triangle at the $n$-th step, $T_{n+1}$ is the orthic triangle of $T_{n}$.
$endgroup$
– Jack D'Aurizio
Jan 18 at 21:37