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Characterizing triangles with integer-degree angles that are similar to any of their iterated orthic triangles

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3

0

$begingroup$

A triangle has integer angles $a,b,c$ (in degrees) and we create the triangle's pedal triangle and call it Pedal-$(1)$. (Here, "pedal triangle" means the specific pedal triangle whose vertices are the feet of the altitudes of the original triangle; in other words, the orthic triangle.)

Pedal-$(n)$'s pedal triangle is Pedal-$(n+1)$.

If the original triangle is similar to any Pedal-$(n)$'s we call it a "x" triangle.

I got if a triangle is a "x" triangle then $a,b,c$ must be divisible by $4$.
But not that if $a,b,c$ are divisible by $4$ then the triangle must be a "x" triangle.

And trying all the trios for $a,b,c$ divisible by $4$ shows that every triangle in this category is indeed a "x" triangle?

Why does this happen?

geometry number-theory

share|cite|improve this question

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's your definition of "pedal triangle"? I'm familiar with the term insofar as it requires an additional point of reference, as in "pedal triangle with respect to $P$". Do you mean "orthic triangle" (a triangle whose vertices are the feet of the altitudes of the original triangle)? The orthic triangle is the pedal triangle with respect to the orthocenter, so it's possible that some authors mix the terms.
  $endgroup$
  – Blue
  Mar 13 at 7:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes i mean the orthic triangle
  $endgroup$
  – Shauryam Akhoury
  Mar 13 at 7:52
  

  
  
  
  

add a comment | 

3

0

$begingroup$

A triangle has integer angles $a,b,c$ (in degrees) and we create the triangle's pedal triangle and call it Pedal-$(1)$. (Here, "pedal triangle" means the specific pedal triangle whose vertices are the feet of the altitudes of the original triangle; in other words, the orthic triangle.)

Pedal-$(n)$'s pedal triangle is Pedal-$(n+1)$.

If the original triangle is similar to any Pedal-$(n)$'s we call it a "x" triangle.

I got if a triangle is a "x" triangle then $a,b,c$ must be divisible by $4$.
But not that if $a,b,c$ are divisible by $4$ then the triangle must be a "x" triangle.

And trying all the trios for $a,b,c$ divisible by $4$ shows that every triangle in this category is indeed a "x" triangle?

Why does this happen?

geometry number-theory

share|cite|improve this question

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What's your definition of "pedal triangle"? I'm familiar with the term insofar as it requires an additional point of reference, as in "pedal triangle with respect to $P$". Do you mean "orthic triangle" (a triangle whose vertices are the feet of the altitudes of the original triangle)? The orthic triangle is the pedal triangle with respect to the orthocenter, so it's possible that some authors mix the terms.
  $endgroup$
  – Blue
  Mar 13 at 7:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes i mean the orthic triangle
  $endgroup$
  – Shauryam Akhoury
  Mar 13 at 7:52
  

  
  
  
  

add a comment | 

$begingroup$

A triangle has integer angles $a,b,c$ (in degrees) and we create the triangle's pedal triangle and call it Pedal-$(1)$. (Here, "pedal triangle" means the specific pedal triangle whose vertices are the feet of the altitudes of the original triangle; in other words, the orthic triangle.)

Pedal-$(n)$'s pedal triangle is Pedal-$(n+1)$.

If the original triangle is similar to any Pedal-$(n)$'s we call it a "x" triangle.

I got if a triangle is a "x" triangle then $a,b,c$ must be divisible by $4$.
But not that if $a,b,c$ are divisible by $4$ then the triangle must be a "x" triangle.

And trying all the trios for $a,b,c$ divisible by $4$ shows that every triangle in this category is indeed a "x" triangle?

Why does this happen?

geometry number-theory

share|cite|improve this question

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

$endgroup$

share|cite|improve this question

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

share|cite|improve this question

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

edited Mar 13 at 8:01

Blue

49.1k870156

edited Mar 13 at 8:01

Blue

49.1k870156

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

asked Mar 13 at 7:06

Shauryam AkhouryShauryam Akhoury

5216

1 Answer
1

active

oldest

votes

2

$begingroup$

If Pedal$(n)$ has angles $(a,b,c)$, then Pedal$(n+1)$ has angles:
$$
cases{
(180°-2a,quad 180°-2b,quad 180°-2c)&if Pedal$(n)$ is an acute triangle,\
(2a-180°,quad2b,quad2c)&if angle $a$ is obtuse.\}
$$
That of course explains why all "x" triangle must have $a$, $b$, $c$ divisible by $4$.

On the other hand, a triangle $T$, with angles $a$, $b$, $c$ divisible by $4$, is not an "x" triangle only if there is no other triangle with angles divisible by $4$ which has $T$ as orthic triangle. But that is never the case, for $T$ is the orthic triangle of a triangle having angles $a'$, $b'$, $c'$ (also divisible by $4$) given by:
$$
cases{
a'=90°-a/2,quad b'=90°-b/2,quad c'=90°-c/2 & if none of $a$, $b$, $c$ is divisible by $8$,\
a'=90°+a/2,quad b'=b/2,quadquadquad c'=c/2 & if $b$ and $c$ are divisible by $8$ while $a$ is not.\
}
$$
No other case is possible, because $a+b+c=180°$ is not divisible by $8$.

In other words: in every chain of nested orthic triangles, if the angles of a triangle are divisible by $4$, then the angles of the triangles preceding and following it in the chain are given uniquely by the relations above. Such a chain cannot then have a first or last triangle, nor branches, and sooner or later a triangle with the same angles as the first one must appear.

share|cite|improve this answer

edited Mar 14 at 14:10

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

$endgroup$

add a comment | 

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2

$begingroup$

If Pedal$(n)$ has angles $(a,b,c)$, then Pedal$(n+1)$ has angles:
$$
cases{
(180°-2a,quad 180°-2b,quad 180°-2c)&if Pedal$(n)$ is an acute triangle,\
(2a-180°,quad2b,quad2c)&if angle $a$ is obtuse.\}
$$
That of course explains why all "x" triangle must have $a$, $b$, $c$ divisible by $4$.

On the other hand, a triangle $T$, with angles $a$, $b$, $c$ divisible by $4$, is not an "x" triangle only if there is no other triangle with angles divisible by $4$ which has $T$ as orthic triangle. But that is never the case, for $T$ is the orthic triangle of a triangle having angles $a'$, $b'$, $c'$ (also divisible by $4$) given by:
$$
cases{
a'=90°-a/2,quad b'=90°-b/2,quad c'=90°-c/2 & if none of $a$, $b$, $c$ is divisible by $8$,\
a'=90°+a/2,quad b'=b/2,quadquadquad c'=c/2 & if $b$ and $c$ are divisible by $8$ while $a$ is not.\
}
$$
No other case is possible, because $a+b+c=180°$ is not divisible by $8$.

In other words: in every chain of nested orthic triangles, if the angles of a triangle are divisible by $4$, then the angles of the triangles preceding and following it in the chain are given uniquely by the relations above. Such a chain cannot then have a first or last triangle, nor branches, and sooner or later a triangle with the same angles as the first one must appear.

share|cite|improve this answer

edited Mar 14 at 14:10

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

$endgroup$

add a comment | 

2

$begingroup$

If Pedal$(n)$ has angles $(a,b,c)$, then Pedal$(n+1)$ has angles:
$$
cases{
(180°-2a,quad 180°-2b,quad 180°-2c)&if Pedal$(n)$ is an acute triangle,\
(2a-180°,quad2b,quad2c)&if angle $a$ is obtuse.\}
$$
That of course explains why all "x" triangle must have $a$, $b$, $c$ divisible by $4$.

On the other hand, a triangle $T$, with angles $a$, $b$, $c$ divisible by $4$, is not an "x" triangle only if there is no other triangle with angles divisible by $4$ which has $T$ as orthic triangle. But that is never the case, for $T$ is the orthic triangle of a triangle having angles $a'$, $b'$, $c'$ (also divisible by $4$) given by:
$$
cases{
a'=90°-a/2,quad b'=90°-b/2,quad c'=90°-c/2 & if none of $a$, $b$, $c$ is divisible by $8$,\
a'=90°+a/2,quad b'=b/2,quadquadquad c'=c/2 & if $b$ and $c$ are divisible by $8$ while $a$ is not.\
}
$$
No other case is possible, because $a+b+c=180°$ is not divisible by $8$.

In other words: in every chain of nested orthic triangles, if the angles of a triangle are divisible by $4$, then the angles of the triangles preceding and following it in the chain are given uniquely by the relations above. Such a chain cannot then have a first or last triangle, nor branches, and sooner or later a triangle with the same angles as the first one must appear.

share|cite|improve this answer

edited Mar 14 at 14:10

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

$endgroup$

add a comment | 

$begingroup$

If Pedal$(n)$ has angles $(a,b,c)$, then Pedal$(n+1)$ has angles:
$$
cases{
(180°-2a,quad 180°-2b,quad 180°-2c)&if Pedal$(n)$ is an acute triangle,\
(2a-180°,quad2b,quad2c)&if angle $a$ is obtuse.\}
$$
That of course explains why all "x" triangle must have $a$, $b$, $c$ divisible by $4$.

On the other hand, a triangle $T$, with angles $a$, $b$, $c$ divisible by $4$, is not an "x" triangle only if there is no other triangle with angles divisible by $4$ which has $T$ as orthic triangle. But that is never the case, for $T$ is the orthic triangle of a triangle having angles $a'$, $b'$, $c'$ (also divisible by $4$) given by:
$$
cases{
a'=90°-a/2,quad b'=90°-b/2,quad c'=90°-c/2 & if none of $a$, $b$, $c$ is divisible by $8$,\
a'=90°+a/2,quad b'=b/2,quadquadquad c'=c/2 & if $b$ and $c$ are divisible by $8$ while $a$ is not.\
}
$$
No other case is possible, because $a+b+c=180°$ is not divisible by $8$.

In other words: in every chain of nested orthic triangles, if the angles of a triangle are divisible by $4$, then the angles of the triangles preceding and following it in the chain are given uniquely by the relations above. Such a chain cannot then have a first or last triangle, nor branches, and sooner or later a triangle with the same angles as the first one must appear.

share|cite|improve this answer

edited Mar 14 at 14:10

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

$endgroup$

share|cite|improve this answer

edited Mar 14 at 14:10

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

answered Mar 14 at 9:51

AretinoAretino

25.4k21445

answered Mar 14 at 9:51

AretinoAretino

25.4k21445