Characterizing triangles with integer-degree angles that are similar to any of their iterated orthic...
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Characterizing triangles with integer-degree angles that are similar to any of their iterated orthic triangles
Proof of “triangles are similar iff corresponding angles are equal”Similar Triangles with proportionsSimilar triangles anglesSimilar Triangles Problem (No corresponding sides)Finding the largest equilateral triangle inside a given triangleIs there a relationship for a triangle's side lengths, altitude/height, and acute/obtuse/right?How are these triangles similar?Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC.Mobius transformation producing a curved triangle with 3 intersecting circlesSAT Math Problem - Corresponding Angles in Similar Triangles
$begingroup$
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
edited Mar 13 at 8:01
Blue
49.1k870156
asked Mar 13 at 7:06
Shauryam AkhouryShauryam Akhoury
5216
$endgroup$
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
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
edited Mar 13 at 8:01
Blue
49.1k870156
asked Mar 13 at 7:06
Shauryam AkhouryShauryam Akhoury
5216
$endgroup$
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
geometry number-theory
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
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
edited Mar 13 at 8:01
Blue
49.1k870156
49.1k870156
asked Mar 13 at 7:06
Shauryam AkhouryShauryam Akhoury
5216
asked Mar 13 at 7:06
Shauryam AkhouryShauryam Akhoury
5216
asked Mar 13 at 7:06
Shauryam AkhouryShauryam Akhoury
5216
5216
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 |
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
1
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
$begingroup$
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
1
$begingroup$
Yes i mean the orthic triangle
$endgroup$
– Shauryam Akhoury
Mar 13 at 7:52
$begingroup$
Yes i mean the orthic triangle
$endgroup$
– Shauryam Akhoury
Mar 13 at 7:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 14 at 9:51
AretinoAretino
25.4k21445
$endgroup$
add a comment |
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$begingroup$
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.
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.
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.
answered Mar 14 at 9:51
AretinoAretino
25.4k21445
$endgroup$
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.
answered Mar 14 at 9:51
AretinoAretino
25.4k21445
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
25.4k21445
add a comment |
add a comment |
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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