Mixing
triangles in $mathbbR^n$ with all vertices in $mathbbQ^n$
Clash Royale CLAN TAG#URR8PPP
up vote
14
down vote
favorite
In the context of this post, a triangle will mean a triple $(a,b,c)$ of positive real numbers which qualify as side lengths of a triangle (i.e., the triangle inequalities are satisfied).
Call triangle $(a,b,c)$ rationally realizable if, for some integer $nge 2$, a triangle with side lengths $a,b,c$ can be placed in $mathbbR^n$ with all vertices in $mathbbQ^n$.
Some examples . . .
- If equilateral triangle $(a,a,a)$ has $ainmathbbQ$, then it's rationally realizable in $mathbbR^6$.$\[6pt]$
Proof:$;$Use vertices
$
(largefraca2,largefraca2,0,0,0,0),,
(0,0,largefraca2,largefraca2,0,0),,
(0,0,0,0,largefraca2,largefraca2)
$. - If right triangle $(a,b,c)$ with legs $a,b$ has $a,binmathbbQ$, then it's rationally realizable in $mathbbR^2$.$\[6pt]$
Proof:$;$Use vertices $(0,0),,(a,0),,(0,b)$.
From the distance formula, for triangle $(a,b,c)$ to be rationally realizable, a necessary condition is $a^2,b^2,c^2in mathbbQ$.
Is this condition also sufficient? More precisely:
Question:$;$If triangle $(a,b,c)$ has $a^2,b^2,c^2in mathbbQ$, must it be rationally realizable?
geometry elementary-number-theory diophantine-equations
asked Jul 20 at 23:04
quasi
33.3k22359
add a comment |Â
up vote
14
down vote
favorite
In the context of this post, a triangle will mean a triple $(a,b,c)$ of positive real numbers which qualify as side lengths of a triangle (i.e., the triangle inequalities are satisfied).
Call triangle $(a,b,c)$ rationally realizable if, for some integer $nge 2$, a triangle with side lengths $a,b,c$ can be placed in $mathbbR^n$ with all vertices in $mathbbQ^n$.
Some examples . . .
- If equilateral triangle $(a,a,a)$ has $ainmathbbQ$, then it's rationally realizable in $mathbbR^6$.$\[6pt]$
Proof:$;$Use vertices
$
(largefraca2,largefraca2,0,0,0,0),,
(0,0,largefraca2,largefraca2,0,0),,
(0,0,0,0,largefraca2,largefraca2)
$. - If right triangle $(a,b,c)$ with legs $a,b$ has $a,binmathbbQ$, then it's rationally realizable in $mathbbR^2$.$\[6pt]$
Proof:$;$Use vertices $(0,0),,(a,0),,(0,b)$.
From the distance formula, for triangle $(a,b,c)$ to be rationally realizable, a necessary condition is $a^2,b^2,c^2in mathbbQ$.
Is this condition also sufficient? More precisely:
Question:$;$If triangle $(a,b,c)$ has $a^2,b^2,c^2in mathbbQ$, must it be rationally realizable?
geometry elementary-number-theory diophantine-equations
asked Jul 20 at 23:04
quasi
33.3k22359
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
In the context of this post, a triangle will mean a triple $(a,b,c)$ of positive real numbers which qualify as side lengths of a triangle (i.e., the triangle inequalities are satisfied).
Call triangle $(a,b,c)$ rationally realizable if, for some integer $nge 2$, a triangle with side lengths $a,b,c$ can be placed in $mathbbR^n$ with all vertices in $mathbbQ^n$.
Some examples . . .
- If equilateral triangle $(a,a,a)$ has $ainmathbbQ$, then it's rationally realizable in $mathbbR^6$.$\[6pt]$
Proof:$;$Use vertices
$
(largefraca2,largefraca2,0,0,0,0),,
(0,0,largefraca2,largefraca2,0,0),,
(0,0,0,0,largefraca2,largefraca2)
$. - If right triangle $(a,b,c)$ with legs $a,b$ has $a,binmathbbQ$, then it's rationally realizable in $mathbbR^2$.$\[6pt]$
Proof:$;$Use vertices $(0,0),,(a,0),,(0,b)$.
From the distance formula, for triangle $(a,b,c)$ to be rationally realizable, a necessary condition is $a^2,b^2,c^2in mathbbQ$.
Is this condition also sufficient? More precisely:
Question:$;$If triangle $(a,b,c)$ has $a^2,b^2,c^2in mathbbQ$, must it be rationally realizable?
geometry elementary-number-theory diophantine-equations
asked Jul 20 at 23:04
quasi
33.3k22359
In the context of this post, a triangle will mean a triple $(a,b,c)$ of positive real numbers which qualify as side lengths of a triangle (i.e., the triangle inequalities are satisfied).
Call triangle $(a,b,c)$ rationally realizable if, for some integer $nge 2$, a triangle with side lengths $a,b,c$ can be placed in $mathbbR^n$ with all vertices in $mathbbQ^n$.
Some examples . . .
- If equilateral triangle $(a,a,a)$ has $ainmathbbQ$, then it's rationally realizable in $mathbbR^6$.$\[6pt]$
Proof:$;$Use vertices
$
(largefraca2,largefraca2,0,0,0,0),,
(0,0,largefraca2,largefraca2,0,0),,
(0,0,0,0,largefraca2,largefraca2)
$. - If right triangle $(a,b,c)$ with legs $a,b$ has $a,binmathbbQ$, then it's rationally realizable in $mathbbR^2$.$\[6pt]$
Proof:$;$Use vertices $(0,0),,(a,0),,(0,b)$.
From the distance formula, for triangle $(a,b,c)$ to be rationally realizable, a necessary condition is $a^2,b^2,c^2in mathbbQ$.
Is this condition also sufficient? More precisely:
Question:$;$If triangle $(a,b,c)$ has $a^2,b^2,c^2in mathbbQ$, must it be rationally realizable?
geometry elementary-number-theory diophantine-equations
asked Jul 20 at 23:04
quasi
33.3k22359
edited Jul 22 at 10:50
asked Jul 20 at 23:04
quasi
33.3k22359
asked Jul 20 at 23:04
quasi
33.3k22359
asked Jul 20 at 23:04
quasi
33.3k22359
33.3k22359
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39
add a comment |Â
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2leq b^2leq c^2$ are all positive even integers.
Case 1: The triangle is not obtuse, i.e., $c^2leq a^2+b^2$.
Let $s:= fraca^2+b^2-c^22in mathbbN$, and let $x:= b^2-sin mathbbN, y:= a^2-sin mathbbN$.
By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers.
Produce vectors of length 4 out of these quartets of numbers, and denote them by $underlinex, underliney, underlines$, respectively.
Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:
$$A=(underlinex,underlines,underline0)$$
$$B=(underline0,underlines,underliney)$$
Then $vert ABvert=c$, $vert BCvert=a$, $vert CAvert=b$.
Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case.
We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= sqrta^2-x^2, b_1:= sqrtb^2-y^2, c_1:=sqrtc^2-z^2$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $lambda>0$ and solve:
$x=lambda a$, $y=lambda b$, $z=x+y$, and
$c^2-z^2= (a^2-x^2)+(b^2-y^2)$.
By substituting $z=x+y$ in the last equation we get:
$$2xy= c^2-a^2-b^2= -2abcos gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=lambda a$, $y=lambda b$ yields the solution $lambda= sqrt-cos gamma$, which satisfies $0<lambda<1$.
As rational numbers are dense in the reals, we can pick rational numbers $xapprox lambda a, yapprox lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.
Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin.
We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $mathbbR^13$.
In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.
Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.
answered Jul 21 at 12:52
A. Pongrácz
2,309221
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
 |Â
show 2 more comments
up vote
5
down vote
I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.
First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2leq b^2leq c^2leq a^2+b^2$.
Let $n:= fraca^2+b^2+c^22$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$.
Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $sqrtn-(a^2+b^2-n)= sqrt2n-(a^2+b^2)= c$.
answered Jul 21 at 0:45
A. Pongrácz
2,309221
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
add a comment |Â
up vote
1
down vote
For the obtuse case, here's a geometric proof . . .
By A. Pongrácz's elegant construction, the claim holds for non-obtuse triangles.
We can reduce the obtuse case to the non-obtuse case as follows . . .
Suppose obtuse triangle $T$ has side lengths $a,b,c$, with $ale b < c$, and $a^2,b^2,c^2inmathbbQ$.
Place a copy of triangle $T$ in $mathbbR^2$ as triangle $ABC$, with $a=|BC|,;b=|AC|,;c=|AB|$.
Reflect $B$ (the vertex of the largest of the two acute angles) over $C$ to a point $C'$, and let $T'$ denote triangle $ABC'$.
By the formula for the length of a median
https://en.wikipedia.org/wiki/Median_(geometry)#Formulas_involving_the_medians%27_lengths
since the squared side lengths of triangle $T$ are rational, so are the squared side lengths of triangle $T'$.
In light of the midpoint formula, if triangle $T'$ is rationally realizable in $mathbbR^n$ (for some integer $nge 2$), so is triangle $T$.
Hence, if $T'$ is non-obtuse, we're done.
Suppose $T'$ is obtuse.
The measure of angle at vertex $B$ is unchanged, so is still acute.
Let the side lengths of triangle $T'$ be $a',b',c'$, where $a'=|BC'|,;b'=|AC'|,;c'=|AB|$.
Then $a'=2a,;c'=c$, and, by the median length formula, we get $(b')^2 = 2a^2+2b^2-c^2$.
Then identically,
$$(b')^2+(c')^2-(a')^2= (2a^2+2b^2-c^2)+(c^2)-(2a)^2=b^2-a^2$$
which is nonnegative, since $bge a$.
Thus, the new angle at $A$ (angle BAC') is non-obtuse, hence, since $T'$ is obtuse, the obtuse angle is at $C'$.
Consider the ratio of the new angle $A$ (angle $BAC'$), to the old one (angle $BAC$) . . .
Since angle $C'$ is obtuse, we have $|AB| > |AC'|$, hence, by the angle bisector theorem,
https://en.wikipedia.org/wiki/Angle_bisector_theorem#Theorem
the angle bisector of angle $BAC'$ meets side $BC'$ strictly between $C$ and $C'$.
It follows that the new angle $A$, while still acute, is more than twice the old one.
In triangle $T'$, the new smallest angle is either angle $BAC'$ or angle $ABC'$.
Thus, either the new smallest angle is more than double the old one, or else the new smallest angle is equal to the larger of the two acute angles of the old one.
Now iterate the process, reflecting the vertex of the largest of the two acute angles of triangle $T'$ over $C'$ to a point $C''$, and let $T''$ denote triangle $ABC''$.
As before, if $T''$ is non-obtuse, we're done.
If not, the smallest angle of $T''$ is more than twice the smallest angle of $T$.
Thus, the process can't yield obtuse triangles forever, since, as long as the triangles are still obtuse, the smallest angle more than doubles after two iterations.
It follows that $T$ is rationally realizable.
This completes the proof.
Remark:$;$With this approach, for the obtuse case, no extra coordinate is needed.
answered Jul 22 at 10:17
quasi
33.3k22359
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
accepted
The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2leq b^2leq c^2$ are all positive even integers.
Case 1: The triangle is not obtuse, i.e., $c^2leq a^2+b^2$.
Let $s:= fraca^2+b^2-c^22in mathbbN$, and let $x:= b^2-sin mathbbN, y:= a^2-sin mathbbN$.
By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers.
Produce vectors of length 4 out of these quartets of numbers, and denote them by $underlinex, underliney, underlines$, respectively.
Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:
$$A=(underlinex,underlines,underline0)$$
$$B=(underline0,underlines,underliney)$$
Then $vert ABvert=c$, $vert BCvert=a$, $vert CAvert=b$.
Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case.
We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= sqrta^2-x^2, b_1:= sqrtb^2-y^2, c_1:=sqrtc^2-z^2$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $lambda>0$ and solve:
$x=lambda a$, $y=lambda b$, $z=x+y$, and
$c^2-z^2= (a^2-x^2)+(b^2-y^2)$.
By substituting $z=x+y$ in the last equation we get:
$$2xy= c^2-a^2-b^2= -2abcos gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=lambda a$, $y=lambda b$ yields the solution $lambda= sqrt-cos gamma$, which satisfies $0<lambda<1$.
As rational numbers are dense in the reals, we can pick rational numbers $xapprox lambda a, yapprox lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.
Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin.
We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $mathbbR^13$.
In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.
Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.
answered Jul 21 at 12:52
A. Pongrácz
2,309221
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
 |Â
show 2 more comments
up vote
2
down vote
accepted
The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2leq b^2leq c^2$ are all positive even integers.
Case 1: The triangle is not obtuse, i.e., $c^2leq a^2+b^2$.
Let $s:= fraca^2+b^2-c^22in mathbbN$, and let $x:= b^2-sin mathbbN, y:= a^2-sin mathbbN$.
By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers.
Produce vectors of length 4 out of these quartets of numbers, and denote them by $underlinex, underliney, underlines$, respectively.
Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:
$$A=(underlinex,underlines,underline0)$$
$$B=(underline0,underlines,underliney)$$
Then $vert ABvert=c$, $vert BCvert=a$, $vert CAvert=b$.
Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case.
We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= sqrta^2-x^2, b_1:= sqrtb^2-y^2, c_1:=sqrtc^2-z^2$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $lambda>0$ and solve:
$x=lambda a$, $y=lambda b$, $z=x+y$, and
$c^2-z^2= (a^2-x^2)+(b^2-y^2)$.
By substituting $z=x+y$ in the last equation we get:
$$2xy= c^2-a^2-b^2= -2abcos gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=lambda a$, $y=lambda b$ yields the solution $lambda= sqrt-cos gamma$, which satisfies $0<lambda<1$.
As rational numbers are dense in the reals, we can pick rational numbers $xapprox lambda a, yapprox lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.
Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin.
We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $mathbbR^13$.
In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.
Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.
answered Jul 21 at 12:52
A. Pongrácz
2,309221
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2leq b^2leq c^2$ are all positive even integers.
Case 1: The triangle is not obtuse, i.e., $c^2leq a^2+b^2$.
Let $s:= fraca^2+b^2-c^22in mathbbN$, and let $x:= b^2-sin mathbbN, y:= a^2-sin mathbbN$.
By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers.
Produce vectors of length 4 out of these quartets of numbers, and denote them by $underlinex, underliney, underlines$, respectively.
Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:
$$A=(underlinex,underlines,underline0)$$
$$B=(underline0,underlines,underliney)$$
Then $vert ABvert=c$, $vert BCvert=a$, $vert CAvert=b$.
Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case.
We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= sqrta^2-x^2, b_1:= sqrtb^2-y^2, c_1:=sqrtc^2-z^2$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $lambda>0$ and solve:
$x=lambda a$, $y=lambda b$, $z=x+y$, and
$c^2-z^2= (a^2-x^2)+(b^2-y^2)$.
By substituting $z=x+y$ in the last equation we get:
$$2xy= c^2-a^2-b^2= -2abcos gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=lambda a$, $y=lambda b$ yields the solution $lambda= sqrt-cos gamma$, which satisfies $0<lambda<1$.
As rational numbers are dense in the reals, we can pick rational numbers $xapprox lambda a, yapprox lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.
Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin.
We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $mathbbR^13$.
In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.
Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.
answered Jul 21 at 12:52
A. Pongrácz
2,309221
The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2leq b^2leq c^2$ are all positive even integers.
Case 1: The triangle is not obtuse, i.e., $c^2leq a^2+b^2$.
Let $s:= fraca^2+b^2-c^22in mathbbN$, and let $x:= b^2-sin mathbbN, y:= a^2-sin mathbbN$.
By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers.
Produce vectors of length 4 out of these quartets of numbers, and denote them by $underlinex, underliney, underlines$, respectively.
Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:
$$A=(underlinex,underlines,underline0)$$
$$B=(underline0,underlines,underliney)$$
Then $vert ABvert=c$, $vert BCvert=a$, $vert CAvert=b$.
Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case.
We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= sqrta^2-x^2, b_1:= sqrtb^2-y^2, c_1:=sqrtc^2-z^2$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $lambda>0$ and solve:
$x=lambda a$, $y=lambda b$, $z=x+y$, and
$c^2-z^2= (a^2-x^2)+(b^2-y^2)$.
By substituting $z=x+y$ in the last equation we get:
$$2xy= c^2-a^2-b^2= -2abcos gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=lambda a$, $y=lambda b$ yields the solution $lambda= sqrt-cos gamma$, which satisfies $0<lambda<1$.
As rational numbers are dense in the reals, we can pick rational numbers $xapprox lambda a, yapprox lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.
Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin.
We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $mathbbR^13$.
In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.
Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.
answered Jul 21 at 12:52
A. Pongrácz
2,309221
edited Jul 21 at 21:21
answered Jul 21 at 12:52
A. Pongrácz
2,309221
answered Jul 21 at 12:52
A. Pongrácz
2,309221
answered Jul 21 at 12:52
A. Pongrácz
2,309221
2,309221
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
 |Â
show 2 more comments
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
I can now push the dimension down to 9. I think with deeper techniques, it can be pushed further down to 5, maybe. I will ask my colleagues, they are experts in Diophantine equations.
– A. Pongrácz
Jul 21 at 16:12
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
In fact, in the right triangle case, even dim=8 is possible. That is how I can improve to 9 in general: at the expense on one extra coordinate, I can always rederive to the right triangle case.
– A. Pongrácz
Jul 21 at 21:17
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
Thanks a lot, I fixed all of these.
– A. Pongrácz
Jul 21 at 21:21
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
How do you know $c_1^2$ is rational?
– quasi
Jul 22 at 3:31
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
$c_1^2=c^2-(x+y)^2$, and $c^2, x, yin mathbbQ$.
– A. Pongrácz
Jul 22 at 8:54
 |Â
show 2 more comments
up vote
5
down vote
I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.
First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2leq b^2leq c^2leq a^2+b^2$.
Let $n:= fraca^2+b^2+c^22$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$.
Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $sqrtn-(a^2+b^2-n)= sqrt2n-(a^2+b^2)= c$.
answered Jul 21 at 0:45
A. Pongrácz
2,309221
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
add a comment |Â
up vote
5
down vote
I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.
First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2leq b^2leq c^2leq a^2+b^2$.
Let $n:= fraca^2+b^2+c^22$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$.
Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $sqrtn-(a^2+b^2-n)= sqrt2n-(a^2+b^2)= c$.
answered Jul 21 at 0:45
A. Pongrácz
2,309221
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
add a comment |Â
up vote
5
down vote
up vote
5
down vote
I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.
First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2leq b^2leq c^2leq a^2+b^2$.
Let $n:= fraca^2+b^2+c^22$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$.
Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $sqrtn-(a^2+b^2-n)= sqrt2n-(a^2+b^2)= c$.
answered Jul 21 at 0:45
A. Pongrácz
2,309221
I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.
First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2leq b^2leq c^2leq a^2+b^2$.
Let $n:= fraca^2+b^2+c^22$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$.
Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $sqrtn-(a^2+b^2-n)= sqrt2n-(a^2+b^2)= c$.
answered Jul 21 at 0:45
A. Pongrácz
2,309221
answered Jul 21 at 0:45
A. Pongrácz
2,309221
answered Jul 21 at 0:45
A. Pongrácz
2,309221
answered Jul 21 at 0:45
A. Pongrácz
2,309221
2,309221
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
add a comment |Â
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
Very nice!$$
– quasi
Jul 21 at 0:51
Very nice!$$
– quasi
Jul 21 at 0:51
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Thank you. I now think that the conjecture is true. When the triangle is obtuse, we can be tricky by the signs in the middle. I mean, in my construction, in the coordinates where both $B$ and $A$ were not zero, I used ones. We can use some $(-1)$ in B, for example, and then when we subtract, instead of cancelling, we increase the length.
– A. Pongrácz
Jul 21 at 0:56
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Are you able to fully handle the obtuse case?
– quasi
Jul 21 at 5:12
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
Also note: Your construction can be modified to show that every non-obtuse triangle $(a,b,c)$ with $a^2,b^2,c^2inmathbbQ$ is rationally realizable in $mathbbR^12$. The idea is that every nonnegative rational number can be expressed as a sum of $4$ rational squares, hence, each of the two non-overlapping segments can be replaced by $4$ rational coordinates, and the overlapping segment can be similarly replaced, so $12$ coordinates will suffice.
– quasi
Jul 21 at 5:24
add a comment |Â
up vote
1
down vote
For the obtuse case, here's a geometric proof . . .
By A. Pongrácz's elegant construction, the claim holds for non-obtuse triangles.
We can reduce the obtuse case to the non-obtuse case as follows . . .
Suppose obtuse triangle $T$ has side lengths $a,b,c$, with $ale b < c$, and $a^2,b^2,c^2inmathbbQ$.
Place a copy of triangle $T$ in $mathbbR^2$ as triangle $ABC$, with $a=|BC|,;b=|AC|,;c=|AB|$.
Reflect $B$ (the vertex of the largest of the two acute angles) over $C$ to a point $C'$, and let $T'$ denote triangle $ABC'$.
By the formula for the length of a median
https://en.wikipedia.org/wiki/Median_(geometry)#Formulas_involving_the_medians%27_lengths
since the squared side lengths of triangle $T$ are rational, so are the squared side lengths of triangle $T'$.
In light of the midpoint formula, if triangle $T'$ is rationally realizable in $mathbbR^n$ (for some integer $nge 2$), so is triangle $T$.
Hence, if $T'$ is non-obtuse, we're done.
Suppose $T'$ is obtuse.
The measure of angle at vertex $B$ is unchanged, so is still acute.
Let the side lengths of triangle $T'$ be $a',b',c'$, where $a'=|BC'|,;b'=|AC'|,;c'=|AB|$.
Then $a'=2a,;c'=c$, and, by the median length formula, we get $(b')^2 = 2a^2+2b^2-c^2$.
Then identically,
$$(b')^2+(c')^2-(a')^2= (2a^2+2b^2-c^2)+(c^2)-(2a)^2=b^2-a^2$$
which is nonnegative, since $bge a$.
Thus, the new angle at $A$ (angle BAC') is non-obtuse, hence, since $T'$ is obtuse, the obtuse angle is at $C'$.
Consider the ratio of the new angle $A$ (angle $BAC'$), to the old one (angle $BAC$) . . .
Since angle $C'$ is obtuse, we have $|AB| > |AC'|$, hence, by the angle bisector theorem,
https://en.wikipedia.org/wiki/Angle_bisector_theorem#Theorem
the angle bisector of angle $BAC'$ meets side $BC'$ strictly between $C$ and $C'$.
It follows that the new angle $A$, while still acute, is more than twice the old one.
In triangle $T'$, the new smallest angle is either angle $BAC'$ or angle $ABC'$.
Thus, either the new smallest angle is more than double the old one, or else the new smallest angle is equal to the larger of the two acute angles of the old one.
Now iterate the process, reflecting the vertex of the largest of the two acute angles of triangle $T'$ over $C'$ to a point $C''$, and let $T''$ denote triangle $ABC''$.
As before, if $T''$ is non-obtuse, we're done.
If not, the smallest angle of $T''$ is more than twice the smallest angle of $T$.
Thus, the process can't yield obtuse triangles forever, since, as long as the triangles are still obtuse, the smallest angle more than doubles after two iterations.
It follows that $T$ is rationally realizable.
This completes the proof.
Remark:$;$With this approach, for the obtuse case, no extra coordinate is needed.
answered Jul 22 at 10:17
quasi
33.3k22359
add a comment |Â
up vote
1
down vote
For the obtuse case, here's a geometric proof . . .
By A. Pongrácz's elegant construction, the claim holds for non-obtuse triangles.
We can reduce the obtuse case to the non-obtuse case as follows . . .
Suppose obtuse triangle $T$ has side lengths $a,b,c$, with $ale b < c$, and $a^2,b^2,c^2inmathbbQ$.
Place a copy of triangle $T$ in $mathbbR^2$ as triangle $ABC$, with $a=|BC|,;b=|AC|,;c=|AB|$.
Reflect $B$ (the vertex of the largest of the two acute angles) over $C$ to a point $C'$, and let $T'$ denote triangle $ABC'$.
By the formula for the length of a median
https://en.wikipedia.org/wiki/Median_(geometry)#Formulas_involving_the_medians%27_lengths
since the squared side lengths of triangle $T$ are rational, so are the squared side lengths of triangle $T'$.
In light of the midpoint formula, if triangle $T'$ is rationally realizable in $mathbbR^n$ (for some integer $nge 2$), so is triangle $T$.
Hence, if $T'$ is non-obtuse, we're done.
Suppose $T'$ is obtuse.
The measure of angle at vertex $B$ is unchanged, so is still acute.
Let the side lengths of triangle $T'$ be $a',b',c'$, where $a'=|BC'|,;b'=|AC'|,;c'=|AB|$.
Then $a'=2a,;c'=c$, and, by the median length formula, we get $(b')^2 = 2a^2+2b^2-c^2$.
Then identically,
$$(b')^2+(c')^2-(a')^2= (2a^2+2b^2-c^2)+(c^2)-(2a)^2=b^2-a^2$$
which is nonnegative, since $bge a$.
Thus, the new angle at $A$ (angle BAC') is non-obtuse, hence, since $T'$ is obtuse, the obtuse angle is at $C'$.
Consider the ratio of the new angle $A$ (angle $BAC'$), to the old one (angle $BAC$) . . .
Since angle $C'$ is obtuse, we have $|AB| > |AC'|$, hence, by the angle bisector theorem,
https://en.wikipedia.org/wiki/Angle_bisector_theorem#Theorem
the angle bisector of angle $BAC'$ meets side $BC'$ strictly between $C$ and $C'$.
It follows that the new angle $A$, while still acute, is more than twice the old one.
In triangle $T'$, the new smallest angle is either angle $BAC'$ or angle $ABC'$.
Thus, either the new smallest angle is more than double the old one, or else the new smallest angle is equal to the larger of the two acute angles of the old one.
Now iterate the process, reflecting the vertex of the largest of the two acute angles of triangle $T'$ over $C'$ to a point $C''$, and let $T''$ denote triangle $ABC''$.
As before, if $T''$ is non-obtuse, we're done.
If not, the smallest angle of $T''$ is more than twice the smallest angle of $T$.
Thus, the process can't yield obtuse triangles forever, since, as long as the triangles are still obtuse, the smallest angle more than doubles after two iterations.
It follows that $T$ is rationally realizable.
This completes the proof.
Remark:$;$With this approach, for the obtuse case, no extra coordinate is needed.
answered Jul 22 at 10:17
quasi
33.3k22359
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For the obtuse case, here's a geometric proof . . .
By A. Pongrácz's elegant construction, the claim holds for non-obtuse triangles.
We can reduce the obtuse case to the non-obtuse case as follows . . .
Suppose obtuse triangle $T$ has side lengths $a,b,c$, with $ale b < c$, and $a^2,b^2,c^2inmathbbQ$.
Place a copy of triangle $T$ in $mathbbR^2$ as triangle $ABC$, with $a=|BC|,;b=|AC|,;c=|AB|$.
Reflect $B$ (the vertex of the largest of the two acute angles) over $C$ to a point $C'$, and let $T'$ denote triangle $ABC'$.
By the formula for the length of a median
https://en.wikipedia.org/wiki/Median_(geometry)#Formulas_involving_the_medians%27_lengths
since the squared side lengths of triangle $T$ are rational, so are the squared side lengths of triangle $T'$.
In light of the midpoint formula, if triangle $T'$ is rationally realizable in $mathbbR^n$ (for some integer $nge 2$), so is triangle $T$.
Hence, if $T'$ is non-obtuse, we're done.
Suppose $T'$ is obtuse.
The measure of angle at vertex $B$ is unchanged, so is still acute.
Let the side lengths of triangle $T'$ be $a',b',c'$, where $a'=|BC'|,;b'=|AC'|,;c'=|AB|$.
Then $a'=2a,;c'=c$, and, by the median length formula, we get $(b')^2 = 2a^2+2b^2-c^2$.
Then identically,
$$(b')^2+(c')^2-(a')^2= (2a^2+2b^2-c^2)+(c^2)-(2a)^2=b^2-a^2$$
which is nonnegative, since $bge a$.
Thus, the new angle at $A$ (angle BAC') is non-obtuse, hence, since $T'$ is obtuse, the obtuse angle is at $C'$.
Consider the ratio of the new angle $A$ (angle $BAC'$), to the old one (angle $BAC$) . . .
Since angle $C'$ is obtuse, we have $|AB| > |AC'|$, hence, by the angle bisector theorem,
https://en.wikipedia.org/wiki/Angle_bisector_theorem#Theorem
the angle bisector of angle $BAC'$ meets side $BC'$ strictly between $C$ and $C'$.
It follows that the new angle $A$, while still acute, is more than twice the old one.
In triangle $T'$, the new smallest angle is either angle $BAC'$ or angle $ABC'$.
Thus, either the new smallest angle is more than double the old one, or else the new smallest angle is equal to the larger of the two acute angles of the old one.
Now iterate the process, reflecting the vertex of the largest of the two acute angles of triangle $T'$ over $C'$ to a point $C''$, and let $T''$ denote triangle $ABC''$.
As before, if $T''$ is non-obtuse, we're done.
If not, the smallest angle of $T''$ is more than twice the smallest angle of $T$.
Thus, the process can't yield obtuse triangles forever, since, as long as the triangles are still obtuse, the smallest angle more than doubles after two iterations.
It follows that $T$ is rationally realizable.
This completes the proof.
Remark:$;$With this approach, for the obtuse case, no extra coordinate is needed.
answered Jul 22 at 10:17
quasi
33.3k22359
For the obtuse case, here's a geometric proof . . .
By A. Pongrácz's elegant construction, the claim holds for non-obtuse triangles.
We can reduce the obtuse case to the non-obtuse case as follows . . .
Suppose obtuse triangle $T$ has side lengths $a,b,c$, with $ale b < c$, and $a^2,b^2,c^2inmathbbQ$.
Place a copy of triangle $T$ in $mathbbR^2$ as triangle $ABC$, with $a=|BC|,;b=|AC|,;c=|AB|$.
Reflect $B$ (the vertex of the largest of the two acute angles) over $C$ to a point $C'$, and let $T'$ denote triangle $ABC'$.
By the formula for the length of a median
https://en.wikipedia.org/wiki/Median_(geometry)#Formulas_involving_the_medians%27_lengths
since the squared side lengths of triangle $T$ are rational, so are the squared side lengths of triangle $T'$.
In light of the midpoint formula, if triangle $T'$ is rationally realizable in $mathbbR^n$ (for some integer $nge 2$), so is triangle $T$.
Hence, if $T'$ is non-obtuse, we're done.
Suppose $T'$ is obtuse.
The measure of angle at vertex $B$ is unchanged, so is still acute.
Let the side lengths of triangle $T'$ be $a',b',c'$, where $a'=|BC'|,;b'=|AC'|,;c'=|AB|$.
Then $a'=2a,;c'=c$, and, by the median length formula, we get $(b')^2 = 2a^2+2b^2-c^2$.
Then identically,
$$(b')^2+(c')^2-(a')^2= (2a^2+2b^2-c^2)+(c^2)-(2a)^2=b^2-a^2$$
which is nonnegative, since $bge a$.
Thus, the new angle at $A$ (angle BAC') is non-obtuse, hence, since $T'$ is obtuse, the obtuse angle is at $C'$.
Consider the ratio of the new angle $A$ (angle $BAC'$), to the old one (angle $BAC$) . . .
Since angle $C'$ is obtuse, we have $|AB| > |AC'|$, hence, by the angle bisector theorem,
https://en.wikipedia.org/wiki/Angle_bisector_theorem#Theorem
the angle bisector of angle $BAC'$ meets side $BC'$ strictly between $C$ and $C'$.
It follows that the new angle $A$, while still acute, is more than twice the old one.
In triangle $T'$, the new smallest angle is either angle $BAC'$ or angle $ABC'$.
Thus, either the new smallest angle is more than double the old one, or else the new smallest angle is equal to the larger of the two acute angles of the old one.
Now iterate the process, reflecting the vertex of the largest of the two acute angles of triangle $T'$ over $C'$ to a point $C''$, and let $T''$ denote triangle $ABC''$.
As before, if $T''$ is non-obtuse, we're done.
If not, the smallest angle of $T''$ is more than twice the smallest angle of $T$.
Thus, the process can't yield obtuse triangles forever, since, as long as the triangles are still obtuse, the smallest angle more than doubles after two iterations.
It follows that $T$ is rationally realizable.
This completes the proof.
Remark:$;$With this approach, for the obtuse case, no extra coordinate is needed.
answered Jul 22 at 10:17
quasi
33.3k22359
edited Jul 23 at 7:45
answered Jul 22 at 10:17
quasi
33.3k22359
answered Jul 22 at 10:17
quasi
33.3k22359
answered Jul 22 at 10:17
quasi
33.3k22359
33.3k22359
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858090%2ftriangles-in-mathbbrn-with-all-vertices-in-mathbbqn%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
How about $a=b=c=1$?
– Henning Makholm
Jul 21 at 0:07
@Henning Makholm: For $a=b=c=1$, use the $R^6$ construction (from my post).
– quasi
Jul 21 at 0:15
Ah, I need to learn to read before posting knee-jerk reactions.
– Henning Makholm
Jul 21 at 0:39