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Is this the most complicated way to construct an equilateral triangle? :)
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Just for fun (inspired by sub-problem described and answered here):
Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:
Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:
Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:
Repeate the same process infinite times.
Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.
euclidean-geometry triangle
asked 3 hours ago
Oldboy
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Just for fun (inspired by sub-problem described and answered here):
Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:
Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:
Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:
Repeate the same process infinite times.
Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.
euclidean-geometry triangle
asked 3 hours ago
Oldboy
3,1931321
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Just for fun (inspired by sub-problem described and answered here):
Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:
Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:
Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:
Repeate the same process infinite times.
Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.
euclidean-geometry triangle
asked 3 hours ago
Oldboy
3,1931321
Just for fun (inspired by sub-problem described and answered here):
Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:
Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:
Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:
Repeate the same process infinite times.
Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.
euclidean-geometry triangle
euclidean-geometry triangle
asked 3 hours ago
Oldboy
3,1931321
asked 3 hours ago
Oldboy
3,1931321
asked 3 hours ago
Oldboy
3,1931321
asked 3 hours ago
Oldboy
3,1931321
asked 3 hours ago
Oldboy
3,1931321
3,1931321
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3 Answers
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Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $vert y-xvert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, x+yover 2, x+yover 2$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$leftvert yover 2-xover 2rightvert=1over 2vert y-xvert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $vert y-xvert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, $1over 2$): if $rin(-1,1)$ then for any $a$ we have
$$lim_nrightarrowinftyar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
answered 2 hours ago
Noah Schweber
112k9142266
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1
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Assume WLOG that the initial triangle is isoceles. Let $alpha$ be the apical angle, and let $beta$ be a remaining angle. Then the transformation in question sends
$$beginbmatrixalpha \ beta endbmatrixmapsto
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix
beginbmatrixalpha \ beta endbmatrixtext.$$
Let $mathsfX$ be the $2times 2$ transformation matrix on the rhs. $mathsfX$ has characteristic polynomial
$x^2-tfrac12x-tfrac12=0.$ By the Cayley–Hamilton theorem,
$$mathsfX^2=tfrac12mathsfX+tfrac12text.$$
Therefore we have a Sylvester formula
$$f(mathsfX)=f(1)left(frac1+2mathsfX3right)+f(-tfrac12)left(frac2-2mathsfX3right)$$
for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus,
$$mathsfX^n=frac1+2mathsfX3+(-tfrac12)^nleft(frac2-2mathsfX3right)text.$$
The second term converges to zero, so
$$beginsplit
lim_ntoinftymathsfX^n&=frac1+2mathsfX3\
&=frac13beginbmatrix1 & 2 \ 1 & 2endbmatrix\
&=frac13beginbmatrix 1\ 1endbmatrixbeginbmatrix1&2endbmatrixtext,
endsplit$$
$$lim_ntoinfty
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix^n
beginbmatrixalpha \ beta endbmatrix=beginbmatrixtfracalpha+2beta3\ tfracalpha+2beta3endbmatrixtext.$$
i.e., the apical and side angles approach equality as the operation is repeated.
answered 2 hours ago
K B Dave
2,744216
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By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $theta_i$, so that the base angles are $frac12(pi - theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $theta_i = frac12(pi-theta_i-1)$. Thus,
$$beginaligntheta_n &= -frac12theta_n-1 + frac12pi \[6pt]
&=frac12left(-frac12(pi-theta_n-2)+piright) = frac14theta_n-2+frac12pi-frac14pi \[6pt]
&= cdots \[6pt]
&= left(-frac12right)^ntheta_0 ;-; sum_i=1^nleft(-frac12right)^npi \[6pt]
lim_ntoinftytheta_n &= 0cdottheta_0 ;-; frac(-1/2)1-(-1/2)pi \
&=fracpi3
endalign$$
Thus, in the limit, the triangle becomes equilateral. $square$
answered 2 hours ago
Blue
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $vert y-xvert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, x+yover 2, x+yover 2$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$leftvert yover 2-xover 2rightvert=1over 2vert y-xvert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $vert y-xvert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, $1over 2$): if $rin(-1,1)$ then for any $a$ we have
$$lim_nrightarrowinftyar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
answered 2 hours ago
Noah Schweber
112k9142266
add a comment |Â
up vote
3
down vote
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $vert y-xvert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, x+yover 2, x+yover 2$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$leftvert yover 2-xover 2rightvert=1over 2vert y-xvert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $vert y-xvert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, $1over 2$): if $rin(-1,1)$ then for any $a$ we have
$$lim_nrightarrowinftyar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
answered 2 hours ago
Noah Schweber
112k9142266
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $vert y-xvert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, x+yover 2, x+yover 2$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$leftvert yover 2-xover 2rightvert=1over 2vert y-xvert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $vert y-xvert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, $1over 2$): if $rin(-1,1)$ then for any $a$ we have
$$lim_nrightarrowinftyar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
answered 2 hours ago
Noah Schweber
112k9142266
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $vert y-xvert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, x+yover 2, x+yover 2$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$leftvert yover 2-xover 2rightvert=1over 2vert y-xvert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $vert y-xvert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, $1over 2$): if $rin(-1,1)$ then for any $a$ we have
$$lim_nrightarrowinftyar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
answered 2 hours ago
Noah Schweber
112k9142266
edited 1 hour ago
answered 2 hours ago
Noah Schweber
112k9142266
answered 2 hours ago
Noah Schweber
112k9142266
answered 2 hours ago
Noah Schweber
112k9142266
112k9142266
add a comment |Â
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up vote
1
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Assume WLOG that the initial triangle is isoceles. Let $alpha$ be the apical angle, and let $beta$ be a remaining angle. Then the transformation in question sends
$$beginbmatrixalpha \ beta endbmatrixmapsto
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix
beginbmatrixalpha \ beta endbmatrixtext.$$
Let $mathsfX$ be the $2times 2$ transformation matrix on the rhs. $mathsfX$ has characteristic polynomial
$x^2-tfrac12x-tfrac12=0.$ By the Cayley–Hamilton theorem,
$$mathsfX^2=tfrac12mathsfX+tfrac12text.$$
Therefore we have a Sylvester formula
$$f(mathsfX)=f(1)left(frac1+2mathsfX3right)+f(-tfrac12)left(frac2-2mathsfX3right)$$
for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus,
$$mathsfX^n=frac1+2mathsfX3+(-tfrac12)^nleft(frac2-2mathsfX3right)text.$$
The second term converges to zero, so
$$beginsplit
lim_ntoinftymathsfX^n&=frac1+2mathsfX3\
&=frac13beginbmatrix1 & 2 \ 1 & 2endbmatrix\
&=frac13beginbmatrix 1\ 1endbmatrixbeginbmatrix1&2endbmatrixtext,
endsplit$$
$$lim_ntoinfty
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix^n
beginbmatrixalpha \ beta endbmatrix=beginbmatrixtfracalpha+2beta3\ tfracalpha+2beta3endbmatrixtext.$$
i.e., the apical and side angles approach equality as the operation is repeated.
answered 2 hours ago
K B Dave
2,744216
add a comment |Â
up vote
1
down vote
Assume WLOG that the initial triangle is isoceles. Let $alpha$ be the apical angle, and let $beta$ be a remaining angle. Then the transformation in question sends
$$beginbmatrixalpha \ beta endbmatrixmapsto
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix
beginbmatrixalpha \ beta endbmatrixtext.$$
Let $mathsfX$ be the $2times 2$ transformation matrix on the rhs. $mathsfX$ has characteristic polynomial
$x^2-tfrac12x-tfrac12=0.$ By the Cayley–Hamilton theorem,
$$mathsfX^2=tfrac12mathsfX+tfrac12text.$$
Therefore we have a Sylvester formula
$$f(mathsfX)=f(1)left(frac1+2mathsfX3right)+f(-tfrac12)left(frac2-2mathsfX3right)$$
for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus,
$$mathsfX^n=frac1+2mathsfX3+(-tfrac12)^nleft(frac2-2mathsfX3right)text.$$
The second term converges to zero, so
$$beginsplit
lim_ntoinftymathsfX^n&=frac1+2mathsfX3\
&=frac13beginbmatrix1 & 2 \ 1 & 2endbmatrix\
&=frac13beginbmatrix 1\ 1endbmatrixbeginbmatrix1&2endbmatrixtext,
endsplit$$
$$lim_ntoinfty
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix^n
beginbmatrixalpha \ beta endbmatrix=beginbmatrixtfracalpha+2beta3\ tfracalpha+2beta3endbmatrixtext.$$
i.e., the apical and side angles approach equality as the operation is repeated.
answered 2 hours ago
K B Dave
2,744216
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume WLOG that the initial triangle is isoceles. Let $alpha$ be the apical angle, and let $beta$ be a remaining angle. Then the transformation in question sends
$$beginbmatrixalpha \ beta endbmatrixmapsto
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix
beginbmatrixalpha \ beta endbmatrixtext.$$
Let $mathsfX$ be the $2times 2$ transformation matrix on the rhs. $mathsfX$ has characteristic polynomial
$x^2-tfrac12x-tfrac12=0.$ By the Cayley–Hamilton theorem,
$$mathsfX^2=tfrac12mathsfX+tfrac12text.$$
Therefore we have a Sylvester formula
$$f(mathsfX)=f(1)left(frac1+2mathsfX3right)+f(-tfrac12)left(frac2-2mathsfX3right)$$
for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus,
$$mathsfX^n=frac1+2mathsfX3+(-tfrac12)^nleft(frac2-2mathsfX3right)text.$$
The second term converges to zero, so
$$beginsplit
lim_ntoinftymathsfX^n&=frac1+2mathsfX3\
&=frac13beginbmatrix1 & 2 \ 1 & 2endbmatrix\
&=frac13beginbmatrix 1\ 1endbmatrixbeginbmatrix1&2endbmatrixtext,
endsplit$$
$$lim_ntoinfty
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix^n
beginbmatrixalpha \ beta endbmatrix=beginbmatrixtfracalpha+2beta3\ tfracalpha+2beta3endbmatrixtext.$$
i.e., the apical and side angles approach equality as the operation is repeated.
answered 2 hours ago
K B Dave
2,744216
Assume WLOG that the initial triangle is isoceles. Let $alpha$ be the apical angle, and let $beta$ be a remaining angle. Then the transformation in question sends
$$beginbmatrixalpha \ beta endbmatrixmapsto
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix
beginbmatrixalpha \ beta endbmatrixtext.$$
Let $mathsfX$ be the $2times 2$ transformation matrix on the rhs. $mathsfX$ has characteristic polynomial
$x^2-tfrac12x-tfrac12=0.$ By the Cayley–Hamilton theorem,
$$mathsfX^2=tfrac12mathsfX+tfrac12text.$$
Therefore we have a Sylvester formula
$$f(mathsfX)=f(1)left(frac1+2mathsfX3right)+f(-tfrac12)left(frac2-2mathsfX3right)$$
for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus,
$$mathsfX^n=frac1+2mathsfX3+(-tfrac12)^nleft(frac2-2mathsfX3right)text.$$
The second term converges to zero, so
$$beginsplit
lim_ntoinftymathsfX^n&=frac1+2mathsfX3\
&=frac13beginbmatrix1 & 2 \ 1 & 2endbmatrix\
&=frac13beginbmatrix 1\ 1endbmatrixbeginbmatrix1&2endbmatrixtext,
endsplit$$
$$lim_ntoinfty
beginbmatrix0 & 1 \ tfrac12 & tfrac12
endbmatrix^n
beginbmatrixalpha \ beta endbmatrix=beginbmatrixtfracalpha+2beta3\ tfracalpha+2beta3endbmatrixtext.$$
i.e., the apical and side angles approach equality as the operation is repeated.
answered 2 hours ago
K B Dave
2,744216
answered 2 hours ago
K B Dave
2,744216
answered 2 hours ago
K B Dave
2,744216
answered 2 hours ago
K B Dave
2,744216
2,744216
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add a comment |Â
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1
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By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $theta_i$, so that the base angles are $frac12(pi - theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $theta_i = frac12(pi-theta_i-1)$. Thus,
$$beginaligntheta_n &= -frac12theta_n-1 + frac12pi \[6pt]
&=frac12left(-frac12(pi-theta_n-2)+piright) = frac14theta_n-2+frac12pi-frac14pi \[6pt]
&= cdots \[6pt]
&= left(-frac12right)^ntheta_0 ;-; sum_i=1^nleft(-frac12right)^npi \[6pt]
lim_ntoinftytheta_n &= 0cdottheta_0 ;-; frac(-1/2)1-(-1/2)pi \
&=fracpi3
endalign$$
Thus, in the limit, the triangle becomes equilateral. $square$
answered 2 hours ago
Blue
44.2k868141
add a comment |Â
up vote
1
down vote
By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $theta_i$, so that the base angles are $frac12(pi - theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $theta_i = frac12(pi-theta_i-1)$. Thus,
$$beginaligntheta_n &= -frac12theta_n-1 + frac12pi \[6pt]
&=frac12left(-frac12(pi-theta_n-2)+piright) = frac14theta_n-2+frac12pi-frac14pi \[6pt]
&= cdots \[6pt]
&= left(-frac12right)^ntheta_0 ;-; sum_i=1^nleft(-frac12right)^npi \[6pt]
lim_ntoinftytheta_n &= 0cdottheta_0 ;-; frac(-1/2)1-(-1/2)pi \
&=fracpi3
endalign$$
Thus, in the limit, the triangle becomes equilateral. $square$
answered 2 hours ago
Blue
44.2k868141
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $theta_i$, so that the base angles are $frac12(pi - theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $theta_i = frac12(pi-theta_i-1)$. Thus,
$$beginaligntheta_n &= -frac12theta_n-1 + frac12pi \[6pt]
&=frac12left(-frac12(pi-theta_n-2)+piright) = frac14theta_n-2+frac12pi-frac14pi \[6pt]
&= cdots \[6pt]
&= left(-frac12right)^ntheta_0 ;-; sum_i=1^nleft(-frac12right)^npi \[6pt]
lim_ntoinftytheta_n &= 0cdottheta_0 ;-; frac(-1/2)1-(-1/2)pi \
&=fracpi3
endalign$$
Thus, in the limit, the triangle becomes equilateral. $square$
answered 2 hours ago
Blue
44.2k868141
By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $theta_i$, so that the base angles are $frac12(pi - theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $theta_i = frac12(pi-theta_i-1)$. Thus,
$$beginaligntheta_n &= -frac12theta_n-1 + frac12pi \[6pt]
&=frac12left(-frac12(pi-theta_n-2)+piright) = frac14theta_n-2+frac12pi-frac14pi \[6pt]
&= cdots \[6pt]
&= left(-frac12right)^ntheta_0 ;-; sum_i=1^nleft(-frac12right)^npi \[6pt]
lim_ntoinftytheta_n &= 0cdottheta_0 ;-; frac(-1/2)1-(-1/2)pi \
&=fracpi3
endalign$$
Thus, in the limit, the triangle becomes equilateral. $square$
answered 2 hours ago
Blue
44.2k868141
answered 2 hours ago
Blue
44.2k868141
answered 2 hours ago
Blue
44.2k868141
answered 2 hours ago
Blue
44.2k868141
44.2k868141
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