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:



enter image description here



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:



enter image description here



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:



enter image description here



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







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    up vote
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    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:



    enter image description here



    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:



    enter image description here



    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:



    enter image description here



    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







    share|cite|improve this question























      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:



      enter image description here



      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:



      enter image description here



      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:



      enter image description here



      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







      share|cite|improve this question













      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:



      enter image description here



      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:



      enter image description here



      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:



      enter image description here



      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






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      asked 3 hours ago









      Oldboy

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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!






          share|cite|improve this answer





























            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.






            share|cite|improve this answer



























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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. $square$






              share|cite|improve this answer




















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                3 Answers
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                3 Answers
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                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!






                share|cite|improve this answer


























                  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!






                  share|cite|improve this answer
























                    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!






                    share|cite|improve this answer














                    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!







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 2 hours ago









                    Noah Schweber

                    112k9142266




                    112k9142266




















                        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.






                        share|cite|improve this answer
























                          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.






                          share|cite|improve this answer






















                            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.






                            share|cite|improve this answer












                            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.







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                            answered 2 hours ago









                            K B Dave

                            2,744216




                            2,744216




















                                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. $square$






                                share|cite|improve this answer
























                                  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. $square$






                                  share|cite|improve this answer






















                                    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. $square$






                                    share|cite|improve this answer












                                    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. $square$







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                                    answered 2 hours ago









                                    Blue

                                    44.2k868141




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                                        Installing NextGIS Connect into QGIS 3?

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                                        Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
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                                        One-line joke

                                        A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
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