Line diving 2 triangles in a plane in equal halves by area











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Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?










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  • Triangles do not overlap, right?
    – Narasimham
    Nov 15 at 9:12






  • 2




    This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
    – Robert Z
    Nov 15 at 9:20












  • Very similar question: math.stackexchange.com/questions/1452917/…
    – Robert Z
    Nov 15 at 9:28












  • @Robert Z It is more than very similar...
    – Jean Marie
    Nov 15 at 21:52















up vote
2
down vote

favorite
3












Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?










share|cite|improve this question






















  • Triangles do not overlap, right?
    – Narasimham
    Nov 15 at 9:12






  • 2




    This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
    – Robert Z
    Nov 15 at 9:20












  • Very similar question: math.stackexchange.com/questions/1452917/…
    – Robert Z
    Nov 15 at 9:28












  • @Robert Z It is more than very similar...
    – Jean Marie
    Nov 15 at 21:52













up vote
2
down vote

favorite
3









up vote
2
down vote

favorite
3






3





Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?










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Can you prove or disprove that for any 2 triangles in a plane, there will always be a line passing through them both which divide each of them into 2 halves of equal area? If said line exists, how can we find it?







geometry






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asked Nov 15 at 9:06









Aashish Rathi

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142












  • Triangles do not overlap, right?
    – Narasimham
    Nov 15 at 9:12






  • 2




    This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
    – Robert Z
    Nov 15 at 9:20












  • Very similar question: math.stackexchange.com/questions/1452917/…
    – Robert Z
    Nov 15 at 9:28












  • @Robert Z It is more than very similar...
    – Jean Marie
    Nov 15 at 21:52


















  • Triangles do not overlap, right?
    – Narasimham
    Nov 15 at 9:12






  • 2




    This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
    – Robert Z
    Nov 15 at 9:20












  • Very similar question: math.stackexchange.com/questions/1452917/…
    – Robert Z
    Nov 15 at 9:28












  • @Robert Z It is more than very similar...
    – Jean Marie
    Nov 15 at 21:52
















Triangles do not overlap, right?
– Narasimham
Nov 15 at 9:12




Triangles do not overlap, right?
– Narasimham
Nov 15 at 9:12




2




2




This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
– Robert Z
Nov 15 at 9:20






This is an application in dimension $2$ of en.wikipedia.org/wiki/Ham_sandwich_theorem
– Robert Z
Nov 15 at 9:20














Very similar question: math.stackexchange.com/questions/1452917/…
– Robert Z
Nov 15 at 9:28






Very similar question: math.stackexchange.com/questions/1452917/…
– Robert Z
Nov 15 at 9:28














@Robert Z It is more than very similar...
– Jean Marie
Nov 15 at 21:52




@Robert Z It is more than very similar...
– Jean Marie
Nov 15 at 21:52










1 Answer
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Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.



We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
$${xover u}+{yover v}=1 ,$$
whereby $u,vin>]0,1[>$ and $uv={1over2}$. The family of these lines can therefore be described in the form
$$2v^2x+y-v=0,qquad0<v<1 .$$
The envelope of this family turns out to be an arc $gamma$ of the hyperbola $xy={1over8}$, whereby the endpoints of this arc are the midpoints $bigl({1over2},{1over4}bigr)$ and $bigl({1over4},{1over2}bigr)$ of the medians through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $ell$ smoothly $180^circ$ along these three arcs so that $ell$ touches (at least) one of the arcs at all times.



enter image description here



If we now are given two arbitrary triangles $T_1$, $T_2$ then each of them has its own "hyperbolic triangle" $H_i$. The rotating $ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of $H_2$ (one would have to check this). In this position the line $ell$ halves both triangle areas simultaneously.






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    Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.



    We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
    $${xover u}+{yover v}=1 ,$$
    whereby $u,vin>]0,1[>$ and $uv={1over2}$. The family of these lines can therefore be described in the form
    $$2v^2x+y-v=0,qquad0<v<1 .$$
    The envelope of this family turns out to be an arc $gamma$ of the hyperbola $xy={1over8}$, whereby the endpoints of this arc are the midpoints $bigl({1over2},{1over4}bigr)$ and $bigl({1over4},{1over2}bigr)$ of the medians through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $ell$ smoothly $180^circ$ along these three arcs so that $ell$ touches (at least) one of the arcs at all times.



    enter image description here



    If we now are given two arbitrary triangles $T_1$, $T_2$ then each of them has its own "hyperbolic triangle" $H_i$. The rotating $ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of $H_2$ (one would have to check this). In this position the line $ell$ halves both triangle areas simultaneously.






    share|cite|improve this answer



























      up vote
      1
      down vote













      Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.



      We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
      $${xover u}+{yover v}=1 ,$$
      whereby $u,vin>]0,1[>$ and $uv={1over2}$. The family of these lines can therefore be described in the form
      $$2v^2x+y-v=0,qquad0<v<1 .$$
      The envelope of this family turns out to be an arc $gamma$ of the hyperbola $xy={1over8}$, whereby the endpoints of this arc are the midpoints $bigl({1over2},{1over4}bigr)$ and $bigl({1over4},{1over2}bigr)$ of the medians through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $ell$ smoothly $180^circ$ along these three arcs so that $ell$ touches (at least) one of the arcs at all times.



      enter image description here



      If we now are given two arbitrary triangles $T_1$, $T_2$ then each of them has its own "hyperbolic triangle" $H_i$. The rotating $ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of $H_2$ (one would have to check this). In this position the line $ell$ halves both triangle areas simultaneously.






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.



        We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
        $${xover u}+{yover v}=1 ,$$
        whereby $u,vin>]0,1[>$ and $uv={1over2}$. The family of these lines can therefore be described in the form
        $$2v^2x+y-v=0,qquad0<v<1 .$$
        The envelope of this family turns out to be an arc $gamma$ of the hyperbola $xy={1over8}$, whereby the endpoints of this arc are the midpoints $bigl({1over2},{1over4}bigr)$ and $bigl({1over4},{1over2}bigr)$ of the medians through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $ell$ smoothly $180^circ$ along these three arcs so that $ell$ touches (at least) one of the arcs at all times.



        enter image description here



        If we now are given two arbitrary triangles $T_1$, $T_2$ then each of them has its own "hyperbolic triangle" $H_i$. The rotating $ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of $H_2$ (one would have to check this). In this position the line $ell$ halves both triangle areas simultaneously.






        share|cite|improve this answer














        Of course one can make appeal to the ham-sandwich-theorem. But for two triangles it should be possible to obtain, at least in principle, an explicit solution.



        We first have to describe the set of lines dividing a single triangle into two parts of equal area. To this end consider the triangle $T$ with vertices $(0,0)$, $(1,0)$ and $(0,1)$. A line cutting off the origin together with half the area of this triangle has an equation of the form
        $${xover u}+{yover v}=1 ,$$
        whereby $u,vin>]0,1[>$ and $uv={1over2}$. The family of these lines can therefore be described in the form
        $$2v^2x+y-v=0,qquad0<v<1 .$$
        The envelope of this family turns out to be an arc $gamma$ of the hyperbola $xy={1over8}$, whereby the endpoints of this arc are the midpoints $bigl({1over2},{1over4}bigr)$ and $bigl({1over4},{1over2}bigr)$ of the medians through the vertices $(1,0)$ and $(0,1)$. Of course these medians themselves are tangent to the arc $gamma$ in these points, since they are halving the area of $T$. By affinity it is not necessary to do a similar computation for cutting off another vertex of $T$: In all we obtain three arcs of hyperbolas that form a "hyperbolic triangle" with cusps in the midpoints of the medians of $T$. It is possible to rotate a line $ell$ smoothly $180^circ$ along these three arcs so that $ell$ touches (at least) one of the arcs at all times.



        enter image description here



        If we now are given two arbitrary triangles $T_1$, $T_2$ then each of them has its own "hyperbolic triangle" $H_i$. The rotating $ell$ of $H_1$ will then at a certain moment be tangent to one of the arcs of $H_2$ (one would have to check this). In this position the line $ell$ halves both triangle areas simultaneously.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 17 at 14:27

























        answered Nov 15 at 11:04









        Christian Blatter

        170k7111325




        170k7111325






























             

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