Necessary/sufficient conditions for certain line arrangement in hexagon

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Are there any necessary or sufficient condition for when in a general convex hexagon, the lines formed by joining the midpoints of opposite sides are concurrent (intersect in a common point)?




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    Are there any necessary or sufficient condition for when in a general convex hexagon, the lines formed by joining the midpoints of opposite sides are concurrent (intersect in a common point)?




    geometry




    share|cite|improve this question























      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      Are there any necessary or sufficient condition for when in a general convex hexagon, the lines formed by joining the midpoints of opposite sides are concurrent (intersect in a common point)?




      geometry




      share|cite|improve this question













      Are there any necessary or sufficient condition for when in a general convex hexagon, the lines formed by joining the midpoints of opposite sides are concurrent (intersect in a common point)?





      geometry





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      M. Winter

      17.5k62664




      17.5k62664









      asked 2 days ago









      dasari naga vijay krishna

      894




      894




















          1 Answer
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          Please take a look at problem 21 in this pdf.




          1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.



          If I recall correctly, the equality of these areas is also a necessary condition.



          The proof uses problem 10 from the same pdf.




          1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.



          Now, let me sketch a proof of 21.



          Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[mathcal F]$ to denote the area of $mathcal F$. Then
          $$[KLS] = frac 14 [ACE] = frac 14 [BDF] = [SNO].$$
          Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.



          So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.






          share|cite|improve this answer























          • Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
            – Blue
            2 days ago











          • Pls can u give proof of this statement,
            – dasari naga vijay krishna
            2 days ago











          • @dasarinagavijaykrishna I have edited my post.
            – timon92
            2 days ago










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          up vote
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          Please take a look at problem 21 in this pdf.




          1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.



          If I recall correctly, the equality of these areas is also a necessary condition.



          The proof uses problem 10 from the same pdf.




          1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.



          Now, let me sketch a proof of 21.



          Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[mathcal F]$ to denote the area of $mathcal F$. Then
          $$[KLS] = frac 14 [ACE] = frac 14 [BDF] = [SNO].$$
          Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.



          So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.






          share|cite|improve this answer























          • Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
            – Blue
            2 days ago











          • Pls can u give proof of this statement,
            – dasari naga vijay krishna
            2 days ago











          • @dasarinagavijaykrishna I have edited my post.
            – timon92
            2 days ago














          up vote
          3
          down vote













          Please take a look at problem 21 in this pdf.




          1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.



          If I recall correctly, the equality of these areas is also a necessary condition.



          The proof uses problem 10 from the same pdf.




          1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.



          Now, let me sketch a proof of 21.



          Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[mathcal F]$ to denote the area of $mathcal F$. Then
          $$[KLS] = frac 14 [ACE] = frac 14 [BDF] = [SNO].$$
          Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.



          So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.






          share|cite|improve this answer























          • Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
            – Blue
            2 days ago











          • Pls can u give proof of this statement,
            – dasari naga vijay krishna
            2 days ago











          • @dasarinagavijaykrishna I have edited my post.
            – timon92
            2 days ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          Please take a look at problem 21 in this pdf.




          1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.



          If I recall correctly, the equality of these areas is also a necessary condition.



          The proof uses problem 10 from the same pdf.




          1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.



          Now, let me sketch a proof of 21.



          Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[mathcal F]$ to denote the area of $mathcal F$. Then
          $$[KLS] = frac 14 [ACE] = frac 14 [BDF] = [SNO].$$
          Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.



          So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.






          share|cite|improve this answer















          Please take a look at problem 21 in this pdf.




          1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.



          If I recall correctly, the equality of these areas is also a necessary condition.



          The proof uses problem 10 from the same pdf.




          1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.



          Now, let me sketch a proof of 21.



          Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[mathcal F]$ to denote the area of $mathcal F$. Then
          $$[KLS] = frac 14 [ACE] = frac 14 [BDF] = [SNO].$$
          Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.



          So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago


























          answered 2 days ago









          timon92

          3,6591724




          3,6591724











          • Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
            – Blue
            2 days ago











          • Pls can u give proof of this statement,
            – dasari naga vijay krishna
            2 days ago











          • @dasarinagavijaykrishna I have edited my post.
            – timon92
            2 days ago
















          • Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
            – Blue
            2 days ago











          • Pls can u give proof of this statement,
            – dasari naga vijay krishna
            2 days ago











          • @dasarinagavijaykrishna I have edited my post.
            – timon92
            2 days ago















          Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
          – Blue
          2 days ago





          Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution.
          – Blue
          2 days ago













          Pls can u give proof of this statement,
          – dasari naga vijay krishna
          2 days ago





          Pls can u give proof of this statement,
          – dasari naga vijay krishna
          2 days ago













          @dasarinagavijaykrishna I have edited my post.
          – timon92
          2 days ago




          @dasarinagavijaykrishna I have edited my post.
          – timon92
          2 days ago












           

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