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)?

geometry

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edited 2 days ago

M. Winter

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dasari naga vijay krishna

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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

edited 2 days ago

M. Winter

17.5k62664

asked 2 days ago

dasari naga vijay krishna

894

add a comment | 

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

edited 2 days ago

M. Winter

17.5k62664

asked 2 days ago

dasari naga vijay krishna

894

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

edited 2 days ago

M. Winter

17.5k62664

asked 2 days ago

dasari naga vijay krishna

894

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

M. Winter

17.5k62664

edited 2 days ago

M. Winter

17.5k62664

edited 2 days ago

M. Winter

17.5k62664

17.5k62664

asked 2 days ago

dasari naga vijay krishna

894

asked 2 days ago

dasari naga vijay krishna

894

asked 2 days ago

dasari naga vijay krishna

894

894

add a comment | 

add a comment | 

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

21. 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.

10. 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

edited 2 days ago

answered 2 days ago

timon92

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
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Please take a look at problem 21 in this pdf.

21. 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.

10. 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

edited 2 days ago

answered 2 days ago

timon92

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
  

  
  
  
  

add a comment | 

up vote
3
down vote

Please take a look at problem 21 in this pdf.

21. 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.

10. 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

edited 2 days ago

answered 2 days ago

timon92

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
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Please take a look at problem 21 in this pdf.

21. 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.

10. 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

edited 2 days ago

answered 2 days ago

timon92

3,6591724

Please take a look at problem 21 in this pdf.

21. 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.

10. 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

edited 2 days ago

answered 2 days ago

timon92

3,6591724

share|cite|improve this answer

share|cite|improve this answer

edited 2 days ago

edited 2 days ago

edited 2 days ago

answered 2 days ago

timon92

3,6591724

answered 2 days ago

timon92

3,6591724

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

 

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