A Quadrilateral (The National Mathematical Olympiad in Bulgaria)

2

$begingroup$

Problem: We have the quadrilateral $ABCD$. The middles of $AB$, $BC$, $CD$ and $DA$, are respectively $M$, $N$, $P$ and $Q$. The centroid of $BNP$ is $F$, and the centroid of $NPD$ is $G$. $MG$ intersects $FQ$ at $K$. We know that $FK = 6$ and we have to show $KQ = 9$.

I made а diagram, but I need some help.

geometry contest-math euclidean-geometry quadrilateral

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The algebra of vectors helps.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you be more specific?
  $endgroup$
  – Nikol Dimitrova
  Feb 2 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'll show it In evening. Just I am very busy now.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 16:25
  

  
  
  
  

add a comment | 

2

$begingroup$

Problem: We have the quadrilateral $ABCD$. The middles of $AB$, $BC$, $CD$ and $DA$, are respectively $M$, $N$, $P$ and $Q$. The centroid of $BNP$ is $F$, and the centroid of $NPD$ is $G$. $MG$ intersects $FQ$ at $K$. We know that $FK = 6$ and we have to show $KQ = 9$.

I made а diagram, but I need some help.

geometry contest-math euclidean-geometry quadrilateral

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The algebra of vectors helps.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you be more specific?
  $endgroup$
  – Nikol Dimitrova
  Feb 2 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'll show it In evening. Just I am very busy now.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 16:25
  

  
  
  
  

add a comment | 

2

2

2

1

$begingroup$

Problem: We have the quadrilateral $ABCD$. The middles of $AB$, $BC$, $CD$ and $DA$, are respectively $M$, $N$, $P$ and $Q$. The centroid of $BNP$ is $F$, and the centroid of $NPD$ is $G$. $MG$ intersects $FQ$ at $K$. We know that $FK = 6$ and we have to show $KQ = 9$.

I made а diagram, but I need some help.

geometry contest-math euclidean-geometry quadrilateral

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

$endgroup$

Problem: We have the quadrilateral $ABCD$. The middles of $AB$, $BC$, $CD$ and $DA$, are respectively $M$, $N$, $P$ and $Q$. The centroid of $BNP$ is $F$, and the centroid of $NPD$ is $G$. $MG$ intersects $FQ$ at $K$. We know that $FK = 6$ and we have to show $KQ = 9$.

I made а diagram, but I need some help.

geometry contest-math euclidean-geometry quadrilateral

geometry contest-math euclidean-geometry quadrilateral

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

share|cite|improve this question

share|cite|improve this question

edited Feb 5 at 8:20

Anirban Niloy

8541319

edited Feb 5 at 8:20

Anirban Niloy

8541319

edited Feb 5 at 8:20

Anirban Niloy

8541319

8541319

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

asked Feb 2 at 16:10

Nikol DimitrovaNikol Dimitrova

337

337

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The algebra of vectors helps.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you be more specific?
  $endgroup$
  – Nikol Dimitrova
  Feb 2 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'll show it In evening. Just I am very busy now.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 16:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The algebra of vectors helps.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you be more specific?
  $endgroup$
  – Nikol Dimitrova
  Feb 2 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'll show it In evening. Just I am very busy now.
  $endgroup$
  – Michael Rozenberg
  Feb 2 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 16:25
  

  
  
  
  

$begingroup$
The algebra of vectors helps.
$endgroup$
– Michael Rozenberg
Feb 2 at 16:38

$begingroup$
The algebra of vectors helps.
$endgroup$
– Michael Rozenberg
Feb 2 at 16:38

$begingroup$
@MichaelRozenberg, can you be more specific?
$endgroup$
– Nikol Dimitrova
Feb 2 at 16:52

$begingroup$
@MichaelRozenberg, can you be more specific?
$endgroup$
– Nikol Dimitrova
Feb 2 at 16:52

$begingroup$
I'll show it In evening. Just I am very busy now.
$endgroup$
– Michael Rozenberg
Feb 2 at 17:05

$begingroup$
I'll show it In evening. Just I am very busy now.
$endgroup$
– Michael Rozenberg
Feb 2 at 17:05

$begingroup$
@MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
$endgroup$
– Nikol Dimitrova
Feb 3 at 16:25

$begingroup$
@MichaelRozenberg, can you show it now because I don't understand how to prove that KQ = 9. Thank you in advance!
$endgroup$
– Nikol Dimitrova
Feb 3 at 16:25

add a comment | 

1 Answer
1

active

oldest

votes

2

$begingroup$

Since $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=frac{1}{3}DB.$$
On the other hand, $$QM||DB$$ and $$QM=frac{1}{2}DB.$$

Thus, $$GF||QM$$ and $$GF=frac{2}{3}QM.$$
Id est, $$frac{FK}{QK}=frac{GF}{QM}=frac{2}{3} $$ or
$$frac{6}{QK}=frac{2}{3},$$ which gives $$QK=9.$$

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 21:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nikol It's the Thales's theorem.
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't studied it. I am going to read about it. Thank you again for your time!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 22:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:13
  

  
  
  
  

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

2

$begingroup$

Since $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=frac{1}{3}DB.$$
On the other hand, $$QM||DB$$ and $$QM=frac{1}{2}DB.$$

Thus, $$GF||QM$$ and $$GF=frac{2}{3}QM.$$
Id est, $$frac{FK}{QK}=frac{GF}{QM}=frac{2}{3} $$ or
$$frac{6}{QK}=frac{2}{3},$$ which gives $$QK=9.$$

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 21:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nikol It's the Thales's theorem.
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't studied it. I am going to read about it. Thank you again for your time!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 22:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:13
  

  
  
  
  

add a comment | 

2

$begingroup$

Since $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=frac{1}{3}DB.$$
On the other hand, $$QM||DB$$ and $$QM=frac{1}{2}DB.$$

Thus, $$GF||QM$$ and $$GF=frac{2}{3}QM.$$
Id est, $$frac{FK}{QK}=frac{GF}{QM}=frac{2}{3} $$ or
$$frac{6}{QK}=frac{2}{3},$$ which gives $$QK=9.$$

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 21:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nikol It's the Thales's theorem.
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't studied it. I am going to read about it. Thank you again for your time!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 22:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:13
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Since $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=frac{1}{3}DB.$$
On the other hand, $$QM||DB$$ and $$QM=frac{1}{2}DB.$$

Thus, $$GF||QM$$ and $$GF=frac{2}{3}QM.$$
Id est, $$frac{FK}{QK}=frac{GF}{QM}=frac{2}{3} $$ or
$$frac{6}{QK}=frac{2}{3},$$ which gives $$QK=9.$$

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

Since $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB},$$ we obtain $$GF||DB$$ and $$GF=frac{1}{3}DB.$$
On the other hand, $$QM||DB$$ and $$QM=frac{1}{2}DB.$$

Thus, $$GF||QM$$ and $$GF=frac{2}{3}QM.$$
Id est, $$frac{FK}{QK}=frac{GF}{QM}=frac{2}{3} $$ or
$$frac{6}{QK}=frac{2}{3},$$ which gives $$QK=9.$$

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

share|cite|improve this answer

share|cite|improve this answer

edited Feb 3 at 20:23

the_fox

2,90231538

edited Feb 3 at 20:23

the_fox

2,90231538

edited Feb 3 at 20:23

the_fox

2,90231538

2,90231538

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

answered Feb 3 at 20:16

Michael RozenbergMichael Rozenberg

111k1896201

111k1896201

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 21:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nikol It's the Thales's theorem.
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't studied it. I am going to read about it. Thank you again for your time!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 22:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 21:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nikol It's the Thales's theorem.
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We haven't studied it. I am going to read about it. Thank you again for your time!
  $endgroup$
  – Nikol Dimitrova
  Feb 3 at 22:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – Michael Rozenberg
  Feb 3 at 22:13
  

  
  
  
  

1

1

$begingroup$
Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
$endgroup$
– Nikol Dimitrova
Feb 3 at 21:59

$begingroup$
Much Appreciate to your help. But why when we know this: $$frac{GT}{DT}=frac{1}{3}=frac{FT}{TB}$$ => GF||DB?
$endgroup$
– Nikol Dimitrova
Feb 3 at 21:59

$begingroup$
@Nikol It's the Thales's theorem.
$endgroup$
– Michael Rozenberg
Feb 3 at 22:08

$begingroup$
@Nikol It's the Thales's theorem.
$endgroup$
– Michael Rozenberg
Feb 3 at 22:08

$begingroup$
We haven't studied it. I am going to read about it. Thank you again for your time!
$endgroup$
– Nikol Dimitrova
Feb 3 at 22:12

$begingroup$
We haven't studied it. I am going to read about it. Thank you again for your time!
$endgroup$
– Nikol Dimitrova
Feb 3 at 22:12

$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Feb 3 at 22:13

$begingroup$
You are welcome!
$endgroup$
– Michael Rozenberg
Feb 3 at 22:13

add a comment | 

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But avoid …

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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Use MathJax to format equations. MathJax reference.

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