Mixing
A Quadrilateral (The National Mathematical Olympiad in Bulgaria)
$begingroup$
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
edited Feb 5 at 8:20
Anirban Niloy
8541319
asked Feb 2 at 16:10
Nikol DimitrovaNikol Dimitrova
337
$endgroup$
add a comment |
$begingroup$
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
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 |
$begingroup$
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
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
edited Feb 5 at 8:20
Anirban Niloy
8541319
asked Feb 2 at 16:10
Nikol DimitrovaNikol Dimitrova
337
edited Feb 5 at 8:20
Anirban Niloy
8541319
asked Feb 2 at 16:10
Nikol DimitrovaNikol Dimitrova
337
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
$begingroup$
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 $$QK=9.$$
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 |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$QK=9.$$
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 |
$begingroup$
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 $$QK=9.$$
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 |
$begingroup$
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 $$QK=9.$$
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 $$QK=9.$$
edited Feb 3 at 20:23
the_fox
2,90231538
answered Feb 3 at 20:16
Michael RozenbergMichael Rozenberg
111k1896201
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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$begingroup$
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