Tom's Mysterious Quadrilateral [Self-Made]
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Phlembac Adib Hasan
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One day Tom decided to have some fun with geometry. So he drew a quadrilateral $ABCD$. Suddenly he noticed that his quadrilateral has a very nice property: $AB\cdot CD=BC\cdot AD$,$AB\cdot BC=CD\cdot AD$.It made him curious and he marked mid-points of side $AB,BC,CD,AD$ as $E,F,G,H$,respectively. \Tom also marked the intersection point of $EG$ and $FH$ as $O$.Tom became surprised seeing that $OE=OF$.If $OE=7cm$,can you find the perimeter of $ABCD$?
Last edited by Phlembac Adib Hasan on Sat Dec 17, 2011 5:06 pm, edited 1 time in total.
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nafistiham
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Re: Tom's Mysterious Quadrilateral [Self-Made]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Tom's Mysterious Quadrilateral [Self-Made]
Not Maybe...
It is 56.
It's not that hard to prove that $EFGH$ is a Rectangle.
Thus $ABCD$ is a rhombus.
$AB$ has the length of $FH=14.$
Thus perimeter of $ABCD$ is $56$.
It is 56.
It's not that hard to prove that $EFGH$ is a Rectangle.
Thus $ABCD$ is a rhombus.
$AB$ has the length of $FH=14.$
Thus perimeter of $ABCD$ is $56$.
Last edited by Labib on Sun Dec 18, 2011 12:15 am, edited 2 times in total.
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nafistiham
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Re: Tom's Mysterious Quadrilateral [Self-Made]
it can be rhombus,too
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Tom's Mysterious Quadrilateral [Self-Made]
It seems to me that there is too much information!
Do not forget that EFGH is a parallelogram and given the condition OE=OF , a rectangle. So ABCD must be a rhombus and O its centroid. Then it directly follows that ABCD has perimeter 56, an it doesn't even needs the facts AB*CD=BC*AD,AB*BC=CD*AD.
Do not forget that EFGH is a parallelogram and given the condition OE=OF , a rectangle. So ABCD must be a rhombus and O its centroid. Then it directly follows that ABCD has perimeter 56, an it doesn't even needs the facts AB*CD=BC*AD,AB*BC=CD*AD.
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nafistiham
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Re: Tom's Mysterious Quadrilateral [Self-Made]
too much information is to confuse people. 
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Phlembac Adib Hasan
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Re: Tom's Mysterious Quadrilateral [Self-Made]
Nope, I think. $EFGH$ is a rectangle. But it implies $ABCD$ a kite, not necessarily a rhombus.
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