BdMO National 2013: Secondary 2, Higher Secondary 1

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BdMO
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BdMO National 2013: Secondary 2, Higher Secondary 1

Unread post by BdMO » Fri Jan 10, 2014 1:33 am

A polygon is called degenerate if one of its vertices falls on a line that joins its neighboring two vertices. In a pentagon $ABCDE$, $AB=AE$, $BC=DE$, $P$ and $Q$ are midpoints of $AE$ and $AB$ respectively. $PQ||CD$, $BD$ is perpendicular to both $AB$ and $DE$. Prove that $ABCDE$ is a degenerate pentagon.

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Fatin Farhan
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

Unread post by Fatin Farhan » Sun Jan 12, 2014 5:32 pm

As $$P$$ and $$Q$$ are the midpoints of $$AB$$ and $$AE$$, $$AB=AE$$.
$$\therefore AP=AQ, PQ||BE$$.
But $$PQ||CD$$.
$$\therefore BE||CD$$.
$$\therefore \angle APQ=\angle AQP=\angle ABE=\angle AEB=\angle ACD$$.
$$\angle ABE+ \angle EBD= 90^\circ= \angle BED + \angle EBD =>\angle ABE=\angle BED $$
In the quadrilateral $$BCDE$$, $$CD||BE, \angle ABE=\angle BED$$.
So $$BCDE$$ is a parallelogram.
$$\angle EDB=\angle DBC=90^\circ$$
Now$$\angle ABD+\angle CBD=180^\circ$$
Thus $$B$$ falls on a line that joins its neighboring two vertices $$A,C$$
So, $$ABCDE$$ is a degenerate pentagon.
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sowmitra
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

Unread post by sowmitra » Sun Jan 12, 2014 6:21 pm

Fatin Farhan wrote:In the quadrilateral $$BCDE$$, $$CD||BE, \angle ABE=\angle BED$$.
So $$BCDE$$ is a parallelogram.
$\angle ABE=\angle BED$ does not necessarily imply that $BCDE$ is a parallelogram. Actually, $BCDE$ can be a Parallelogram, OR, an Isosceles-Trapezium.

**Note: This is actually a hole in the problem, as there aren't any additional conditions to determine whether $BCDE$ is a Parallelogram or an Isosceles Trapezium. So, the given construction doesn't always produce a degenerate pentagon.
As you can see below, both the pentagons $ABCDE$ and $ABC'DE$ satisfy all the given conditions. But, $ABCDE$ is non-degenerate, whereas, $ABC'DE$ is degenerate.
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Pentagons- Not So Degenerate.png
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*Mahi*
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1

Unread post by *Mahi* » Sun Jan 12, 2014 6:37 pm

sowmitra wrote: **Note: This is actually a hole in the problem, as there aren't any additional conditions to determine whether $BCDE$ is a Parallelogram or an Isosceles Trapezium. So, the given construction doesn't always produce a degenerate pentagon.
Yep, there actually is one. This question statement was false, even if it was given as a no. 1 in higher secondary.
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