BdMO National 2013: Secondary 2, Higher Secondary 1
BdMO National 2013: Secondary 2, Higher Secondary 1
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. $PQCD$, $BD$ is perpendicular to both $AB$ and $DE$. Prove that $ABCDE$ is a degenerate pentagon.
 Fatin Farhan
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1
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Re: BdMO National 2013: Secondary 2, Higher Secondary 1
$\angle ABE=\angle BED$ does not necessarily imply that $BCDE$ is a parallelogram. Actually, $BCDE$ can be a Parallelogram, OR, an IsoscelesTrapezium.Fatin Farhan wrote:In the quadrilateral $$BCDE$$, $$CDBE, \angle ABE=\angle BED$$.
So $$BCDE$$ is a parallelogram.
**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 nondegenerate, whereas, $ABC'DE$ is degenerate.
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 Pentagons Not So Degenerate.png (45.23 KiB) Viewed 2459 times
Re: BdMO National 2013: Secondary 2, Higher Secondary 1
Yep, there actually is one. This question statement was false, even if it was given as a no. 1 in higher secondary.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.
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