locus of center parallelogram

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locus of center parallelogram

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Source: 31-th Vietnamese Mathematical Olympiad 1993

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#1 h Feb 17, 2007, 3:57 PM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is a quadrilateral such that $AB$ is not parallel to $CD$ , and $BC$ is not parallel to $AD$ . Variable points $P, Q, R, S$ are taken on $AB, BC, CD, DA$ respectively so that $PQRS$ is a parallelogram. Find the locus of its center.

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#2 h Dec 5, 2008, 9:08 AM • 2 Y

Y by Adventure10, Mango247

It is easy to construct a parallelogram $PQRS$ as the problem states, beginning from an arbitrary point $P$ on the side-segment $AB.$

Through the point $P,$ we draw two lines parallel to $AC,\ BD,$ which intersect the side-segments $BC,\ AD$ of the given quadrilateral $ABCD,$ at points $Q,\ S,$ respectively.

If now, $R$ is the point on $CD,$ such that $QR\parallel BD\parallel PS,$ it is easy to show that $RS\parallel AC\parallel PQ,$ from $\frac {AS}{SD} = \frac {AP}{PB} = \frac {CQ}{QB} = \frac {CR}{RD}.$

$\bullet$ In our problem now, if we consider the non-convex quadrilateral $BCAD,$ it is easy to show that the midpoint $M$ of $QS,$ as the center of the parallelogram $PQRS,$ lies on the segment $XY,$ where $X,\ Y$ are the midpoints of $AC,\ BD$ respectively,

as a direct application of the ERIQ theorem ( = Equal Ratios In Quadrilateral ), because of $\frac {BQ}{QC} = \frac {DS}{SA}$ and $\frac {BY}{YD} = \frac {QM}{MS} = \frac {CX}{XA}.$

Hence, the wanted locus is the segment $XY$ and the solution is completed.

Kostas Vittas.

PS. When the point $P$ is taken outwardly to the side-segment $AB$ then, all the vertices of the parallelogram $PQRS,$ are also outwardly to the side-segments of $ABCD$ and in this general case, we can say that the locus of $M$ is the line segment $XY.$

Attachments:

t=134116.pdf (5kb)

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#3 h Jan 2, 2009, 1:31 PM • 2 Y

Y by Adventure10, Mango247

vittasko wrote:

PS. When the point $P$ is taken outwardly to the side-segment $AB$ then, all the vertices of the parallelogram $PQRS,$ are also outwardly to the side-segments of $ABCD$ and in this general case, we can say that the locus of $M$ is the line segment $XY.$

See the schema t=134116(a).

But, we can easy to construct a parallelogram $PQRS$ as the problem states, without $PQ\parallel AC\parallel RS$ and $QR\parallel BD\parallel PS,$ as it seems in the schema t=134116(b).

In this general case, the problem has a meaning only when the points $P,\ Q,\ R,\ S,$ lie on the side-segments of $ABCD$ and then, the wanted locus is the area of the parallelogram $XX'YY'$ with $XX'\parallel YY'\parallel CD$ and $XY'\parallel X'Y\parallel AD,$ including its side-segments.

I am surprised because such that problem was a contest problem and I wonder about its official solution.

Kostas Vittas.

Attachments:

t=134116(b).pdf (4kb)

t=134116(a).pdf (6kb)

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