1996 Paper 2 Q7

Year: 1996
Paper: 2
Question Number: 7

Course: LFM Pure
Section: Coordinate Geometry

Difficulty: 1600.0 Banger: 1484.0

coordinate geometry locus circle perpendicular distance square geometric proof straight lines

Problem

Consider a fixed square \(ABCD\) and a variable point \(P\) in the plane of the square. We write the perpendicular distance from \(P\) to \(AB\) as \(p\), from \(P\) to \(BC\) as \(q\), from \(P\) to \(CD\) as \(r\) and from \(P\) to \(DA\) as \(s\). (Remember that distance is never negative, so \(p,q,r,s\geqslant 0\).) If \(pr=qs\), show that the only possible positions of \(P\) lie on two straight lines and a circle and that every point on these two lines and a circle is indeed a possible position of \(P\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source

Consider a fixed square $ABCD$ and a variable
point $P$ in the plane of the square. We write the perpendicular
distance from $P$ to $AB$ as $p$, from $P$ to $BC$ as $q$,
from $P$ to $CD$ as $r$ and from $P$ to $DA$ as $s$.
(Remember that distance is never negative, so $p,q,r,s\geqslant 0$.)
If $pr=qs$, show that the only possible positions of $P$
lie on two straight lines and a circle and that every point
on these two lines and a circle is indeed a possible position
of $P$.

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