Hungary Mathematical Competition 2001

[MATHPRITOM]

Points A,B,C,D lie on the line l in that order.Find the locus of points P in the plane for which $ \angle APB=\angle CPD $.

[nafistiham]

let us found a point $K$ in $l$ such that, it lies between $B,C$ such that,
\[\frac {CD}{AB}= \frac {BK}{CK}\]
$P$ will be any point in the perpendicular line to $l$ standing on $K$

[sourav das]

Nice Problem.
Solution:

Take point $P$ on the locus,
Using sine law:
$\frac{AB}{BD}=\frac{sin \angle APB}{sin \angle BPD} \frac{PA}{PD}$
Same way, $\frac{AC}{CD}=\frac{sin \angle DPB}{sin \angle BPA} \frac{PA}{PD}$
So $\frac{PA^2}{PD^2}=\frac{AB}{BD}\frac{AC}{CD}=k^2$; $k$ is a constant ratio that is known as $A,B,C,D$ are known. So $\frac{PA}{PD}=k$. If we take 2 points $M,M'$ ON l so that $\frac{MA}{MD}=-\frac{M'A}{M'D}=k$ it follows from Appolonious theorem that the locus will be a circle with diameter $MM'$ and center midpoint of $MM'$.

EDITED. SORRY FOR THE TYPO. THANKS TO *Mahi*

[sourav das]

let us found a point $K$ in $l$ such that, it lies between $B,C$ such that,
\[\frac {CD}{AB}= \frac {BK}{CK}\]
$P$ will be any point in the perpendicular line to $l$ standing on $K$

A Proof would be much better...

[*Mahi*]

Nice Problem.
Solution:

Take point $P$ on the locus,
Using sine law:
$\frac{AB}{BD}=\frac{sin \angle APB}{sin \angle BPD} \frac{PA}{PD}$
Same way, $\frac{AC}{CD}=\frac{sin \angle DPB}{sin \angle BPA} \frac{PA}{PD}$
So $\frac{PA^2}{PD^2}=\frac{AB}{BD}\frac{AC}{CD}=k^2$; $k$ is a constant ratio that is known as $A,B,C,D$ are known. So $\frac{PA}{PD}=k$. If we take 2 points $M,M'$ so that $\frac{MA}{MD}=-\frac{M'A}{M'D}=k$ it follows from Appolonious theorem that the locus will be a circle with diameter $MM'$ and center midpoint of $MM'$.

And , of course , the segment $BC$.