Two nonperpendicular lines throught the point $A$ and a point $F$ on one of these lines different from $A$ are given. Let $P_{G}$ be the intersection point of tangent lines at $G$ and $F$ to the circle through the point $A$, $F$ and $G$ where $G$ is a point on the given line different from the line $FA$. What is the locus of $P_{G}$ as $G$ varies.
Problem
Source: Turkey NMO 2001 Problem 5
Tags: trigonometry, geometry, circumcircle, perpendicular bisector, geometric transformation, geometry proposed
Luis González
30.09.2011 18:22
Let the perpendicular bisector of $\overline{AF}$ cut $AG$ at $E.$ The isosceles triangles $\triangle EAF$ and $\triangle P_GGF$ are similar due to $\angle EFA=\angle EAF=\angle P_GGF=\angle P_GFG.$ Therefore, $\angle AEF=\angle GP_GF$ $\Longrightarrow$ The points $E,F,G,P_G$ are concyclic $\Longrightarrow$ $\angle FEP_G=\angle FGP_G=\angle EFA$ $\Longrightarrow$ $EP_G \parallel AF,$ i.e. Locus of $P_G$ is the parallel line to $AF$ through $E.$
Number1
02.10.2011 19:16
Let $M$ be midpoint of $GF$. Then as $G$ varies, $M$ descirbes line $\ell$ parallel to $AF$, since: $d(M,AF) = {1\over 2} d(G,AF) = const.$ Now $\angle GFP_G = \angle (FA,GA)=\alpha $ is also constant, since lines $FA$ and $GA$ are fixed. Finally, map $M\mapsto P_G$ is spiral similarity with center at $F$, and since $M$ describes line $\ell$, then $P_G$ describes line $\ell '$ which we get with rotating $\ell$ by angle $\alpha$ around $F$. So $\ell '$ is parallel to $AG$.
xeroxia
11.01.2013 23:34
Trigonometric Approach: $\angle FAG = \angle P_GFG = \angle P_GGF$. $\frac {AF}{FG} = \frac {\sin \angle AGF} { \sin \angle FAG}$. $ \frac {FG}{P_GF} = \frac {\sin \angle FP_GG }{\sin \angle P_GFG} = \frac {\sin 2\cdot \angle FAG}{\sin \angle FAG} = 2\cos \angle FAG$ $\frac {AF}{FG} \cdot \frac {FG}{P_GF} = \frac {AF}{P_GF} = 2 \cot \angle FAG \cdot \sin \angle AGF $. $\angle (FA, FP_G) = \angle AGF$ or $\angle (FA, FP_G) = 180^\circ - \angle AGF$. So the distance of $P_G$ to $AF$ is $d(P_G, AF) = P_GF \cdot \sin \angle AGF = \frac {AF \cdot \tan \angle FAG}{2}$ which is constant. So locus of $P_G$ is a a line parallel to $AF$ with distance $\frac {AF \cdot \tan \angle FAG}{2}$. $\blacksquare$ Now, we will show for every point on the line parallel to $AF$ with distance $\frac {AF \cdot \tan \angle FAG}{2}$, we can find $G$ on the given line. Let the line perpendicular to $AF$ at $F$ meet the line $AG$ (Here $G$ is not defined, but the line $AG$ is constant) at $Q$. Let $H$ be the foot of perpendicular from $P_G$ to $QF$. Since $HF = \frac {AF \cdot \tan \angle FAG}{2}$ and $QF = AF \tan \alpha$, $H$ is the midpoint of $QF$. So $P_GQ=P_GF$. Let the circle with center $P_G$ and radius $P_GF=P_GQ$ cut $AQ$ at $G$. Since $\angle GP_GF = 2\cdot \angle FQA = 180^\circ - 2\cdot \angle FAG$, and $P_GF = P_GG$, then $\angle P_GFG = \angle P_GGF = \angle FAG$. So $P_GF$ and $P_GG$ are tangent to the circumcircle of $\triangle AFG$.