Point $P$ lies inside a triangle $ABC$. Let $D,E$ and $F$ be reflections of the point $P$ in the lines $BC,CA$ and $AB$, respectively. Prove that if the triangle $DEF$ is equilateral, then the lines $AD,BE$ and $CF$ intersect in a common point.
Problem
Source: Pan African Olympiad 2009
Tags: geometry, geometric transformation, reflection, geometry proposed
Luis González
01.10.2011 19:17
Clearly, if DEF is equilateral, then the pedal triangle of P WRT ABC is also equilateral, i.e. P is an isodynamic point of ABC. Thus, according to the problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=304719, AD,BE,CF concur at the isogonal conjugate of P, namely a Fermat point of ABC. P.S. For a generalization, see Reflections of a point M.
r1234
04.05.2012 20:01
Since $\triangle DEF$ is equilateral we conclude that $P$ is the first Isodyanamic point of $\triangle ABC$.We will prove that $AD,BE,CF$ concur at the Fermat point of $\triangle ABC$ i.e at $P'$, the isogonal conjugate of $P$ wrt $\triangle ABC$.For this note that $D$ is isogonal conjugate of $A$ wrt $\triangle P'BC$.We know that $AP'$ bisects $\angle BP'C$.Hence $A,P'D$ are collinear. So done.
jred
13.03.2015 09:32
Wow, I don't know what it is " isodynamic point ", however this problem can be solved by some basic knowledge. Easily we see $AE=AP=AF$, and also $DE=DF$, thus $\triangle ADE$ is congruent to $\triangle ADF$, so we have $DA$ is the bisector of $\angle EDF$, similarily we get $BE, CF$ are bisectors of $\angle DEF, \angle DFE$ respectively. hence we know lines $AD,BE$ and $CF$ are concurrent at the incenter of $\triangle DEF$.