Problem E : 17 499

 Sunt noua zile de cand ALINA mi-a spus acel hotarat...ADIO. De atunci nu cred ca am mai vazut-o pe chat. Poza pe care am pus-o in postare voia sa sugereze ca ma voi lasa de fumat (voiam sa fie o ULTIMA tigara ; EA imi spunea ca nu fumeaza...eu de ce n-as putea ?? Hm, inca nu am reusit...)

<oh, dar am vazut-o pe "sanda"(sanda@3C49464F.86802563.71AA9C83.IP) la 17:35...>

In the meantime, we console ourselves with another beauty, Geometry (in Algebra I missed a simple problem)

   "E:17499.   Let  $ABCD$  be a square. We construct the equilateral triangle

           $ABE$  with  $E$  and  $C$  on either side of the line  $AB$  and denote by  $G$  the

 intersection of the lines  $AB\;\; and\;\; CE$ , and by  $F$  the intersection of the 

lines $BD\;\; and\;\; CE$.    We also construct the equilateral triangle  

$DGH$  with $H$  and   $C$ on either side of the line  $DG.$  Show that :

 a) the points  $H,\;\;A\;\; and\;\; F$  are collinear .           b)  $HG=HE.$

{Author : } Adrian BUD, Negrești-Oaș"

Solution CiP

< Figure of the statement>

          To solve, let's first FORGET the green equilateral triangle  $DGH$. We will add a few lines and, most importantly, draw the circle  $\mathscr{C}(DFG)$  through the three points  $D,\;F,\;G$.

Let  $K$  be the point of intersection between the circle  $\mathscr{C}(DFG)$  and the ray  $[FA)$.

According to this Post, we have  $\widehat{AFD}=60^{\circ}$  so with Inscrible Angle Theorem we have  $\overset{\frown}{DK}=120^{\circ}$. Similarly we have  $\widehat{AFE}=60^{\circ}\Rightarrow \overset{\frown}{KG}=120^{\circ}$. Hence

$\overset{\frown}{DG}=360^{\circ}-\overset{\frown}{DK}-\overset{\frown}{KG}=360^{\circ}-120^{\circ}-120^{\circ}=120^{\circ}$

It follows that triangle  $DGK$  is equilateral and therefore coincides with the one in the original statement. So  $K=H$.

          But we have that the points  $F\;,\;A\;and\; K$ are, by construction, collinear, so we obtain the conclusion  a). Also, in the aforementioned post I showed that the line  $AF$ is the perpendicular bisector of  $[DE]$ , so  $KD=KE$ , but  $KD=KG$ and therefore  $KG=KE$  which proves b). 

$\blacksquare$