(If $p'$ is reflection of point $p$ over segment $xy$ the we will write that statement thus $p'=R(p,xy)$)
- HEX 1.JPG (32.65KiB)Viewed 1806 times
$G \in AB$ , $G'=R(G,BC)$
Now $(IH+HG)_{min}=IG'$ .
$G''=R(G',CD)$
So $(IH+HG+IJ)_{min}=(IJ+IG')_{min}=JG"$
..
Thus $GM'=h_{min}$ .
So Its enough to prove that $GM'=3 \sqrt 3$
- HEX 2.JPG (29.15KiB)Viewed 1806 times
In Hex2
$X'=R(X,AF),X=R(A''',EF),A"'=R(A'',DE),A''=R(A',CD),A'=R(A,BC)$ ,
$M'=R(M,AF),M=R(G''',EF),G"'=R(G'',DE),G''=R(G',CD),G'=R(G,BC)$ .
Now Using The properties of reflection Its easy to show that ,
$AGM'X'$ is a parellogram .
So for an arbitrary $G$ , $GM'=AX'$ .
Note that $E,C,A" ; A,E,X$ are collinear .$ \triangle CAA" ,\triangle EA"A''',\triangle EA'''X,\triangle AXX'$ are equilateral .
Now $AX'=AX=\sqrt3 +EX=\sqrt3 +EA''=2\sqrt3 +CA'=3\sqrt3$
And using triangle inequality its must that $h$ will be max iff the insscribed hexagon be $ABCDEF$ itself .
