By Phytagoras theorem it can be prove; a^2+b^2+c^2=q^2+r^2+s^2 (a^2-q^2)+(b^2-r^2)+(c^2-s^2)=0 Let a+q = b+r = s+c = x (a-q)x+(b-r)x+(c-s)x=0 a+b+c=q+r+s
Very good! But what about "just drawing lines in the sand" ? There is a geometric solution without any equation - or should say with simple additions...
draw lines from point P, parallel to three sides. additional equilaterals will be formed. simple equalities proves the relationship. a nice problems! thanks.
By Phytagoras theorem it can be prove;
ReplyDeletea^2+b^2+c^2=q^2+r^2+s^2
(a^2-q^2)+(b^2-r^2)+(c^2-s^2)=0
Let
a+q = b+r = s+c = x
(a-q)x+(b-r)x+(c-s)x=0
a+b+c=q+r+s
Very good! But what about "just drawing lines in the sand" ? There is a geometric solution without any equation - or should say with simple additions...
ReplyDeleteI am trying
ReplyDeletedraw lines from point P, parallel to three sides. additional equilaterals will be formed. simple equalities proves the relationship. a nice problems! thanks.
ReplyDelete-binary