Michael Reid's proof that every Heronian triangle is embeddable in the plane with integer coordinates for vertices [posted here 17 Nov 2011]... actually (more strongly) shows that for every triangle with 1. Integer edge lengths a,b,c 2. Rational area. There is no requirement for integer area. Now the Heron area formula Area=sqrt(s*(s-a)*(s-b)*(s-c)) where 2s=a+b+c shows that any triangle with integer edge lengths a,b,c either has 1. Area that is an integer/4 2. Irrational area. This suggests that a more natural class than the "Heronian" triangles are the "quarter-Heroic" triangles whose area is a quarter-integer. But... a funny thing happens. The only quarter-Heroic triangles actually ARE Heronian, i.e. quarter-integer area implies integer area. Equivalently: every quarter-Heroic triangle has EVEN perimeter so that s automatically is an integer. Equivalently: every primitive quarter-Heroic triangle has two odd sides and one even side. Equivalently: THEOREM: The "Heronian" triangles can also be characterized as the triangles with integer sides and rational area (they then automatically have integer area). PROOF: For a primitive, (a,b,c)=(even,even,even) is impossible (not primitive). (a,b,c)=(even,even,odd) and (a,b,c)=(odd,odd,odd) both are impossible since then we'd have 16*area^2=p*(p-2a)*(p-2b)*(p-2c)=odd*odd*odd*odd=odd. Hence (a,b,c)=(odd,odd,even) which causes area and semiperimeter both to be integer. QED.