△ABC面积是根号三,
∴△APQ面积是二分之根号三,
令AP=a,AQ=b,1/2*a*b*sin60°=二分之根号三,
∴a*b=2,
设PQ=c（a<=2,b<=2)
根据余弦定理,
c^2=a^2+b^2-2*ab*cos60°
=a^2+b^2-aba^2+b^2-ab
>=2ab-ab=ab=2,
当ab=2（定值）,
a+b有最小值二倍根号二,最大值3（极端为最大值）a^2+b^2-ab=(a+b)^2-3ab<=3^2-6=3
综上,
2<=a^2+b^2-ab<=3,
∴2<=c^2<=3
∴根号二<=c=PQ<=根号三