China Western Mathematical Olympiad

We shall prove that no such a triangle satisfies the condition. If $△ABC$ satisfies the condition, let $\angle A\le \angle B\le \angle C$, then $\angle C=2\angle A$, and $a=2007$. Draw the bisector of $\angle ACB$ and let it intersect $AB$ at point $D$. Then $\angle BCD=\angle A$, so $△CDB\sim △ACB$, it follows that $$\frac{CB}{AB}=\frac{BD}{BC}=\frac{CD}{AC}=\frac{BD+CD}{BC+AC}=\frac{AB}{BC+AC}.$$ Thus $$c^2=a(a+b)=2007(2007+b),\text{①}$$ where $2007\le b\le c<2007+b$. Since $a,b,c$ are integers, so $2007|c^2$, then $3\cdot 223|c^2$. We can let $c=669m$, from ①, we get $223m^2=2007+b$. Thus $b=223m^2-2007\ge 2007$, so $m\ge 5$. But $c\ge b$, so $669m\ge 223m^2-2007$, this implies $m<5$, contradiction.