Let $\triangle ABC$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle? $\textbf{(A) }13 \qquad$ $ \textbf{(B) }14 \qquad$ $ \textbf{(C) }15 \qquad$ $ \textbf{(D) }16 \qquad$ $ \textbf{(E) }17 \qquad$
Solution
Let $a=BC$, $b=CA$, $c=AB$. By the law of sines, $\dfrac ba=\dfrac{\sin2A}{\sin A}=2\cos A$, so $\cos A=\dfrac b{2a}$.
By the law of cosines, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. Equating, $b^2c=a(b^2+c^2-a^2)$, i.e. $b^2(c-a)=a(c-a)(c+a)$. If $c=a$ then $C=A$ and $4A=180^\circ$, so $B=90^\circ$ and $b^2=2a^2$, impossible for integers. Hence $c\ne a$ and $$b^2=a(a+c).$$
Since $3A<180^\circ$, $A<60^\circ$, so $\frac12<\cos A<1$, i.e. $a<b<2a$.
$c=\dfrac{b^2}a-a$, so the perimeter is $a+b+c=b+\dfrac{b^2}a>2a$. A perimeter at most $15$ therefore needs $a\le7$. We need $a\mid b^2$ with $a<b<2a$: for $a=1,2,3$ there is none; for $a=4$ only $b=6$ works (giving $c=5$); for $a=5,6,7$ no $b$ in $(a,2a)$ has $a\mid b^2$ (for $5$ and $7$ this would need $a\mid b$; for $6$ it would need $6\mid b$).
So the smallest triangle is $(a,b,c)=(4,6,5)$, with perimeter $15$, and every other triangle has $a\ge8$ and perimeter greater than $16$. (Check: $\cos A=\frac34$, $\cos B=\frac18=2\cdot\frac9{16}-1$.)
Let $\triangle ABC$ be a triangle with integer side lengths and the property that $BC=9$ and $\angle B=2\angle A$. What is the sum of all possible perimeters of such triangles?
