Question

Suppose that the side lengths of a triangle are three consecutive integers and one of the angles is twice another. The perimeter of such a unique triangle is

Answer 1

Let $B=2A$ and $BD$ be bisector of the angle $B$, then
$CD=\frac{ab}{a+c}$ and $AD=\frac{cb}{a+c}$

Now,

$\Delta ABC$ and $\Delta BDC$ are similar
$\frac{BC}{AC}=\frac{CD}{BC}$

$\Rightarrow a^2=\frac{ab}{a+c}b$

$\Rightarrow b^2=a(a+c)$ ...(1)

Since, $b>a\Rightarrow$ Either $b=a+1$ or $b=a+2$,

if $b=a+1$, then from (1)
${(a+1)}^2=(a+c)a\Rightarrow c=2+\frac{1}{a}$
$\because$ c is integer $\Rightarrow a=1,b=2,c=3,$ but then no triangle will form.
If $b=a+2$, then obviously $c=a+1$, and then from (1)

$(a+2{)}^2=a(2a+1)\Rightarrow a^2-3a-4=0$ or $a=4$
$\therefore a=4,b=6,c=5$ is the only possible solution.

Hence perimeter is $15$