Elegant way of finding the least perimeter of triangle

Question

A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?

We have $\angle A = 2\angle B$

$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $

$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\sin^3B$

Using $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$a=2b\cos B $

$c=b(3-4\sin^2 B)$

From above two equations $a^2=b(c+b) $

NOTE: similar questions have been asked earlier but I want to do this question analysing with $a^2=b(c+b)$ and no fancy inequalities

Answer 1

First of all, it is consider very bad style to just throw random equations around. Write down in English what exactly you are doing --- is it an assumption you make? A given condition? Some logical deduction from earlier? And every sentence should start with an English word not an equation, unless you absolutely have to.

Note that it suffices to consider $a,b,c$ coprime (otherwise you get a smaller triangle by scaling). Now $a^2=b^2+bc$ is equivalent to $c^2+4a^2=(2b+c)^2$. If $c$ is even we have $a^2+(\frac12c)^2=(b+\frac12c)^2$, and we must have $a$ odd (otherwise $a,c$ even gives $b$ also even, and so not primitive). So applying the classification of primitive Pythagorean triples, we have: \begin{align*} (c,2a,2b+c)&=(m^2-n^2,2mn,m^2+n^2)\text{ or }\\ (\tfrac12c,a,b+\tfrac12c)&=(2mn,m^2-n^2,m^2+n^2)\\ \end{align*} for some $m,n$ coprime, $m>n$ opposite parity. Now analyse each case separately:

Case 1: $(c,2a,2b+c)=(m^2-n^2,2mn,m^2+n^2)$, so $(a,b,c)=(mn,n^2,m^2-n^2)$. We want $c^2>a^2+b^2$, so $m^2>3n^2$. Also triangle inequality $c<a+b$ gives $(m+n)(m-2n)<0$. So seek $\sqrt3<\frac{m}n<2$ (it is probably not a good idea to bring in $\sqrt3$ but you know what I mean) and we will have perimeter $m(m+n)$. $\frac{m}n=\frac74$ is obviously the best candidate here with least $n$ and least $m$. So we have $(a,b,c)=(28,16,33)$ and perimeter $77$.

Case 2: $(\tfrac12c,a,b+\tfrac12c)=(2mn,m^2-n^2,m^2+n^2)$ so $(a,b,c)=(m^2-n^2,(m-n)^2,4mn)$. Then $c^2>a^2+b^2$ gives $16m^2n^2>(m+n)^2(m-n)^2+(m-n)^4$, i.e., $(m+n)^2 (m^2-4mn+n^2)<0$ and $c<a+b$ gives $m>3n$, so $3<\frac{m}{n}<2+\sqrt3$ and the perimeter is $2m(m+n)$. The best candidate here is the choice $\frac{m}n=\frac72$, which gives $(a,b,c)=(45,25,56)$ and perimeter $126$.

So the least perimeter is 77.

Answer 2

Found an another solution:

we have $a^2=b(c+b)$

A triangle of smallest perimeter means $gcd(a,b,c)=1$

In fact $gcd(b,c)=1$ since any common factor of $b,c$ would be a factor of $a$ as well.

A perfect square $a^2$ is being expressed as the product of two relatively prime integers $b$ and $c$.

it must be the case that both $b$ & $b + c$ are perfect squares. Thus for some integer $m$ & $n$ with $gcd(m,n)= 1$ we have

$b=m^2$ and $b+c=n^2$,$a=mn$

$2\ cosB=\frac{n}{m}=\frac{a}{b}$

As $\angle C >\frac{\pi}{2} \Rightarrow 0<\angle B<\frac{\pi}{6}$

$\Rightarrow \sqrt{3}<2\ cosB=\frac{n}{m}<2$

The smallest value of $(m, n)$ that satisfies the above mentioned conditions are $4$ and $7$ $\blacksquare$