The angles of the triangle ABC satisfy A = 3B. What is the least possible perimeter of ABC assuming its sides are integers. The angles of the triangle $\bigtriangleup ABC$ satisfy $\measuredangle A = 3* \measuredangle B$. What is the least possible perimeter of $ \bigtriangleup ABC$ assuming its lengths are integers.
This is a problem that was in a packet we received in our problem-solving club at school. Here is what I have so far
The smallest possible triangle whose sides are all integers would be $(1,1,1)$ with perimeter, $P = 3$. The smallest right triangle would be $(3,4,5)$ with $P = 12$. Since the right triangle has angles $(30,60,90)$, it meets the conditions to be the triangle we need. From this, I've deduced that,
$$3 < P_{ABC} \leq 12$$
From here I have gone case by case, $P = 4$, $P= 5$, and so on to see if there is a sum of three integers that would make a triangle. I used the triangle inequality to get rid of any sums that don't make a triangle and if it was possible to make I triangle I calculated the angles.
Going through these cases I have not found a triangle that meets the condition, so I believe the smallest possible perimeter is 12. I was curious if anyone has another way to do this problem, thanks
EDIT: As Oscar pointed out I am wrong with my assumptions so I am back to the drawing board
 A: Let $D$ be on the segment $\overline{BC}$ such that $\angle DAB = \angle B$. Then clearly: (1) $\overline{DB} = \overline{DA}$ and $\overline{DC} = \overline{AC}$.

Assume that $\angle B = \alpha$. Then $\angle CDA = 2\alpha$. Also, assume $\overline{CD} = x$; $\overline{AD} = y$ and $\overline{AB} = z$.
We then have
$$
\begin{aligned}
y &= 2x\cos 2\alpha = 2x(2\cos^2\alpha - 1);\\
z &= 2y\cos \alpha.
\end{aligned}
$$
Obviously, the three sides are all integers if and only if $x, y$ and $z$ are all integers. Thus $\cos \alpha \in \mathbb Q$. Also we know that $\alpha < 45^\circ$ since the sum of degree of the three angles are only $180^\circ$. This implies $\cos \alpha > \frac{\sqrt2}2$.
Suppose $\cos \alpha = \frac{p}{q}$ where $p$ and $q$ are coprime integers. Then $q \geq 4$.
Note that when $\cos \alpha = \frac 34$, you get $\cos 2\alpha = \frac 18$ and hence $x = 8, y=2$ and $z=3$ is a solution, giving a perimeter of 21. I claim that this is optimal. To show this you need to find a way to "exhaustively search all hopeful $p$ and $q$" and argue that the perimeter cannot be smaller.
A: Hw Chu's conjecture is correct, the triangle with sides $x=8,x+y=10,z=3$, perimeter= $21$ is minimal.
Relabel $x+y$ as $w$. Then from the Law of Sines
$x=D\sin \alpha$
$w=D\sin 3\alpha$
$z=D\sin 4\alpha$
Using multiple angle formulas we then obtain the ratios
$w/x=(\sin 3\alpha)/(\sin \alpha)=4\cos^2 \alpha-1$
$z/x=(\sin 4\alpha)/(\sin \alpha)=4\cos\alpha(2\cos^2 \alpha-1)$
These are rational if $\cos\alpha$ is rational, following Hw's requirement, and positive sides require $\cos\alpha>(\sqrt{2})/2$.  Plug in $\cos\alpha=(p/q)$ with $p,q$ a pair of relatively prime whole numbers and $q\ge 4$ to meet the positive sides requirement.  The perimeter $\Pi$ is then found from the above to be:
$\Pi=\frac{4xp(2p-q)(p+q)}{q^3}$
The minimal whole number perimeter (which turns out to give whole number sides) then requires
$x=(q^3)/g$
$g=gcd(4p(2p-q)(p+q),q^3)$
Then
$\Pi=\frac{4p(2p-q)(p+q)}{g}>(2q^3)/g$
where the inequality comes from putting $p/q>(\sqrt{2})/2$.
Using polynomial resultants $g$ is found to be a divisor of $32$. Then if $q$ is odd, $g=1$ and $\Pi>2q^3$.  With the odd $q$ having to be at least $5$ this means $\Pi>250$.  The actual minimum for this case is obtained with $\cos\alpha=(4/5),x=125,w=195,z=112,\Pi=432$.
If $q$ is even then $g$ is greater than $1$.  However, note that $p$ is odd, and then:
1)  If $q$ is a multiple of $4$ then $4p(2p-q)(p+q)$ is a multiple of $8$ but not of $16$.  Then $g=8$.
2)  If $q$ is twice an odd number then $4p(2p-q)(p+q)$ is a multiple of $16$ but(among powers of $2$) $q^3$ is only a multiple of $8$.  Again $g=8$.
So for even $q$, $\Pi>(q^3)/4$ meaning for the minimum $q=4$, $\Pi>16$ consistent with Hw's triangle where $\Pi=21$ (which is the only solution with positive sides for $q=4$).  For larger even $q$, $\Pi>(6^3)/4=54$ proving Hw's triangle minimal.
