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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

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  • $\begingroup$ Do you know the law of sines? $A = 3B$ and $C = 180 - 4B$ is a huge restriction. $\endgroup$ – fleablood Nov 30 '17 at 2:31
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    $\begingroup$ No, the 3-4-5 right triangle does not have acute angles of 30 and 60 degrees. No right triangle with integer sides has rational degree measures for the acute angles. $\endgroup$ – Oscar Lanzi Nov 30 '17 at 2:33
  • $\begingroup$ @fleablood Ill look into that thanks, its been awhile since I heard of that $\endgroup$ – Hank Manks Nov 30 '17 at 2:41
  • $\begingroup$ @OscarLanzi Ohh crap youre right, back to the drawing board $\endgroup$ – Hank Manks Nov 30 '17 at 2:43
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Let $D$ be on the segment $\overline{BC}$ such that $\angle DAB = \angle B$. Then clearly: (1) $\overline{DB} = \overline{DA}$ and $\overline{DC} = \overline{AC}$.

a brief graph

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.

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  • $\begingroup$ Well, it does beat the perimeter of 1296 I had come up with by a different method. Impressive, +1! $\endgroup$ – Oscar Lanzi Nov 30 '17 at 10:41
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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.

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