In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?

However, I can't think even of the length of the sides related with the possible values for all angle $\angle A, \angle B$ and $\angle C$. How can I construct the triangle and then get all the side having a length belonging to the positive integers? The problem was very weird for me and all of my effort can be hardly shown or described. And how can I get the minimum possible perimeter?

Thanks in advance.

  • $\begingroup$ Is the problem statement missing a condition? As it is, given any triangle satisfying the conditions, one can scale the triangle to get another, smaller triangle also satisfying the conditions. $\endgroup$ – Travis Mar 8 at 6:03
  • $\begingroup$ @Travis I haven't found any condition yet and I don't know about how to get the best possible triangle with given so little information satisfying that condition. Is it possible? Sorry, I have made a mistake. I corrected that. $\endgroup$ – Anirban Niloy Mar 8 at 6:08
  • 2
    $\begingroup$ No. Like I said in my previous comment, if you have a triangle satisfying the conditions as written, you can scale to produce another triangle also satisfying the conditions but with a smaller perimeter. Therefore there is no "best possible triangle". $\endgroup$ – Travis Mar 8 at 6:11
  • $\begingroup$ Is it possible that the lengths are supposed to be integers, rather than real numbers? $\endgroup$ – Travis Mar 8 at 6:11

In the standard notation we have: $$a+b>c,$$ $$c^2>a^2+b^2$$ and $$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}>\frac{\sin2\beta}{a}=\frac{2\sin\beta\cos\beta}{a},$$ which gives $$\frac{a}{b}>\frac{a^2+c^2-b^2}{ac}$$ or $$a^2(c-b)>b(c^2-b^2)$$ or $$a^2>bc+b^2.$$ If $b=1$ we obtain $$c<a+1$$ or $$c-a<1,$$ which is impossible.

Thus, $b\geq2$.

Now, $$a<c<\frac{a^2-b^2}{b},$$ which gives $$a+1\leq c\leq\frac{a^2}{b}-b-1$$ and $$a^2-ab-b^2-2b\geq0,$$ which gives $$a\geq\frac{b+\sqrt{5b^2+8b}}{2}\geq4$$ and since $$c^2>4^2+2^2,$$ we obtain $$c\geq5$$ and $$a+b+c\geq11.$$ Can you end it now?

  • $\begingroup$ @Anirban Niloy I added something. See now. $\endgroup$ – Michael Rozenberg Mar 8 at 7:12

So, how about a really simple approach: run through positive integer triples $b<a<c$. If they're sides of a triangle, the angles will be in order $B<A<C$.

  • $(1,2,3)$: Not a triangle. In fact, we can't have a triangle if the small side is $1$.
  • $(2,3,4)$: Since $4^2=16>13=2^2+3^2$, it's obtuse. Now, the smaller angles: $\cos B = \frac{4^2+3^2-2^2}{2\cdot 4\cdot 3}=\frac{7}{8}$ and $\cos A = \frac{4^2+2^2-3^2}{2\cdot 4\cdot 2}=\frac{11}{16}$. By the double-angle formula, $\cos 2B = 2\cdot\frac{7^2}{8^2}-1=\frac{98-64}{64}=\frac{17}{32}<\frac{11}{16}$. Not this one.

  • $(2,3,5)$: Not a triangle.

  • $(2,4,5)$: $25>20=16+4$, so it's obtuse. $\sin A=2\sin B>\sin 2B$, so it satisfies the second condition as well.

    There it is - the second smallest triangle with three different integer sides. Sometimes, the simple approach pays off.


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