# Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle

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?

• 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. – Travis Mar 8 at 6:03
• @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. – Anirban Niloy Mar 8 at 6:08
• 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". – Travis Mar 8 at 6:11
• Is it possible that the lengths are supposed to be integers, rather than real numbers? – 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 $$c or $$c-a<1,$$ which is impossible.

Thus, $$b\geq2$$.

Now, $$a which gives $$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?

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

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

• $$(1,2,3)$$: 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.