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

contest-mathgeometric-constructiongeometryoptimizationtriangles

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.

Best Answer

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?

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[Math] Triangle with integral side lengths and $\angle A=3\angle B$

Such triangles exist. I found one 3:8:10, where the angle opposite 8 is 1/3 of that opposite 10. The trick here is to pick really small primes in the form of $x+y\sqrt{-n}$, and cube the result. Note here that we're using $n=7$, and the prime is $1\frac 12 + \frac 12\sqrt{-7}$. This is a pretty tiny cube, one gets then a matrix

   ( 3  -7 )  (3)  (2)  (36)
   ( 1   3 )  (1)  (6)  (20)
               A         B = A^3

You then divide through by common factors, to get coordinates at $A =0,0$ $B=+6,0$, and $C=-9,5\sqrt{7}$. The three sides are AB=6, AC=16, and BC=20, which gives the indicated triangle.

The perimeter of an isosceles triangle $\triangle ABC$

The first case.

Let $BK$ be an altitude of $\Delta ABC$.

Thus, $$BK=8,$$ $$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and $$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$ $$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$ which gives the answer: $32+16\sqrt{2-\sqrt3}.$

We can solve the second problem by the similar way.

Best Answer

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[Math] Triangle with integral side lengths and $\angle A=3\angle B$

The perimeter of an isosceles triangle $\triangle ABC$

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