If $a\ge b\ge c>0$ be real numbers such that $\mathrm{\forall}n\in N$, there exists triangles of side lengths ${a}^{n},{b}^{n},{c}^{n}$. Then prove that those triangles must be Isosceles.

icedagecs
2022-07-06
Answered

If $a\ge b\ge c>0$ be real numbers such that $\mathrm{\forall}n\in N$, there exists triangles of side lengths ${a}^{n},{b}^{n},{c}^{n}$. Then prove that those triangles must be Isosceles.

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Kiana Cantu

Answered 2022-07-07
Author has **22** answers

Suppose the triangle is not isosceles. Then $a>b>c>0$.

We need to show that for sufficiently large $n$, ${a}^{n}>{b}^{n}+{c}^{n}$, which violates the triangle inequality.

Note that $\frac{{a}^{n}}{{b}^{n}+{c}^{n}}}\ge {\displaystyle \frac{{a}^{n}}{2{b}^{n}}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{a}{b}}\right)}^{n}\to \mathrm{\infty$ as $\frac{a}{b}}>1$.

Thus there exists some large $N$ such that the quotient is larger than 1, and this completes the proof by contrapositive.

We need to show that for sufficiently large $n$, ${a}^{n}>{b}^{n}+{c}^{n}$, which violates the triangle inequality.

Note that $\frac{{a}^{n}}{{b}^{n}+{c}^{n}}}\ge {\displaystyle \frac{{a}^{n}}{2{b}^{n}}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{a}{b}}\right)}^{n}\to \mathrm{\infty$ as $\frac{a}{b}}>1$.

Thus there exists some large $N$ such that the quotient is larger than 1, and this completes the proof by contrapositive.

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