Let, \( a \geq b \geq c > 0 \) be real numbers such that for all natural number n, there exist triangles of side lengths \( a^n,b^n,c^n \) Prove that the triangles are isosceles.

If a, b, c are sides of a triangle,

*triangular inequality assures that difference of two sides is lesser than the third side.*Since \( a \ge b \ge c > 0 \), hence using triangular inequality we have a – b < c. Infact for all n, \( a^n – b^n < c^n \)We have \( a^n – b^n = (a-b)(a^{n-1} + a^{n-2}b + … + b^{n-1} ) < c^n \) Replacing every a by b in the left hand side, we make the expression to the left even smaller. i.e. \((a-b)(b^{n-1} + b^{n-2}b + … + b^{n-1} ) \le (a-b)(a^{n-1} + a^{n-2}b + … + b^{n-1} ) < c^n \) Hence \( (a-b) \times n \times b^{n-1} < c^n \)

Now notice \( (a-b) < \frac {c^n}{n\times b^{n-1}} = \frac{c}{n} \times \frac {c^{n-1}}{b^{n-1}} = \frac {c}{n} (\frac{c}{b})^{n-1}\) Clearly \( \frac{c}{b} \le 1 \) by given hypothesis. Hence \( a-b \le \frac{c}{n} \) for all n. But letting n go to infinity, we see that a and b can be made

**arbitrarily**close to each other. This implies a=b. Hence each triangle in the sequence is isosceles# Get Started with I.S.I. Entrance Program

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