A sequence of triangles is defined as follows:

\( T_1\) is an isosceles triangles inscribed in a given circle. With one of the equal sides of \(T_1\) as the base , \( T_2\) is an isosceles triangles inscribed in the circle so that the non-coincident vertices of \( T_1\) and \(T_2\) are on the same arc. Similarly successive isosceles triangles are drawn with preceding triangle's equal side as base . Proceeded this way till infinitum . Prove that as \( n \to \infty\) , \(T_n \to \) an equilateral triangle.

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## Comments

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TopNewestThis problem can benefit from having a "good interpretation".

Let the circle have perimeter 1. Pick a Starting point on the circle (like the top most point). For any other point, define the value as the clockwise distance along the perimeter from the Starting point. Suppose the vertices of \(T_1 \) are values \( a, b, c \).

Hence, prove that we approach an equilateral triangle.

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I didn't completely understand how to relate the stuff. The perimeter of the arc cut by equal sides of triangle must be same. So if the triangle is becoming an equilateral triangle then it must have all the perimeter of the three arc same . Right ? I tried to make so relation but didn't got too far.

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Generalize and simplify this expression. Hint: Work modulo 1.

The condition for an equilaterial triangle is \( c = b + \frac{1}{3} = a + \frac{2}{3} \pmod{1} \) (but be careful of orientation). Show that these equations hold true in the limit.

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If you consider the base lengths as a sequence of numbers then you can form a recurrence relation by pythagoras' theorem. The problem is equivalent to proving that the limit tends to \(\sqrt3 r\) which an equilateral triangle has.

I'm not sure how to solve the final equation though. I got something like \(k_{n+1}=\sqrt{2-k}\) and it must be proven that its limit is 1.

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