Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

A sequence a1,a2,a3,a_{1}, a_{2}, a_{3}, \ldots of positive reals satisfies an+1=1+an2a_{n+1}=\sqrt{\frac{1+a_{n}}{2}}. Determine all a1a_{1} such that ai=6+24a_{i}=\frac{\sqrt{6}+\sqrt{2}}{4} for some positive integer ii.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Clearly a1<1a_{1}<1, or else 1a1a2a31 \leq a_{1} \leq a_{2} \leq a_{3} \leq \ldots We can therefore write a1=cosθa_{1}=\cos \theta for some 0<θ<900<\theta<90^{\circ}. Note that cosθ2=1+cosθ2\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}, and cos15=\cos 15^{\circ}= 6+24\frac{\sqrt{6}+\sqrt{2}}{4}. Hence, the possibilities for a1a_{1} are cos15,cos30\cos 15^{\circ}, \cos 30^{\circ}, and cos60\cos 60^{\circ}, which are 2+62,32\frac{\sqrt{2}+\sqrt{6}}{2}, \frac{\sqrt{3}}{2}, and 12\frac{1}{2}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.