{
    "problem": "Let $a_0$, $a_1$, $a_2$, $\\dots$ be an infinite sequence of real numbers such that $a_0 = \\frac{5}{13}$ and\n\\[\n  a_{n} = 2 a_{n-1}^2 - 1\n\\]for every positive integer $n$.  Let $c$ be the smallest number such that for every positive integer $n$, the product of the first $n$ terms satisfies the inequality\n\\[|a_0 a_1 \\dotsm a_{n - 1}| \\le \\frac{c}{2^n}.\\]What is the value of $100c$, rounded to the nearest integer?",
    "level": "Level 5",
    "type": "Precalculus",
    "solution": "Define the sequence $(\\theta_n)$ by $\\theta_0 = \\arccos \\frac{5}{13}$ and\n\\[\\theta_n = 2 \\theta_{n - 1}.\\]Then $\\cos \\theta_0 = \\frac{5}{13},$ and\n\\begin{align*}\n\\cos \\theta_n &= \\cos (2 \\theta_{n - 1}) \\\\\n&= 2 \\cos^2 \\theta_{n - 1} - 1.\n\\end{align*}Since the sequences $(a_n)$ and $(\\cos \\theta_n)$ have the same initial term, and the same recursion, they coincide.\n\nWe have that\n\\[\\sin^2 \\theta_0 = 1 - \\cos^2 \\theta_0 = \\frac{144}{169}.\\]Since $\\theta_0$ is acute, $\\sin \\theta_0 = \\frac{12}{13}.$\n\nNow,\n\\begin{align*}\na_0 a_1 \\dotsm a_{n - 1} &= \\cos \\theta_0 \\cos \\theta_1 \\dotsm \\cos \\theta_{n - 1} \\\\\n&= \\cos \\theta_0 \\cos 2 \\theta_0 \\dotsm \\cos 2^{n - 1} \\theta_0.\n\\end{align*}Multiplying both sides by $\\sin \\theta_0 = \\frac{12}{13},$ we get\n\\begin{align*}\n\\frac{12}{13} a_0 a_1 \\dotsm a_{n - 1} &= \\sin \\theta_0 \\cos \\theta_0 \\cos 2 \\theta_0 \\cos 4 \\theta_0 \\dotsm \\cos 2^{n - 1} \\theta_0 \\\\\n&= \\frac{1}{2} \\sin 2 \\theta_0 \\cos 2 \\theta_0 \\cos 4 \\theta_0 \\dotsm \\cos 2^{n - 1} \\theta_0 \\\\\n&= \\frac{1}{4} \\sin 4 \\theta_0 \\dotsm \\cos 2^{n - 1} \\theta_0 \\\\\n&= \\dotsb \\\\\n&= \\frac{1}{2^n} \\sin 2^n \\theta_0.\n\\end{align*}Hence,\n\\[|a_0 a_2 \\dotsm a_{n - 1}| = \\frac{1}{2^n} \\cdot \\frac{13}{12} |\\sin 2^n \\theta_0| \\le \\frac{1}{2^n} \\cdot \\frac{13}{12}.\\]This tells us $c \\le \\frac{13}{12}.$\n\nWe can compute that $a_1 = 2a_0^2 - 1 = 2 \\left( \\frac{5}{13} \\right)^2 - 1 = -\\frac{119}{169},$ so\n\\[\\frac{5}{13} \\cdot \\frac{119}{169} \\le \\frac{c}{4}.\\]Then $c \\ge \\frac{2380}{2197}.$  The bound\n\\[\\frac{2380}{2197} \\le c \\le \\frac{13}{12}\\]tells us that the integer closest to $100c$ is $\\boxed{108}.$"
}