https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Chainsmokers&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:03:57ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=1189762020 AMC 12B Problems/Problem 222020-03-10T00:28:22Z<p>Chainsmokers: /* Solution 4 (AM-GM) */</p>
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<div>==Problem 22==<br />
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What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math><br />
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<math>\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}</math><br />
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==Solution 1==<br />
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Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .</cmath> It is easy to see that <math>u =\frac{1}{6}</math> is attained for some value of <math>t</math> between <math>t = 0</math> and <math>t = 1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>.<br />
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==Solution 2 (Calculus Needed)==<br />
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We want to maximize <math>f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}</math>. We can use the first derivative test. Use quotient rule to get the following:<br />
<cmath><br />
\frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br />
</cmath><br />
<cmath><br />
\implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br />
</cmath><br />
<cmath><br />
\implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br />
</cmath>Therefore, we plug this back into the original equation to get <math>\boxed{\textbf{(C)} \frac{1}{12}}</math><br />
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~awesome1st<br />
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==Solution 3==<br />
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First, substitute <math>2^t = x (\log_2{x} = t)</math> so that <br />
<cmath><br />
\frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br />
</cmath><br />
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Notice that <br />
<cmath><br />
\frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br />
</cmath><br />
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When seen as a function, <math>\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2</math> is a synthesis function that has <math>\frac{\log_2{x}}{x}</math> as its inner function.<br />
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If we substitute <math>\frac{\log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>.<br />
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Now we need to check if <math>\frac{\log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers.<br />
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In the range of (positive) real numbers, function <math>\frac{\log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{\log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</math>.<br />
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Therefore, we can assume that <math>\frac{\log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.<br />
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==See Also==<br />
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{{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Chainsmokershttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_14&diff=942292018 AIME II Problems/Problem 142018-04-28T20:47:02Z<p>Chainsmokers: /* Solution 1 */</p>
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<div>==Problem==<br />
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The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
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==Solution 1==<br />
Let sides <math>\overline{AB}</math> and <math>\overline{AC}</math> be tangent to <math>\omega</math> at <math>Z</math> and <math>W</math>, respectively. Let <math>\alpha = \angle BAX</math> and <math>\beta = \angle AXC</math>. Because <math>\overline{PQ}</math> and <math>\overline{BC}</math> are both tangent to <math>\omega</math> and <math>\angle YXC</math> and <math>\angle QYX</math> subtend the same arc of <math>\omega</math>, it follows that <math>\angle AYP = \angle QYX = \angle YXC = \beta</math>. By equal tangents, <math>PZ = PY</math>. Applying the Law of Sines to <math>\triangle APY</math> yields <cmath>\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.</cmath>Similarly, applying the Law of Sines to <math>\triangle ABX</math> gives <cmath>\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfrac{21}5</math>. Applying the same argument to <math>\triangle AQY</math> yields <cmath>2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),</cmath>from which <math>AQ = \tfrac{168}{59}</math>. The requested sum is <math>168 + 59 = \boxed{227}</math>.<br />
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==Solution 2 (Projective)==<br />
Let the incircle of <math>ABC</math> be tangent to <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math>. By Brianchon's theorem on tangential hexagons <math>QNCBMP</math> and <math>PYQCXB</math>, we know that <math>MN,CP,BQ</math> and <math>XY</math> are concurrent at a point <math>O</math>. Let <math>PQ \cap BC = Z</math>. Then by La Hire's <math>A</math> lies on the polar of <math>Z</math> so <math>Z</math> lies on the polar of <math>A</math>. Therefore, <math>MN</math> also passes through <math>Z</math>. Then projecting through <math>Z</math>, we have<br />
<cmath> -1 = (A,O;Y,X) \stackrel{Z}{=} (A,M;P,B) \stackrel{Z}{=} (A,N;Q,C).</cmath>Therefore, <math>\frac{AP \cdot MB}{MP \cdot AB} = 1 \implies \frac{3 \cdot MB}{MP \cdot 7} = 1</math>. Since <math>MB+MP=4</math> we know that <math>MB = \frac{6}{5}</math> and <math>MB = \frac{14}{5}</math>. Therefore, <math>AN = AM = \frac{21}{5}</math> and <math>NC = 8 - \frac{21}{5} = \frac{19}{5}</math>. Since <math>(A,N;Q,C) = -1</math>, we also have <math>\frac{AQ \cdot NC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1</math>. Solving for <math>AQ</math>, we obtain <math>AQ = \frac{168}{59} \implies m+n = \boxed{227}</math>.<br />
-Vfire<br />
{{AIME box|year=2018|n=II|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Chainsmokers