https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=MathLearner01&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-23T15:36:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_2&diff=86389 1995 IMO Problems/Problem 2 2017-07-15T13:27:05Z <p>MathLearner01: better LaTeX, very small edit</p> <hr /> <div>== Problem ==<br /> <br /> (''Nazar Agakhanov, Russia'')<br /> Let &lt;math&gt;a, b, c&lt;/math&gt; be positive real numbers such that &lt;math&gt;abc = 1&lt;/math&gt;. Prove that<br /> &lt;cmath&gt; \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. &lt;/cmath&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> We make the substitution &lt;math&gt;x= 1/a&lt;/math&gt;, &lt;math&gt;y=1/b&lt;/math&gt;, &lt;math&gt;z=1/c&lt;/math&gt;. Then<br /> &lt;cmath&gt; \begin{align*}<br /> \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} &amp;= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\<br /> &amp;= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} .<br /> \end{align*} &lt;/cmath&gt;<br /> Since &lt;math&gt;(x^2,y^2,z^2)&lt;/math&gt; and &lt;math&gt;\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \bigr)&lt;/math&gt; are similarly sorted sequences, it follows from the [[Rearrangement Inequality]] that<br /> &lt;cmath&gt; \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{2} \left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) . &lt;/cmath&gt;<br /> By the [[Power Mean Inequality]],<br /> &lt;cmath&gt; \frac{y^2+z^2}{y+z} \ge \frac{(y+z)^2}{2(x+y)} = \frac{x+y}{2} . &lt;/cmath&gt;<br /> Symmetric application of this argument yields<br /> &lt;cmath&gt; \frac{1}{2}\left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) \ge \frac{1}{2}(x+y+z) . &lt;/cmath&gt;<br /> Finally, [[AM-GM]] gives us<br /> &lt;cmath&gt; \frac{1}{2}(x+y+z) \ge \frac{3}{2}xyz = \frac{3}{2}, &lt;/cmath&gt;<br /> as desired. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> We make the same substitution as in the first solution. We note that in general,<br /> &lt;cmath&gt; \frac{p}{q+r} = \frac{(p+q+r)}{(p+q+r)-p} - 1 . &lt;/cmath&gt;<br /> It follows that &lt;math&gt;(x,y,z)&lt;/math&gt; and &lt;math&gt;\bigl(x/(y+z), y/(z+x), z/(x+y)\bigr)&lt;/math&gt; are similarly sorted sequences. Then by [[Chebyshev's Inequality]],<br /> &lt;cmath&gt; \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{3}(x+y+z) \left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \right) . &lt;/cmath&gt;<br /> By AM-GM, &lt;math&gt;\frac{x+y+z}{3} \ge \sqrt{xyz}=1&lt;/math&gt;, and by [[Nesbitt's Inequality]],<br /> &lt;cmath&gt; \frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \ge \frac{3}{2}. &lt;/cmath&gt;<br /> The desired conclusion follows. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Without clever substitutions:<br /> By Cauchy-Schwarz, &lt;cmath&gt;\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2&lt;/cmath&gt; Dividing by &lt;math&gt;2(ab+bc+ac)&lt;/math&gt; gives &lt;cmath&gt;\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}&lt;/cmath&gt; by AM-GM.<br /> <br /> === Solution 3b ===<br /> Without clever notation:<br /> By Cauchy-Schwarz, &lt;cmath&gt;\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;= (ab + bc + ac)^2&lt;/cmath&gt;<br /> <br /> Dividing by &lt;math&gt;2(ab + bc + ac)&lt;/math&gt; and noting that &lt;math&gt;ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3&lt;/math&gt; by AM-GM gives<br /> &lt;cmath&gt;\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},&lt;/cmath&gt;<br /> as desired.<br /> <br /> === Solution 4 ===<br /> Proceed as in Solution 1, to arrive at the equivalent inequality<br /> &lt;cmath&gt; \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . &lt;/cmath&gt;<br /> But we know that &lt;cmath&gt;x + y + z \ge 3xyz \ge 3&lt;/cmath&gt; by AM-GM. Furthermore,<br /> &lt;cmath&gt; (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 &lt;/cmath&gt;<br /> by Cauchy-Schwarz, and so dividing by &lt;math&gt;2(x + y + z)&lt;/math&gt; gives<br /> &lt;cmath&gt; \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &amp;\ge \frac{(x + y + z)}{2} \\ &amp;\ge \frac{3}{2} \end{align*},&lt;/cmath&gt;<br /> as desired.<br /> <br /> {{alternate solutions}}<br /> <br /> == Resources ==<br /> <br /> * [[1995 IMO Problems]]<br /> * [http://www.mathlinks.ro/Forum/viewtopic.php?p=365178#p365178 Discussion on AoPS/MathLinks]<br /> <br /> <br /> [[Category:Olympiad Algebra Problems]]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_9&diff=84509 2017 AIME I Problems/Problem 9 2017-03-08T23:21:50Z <p>MathLearner01: fixed typo</p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;a_{10} = 10&lt;/math&gt;, and for each integer &lt;math&gt;n &gt;10&lt;/math&gt; let &lt;math&gt;a_n = 100a_{n - 1} + n&lt;/math&gt;. Find the least &lt;math&gt;n &gt; 10&lt;/math&gt; such that &lt;math&gt;a_n&lt;/math&gt; is a multiple of &lt;math&gt;99&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Writing out the recursive statement for &lt;math&gt;a_n, a_{n-1}, \dots, a_{10}&lt;/math&gt; and summing them gives &lt;cmath&gt;a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10&lt;/cmath&gt;<br /> Which simplifies to &lt;cmath&gt;a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;a_n&lt;/math&gt; is divisible by 99 if and only if &lt;math&gt;\frac{1}{2}(n+10)(n-9)&lt;/math&gt; is divisible by 99, so &lt;math&gt;(n+10)(n-9)&lt;/math&gt; needs to be divisible by 9 and 11. Assume that &lt;math&gt;n+10&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=12, 23, 34, 45&lt;/math&gt;, we see that &lt;math&gt;n=45&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. Next, assume that &lt;math&gt;n-9&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=20, 31, 42, 53&lt;/math&gt;, we see that &lt;math&gt;n=53&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. The smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{45}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;cmath&gt;a_n \equiv a_{n-1} + n \pmod {99} &lt;/cmath&gt;<br /> By looking at the first few terms, we can see that <br /> &lt;cmath&gt;a_n \equiv 10+11+12+ \dots + n \pmod {99} &lt;/cmath&gt;<br /> This implies<br /> &lt;cmath&gt;a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> Since &lt;math&gt;a_n \equiv 0 \pmod {99}&lt;/math&gt;, we can rewrite the equivalence, and simplify <br /> &lt;cmath&gt;0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv n(n+1) - 90 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv n^2+n+9 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv 4n^2+4n+36 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv (2n+1)^2+35 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;64 \equiv (2n+1)^2 \pmod {99} &lt;/cmath&gt;<br /> The only squares that are congruent to &lt;math&gt;64 \pmod {99}&lt;/math&gt; are &lt;math&gt;(\pm 8)^2&lt;/math&gt; and &lt;math&gt;(\pm 19)^2&lt;/math&gt;, so <br /> &lt;cmath&gt;2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}&lt;/cmath&gt;<br /> &lt;math&gt;2n+1 \equiv -8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=45&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv 8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=53&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv -19 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=89&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv -8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=9&lt;/math&gt; as the smallest integer solution. However, &lt;math&gt;n&lt;/math&gt; must be greater than &lt;math&gt;10&lt;/math&gt;.<br /> <br /> The smallest positive integer solution greater than &lt;math&gt;10&lt;/math&gt; is &lt;math&gt;n=\boxed{045}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2017|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_9&diff=84507 2017 AIME I Problems/Problem 9 2017-03-08T23:19:55Z <p>MathLearner01: added solution</p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;a_{10} = 10&lt;/math&gt;, and for each integer &lt;math&gt;n &gt;10&lt;/math&gt; let &lt;math&gt;a_n = 100a_{n - 1} + n&lt;/math&gt;. Find the least &lt;math&gt;n &gt; 10&lt;/math&gt; such that &lt;math&gt;a_n&lt;/math&gt; is a multiple of &lt;math&gt;99&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Writing out the recursive statement for &lt;math&gt;a_n, a_{n-1}, \dots, a_{10}&lt;/math&gt; and summing them gives &lt;cmath&gt;a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10&lt;/cmath&gt;<br /> Which simplifies to &lt;cmath&gt;a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;a_n&lt;/math&gt; is divisible by 99 if and only if &lt;math&gt;\frac{1}{2}(n+10)(n-9)&lt;/math&gt; is divisible by 99, so &lt;math&gt;(n+10)(n-9)&lt;/math&gt; needs to be divisible by 9 and 11. Assume that &lt;math&gt;n+10&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=12, 23, 34, 45&lt;/math&gt;, we see that &lt;math&gt;n=45&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. Next, assume that &lt;math&gt;n-9&lt;/math&gt; is a multiple of 11. Writing out a few terms, &lt;math&gt;n=20, 31, 42, 53&lt;/math&gt;, we see that &lt;math&gt;n=53&lt;/math&gt; is the smallest &lt;math&gt;n&lt;/math&gt; that works in this case. The smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{45}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;cmath&gt;a_n \equiv a_{n-1} + n \pmod {99} &lt;/cmath&gt;<br /> By looking at the first few terms, we can see that <br /> &lt;cmath&gt;a_n \equiv 10+11+12+ \dots + n \pmod {99} &lt;/cmath&gt;<br /> This implies<br /> &lt;cmath&gt;a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> Since &lt;math&gt;a_n \equiv 0 \pmod {99}&lt;/math&gt;, we can rewrite the equivalence, and simplify <br /> &lt;cmath&gt;0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv n(n+1) - 90 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv n^2+n+9 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv 4n^2+4n+36 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;0 \equiv (2n+1)^2+35 \pmod {99} &lt;/cmath&gt;<br /> &lt;cmath&gt;64 \equiv (2n+1)^2 \pmod {99} &lt;/cmath&gt;<br /> The smallest squares that are congruent to &lt;math&gt;64 \pmod {99}&lt;/math&gt; are &lt;math&gt;(\pm 8)^2&lt;/math&gt; and &lt;math&gt;(\pm 19)^2&lt;/math&gt;, so <br /> &lt;cmath&gt;2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}&lt;/cmath&gt;<br /> &lt;math&gt;2n+1 \equiv -8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=45&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv 8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=53&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv -19 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=89&lt;/math&gt; as the smallest integer solution.<br /> <br /> &lt;math&gt;2n+1 \equiv -8 \pmod {99}&lt;/math&gt; yields &lt;math&gt;n=9&lt;/math&gt; as the smallest integer solution. However, &lt;math&gt;n&lt;/math&gt; must be greater than &lt;math&gt;10&lt;/math&gt;.<br /> <br /> The smallest positive integer solution greater than &lt;math&gt;10&lt;/math&gt; is &lt;math&gt;n=\boxed{045}&lt;/math&gt;.<br /> <br /> &lt;i&gt; Solution by MathLearner01 &lt;/i&gt;<br /> ==See also==<br /> {{AIME box|year=2017|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems&diff=84192 2017 AMC 12B Problems 2017-02-25T02:06:32Z <p>MathLearner01: added missing solution link for #22</p> <hr /> <div>WORK IN PROGRESS<br /> <br /> {{AMC12 Problems|year=2017|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> Kymbrea's comic book collection currently has &lt;math&gt;30&lt;/math&gt; comic books in it, and she is adding to her collection at the rate of &lt;math&gt;2&lt;/math&gt; comic books per month. LaShawn's collection currently has &lt;math&gt;10&lt;/math&gt; comic books in it, and he is adding to his collection at the rate of &lt;math&gt;6&lt;/math&gt; comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; satify the inequalities<br /> &lt;math&gt;0&lt;x&lt;1&lt;/math&gt;, &lt;math&gt;-1&lt;y&lt;0&lt;/math&gt;, and &lt;math&gt;1&lt;z&lt;2&lt;/math&gt;.<br /> Which of the following numbers is necessarily positive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Supposed that &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are nonzero real numbers such that &lt;math&gt;\frac{3x+y}{x-3y}=-2&lt;/math&gt;. What is the value of &lt;math&gt;\frac{x+3y}{3x-y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 4==<br /> Samia set off on her bicycle to visit her friend, traveling at an average speed of &lt;math&gt;17&lt;/math&gt; kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at &lt;math&gt;5&lt;/math&gt; kilometers per hour. In all it took her &lt;math&gt;44&lt;/math&gt; minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> The data set &lt;math&gt;[6,19,33,33,39,41,41,43,51,57]&lt;/math&gt; has median &lt;math&gt;Q_2 = 40&lt;/math&gt;, first quartile &lt;math&gt;Q_1 = 33&lt;/math&gt;, and third quartile &lt;math&gt;Q_3=43&lt;/math&gt;. An outlier in a data set is a value that is more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range below the first quartile &lt;math&gt;(Q_1)&lt;/math&gt; or more than &lt;math&gt;1.5&lt;/math&gt; times the interquartile range above the third quartile &lt;math&gt;(Q_3)&lt;/math&gt;, where the interquartile range is defined as &lt;math&gt;Q_3 - Q_1&lt;/math&gt;. How many outliers does this data set have?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The circle having &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(8,6)&lt;/math&gt; as the endpoints of a diameter intersects the &lt;math&gt;x&lt;/math&gt;-axis at a second point. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of this point? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\sqrt{2} \qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 6\sqrt{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The functions &lt;math&gt;\sin(x)&lt;/math&gt; and &lt;math&gt;\cos(x)&lt;/math&gt; are periodic with least period &lt;math&gt;2\pi&lt;/math&gt;. What is the least period of the function &lt;math&gt;\cos(\sin(x))&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\pi}{2}\qquad\textbf{(B)}\ \pi\qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ 4\pi \qquad\textbf{(E)}&lt;/math&gt; It's not periodic.<br /> <br /> [[2017 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{3}-1}{2}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{2} \qquad\textbf{(E)}\ \frac{\sqrt{6}-1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A circle has center &lt;math&gt;(-10,-4)&lt;/math&gt; and radius &lt;math&gt;13&lt;/math&gt;. Another circle has center &lt;math&gt;(3,9)&lt;/math&gt; and radius &lt;math&gt;\sqrt{65}&lt;/math&gt;. The line passing through the two points of intersection of the two circles has equation &lt;math&gt;x + y = c&lt;/math&gt;. What is &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> At Typico High School, &lt;math&gt;60\%&lt;/math&gt; of the students like dancing, and the rest dislike it. Of those who like dancing, &lt;math&gt;80\%&lt;/math&gt; say that they like it, and the rest say that they dislike it. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt; say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Call a positive integer &lt;math&gt;monotonous&lt;/math&gt; if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;23578&lt;/math&gt;, and &lt;math&gt;987620&lt;/math&gt; are monotonous, but &lt;math&gt;88&lt;/math&gt;, &lt;math&gt;7434&lt;/math&gt;, and &lt;math&gt;23557&lt;/math&gt; are not. How many monotonous positive integers are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048&lt;/math&gt;<br /> <br /> [[2017 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 12==<br /> What is the sum of the roots of &lt;math&gt;z^{12}=64&lt;/math&gt; that have a positive real part? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> In the figure below, &lt;math&gt;3&lt;/math&gt; of the &lt;math&gt;6&lt;/math&gt; disks are to be painted blue, &lt;math&gt;2&lt;/math&gt; are to be painted red, and &lt;math&gt;1&lt;/math&gt; is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?<br /> <br /> &lt;asy&gt;<br /> size(100);<br /> pair A, B, C, D, E, F;<br /> A = (0,0);<br /> B = (1,0);<br /> C = (2,0);<br /> D = rotate(60, A)*B;<br /> E = B + D;<br /> F = rotate(60, A)*C;<br /> draw(Circle(A, 0.5));<br /> draw(Circle(B, 0.5));<br /> draw(Circle(C, 0.5));<br /> draw(Circle(D, 0.5));<br /> draw(Circle(E, 0.5));<br /> draw(Circle(F, 0.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The number &lt;math&gt;21!=51,090,942,171,709,440,000&lt;/math&gt; has over &lt;math&gt;60,000&lt;/math&gt; positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> A coin is biased in such a way that on each toss the probability of heads is &lt;math&gt;\frac{2}{3}&lt;/math&gt; and the probability of tails is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; less than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; The probabilities are the same.<br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{2}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; The probability of winning Game A is &lt;math&gt;\frac{4}{81}&lt;/math&gt; greater than the probability of winning Game B.<br /> <br /> [[2017 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> The diameter &lt;math&gt;AB&lt;/math&gt; of a circle of radius &lt;math&gt;2&lt;/math&gt; is extended to a point &lt;math&gt;D&lt;/math&gt; outside the circle so that &lt;math&gt;BD=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is chosen so that &lt;math&gt;ED=5&lt;/math&gt; and line &lt;math&gt;ED&lt;/math&gt; is perpendicular to line &lt;math&gt;AD&lt;/math&gt;. Segment &lt;math&gt;AE&lt;/math&gt; intersects the circle at a point &lt;math&gt;C&lt;/math&gt; between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. What is the area of &lt;math&gt;\triangle <br /> ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are chosen independently and uniformly at random from the interval &lt;math&gt;(0,1)&lt;/math&gt;. What is the probability that &lt;math&gt;\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor&lt;/math&gt;, where &lt;math&gt;\lfloor r\rfloor&lt;/math&gt; denotes the greatest integer less than or equal to the real number &lt;math&gt;r&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Last year Isabella took &lt;math&gt;7&lt;/math&gt; math tests and received &lt;math&gt;7&lt;/math&gt; different scores, each an integer between &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was &lt;math&gt;95&lt;/math&gt;. What was her score on the sixth test?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{7}{576} \qquad \textbf{(B)}\ \frac{5}{192} \qquad \textbf{(C)}\ \frac{1}{36} \qquad \textbf{(D)}\ \frac{5}{144} \qquad\textbf{(E)}\ \frac{7}{48}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> &lt;math&gt;\textbf{(A)}\ -2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{24}{5} \qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; has right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;\triangle ABC \sim \triangle BCD&lt;/math&gt;, and &lt;math&gt;AB &gt; BC&lt;/math&gt;. There is a point &lt;math&gt;E&lt;/math&gt; in the interior of &lt;math&gt;ABCD&lt;/math&gt; such that &lt;math&gt;\triangle ABC \sim \triangle CEB&lt;/math&gt; and the area of &lt;math&gt;\triangle AED&lt;/math&gt; is &lt;math&gt;17&lt;/math&gt; times the area of &lt;math&gt;\triangle CEB&lt;/math&gt;. What is &lt;math&gt;\frac{AB}{BC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2} \qquad \textbf{(B)}\ 2 + \sqrt{2} \qquad \textbf{(C)}\ \sqrt{17} \qquad \textbf{(D)}\ 2 + \sqrt{5} \qquad\textbf{(E)}\ 1 + 2\sqrt{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> A set of &lt;math&gt;n&lt;/math&gt; people participate in an online video basketball tournament. Each person may be a member of any number of &lt;math&gt;5&lt;/math&gt;-player teams, but no teams may have exactly the same &lt;math&gt;5&lt;/math&gt; members. The site statistics show a curious fact: The average, over all subsets of size &lt;math&gt;9&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those 9 people is equal to the reciprocal of the average, over all subsets of size &lt;math&gt;8&lt;/math&gt; of the set of &lt;math&gt;n&lt;/math&gt; participants, of the number of complete teams whose members are among those &lt;math&gt;8&lt;/math&gt; people. How many values &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;9 \leq n \leq 2017&lt;/math&gt;, can be the number of participants?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 477 \qquad \textbf{(B)}\ 482 \qquad \textbf{(C)}\ 487 \qquad \textbf{(D)}\ 557 \qquad\textbf{(E)}\ 562&lt;/math&gt;<br /> <br /> [[2017 AMC 12B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2017|ab=B|before=[[2017 AMC 12A Problems]]|after=[[2018 AMC 12A Problems]]}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=82977 2017 AMC 10A Problems/Problem 4 2017-02-08T22:15:30Z <p>MathLearner01: Fixed minor LaTeX error</p> <hr /> <div>==Problem==<br /> Mia is “helping” her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> Every &lt;math&gt;30&lt;/math&gt; seconds &lt;math&gt;3-2&lt;/math&gt; toys are put in the box, so after &lt;math&gt;27\cdot30&lt;/math&gt; seconds there will be &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into to the box and we have our total amount of time to be &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, which equals &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(B) 14}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=82969 2017 AMC 10A Problems/Problem 20 2017-02-08T22:13:39Z <p>MathLearner01: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1265}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems&diff=79769 1997 AIME Problems 2016-07-30T18:53:51Z <p>MathLearner01: fixed typo</p> <hr /> <div>{{AIME Problems|year=1997}}<br /> <br /> == Problem 1 ==<br /> How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?<br /> <br /> [[1997 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> The nine horizontal and nine vertical lines on an &lt;math&gt;8\times8&lt;/math&gt; checkerboard form &lt;math&gt;r&lt;/math&gt; rectangles, of which &lt;math&gt;s&lt;/math&gt; are squares. The number &lt;math&gt;s/r&lt;/math&gt; can be written in the form &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[1997 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?<br /> <br /> [[1997 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Circles of radii 5, 5, 8, and &lt;math&gt;m/n&lt;/math&gt; are mutually externally tangent, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[1997 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> The number &lt;math&gt;r&lt;/math&gt; can be expressed as a four-place decimal &lt;math&gt;0.abcd,&lt;/math&gt; where &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; represent digits, any of which could be zero. It is desired to approximate &lt;math&gt;r&lt;/math&gt; by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\frac 27.&lt;/math&gt; What is the number of possible values for &lt;math&gt;r&lt;/math&gt;?<br /> <br /> [[1997 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Point &lt;math&gt;B&lt;/math&gt; is in the exterior of the regular &lt;math&gt;n&lt;/math&gt;-sided polygon &lt;math&gt;A_1A_2\cdots A_n&lt;/math&gt;, and &lt;math&gt;A_1A_2B&lt;/math&gt; is an equilateral triangle. What is the largest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;A_n&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; are consecutive vertices of a regular polygon?<br /> <br /> [[1997 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> A car travels due east at &lt;math&gt;\frac 23&lt;/math&gt; mile per minute on a long, straight road. At the same time, a circular storm, whose radius is &lt;math&gt;51&lt;/math&gt; miles, moves southeast at &lt;math&gt;\frac 12\sqrt{2}&lt;/math&gt; mile per minute. At time &lt;math&gt;t=0&lt;/math&gt;, the center of the storm is &lt;math&gt;110&lt;/math&gt; miles due north of the car. At time &lt;math&gt;t=t_1&lt;/math&gt; minutes, the car enters the storm circle, and at time &lt;math&gt;t=t_2&lt;/math&gt; minutes, the car leaves the storm circle. Find &lt;math&gt;\frac 12(t_1+t_2)&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> How many different &lt;math&gt;4\times 4&lt;/math&gt; arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?<br /> <br /> [[1997 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Given a nonnegative real number &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;\langle x\rangle&lt;/math&gt; denote the fractional part of &lt;math&gt;x&lt;/math&gt;; that is, &lt;math&gt;\langle x\rangle=x-\lfloor x\rfloor&lt;/math&gt;, where &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;. Suppose that &lt;math&gt;a&lt;/math&gt; is positive, &lt;math&gt;\langle a^{-1}\rangle=\langle a^2\rangle&lt;/math&gt;, and &lt;math&gt;2&lt;a^2&lt;3&lt;/math&gt;. Find the value of &lt;math&gt;a^{12}-144a^{-1}&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:<br /> <br /> i. Either each of the three cards has a different shape or all three of the card have the same shape.<br /> <br /> ii. Either each of the three cards has a different color or all three of the cards have the same color.<br /> <br /> iii. Either each of the three cards has a different shade or all three of the cards have the same shade.<br /> <br /> How many different complementary three-card sets are there?<br /> <br /> [[1997 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> [[1997 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The function &lt;math&gt;f&lt;/math&gt; defined by &lt;math&gt;f(x)= \frac{ax+b}{cx+d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;,&lt;math&gt;b&lt;/math&gt;,&lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are nonzero real numbers, has the properties &lt;math&gt;f(19)=19&lt;/math&gt;, &lt;math&gt;f(97)=97&lt;/math&gt; and &lt;math&gt;f(f(x))=x&lt;/math&gt; for all values except &lt;math&gt;\frac{-d}{c}&lt;/math&gt;. Find the unique number that is not in the range of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points in the Cartesian plane that satisfy &lt;center&gt;&lt;math&gt;\Big|\big| |x|-2\big|-1\Big|+\Big|\big| |y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; be distinct, randomly chosen roots of the equation &lt;math&gt;z^{1997}-1=0&lt;/math&gt;. Let &lt;math&gt;m/n&lt;/math&gt; be the probability that &lt;math&gt;\sqrt{2+\sqrt{3}}\le |v+w|&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The sides of rectangle &lt;math&gt;ABCD&lt;/math&gt; have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt;. An equilateral triangle is drawn so that no point of the triangle lies outside &lt;math&gt;ABCD&lt;/math&gt;. The maximum possible area of such a triangle can be written in the form &lt;math&gt;p\sqrt{q}-r&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are positive integers, and &lt;math&gt;q&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;p+q+r&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_12&diff=79109 2011 AIME I Problems/Problem 12 2016-06-28T16:41:43Z <p>MathLearner01: Fixed small error at end</p> <hr /> <div>== Problem ==<br /> Six men and some number of women stand in a line in random order. Let &lt;math&gt;p&lt;/math&gt; be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that &lt;math&gt;p&lt;/math&gt; does not exceed 1 percent.<br /> <br /> == Solution ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where &lt;math&gt;(k)&lt;/math&gt; refers to a consecutive group of &lt;math&gt;k&lt;/math&gt; men:<br /> <br /> _(2)_(2)_(2)_<br /> <br /> _(3)_(3)_<br /> <br /> _(2)_(4)_<br /> <br /> _(4)_(2)_<br /> <br /> _(6)_<br /> <br /> For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the &lt;math&gt;n&lt;/math&gt; women. Since there are &lt;math&gt;n+1&lt;/math&gt; possible places to insert the dividers, and we need to choose any three of these locations, we have &lt;math&gt;\dbinom{n+1}{3}&lt;/math&gt; ways.<br /> <br /> The second, third, and fourth cases are like the first, only that we need to insert two dividers among the &lt;math&gt;n+1&lt;/math&gt; possible locations. Each gives us &lt;math&gt;\dbinom{n+1}{2}&lt;/math&gt; ways, for a total of &lt;math&gt;3\dbinom{n+1}{2}&lt;/math&gt; ways.<br /> <br /> The last case gives us &lt;math&gt;\dbinom{n+1}{1}=n+1&lt;/math&gt; ways.<br /> <br /> Therefore, the total number of possible ways where there are no isolated men is<br /> <br /> &lt;cmath&gt;\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).&lt;/cmath&gt;<br /> <br /> The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or<br /> <br /> &lt;cmath&gt;2\dbinom{n+1}{2}+(n+1).&lt;/cmath&gt;<br /> <br /> Thus, we want to find the minimum possible value of &lt;math&gt;n&lt;/math&gt; where &lt;math&gt;n&lt;/math&gt; is a positive integer such that<br /> <br /> &lt;cmath&gt;\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.&lt;/cmath&gt;<br /> <br /> The numerator is equal to<br /> <br /> &lt;cmath&gt;2\cdot\dfrac{(n+1)!}{2!(n-1)!}+(n+1)=2\cdot\dfrac{(n+1)(n)}{2}+(n+1)=n(n+1)+1(n+1)=(n+1)^2.&lt;/cmath&gt;<br /> <br /> For the denominator, we get<br /> <br /> &lt;cmath&gt;\begin{align*}\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)&amp;=\dfrac{(n+1)!}{3!(n-2)!}+3\dfrac{(n+1)!}{2!(n-1)!}+(n+1)\\<br /> &amp;=\dfrac{(n+1)(n)(n-1)}{6}+3\dfrac{(n+1)(n)}{2}+(n+1)\\<br /> &amp;=(n+1)\left[\dfrac{n^2-n}{6}+\dfrac{3n}{2}+1\right]\\<br /> &amp;=\dfrac{1}{6}(n+1)(n^2-n+9n+6)\\<br /> &amp;=\dfrac{1}{6}(n+1)(n^2+8n+6).<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So, we get<br /> <br /> &lt;cmath&gt;\dfrac{(n+1)^2}{\frac{1}{6}(n+1)(n^2+8n+6)}=\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.&lt;/cmath&gt;<br /> <br /> We know that &lt;math&gt;100(n^2+8n+6)&lt;/math&gt; is positive since &lt;math&gt;n&lt;/math&gt; must be positive. So, when multiplying both sides of the inequality by that expression, it will not change the inequality sign. After multiplying by it, we get<br /> <br /> &lt;cmath&gt;\begin{align*}100\cdot6(n+1)&amp;\le n^2+8n+6\\<br /> 600n+600&amp;\le n^2+8n+6\\<br /> 0&amp;\le n^2-592n-594.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus we seek the smallest positive integer value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;n^2-592n-594\ge0&lt;/math&gt;. Since the quadratic function's discriminant, or &lt;math&gt;\sqrt{592^2-4(-594)}=\sqrt{592^2+4\cdot594}&lt;/math&gt;, is positive, the polynomial has two distinct real roots.<br /> <br /> Also, since the polynomial has a positive leading coefficient, the graph of the polynomial is concave up, and the value of &lt;math&gt;n&lt;/math&gt; we want must be either slightly larger than the positive root (if the other, smaller root is less than &lt;math&gt;1&lt;/math&gt;) or equal to &lt;math&gt;1&lt;/math&gt; (if the smaller root is greater than or equal to &lt;math&gt;1&lt;/math&gt;). We see that &lt;math&gt;n=1&lt;/math&gt; does not satisfy the inequality, so the smaller root is irrelevant.<br /> <br /> The solution to the polynomial is<br /> <br /> &lt;cmath&gt;\dfrac{592\pm\sqrt{592^2+4\cdot594}}{2}.&lt;/cmath&gt;<br /> <br /> We want the larger solution and to find the smallest integer greater than that solution. So, we will look only at the &lt;math&gt;+&lt;/math&gt; case, not the &lt;math&gt;-&lt;/math&gt;.<br /> <br /> Let's look at the discriminant:<br /> <br /> &lt;cmath&gt;\begin{align*}\sqrt{592^2+4\cdot594}&amp;=\sqrt{592^2+4(592+2)}\\<br /> &amp;=\sqrt{592(592)+4(592)+8}\\<br /> &amp;=\sqrt{592(592+4)+8}\\<br /> &amp;=\sqrt{(594-2)(594+2)+8}\\<br /> &amp;=\sqrt{594^2-4+8}\\<br /> &amp;=\sqrt{594^2+4}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;594&lt;\text{discriminant}&lt;595&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;k=\dfrac{592+\sqrt{592^2+4\cdot594}}{2}&lt;/math&gt;.<br /> <br /> Therefore, we're looking for<br /> <br /> &lt;cmath&gt;n=\left\lceil\dfrac{592+\sqrt{592^2+4\cdot594}}{2}\right\rceil=\lceil k\rceil.&lt;/cmath&gt;<br /> <br /> Since we have &lt;math&gt;594&lt;\sqrt{592^2+4\cdot594}&lt;595&lt;/math&gt;, we get<br /> <br /> &lt;cmath&gt;\begin{align*}\dfrac{592+594}{2}&lt;&amp;k&lt;\dfrac{592+595}{2}\\<br /> \implies593&lt;&amp;k&lt;593.5<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;n=\lceil k\rceil&lt;/math&gt;, &lt;math&gt;n=\boxed{594}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=User:MathLearner01&diff=74392 User:MathLearner01 2016-01-09T19:06:42Z <p>MathLearner01: woops old link doesn't work</p> <hr /> <div>MathLearner01's user page has moved. The new user page can be found here: http://www.artofproblemsolving.com/wiki/index.php?title=User:MathLearner01</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=71511 AoPS Wiki:FAQ 2015-08-10T16:31:17Z <p>MathLearner01: /* I believe a post needs corrective action. What should I do? */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&amp;t=414129 here].<br /> <br /> == General==<br /> <br /> <br /> ==== Can I change my user name? ====<br /> <br /> :As indicated during the time of your registration, you are unable to change your username.<br /> <br /> ==== Can I make more than one account? ====<br /> <br /> It is preferred that you have one and only one account. 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Your problems may eventually be submitted into the Contest page.<br /> <br /> ==== Who can I ask to add posts to the contests section? ====<br /> :Any one of the members in the the [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=group&amp;g=417 RManagers] group.<br /> <br /> ==== What are the guidelines for posting problems to be added to the contests section? ====<br /> :Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&amp;t=195579 guidelines in this post].<br /> <br /> ==== Why is the wiki missing many contest questions? ====<br /> :Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations. See, for example, [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1391106#p1391106 this post].<br /> <br /> ==== What if I find an error on a problem? ====<br /> Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=426693 this thread]<br /> <br /> == LaTeX, Asymptote, GeoGebra ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\LaTeX&lt;/math&gt; is a typesetting markup language that is useful to produce properly formatted mathematical and scientific expressions.<br /> <br /> ==== How can I download LaTeX to use on the forums? ====<br /> <br /> :There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br /> <br /> ==== How can I download LaTeX for personal use? ====<br /> :You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br /> <br /> ==== Where can I find a list of LaTeX commands? ====<br /> :See [[LaTeX:Symbols|here]].<br /> <br /> ==== Where can I test LaTeX commands? ====<br /> <br /> :[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. <br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br /> <br /> ==== How can I draw 3D diagrams? ====<br /> <br /> :See [[Asymptote: 3D graphics]].<br /> <br /> ==== What is the cse5 package? ==== <br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br /> <br /> ==== Can I convert diagrams from GeoGebra to other formats? ====<br /> :It is possible to export GeoGebra to [[Asymptote (Vector Graphics Language)|Asymptote]] (see [[User:Azjps/geogebra|here]]), PsTricks, and PGF/TikZ; and GeoGebra animations into .gif or video files. <br /> <br /> == AoPSWiki ==<br /> ==== Is there a guide for wiki syntax? ====<br /> <br /> :See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br /> <br /> ==== What do I do if I see a mistake in the wiki? ====<br /> <br /> :Edit the page and correct the error! You can edit most pages on the wiki. Click the &quot;edit&quot; button on the right sidebar to edit a page.<br /> <br /> ==== Why can't I edit the wiki? ====<br /> <br /> You must be a registered user to edit. To be registered, make sure you give a correct email, and activate your account.<br /> <br /> == Miscellaneous ==<br /> ==== Is it possible to join the AoPS Staff? ====<br /> <br /> :Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br /> <br /> ==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br /> <br /> :You must have graduated from high school, or at least be in the second term of your senior year.<br /> <br /> ==What do some of the acronyms such as &quot;OP&quot; stand for?==<br /> *'''AFK'''- Away from keyboard<br /> *'''AoPS'''- Art of Problem Solving, the website you're on right now!<br /> *'''AIME'''- American Invitational Mathematics Examination<br /> *'''AMC'''- American Math Competititions<br /> *'''ATM'''- At the Moment<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''ID(R)K'''-I Don't (Really) Know<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''MOP'''- Mathematical Olympiad (Summer) Program<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh.<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''QED'''- Quod erat demonstrandum, Latin for Which was to be proven; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''rotfl''' - Rolling on the floor laughing<br /> *'''sa''' - sa<br /> *'''smh''' - Shaking my head<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WLOG'''- Without loss of generality<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''xD'''- Bursting Laugh<br /> <br /> == FTW! ==<br /> <br /> ==== How do you access FTW? ====<br /> You can access FTW by clicking FTW! on the green bar at the top of the page.<br /> <br /> <br /> ==== Did FTW miscount my number of games?====<br /> <br /> No! However, the (Overall) rating statistics do not count games with less than 6 problems or less than 15 seconds.<br /> <br /> For example, if you have played 30 games, but not all of them were 6 problems or higher, then you will still be muted.<br /> <br /> == School ==<br /> <br /> ==== What if I want to drop out of a class? ====<br /> :For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br /> <br /> ==== What if I miss a class? ====<br /> :There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board.<br /> <br /> ==== Is there audio or video in class? ====<br /> There is no audio or video in the class. The classes are completely text based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video. <br /> <br /> ====I feel like joining! What are my class choices? ====<br /> :[http://www.artofproblemsolving.com/School/classlist.php Class List] [http://www.artofproblemsolving.com/School/index.php?page=school.instructors Instructors List]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=71510 AoPS Wiki:FAQ 2015-08-10T16:30:10Z <p>MathLearner01: /* What is my post rating? */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&amp;t=414129 here].<br /> <br /> == General==<br /> <br /> <br /> ==== Can I change my user name? ====<br /> <br /> :As indicated during the time of your registration, you are unable to change your username.<br /> <br /> ==== Can I make more than one account? ====<br /> <br /> It is preferred that you have one and only one account. Having more leads to issues of not remembering on what account you did what. Using multiple accounts to &quot;game&quot; the system, (e.g., increase rating for posts or in online games) will lead to bans on all accounts associated to you. If you have already made additional accounts, please choose one account and stop using the others.<br /> <br /> ====What software does Art of Problem Solving use to run the website?====<br /> <br /> :* Forums: Custom<br /> :* Blog: Custom<br /> :* Search: Solr<br /> :* Wiki: MediaWiki<br /> :* Asymptote, Latex, and Geogebra are generated through their respective binary packages<br /> :* Videos: YouTube<br /> <br /> == Forums ==<br /> <br /> ==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br /> <br /> :AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br /> <br /> ==== I got the message &quot;You can not post at this time&quot; when trying to post, why? ====<br /> <br /> :New users are not allowed to post messages with URLs and various other things. Once you have five posts you can post normally.<br /> <br /> ====I got the message &quot;Too many messages.&quot; when trying to send a private message, why?====<br /> :To prevent PM spam abuse, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br /> <br /> ==== If I make more posts, it means I'm a better user, right? ====<br /> <br /> :Absolutely not. Post quality is far more important than post quantity. Users making a lot of senseless posts are often considered worse users, or spammers.<br /> <br /> ==== I have made some posts but my post count did not increase. Why? ====<br /> <br /> :When you post in some of the forums, such as the Test Forum, Mafia Forum, and the Fun Factory, it does not count towards your post count.<br /> <br /> ==== What is my post rating? ====<br /> <br /> :On certain forums, the posts you make can be rated by other users on a scale from 1-6 in terms of how useful or helpful it is (1 = :(, 6 = :D). Your post rating is a weighted average of the ratings you have received, and will only show once you have accumulated a certain number of ratings. This feature existed in the old AoPS and does not exist today. Today, users can upvote and downvote posts.<br /> <br /> ==== When can I rate posts? ====<br /> <br /> :You will be able to rate posts after posting ten messages.<br /> <br /> ==== Who can see my post rating? ====<br /> <br /> :Only moderators, and administrators.<br /> <br /> ==== I rated a post but the post rating does not appear on the post. Why?====<br /> :One or two ratings is not enough to determine a post's overall rating. Therefore, a post has to receive a preset number of ratings before the overall rating of the post appears. Ratings will continue to affect the user's overall post rating even if they are not yet displayed on the post.<br /> <br /> ==== How does AoPS select moderators? ====<br /> <br /> :When a new moderator is needed in the forums, AoPS administrators first check if any current moderators could serve as a moderator of the forum which needs a moderator. Should none be found, AoPS administrators and/or other moderators scour the forum looking for productive users. They may also ask for suggestions from other moderators or trusted users on the site. Once they have pinpointed a possible candidate based on their long term usage of the site, productive posts in the forum, and having no recent behavioral issues, that user is asked if he or she would like to moderate the forum. <br /> <br /> :Less active forums often have no moderator. Inappropriate posts should be reported by users and administrators will take appropriate action.<br /> <br /> :AoPS receives MANY requests to be a moderator. As they receive so many, it is possible that you won't get a response should you request to be one. Also, AoPS very rarely makes someone a mod for asking to be one, so '''please do not ask'''.<br /> <br /> ==== I believe a post needs corrective action. What should I do? ====<br /> <br /> :If you believe a post needs moderative action, you may report it by clicking the &quot;!&quot; icon on the bottom-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called &quot;backseat moderation&quot; and is considered rude.<br /> <br /> ==== How long of a non-commented thread is considered reviving? ====<br /> <br /> :If any post is still on-topic and isn't spammy or anything, it isn't considered reviving. The definition of reviving in the Games forum is 1 month. However, everyone has a different period of time that they consider reviving. In general, apply common sense.<br /> <br /> ==== Someone is marking all my posts as spam, what should I do? ====<br /> <br /> :It happens to everyone. There's really not much you can do.<br /> <br /> ==== Are posts marked spam more often than good? ====<br /> <br /> :No. The most common rating is 6 cubes. We understand that many posts are rated 1 when they shouldn't be. We also know that many posts are rated a 6 when they shouldn't be. It pretty much all averages out in the end. The best way to safeguard yourself is not to complain about it! In fact, most other members cannot see your rating. If you want to make a mark here, let your post quality do the talking.<br /> <br /> ==== How do I post images? ====<br /> :There are limited attachment options for posts. Attachments have an overall size limit, and will be deleted as they get old. Attachments also may be deleted during any server move or software upgrade or change. You may instead wish to host images on another site and embed them in to your post using the [img] tags. The general format is [img]{url to image}[/img], excluding the braces. There are a number of image hosting sites, including:<br /> :* [http://imgur.com/ Imgur]<br /> :* [http://photobucket.com Photobucket]<br /> ::*In thumbnail view<br /> :::*Hover over image and click the text box labeled IMG code. It will automatically copy to your clipboard<br /> :::*Paste to your message<br /> ::*In image view<br /> :::*Look for the '''Links''' box which should appear at the right side of your screen<br /> :::*Click the box labeled IMG Code<br /> :::*Copy the text<br /> :::*Paste to your message<br /> :* [http://imageshack.com ImageShack]<br /> :* [http://minus.com minus.com]<br /> ::*Go to image you wish to embed<br /> ::*Click the share tab<br /> ::*Copy the contents of the Forum Code text<br /> ::*Paste to your message<br /> :* [http://bayfiles.com bayfiles.com]<br /> :* [http://picasaweb.google.com Picasa]<br /> :** Go to the image you want to embed<br /> :** Right click on the image<br /> :** Select Copy Image Location on Firefox, Copy Image URL on Chrome, or similar on IE/Opera &lt;!-- NEEDS VERIFICATION ABOUT THE OPERA/IE PART! --&gt;<br /> :** Paste into your message, surrounding the pasted text with [img] and [/img] tags<br /> :* [http://www.flickr.com Flickr]<br /> :* [http://www.tinypic.com TinyPic]<br /> <br /> See also:<br /> [http://www.artofproblemsolving.com/Wiki/index.php/Direct_Image_Link Direct Image Link]<br /> <br /> == Blogs ==<br /> ==== How come I can't create a blog? ====<br /> :One needs to have at least 5 posts in order to make a blog.<br /> ==== How do I make my blog look nice? ====<br /> :Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br /> <br /> == Contests ==<br /> ==== Where can I find past contest questions and solutions? ====<br /> :In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br /> <br /> ==== How do I get problems onto the contest page? ====<br /> <br /> :Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br /> <br /> ==== Who can I ask to add posts to the contests section? ====<br /> :Any one of the members in the the [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=group&amp;g=417 RManagers] group.<br /> <br /> ==== What are the guidelines for posting problems to be added to the contests section? ====<br /> :Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&amp;t=195579 guidelines in this post].<br /> <br /> ==== Why is the wiki missing many contest questions? ====<br /> :Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations. See, for example, [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1391106#p1391106 this post].<br /> <br /> ==== What if I find an error on a problem? ====<br /> Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=426693 this thread]<br /> <br /> == LaTeX, Asymptote, GeoGebra ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\LaTeX&lt;/math&gt; is a typesetting markup language that is useful to produce properly formatted mathematical and scientific expressions.<br /> <br /> ==== How can I download LaTeX to use on the forums? ====<br /> <br /> :There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br /> <br /> ==== How can I download LaTeX for personal use? ====<br /> :You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br /> <br /> ==== Where can I find a list of LaTeX commands? ====<br /> :See [[LaTeX:Symbols|here]].<br /> <br /> ==== Where can I test LaTeX commands? ====<br /> <br /> :[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. <br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br /> <br /> ==== How can I draw 3D diagrams? ====<br /> <br /> :See [[Asymptote: 3D graphics]].<br /> <br /> ==== What is the cse5 package? ==== <br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br /> <br /> ==== Can I convert diagrams from GeoGebra to other formats? ====<br /> :It is possible to export GeoGebra to [[Asymptote (Vector Graphics Language)|Asymptote]] (see [[User:Azjps/geogebra|here]]), PsTricks, and PGF/TikZ; and GeoGebra animations into .gif or video files. <br /> <br /> == AoPSWiki ==<br /> ==== Is there a guide for wiki syntax? ====<br /> <br /> :See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br /> <br /> ==== What do I do if I see a mistake in the wiki? ====<br /> <br /> :Edit the page and correct the error! You can edit most pages on the wiki. Click the &quot;edit&quot; button on the right sidebar to edit a page.<br /> <br /> ==== Why can't I edit the wiki? ====<br /> <br /> You must be a registered user to edit. To be registered, make sure you give a correct email, and activate your account.<br /> <br /> == Miscellaneous ==<br /> ==== Is it possible to join the AoPS Staff? ====<br /> <br /> :Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br /> <br /> ==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br /> <br /> :You must have graduated from high school, or at least be in the second term of your senior year.<br /> <br /> ==What do some of the acronyms such as &quot;OP&quot; stand for?==<br /> *'''AFK'''- Away from keyboard<br /> *'''AoPS'''- Art of Problem Solving, the website you're on right now!<br /> *'''AIME'''- American Invitational Mathematics Examination<br /> *'''AMC'''- American Math Competititions<br /> *'''ATM'''- At the Moment<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''ID(R)K'''-I Don't (Really) Know<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''MOP'''- Mathematical Olympiad (Summer) Program<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh.<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''QED'''- Quod erat demonstrandum, Latin for Which was to be proven; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''rotfl''' - Rolling on the floor laughing<br /> *'''sa''' - sa<br /> *'''smh''' - Shaking my head<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WLOG'''- Without loss of generality<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''xD'''- Bursting Laugh<br /> <br /> == FTW! ==<br /> <br /> ==== How do you access FTW? ====<br /> You can access FTW by clicking FTW! on the green bar at the top of the page.<br /> <br /> <br /> ==== Did FTW miscount my number of games?====<br /> <br /> No! However, the (Overall) rating statistics do not count games with less than 6 problems or less than 15 seconds.<br /> <br /> For example, if you have played 30 games, but not all of them were 6 problems or higher, then you will still be muted.<br /> <br /> == School ==<br /> <br /> ==== What if I want to drop out of a class? ====<br /> :For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br /> <br /> ==== What if I miss a class? ====<br /> :There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board.<br /> <br /> ==== Is there audio or video in class? ====<br /> There is no audio or video in the class. The classes are completely text based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video. <br /> <br /> ====I feel like joining! What are my class choices? ====<br /> :[http://www.artofproblemsolving.com/School/classlist.php Class List] [http://www.artofproblemsolving.com/School/index.php?page=school.instructors Instructors List]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=Asymptote:_Drawing_part_2&diff=71006 Asymptote: Drawing part 2 2015-07-06T17:27:05Z <p>MathLearner01: fixed code for filldraw (how did this go unnoticed for such a long time?), and fixed typoes</p> <hr /> <div>{{asymptote}}<br /> <br /> In this article, we create circular objects.<br /> <br /> &lt;tt&gt;draw(circle((0,0),5));&lt;/tt&gt;<br /> <br /> We see that the first '''draw()''' command creates the circle, which uses the '''circle()''' command. Within the circle command, we see the center point is located at the cartesian plane point (0,0), and it has a radius of 5.<br /> <br /> This code produces:<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),5));<br /> &lt;/asy&gt;<br /> <br /> Once again, we can fix certain attributes to this code:<br /> <br /> &lt;tt&gt;draw(circle((0,0),5),red+linewidth(1));&lt;/tt&gt;<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),5),red+linewidth(1));<br /> &lt;/asy&gt;<br /> <br /> And we can fill the inside:<br /> <br /> &lt;tt&gt;filldraw(circle((0,0),5),green,red+linewidth(1));&lt;/tt&gt;<br /> <br /> &lt;asy&gt;<br /> filldraw(circle((0,0),5),green,red+linewidth(1));<br /> &lt;/asy&gt;<br /> <br /> Another rounded figure we can create is the ellipse. <br /> <br /> &lt;tt&gt;draw(ellipse((0,0),5,3));&lt;/tt&gt;<br /> <br /> In this case, the (0,0) is the center of the ellipse, the 5 is the length of the major axis and the 3 is the length of the minor axis. This results in:<br /> <br /> &lt;asy&gt;<br /> draw(ellipse((0,0),5,3));<br /> &lt;/asy&gt;<br /> <br /> Once again, we can fix attributes and fill the inside.<br /> <br /> &lt;asy&gt;<br /> filldraw(ellipse((0,0),5,3),green,red+linewidth(1));<br /> &lt;/asy&gt;</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_14&diff=70930 1985 AIME Problems/Problem 14 2015-06-29T15:56:52Z <p>MathLearner01: /* Problem */</p> <hr /> <div>== Problem ==<br /> In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded &lt;math&gt;1&lt;/math&gt; point, the loser got &lt;math&gt;0&lt;/math&gt; points, and each of the two players earned &lt;math&gt;\frac{1}{2}&lt;/math&gt; point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?<br /> <br /> == Solution ==<br /> Let us suppose for convenience that there were &lt;math&gt;n + 10&lt;/math&gt; players over all. Among the &lt;math&gt;n&lt;/math&gt; players not in the weakest 10 there were &lt;math&gt;n \choose 2&lt;/math&gt; games played and thus &lt;math&gt;n \choose 2&lt;/math&gt; points earned. By the givens, this means that these &lt;math&gt;n&lt;/math&gt; players also earned &lt;math&gt;n \choose 2&lt;/math&gt; points against our weakest 10. Now, the 10 weakest players playing amongst themselves played &lt;math&gt;{10 \choose 2} = 45&lt;/math&gt; games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger &lt;math&gt;n&lt;/math&gt; players. Since every point earned falls into one of these categories, It follows that the total number of points earned was &lt;math&gt;2{n \choose 2} + 90 = n^2 - n + 90&lt;/math&gt;. However, there was one point earned per game, and there were a total of &lt;math&gt;{n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}&lt;/math&gt; games played and thus &lt;math&gt;\frac{(n + 10)(n + 9)}{2}&lt;/math&gt; points earned. So we have &lt;math&gt;n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}&lt;/math&gt; so &lt;math&gt;2n^2 - 2n + 180 = n^2 + 19n + 90&lt;/math&gt; and &lt;math&gt;n^2 -21n + 90 = 0&lt;/math&gt; and &lt;math&gt;n = 6&lt;/math&gt; or &lt;math&gt;n = 15&lt;/math&gt;. Now, note that the top &lt;math&gt;n&lt;/math&gt; players got &lt;math&gt;n(n - 1)&lt;/math&gt; points in total (by our previous calculation) for an average of &lt;math&gt;n - 1&lt;/math&gt;, while the bottom 10 got 90 points total, for an average of 9. Thus we must have &lt;math&gt;n &gt; 10&lt;/math&gt;, so &lt;math&gt;n = 15&lt;/math&gt; and the answer is &lt;math&gt;15 + 10 = \boxed{025}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1985|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=70929 1983 AIME Problems/Problem 13 2015-06-29T15:51:03Z <p>MathLearner01: fixed typo</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;For &lt;math&gt;\{1, 2, 3, \ldots, n\}&lt;/math&gt; and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for &lt;math&gt;\{1, 2, 4, 6,9\}&lt;/math&gt; is &lt;math&gt;9-6+4-2+1=6&lt;/math&gt; and for &lt;math&gt;\{5\}&lt;/math&gt; it is simply &lt;math&gt;5&lt;/math&gt;. Find the sum of all such alternating sums for &lt;math&gt;n=7&lt;/math&gt;.&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;S&lt;/math&gt; be a non-[[empty set | empty]] [[subset]] of &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt;. <br /> <br /> Then the alternating sum of &lt;math&gt;S&lt;/math&gt; plus the alternating sum of &lt;math&gt;S&lt;/math&gt; with 7 included is 7. In mathematical terms, &lt;math&gt;S+ (S\cup 7)=7&lt;/math&gt;. This is true because when we take an alternating sum, each term of &lt;math&gt;S&lt;/math&gt; has the opposite sign of each corresponding term of &lt;math&gt;S\cup 7&lt;/math&gt;.<br /> <br /> Because there are &lt;math&gt;63&lt;/math&gt; of these pairs, the sum of all possible subsets of our given set is &lt;math&gt;63*7&lt;/math&gt;. However, we forgot to include the subset that only contains &lt;math&gt;7&lt;/math&gt;, so our answer is &lt;math&gt;64\cdot 7=\boxed{448}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Consider a given subset &lt;math&gt;T&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt; that contains 7; then there is a subset &lt;math&gt;T'&lt;/math&gt; which contains all the elements of &lt;math&gt;T&lt;/math&gt; except for 7, and only those. Since each element of &lt;math&gt;T'&lt;/math&gt; has one element fewer preceding it than it does in &lt;math&gt;T&lt;/math&gt;, their signs are opposite; so the sum of the alternating sums of &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt; is equal to 7. There are &lt;math&gt;2^6&lt;/math&gt; subsets containing 7, so our answer is &lt;math&gt;7 * 2^6 = \boxed{448}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=Riemann_zeta_function&diff=69789 Riemann zeta function 2015-04-01T00:50:32Z <p>MathLearner01: /* Zeros of the Zeta Function */</p> <hr /> <div>The '''Riemann zeta function''' is a function very important in<br /> [[number theory]]. In particular, the [[Riemann Hypothesis]] is a conjecture<br /> about the roots of the zeta function.<br /> <br /> The function is defined by <br /> &lt;cmath&gt;\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=<br /> 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots&lt;/cmath&gt;<br /> when the [[real part]] &lt;math&gt;\Re(s)&lt;/math&gt; is greater than 1. (When &lt;math&gt;\Re(s) \le<br /> 1&lt;/math&gt; the [[series]] '''does not''' converge, but it can be extended to all<br /> [[complex number]]s except &lt;math&gt;s = 1&lt;/math&gt;&amp;mdash;see<br /> [[#Extending_the_zeta_function | below]].)<br /> <br /> [[Leonhard Euler]] showed that when &lt;math&gt;s=2&lt;/math&gt;, the sum is equal to<br /> &lt;math&gt;\frac{\pi^2}{6}&lt;/math&gt;. Euler also found that since every number is the product<br /> of a unique combination of [[prime number]]s, the zeta function can be<br /> expressed as an infinite product: <br /> &lt;cmath&gt;\zeta(s) = \left(\frac{1}{(2^0)^s} + \frac{1}{(2^1)^s}+<br /> \frac{1}{(2^2)^s} + \cdots\right) \left(\frac{1}{(3^0)^s} + \frac{1}<br /> {(3^1)^s} + \frac{1}{(3^2)^s} + \cdots\right) \left(\frac{1}{(5^0)^s}<br /> + \frac{1}{(5^1)^s} + \frac{1}{(5^2)^s} + \cdots\right) \cdots.&lt;/cmath&gt;<br /> By summing up each of these [[geometric series]] in parentheses, we arrive<br /> at the following identity (the [[Euler Product]]): <br /> &lt;cmath&gt;\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p \text{ prime}}<br /> (1-p^{-s})^{-1}.&lt;/cmath&gt;<br /> <br /> This gives a hint of why an [[analysis | analytic]] object like the<br /> zeta function could be related to number theoretic results.<br /> <br /> <br /> == Extending the zeta function ==<br /> <br /> The most important properties of the zeta function are based on the<br /> fact that it extends to a [[meromorphic]] function on the full<br /> [[complex plane]] which is [[holomorphic]] except at &lt;math&gt;s=1&lt;/math&gt;, where<br /> there is a [[simple pole]] of [[residue]] 1. Let us see how this is done.<br /> <br /> First, we wish to extend &lt;math&gt;\zeta(s)&lt;/math&gt; to the strip &lt;math&gt;\Re(s)&gt;0&lt;/math&gt;. To do this,<br /> we introduce the ''alternating zeta function''<br /> &lt;cmath&gt;\zeta_a(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} .&lt;/cmath&gt;<br /> For &lt;math&gt;\Re(s) &gt; 1&lt;/math&gt;, we have<br /> &lt;cmath&gt;\zeta(s) = \zeta_a(s) + \frac{2}{2^s} + \frac{2}{4^s} + \frac{2}{6^s}<br /> + \cdots = \zeta_a(s) + 2^{1-s}{\zeta(s)}, &lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt; \zeta(s) = \frac{1}{1- 2^{1-s}} \zeta_a(s) . &lt;/cmath&gt;<br /> We may thus use the alternating zeta function to extend the zeta<br /> function.<br /> <br /> '''Proposition.''' The series &lt;math&gt;\zeta_a(s)&lt;/math&gt; converges whenever<br /> &lt;math&gt;\Re(s) \ge 0&lt;/math&gt;.<br /> <br /> ''Proof.'' We have<br /> &lt;cmath&gt; \zeta_a(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}<br /> = \sum_{n=1}^{\infty} \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} .&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt; \lvert d(x^{-s})/dx \rvert = \lvert s x^{-s-1} \rvert \le<br /> \left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert &lt;/cmath&gt;<br /> for &lt;math&gt;x \in [2n-1, 2n]&lt;/math&gt;, it follows that<br /> &lt;cmath&gt; \left\lvert \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} \right\rvert<br /> \le \left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert . &lt;/cmath&gt;<br /> Since &lt;math&gt;\Re(s+1) &gt; 1&lt;/math&gt;, the series in question converges.<br /> &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Now we can extend the zeta function.<br /> <br /> '''Theorem 1.''' The function &lt;math&gt;\zeta(s)&lt;/math&gt; has a meromorphic extension<br /> to &lt;math&gt;\Re(s) &gt; 0&lt;/math&gt;, and it is holomorphic there except at &lt;math&gt;s=1&lt;/math&gt;, where<br /> it has a simple pole of residue 1.<br /> <br /> ''Proof.'' For &lt;math&gt;s \neq 1&lt;/math&gt;, we have the extension<br /> &lt;cmath&gt; \zeta(s) - \frac{1}{s-1} = \frac{1}{1 - 2^{1-s}}\zeta_a(s) -<br /> \frac{1}{s-1} . &lt;/cmath&gt;<br /> For &lt;math&gt;s= 1&lt;/math&gt;, we have<br /> &lt;cmath&gt; \lim_{s\to 1} \frac{(s-1) \zeta_a(s)}{1- 2^{1-s}} = \lim_{s\to1}<br /> \frac{\zeta_a(s)}{\log 2 \cdot 2^{1-s}} = \frac{\zeta_a(1)}{<br /> \log 2} ,&lt;/cmath&gt;<br /> by [[l'Hôpital's Rule]], so the pole at &lt;math&gt;s=1&lt;/math&gt; is simple, and its<br /> residue is &lt;math&gt;\zeta_a(1) / \log 2&lt;/math&gt;.<br /> <br /> Now, in general,<br /> &lt;cmath&gt; \frac{d^n(\log t)}{(dt)^n} = \frac{(-1)^{n-1}(n-1)!}{t^n} . &lt;/cmath&gt;<br /> It follows that the [[Taylor series]] expansion of &lt;math&gt;\log x&lt;/math&gt;<br /> about &lt;math&gt;x=1&lt;/math&gt; is<br /> &lt;cmath&gt; \sum_{k=0}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} . &lt;/cmath&gt;<br /> It follows that &lt;math&gt;\zeta_a(1) = \log 2&lt;/math&gt;. Thus the residue of the<br /> pole is 1. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> The next step is the<br /> [[functional equation for the zeta function|functional equation]]:<br /> Let<br /> &lt;cmath&gt;\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s).&lt;/cmath&gt;<br /> Then &lt;math&gt;\xi(s)=\xi(1-s)&lt;/math&gt;. This gives us a meromorphic continuation<br /> of &lt;math&gt;\zeta(s)&lt;/math&gt; to all of &lt;math&gt;\mathbb{C}&lt;/math&gt;.<br /> <br /> == Zeroes of the Zeta Function ==<br /> <br /> Using the Euler product, it is not too difficult to show that<br /> &lt;math&gt;\zeta(s)&lt;/math&gt; has no zeros for &lt;math&gt;\Re s &gt; 1&lt;/math&gt;. Indeed, suppose this<br /> is the case; let &lt;math&gt;x = \Re s&lt;/math&gt;. Then<br /> &lt;cmath&gt; \begin{align*}<br /> \sum_{p} \bigl\lvert \log \lvert (1-p^{-s})^{-1} \rvert \bigr\rvert<br /> &amp;= \sum_p \log \lvert p^s \rvert - \log \lvert p^s -1 \rvert<br /> = \sum_p \int\limits_{\lvert p^s - 1 \rvert}^{\lvert p^s \rvert}<br /> \frac{dt}{t} \\<br /> &amp;\approx \sum_p \frac{1}{\lvert p^s - 1 \rvert} \\<br /> &amp;&lt; \sum_p 1/p^s,<br /> \end{align*} &lt;/cmath&gt;<br /> which converges. It follows that<br /> &lt;cmath&gt; \prod_p (1 - p^{-s})^{-1} \neq 0 . &lt;/cmath&gt;<br /> <br /> From the functional equation<br /> &lt;cmath&gt; \zeta(1-s) = (2\pi)^{-s} 2 \cos(\pi s/2) \Gamma(s) \zeta(s), &lt;/cmath&gt;<br /> it is evident that the zeta function has zeroes at &lt;math&gt;s= -2n&lt;/math&gt;, for<br /> &lt;math&gt;n&lt;/math&gt; a postive integer. These are called the trivial zeros.<br /> Since the [[gamma function]] has no zeros, it follows that these<br /> are the only zeros with real part less than 0.<br /> <br /> In 1859, Georg Friedrich Bernhard Riemann, after whom the<br /> function is named, established the functional equation and<br /> proved that &lt;math&gt;\zeta(s)&lt;/math&gt; has infinitely many zeros in the strip<br /> &lt;math&gt;0 \le \Re(s) \le 1&lt;/math&gt;. He conjectured that they all lie on the<br /> line &lt;math&gt;\Re s = 1/2&lt;/math&gt;. This is the famous [[Riemann Hypothesis]],<br /> and to this day it remains one of the great unsolved problems<br /> of mathematics. Recently it has been proven that the function's<br /> first ten trillion zeros lie on the line<br /> &lt;math&gt;\Re s = 1/2&lt;/math&gt;[http://mathworld.wolfram.com/RiemannHypothesis.html], but<br /> proof of the Riemann hypothesis still eludes us.<br /> <br /> In 1896, Jacque Hadamard and Charles-Jean de la Vallée Poussin<br /> independently proved that &lt;math&gt;\zeta(s)&lt;/math&gt; has no zeros on the line<br /> &lt;math&gt;\Re(s) = 1&lt;/math&gt;. From this they proved the [[prime number theorem]].<br /> We prove this result here.<br /> <br /> We first define the phi function,<br /> &lt;cmath&gt; \phi(s) = \sum_{p \text{ prime}} \frac{\log p}{p^s} . &lt;/cmath&gt;<br /> <br /> '''Theorem 2.''' The function &lt;math&gt;\phi(s)&lt;/math&gt; has a meromorphic<br /> continuation to &lt;math&gt;\Re(s) &gt; 1/2&lt;/math&gt; with simple poles at<br /> the poles and zeros of &lt;math&gt;\zeta(s)&lt;/math&gt;, and with no other poles.<br /> The continuation is<br /> &lt;cmath&gt; \phi(s) = - \frac{\zeta'(s)}{\zeta(s)} - \sum_p<br /> \frac{\log p}{p^s(p^s-1)} .&lt;/cmath&gt;<br /> <br /> ''Proof.'' It follows from the Euler product formula that for<br /> &lt;math&gt;\Re(s) &gt; 1&lt;/math&gt;,<br /> &lt;cmath&gt; \begin{align*}<br /> \frac{\zeta'(s)}{\zeta(s)} &amp;= \sum_p \frac{d (1- p^{-s})^{-1}/ds}<br /> {(1-p^{-s})^{-1}} = -\sum_p \frac{d(1-p^{-s})/ds}{(1-p^{-s})} \\<br /> &amp;= -\sum_p \frac{\log p \cdot p^{-s}}{1-p^{-s}} \\<br /> &amp;= -\sum_p \frac{\log p}{p^s -1 } <br /> = - \phi(s) - \sum_p \frac{\log p}{p^s (p^s - 1)}.<br /> \end{align*} &lt;/cmath&gt;<br /> Since &lt;math&gt;\sum_p \frac{\log p}{p^s (p^s- 1)}&lt;/math&gt; converges when<br /> &lt;math&gt;\Re s &gt; 1/2&lt;/math&gt;, the theorem statement follows. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> Now we proceed to the main result.<br /> <br /> '''Theorem 3.''' The zeta function has no zeros on the<br /> line &lt;math&gt;\Re(s) = 1&lt;/math&gt;.<br /> <br /> ''Proof.'' We use the fact that &lt;math&gt;\zeta(\bar s) = \overline{<br /> \zeta(s)}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g(s) = 1/ \zeta(s)&lt;/math&gt;. Then 1 is a zero of &lt;math&gt;g&lt;/math&gt; of order 1.<br /> Thus<br /> &lt;cmath&gt;<br /> \lim_{\epsilon \to 0} \epsilon \phi(1+\epsilon)<br /> = \lim_{\epsilon \to 0} -\frac{\epsilon(1/g(1+\epsilon))'}{1/g(1+<br /> \epsilon)}<br /> = \lim_{\epsilon \to 0} \frac{\epsilon g'(1+\epsilon)}{g(1+\epsilon)}<br /> = 1 . &lt;/cmath&gt;<br /> <br /> Suppose now that &lt;math&gt;1+ki&lt;/math&gt; and &lt;math&gt;1+2ki&lt;/math&gt; are zeros of &lt;math&gt;\zeta(s)&lt;/math&gt; of<br /> &lt;math&gt;\zeta(s)&lt;/math&gt; of order &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, respectively. (Note that &lt;math&gt;m&lt;/math&gt;<br /> and &lt;math&gt;n&lt;/math&gt; may be zero.)<br /> Then<br /> &lt;cmath&gt; \begin{align*}<br /> \lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm ki)<br /> &amp;= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm ki)}{<br /> \zeta(1+\epsilon \pm ki)} = -m , \\<br /> \lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm 2ki)<br /> &amp;= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm 2ki)}{<br /> \zeta(1+\epsilon \pm 2ki)} = -n .<br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Now for real, positive &lt;math&gt;\epsilon&lt;/math&gt;,<br /> &lt;cmath&gt; \sum_{a=0}^{4} \binom{4}{a} \phi(1+\epsilon - 2ki+4kai)<br /> = \sum_p \frac{\log p}{p^{1+\epsilon}} (p^{ki/2} + p^{-ki/2})^4<br /> \ge 0, &lt;/cmath&gt;<br /> since &lt;math&gt;p^{-ki/2} = \overline{p^{ki/2}}&lt;/math&gt;. It follows that<br /> &lt;cmath&gt; -2n - 8m + 6 \ge 0 . &lt;/cmath&gt;<br /> Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; must be nonnegative integers, it follows that &lt;math&gt;m=0&lt;/math&gt;.<br /> Thus &lt;math&gt;\zeta(1+ki) \neq 0&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; was arbitrary, it follows<br /> that &lt;math&gt;\zeta(s)&lt;/math&gt; has no zeros on the line &lt;math&gt;\Re s = 1&lt;/math&gt;.<br /> &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> == Resources ==<br /> <br /> * Koch, Helmut (trans. David Kramer), ''Number Theory: Algebraic Numbers and Functions.'' AMS 2000, ISBN 0-8218-2054-0.<br /> <br /> == See also ==<br /> <br /> * [[Riemann Hypothesis]]<br /> * [[Prime number theorem]]<br /> <br /> [[Category:Number theory]]<br /> [[Category:Analytic number theory]]<br /> [[Category:Complex analysis]]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems&diff=66090 2005 AMC 10A Problems 2014-11-26T18:48:02Z <p>MathLearner01: /* Problem 8 */</p> <hr /> <div>==Problem 1==<br /> While eating out, Mike and Joe each tipped their server &lt;math&gt;2&lt;/math&gt; dollars. Mike tipped &lt;math&gt;10\%&lt;/math&gt; of his bill and Joe tipped &lt;math&gt;20\%&lt;/math&gt; of his bill. What was the difference, in dollars between their bills? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> For each pair of real numbers &lt;math&gt;a\neq b&lt;/math&gt;, define the operation &lt;math&gt;\star&lt;/math&gt; as<br /> <br /> &lt;math&gt; (a \star b) = \frac{a+b}{a-b} &lt;/math&gt;.<br /> <br /> What is the value of &lt;math&gt; ((1 \star 2) \star 3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> The equations &lt;math&gt; 2x + 7 = 3 &lt;/math&gt; and &lt;math&gt; bx - 10 = -2 &lt;/math&gt; have the same solution &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;b&lt;/math&gt;? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } -8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> A rectangle with a diagonal of length &lt;math&gt;x&lt;/math&gt; is twice as long as it is wide. What is the area of the rectangle? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A store normally sells windows at &lt;dollar/&gt;100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> The average (mean) of &lt;math&gt;20&lt;/math&gt; numbers is &lt;math&gt;30&lt;/math&gt;, and the average of &lt;math&gt;30&lt;/math&gt; other numbers is &lt;math&gt;20&lt;/math&gt;. What is the average of all &lt;math&gt;50&lt;/math&gt; numbers?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Josh and Mike live &lt;math&gt;13&lt;/math&gt; miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> In the figure, the length of side &lt;math&gt;AB&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\sqrt{50}&lt;/math&gt; and &lt;math&gt;BE=1&lt;/math&gt;. What is the area of the inner square &lt;math&gt;EFGH&lt;/math&gt;?<br /> <br /> [[File:AMC102005Aq.png]]<br /> <br /> &lt;math&gt; \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Three tiles are marked &lt;math&gt;X&lt;/math&gt; and two other tiles are marked &lt;math&gt;O&lt;/math&gt;. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads &lt;math&gt;XOXOX&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> A wooden cube &lt;math&gt;n&lt;/math&gt; units on a side is painted red on all six faces and then cut into &lt;math&gt;n^3&lt;/math&gt; unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length &lt;math&gt;2&lt;/math&gt;?<br /> <br /> [[Image:2005amc10a12.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy the following condition:<br /> <br /> &lt;math&gt; (130n)^{50} &gt; n^{100} &gt; 2^{200} &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> How many positive cubes divide &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; ?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is &lt;math&gt;6&lt;/math&gt;. How many two-digit numbers have this property? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> In the five-sided star shown, the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are replaced by the numbers &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;9&lt;/math&gt;, although not necessarily in this order. The sums of the numbers at the ends of the line segments &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt;, and &lt;math&gt;EA&lt;/math&gt; form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence? <br /> <br /> [[Image:2005amc10a17.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point &lt;math&gt;B&lt;/math&gt; from the line on which the bases of the original squares were placed?<br /> <br /> &lt;asy&gt;<br /> unitsize(1inch);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> draw((0,0)--((1/3) + 3*(1/2),0));<br /> fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7));<br /> draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle);<br /> draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle);<br /> draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle);<br /> draw((2,0)--(2 + (1/3) + (3/2),0));<br /> draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle);<br /> draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle);<br /> label(&quot;$B$&quot;,((1/6) + (1/2),(1/2)),NW);<br /> label(&quot;$B$&quot;,((2/3) + 2 + (1/4),(29/30)),NNE);<br /> draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3));<br /> fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7));<br /> draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> An equiangular octagon has four sides of length 1 and four sides of length &lt;math&gt;\frac{\sqrt{2}}{2}&lt;/math&gt;, arranged so that no two consecutive sides have the same length. What is the area of the octagon?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> For how many positive integers &lt;math&gt;n&lt;/math&gt; does &lt;math&gt; 1+2+...+n &lt;/math&gt; evenly divide &lt;math&gt;6n&lt;/math&gt;? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of the &lt;math&gt;2005&lt;/math&gt; smallest positive multiples of &lt;math&gt;4&lt;/math&gt;, and let &lt;math&gt;T&lt;/math&gt; be the set of the &lt;math&gt;2005&lt;/math&gt; smallest positive multiples of &lt;math&gt;6&lt;/math&gt;. How many elements are common to &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 166\qquad \mathrm{(B) \ } 333\qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ } 668\qquad \mathrm{(E) \ } 1001 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> Let &lt;math&gt;AB&lt;/math&gt; be a diameter of a circle and let &lt;math&gt;C&lt;/math&gt; be a point on &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;2\cdot AC=BC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be points on the circle such that &lt;math&gt;DC \perp AB&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; is a second diameter. What is the ratio of the area of &lt;math&gt;\triangle DCE&lt;/math&gt; to the area of &lt;math&gt;\triangle ABD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2.5cm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=3;<br /> pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);<br /> pair D=dir(aCos(C.x)), E=(-D.x,-D.y);<br /> draw(A--B--D--cycle);<br /> draw(D--E--C);<br /> draw(unitcircle,white);<br /> drawline(D,C);<br /> dot(O);<br /> clip(unitcircle);<br /> draw(unitcircle);<br /> label(&quot;$E$&quot;,E,SSE);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$D$&quot;,D,NNW);<br /> label(&quot;$C$&quot;,C,SW);<br /> draw(rightanglemark(D,C,B,2));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{6} \qquad \mathrm{(B) \ } \frac{1}{4} \qquad \mathrm{(C) \ } \frac{1}{3} \qquad \mathrm{(D) \ } \frac{1}{2} \qquad \mathrm{(E) \ } \frac{2}{3} &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> For each positive integer &lt;math&gt; m &gt; 1 &lt;/math&gt;, let &lt;math&gt;P(m)&lt;/math&gt; denote the greatest prime factor of &lt;math&gt;m&lt;/math&gt;. For how many positive integers &lt;math&gt;n&lt;/math&gt; is it true that both &lt;math&gt; P(n) = \sqrt{n} &lt;/math&gt; and &lt;math&gt; P(n+48) = \sqrt{n+48} &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> In &lt;math&gt;ABC&lt;/math&gt; we have &lt;math&gt; AB = 25 &lt;/math&gt;, &lt;math&gt; BC = 39 &lt;/math&gt;, and &lt;math&gt;AC=42&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively, with &lt;math&gt; AD = 19 &lt;/math&gt; and &lt;math&gt; AE = 14 &lt;/math&gt;. What is the ratio of the area of triangle &lt;math&gt;ADE&lt;/math&gt; to the area of the quadrilateral &lt;math&gt;BCED&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 &lt;/math&gt;<br /> <br /> [[2005 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2005|ab=A|before=[[2004 AMC 10B Problems]]|after=[[2005 AMC 10B Problems]]}}<br /> *[[AMC 10 Problems and Solutions]]<br /> * [[AMC Problems and Solutions]]<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8&diff=66079 2014 AMC 8 2014-11-26T14:15:55Z <p>MathLearner01: deadline has been reached: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=614431</p> <hr /> <div>*[[2014 AMC 8 Problems]]<br /> * [[2014 AMC 8 Answer Key]]<br /> ** [[2014 AMC 8 Problems/Problem 1]]<br /> ** [[2014 AMC 8 Problems/Problem 2]]<br /> ** [[2014 AMC 8 Problems/Problem 3]]<br /> ** [[2014 AMC 8 Problems/Problem 4]]<br /> ** [[2014 AMC 8 Problems/Problem 5]]<br /> ** [[2014 AMC 8 Problems/Problem 6]]<br /> ** [[2014 AMC 8 Problems/Problem 7]]<br /> ** [[2014 AMC 8 Problems/Problem 8]]<br /> ** [[2014 AMC 8 Problems/Problem 9]]<br /> ** [[2014 AMC 8 Problems/Problem 10]]<br /> ** [[2014 AMC 8 Problems/Problem 11]]<br /> ** [[2014 AMC 8 Problems/Problem 12]]<br /> ** [[2014 AMC 8 Problems/Problem 13]]<br /> ** [[2014 AMC 8 Problems/Problem 14]]<br /> ** [[2014 AMC 8 Problems/Problem 15]]<br /> ** [[2014 AMC 8 Problems/Problem 16]]<br /> ** [[2014 AMC 8 Problems/Problem 17]]<br /> ** [[2014 AMC 8 Problems/Problem 18]]<br /> ** [[2014 AMC 8 Problems/Problem 19]]<br /> ** [[2014 AMC 8 Problems/Problem 20]]<br /> ** [[2014 AMC 8 Problems/Problem 21]]<br /> ** [[2014 AMC 8 Problems/Problem 22]]<br /> ** [[2014 AMC 8 Problems/Problem 23]]<br /> ** [[2014 AMC 8 Problems/Problem 24]]<br /> ** [[2014 AMC 8 Problems/Problem 25]]</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=User:MathLearner01&diff=63442 User:MathLearner01 2014-09-22T14:13:14Z <p>MathLearner01: </p> <hr /> <div>MathLearner01's user page has moved. The new user page can be found here: http://www.artofproblemsolving.com/Wiki/index.php?title=User:MathLearner01.</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1998_PMWC_Problems/Problem_I2&diff=62646 1998 PMWC Problems/Problem I2 2014-07-23T19:06:51Z <p>MathLearner01: Created page with &quot;==Problem I2== Triangular numbers and Square numbers can be represented in the following manner: &lt;asy&gt; int triangle(pair z, int n){ for(int i = 0; i &lt; n; ++i){ for(int j = n-i; ...&quot;</p> <hr /> <div>==Problem I2==<br /> Triangular numbers and Square numbers can be represented in the following manner:<br /> <br /> &lt;asy&gt;<br /> int triangle(pair z, int n){<br /> for(int i = 0; i &lt; n; ++i){<br /> for(int j = n-i; j &gt; 0; --j){<br /> dot((z.x+j -1 + i/2 ,z.y + i*sqrt(3)/2));<br /> }<br /> }<br /> return 0;<br /> }<br /> triangle((0,0),2);<br /> label(&quot;3&quot;,(0.5,0),2S);<br /> triangle((5,0),3);<br /> label(&quot;6&quot;,(6,0),2S);<br /> triangle((11,0),4);<br /> label(&quot;10&quot;,(12.5,0),2S);<br /> int squ(pair z, int n){<br /> for(int i = 0; i &lt; n; ++i){<br /> for(int j = 0; j &lt; n; ++j){<br /> dot((z.x + i , z.y + j));<br /> }<br /> }<br /> return 0;<br /> }<br /> squ((0,-6),2);<br /> label(&quot;4&quot;,(0.5,-6),2S);<br /> squ((5,-6),3);<br /> label(&quot;9&quot;,(6,-6),2S);<br /> squ((11,-6),4);<br /> label(&quot;16&quot;,(12.5,-6),2S);<br /> //Credit to chezbgone2 for the diagram&lt;/asy&gt;<br /> <br /> Find a pair of consecutive Triangular Numbers and the difference between a pair of consecutive Square Numbers whose difference are both 11. What is the sum of these four numbers?<br /> <br /> ==Solution==<br /> <br /> We want &lt;math&gt;n^2-(n-1)^2=11&lt;/math&gt; or &lt;math&gt;2n-1=11&lt;/math&gt; for &lt;math&gt;n=6&lt;/math&gt;. Therefore the two square numbers are &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;36&lt;/math&gt;.<br /> <br /> The &lt;math&gt;n&lt;/math&gt;th triangular number is equal to &lt;math&gt;n(n+1)/2&lt;/math&gt;. Therefore we want &lt;math&gt;\frac{n(n+1)}{2}=\frac{(n-1)n}{2}+11&lt;/math&gt; or &lt;math&gt;n(n+1)=n(n-1)+22&lt;/math&gt;. Solving gives &lt;math&gt;n=11&lt;/math&gt; so our two numbers are &lt;math&gt;55&lt;/math&gt; and &lt;math&gt;66&lt;/math&gt;.<br /> <br /> The sum of all our numbers is therefore &lt;math&gt;25+36+55+66=\boxed{182}&lt;/math&gt;.<br /> <br /> (Credit to 54math for the solution [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=586102 here])</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_2&diff=62620 1991 AHSME Problems/Problem 2 2014-07-22T18:42:09Z <p>MathLearner01: Added solution and changed format of answer choices.</p> <hr /> <div>==Problem==<br /> &lt;math&gt;|3-\pi|=&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)\ }\frac{1}{7}\qquad\textbf{(B)\ }0.14\qquad\textbf{(C)\ }3-\pi\qquad\textbf{(D)\ }3+\pi\qquad\textbf{(E)\ }\pi-3 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since &lt;math&gt;\pi&gt;3&lt;/math&gt;, the value of &lt;math&gt;\abs{3-\pi}&lt;/math&gt; is negative. The absolute value of a negative quantity is the negative quantity multiplied by &lt;math&gt;-1&lt;/math&gt;, or the negative of that quantity. Therefore &lt;math&gt;|3-\pi|=-(3-\pi)=\pi-3&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{E}}&lt;/math&gt;<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_1&diff=62619 1991 AHSME Problems/Problem 1 2014-07-22T17:51:21Z <p>MathLearner01: Added solution and changed format of answer choices.</p> <hr /> <div>==Problem==<br /> If for any three distinct numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; we define &lt;math&gt;f(a,b,c)=\frac{c+a}{c-b}&lt;/math&gt;, then &lt;math&gt;f(1,-2,-3)&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf {(A) } -2 \qquad \textbf {(B) } -\frac{2}{5} \qquad \textbf {(C) } -\frac{1}{4} \qquad \textbf {(D) } \frac{2}{5} \qquad \textbf {(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If we plug in &lt;math&gt;1&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;-2&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;-3&lt;/math&gt; as &lt;math&gt;c&lt;/math&gt; in the expression &lt;math&gt;\frac{c+a}{c-b}&lt;/math&gt;, then we get &lt;math&gt;\frac{-3+1}{-3-(-2)}=\frac{-2}{-1}=2&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{A}}&lt;/math&gt;.<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_5&diff=62495 1950 AHSME Problems/Problem 5 2014-07-13T23:50:05Z <p>MathLearner01: /* Problem */</p> <hr /> <div>== Problem==<br /> <br /> If five geometric means are inserted between &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;5832&lt;/math&gt;, the fifth term in the geometric series:<br /> <br /> &lt;math&gt; \textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> We can let the common ratio of the geometric sequence be &lt;math&gt;r&lt;/math&gt;. &lt;math&gt;5832&lt;/math&gt; is given to be the seventh term in the geometric sequence as there are five terms between it and &lt;math&gt;a_1&lt;/math&gt; if we consider &lt;math&gt;a_1=8&lt;/math&gt;.<br /> By the formula for each term in a geometric sequence, we find that &lt;math&gt;a_n=a_1r^{n-1}&lt;/math&gt; or &lt;math&gt;(5382)=(8)r^6&lt;/math&gt;<br /> We divide by eight to find:<br /> &lt;math&gt;r^6=729&lt;/math&gt;<br /> &lt;math&gt;r=\pm 3&lt;/math&gt;<br /> <br /> Because &lt;math&gt;a_2&lt;/math&gt; will not be between &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;5832&lt;/math&gt; if &lt;math&gt;r=-3&lt;/math&gt; we can discard it as an extraneous solution. We find &lt;math&gt;r=3&lt;/math&gt; and &lt;math&gt;a_5=a_1r^4=(8)(3)^4=\boxed{\textbf{(A)}\ 648}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME 50p box|year=1950|num-b=4|num-a=6}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_37&diff=62460 1950 AHSME Problems/Problem 37 2014-07-07T01:45:59Z <p>MathLearner01: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> If &lt;math&gt; y \equal{} \log_{a}{x}&lt;/math&gt;, &lt;math&gt; a &gt; 1&lt;/math&gt;, which of the following statements is incorrect?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\<br /> \textbf{(B)}\ \text{If }x=a,y=1 \qquad\\<br /> \textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\<br /> \textbf{(D)}\ \text{If }0&lt;x&lt;a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\<br /> \textbf{(E)}\ \text{Only some of the above statements are correct}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let us first check <br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{If }x=1,y=0&lt;/math&gt;. Rewriting into exponential form gives &lt;math&gt;a^0=1&lt;/math&gt;. This is certainly correct.<br /> <br /> &lt;math&gt;\textbf{(B)}\ \text{If }x=a,y=1&lt;/math&gt;. Rewriting gives &lt;math&gt;a^1=a&lt;/math&gt;. This is also certainly correct.<br /> <br /> &lt;math&gt;\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)}&lt;/math&gt;. Rewriting gives &lt;math&gt;a^{\text{complex number}}=-1&lt;/math&gt;. Because &lt;math&gt;a&gt;1&lt;/math&gt;, therefore positive, there is no real solution to &lt;math&gt;y&lt;/math&gt;, but there is imaginary.<br /> <br /> &lt;math&gt;\textbf{(D)}\ \text{If }0&lt;x&lt;a,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero}&lt;/math&gt;. Rewriting: &lt;math&gt;a^y=x&lt;/math&gt; such that &lt;math&gt;x&lt;a&lt;/math&gt;. Well, a power of &lt;math&gt;a&lt;/math&gt; can be less than &lt;math&gt;a&lt;/math&gt; only if &lt;math&gt;y&lt;1&lt;/math&gt;. And we observe, &lt;math&gt;y&lt;/math&gt; has no lower asymptote, because it is perfectly possible to have &lt;math&gt;y&lt;/math&gt; be &lt;math&gt;-100000000&lt;/math&gt;; in fact, the lower &lt;math&gt;y&lt;/math&gt; gets, &lt;math&gt;x&lt;/math&gt; approaches &lt;math&gt;0&lt;/math&gt;. This is also correct.<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Only some of the above statements are correct}&lt;/math&gt;. This is the last option, so it follows that our answer is &lt;math&gt;\boxed{\textbf{(E)}\ \text{Only some of the above statements are correct}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AHSME 50p box|year=1950|num-b=36|num-a=38}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> MathLearner01 https://artofproblemsolving.com/wiki/index.php?title=User:MathLearner01&diff=62420 User:MathLearner01 2014-06-30T19:13:58Z <p>MathLearner01: Created page with &quot;http://www.artofproblemsolving.com/Wiki/index.php?title=User:MathLearner01&quot;</p> <hr /> <div>http://www.artofproblemsolving.com/Wiki/index.php?title=User:MathLearner01</div> MathLearner01