https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Programjames1&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-20T10:24:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_USAJMO_Problems/Problem_1&diff=119374 2015 USAJMO Problems/Problem 1 2020-03-14T14:34:35Z <p>Programjames1: /* Solution 2 (INCORRECT) */</p> <hr /> <div>==Problem==<br /> Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of &lt;math&gt;2015&lt;/math&gt; distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.<br /> <br /> ==Solution==<br /> Let the set be &lt;math&gt;{-1007, -1006, ...0,1,2...1006, 1007}&lt;/math&gt;, namely all the consecutive integers from &lt;math&gt;-1007&lt;/math&gt; to &lt;math&gt;1007&lt;/math&gt;. Notice that the operation we are applying in this problem does not change the sum or the mean of the set, which is &lt;math&gt;0&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;1007&lt;/math&gt; pairs of opposite integers &lt;math&gt;\{a,-a\}&lt;/math&gt;. After the first two elements are chosen, there are at least &lt;math&gt;1005&lt;/math&gt; such pairs. For each such pair we perform the operation of average, hence reducing these &lt;math&gt;2010&lt;/math&gt; elements to &lt;math&gt;0&lt;/math&gt;. Then use the other &lt;math&gt;5&lt;/math&gt; elements together with three &lt;math&gt;0&lt;/math&gt;'s produced to form the group of eight: &lt;math&gt;{a_1,a_2,a_3,a_4,a_5,a_6=0,a_7=0,a_8=0}&lt;/math&gt;, and perform the operation in the following order: &lt;cmath&gt;(a_1,a_2)\to(m_1,m_1), (a_3,a_4)\to(m_2,m_2), (a_5,a_6)\to(m_3,m_3), (a_7,a_8)\to(m_4,m_4),&lt;/cmath&gt; where &lt;math&gt;m_i=\frac{a_i+a_{i+1}}{2}&lt;/math&gt;. Then, &lt;math&gt;(m_1,m_2)\to(m_{11}, m_{11})&lt;/math&gt; for two groups, &lt;math&gt;(m_3,m_4)\to(m_{12}, m_{12})&lt;/math&gt; for the other two groups, and finally &lt;math&gt;(m_{11},m_{12})\to(m_{111}, m_{111})&lt;/math&gt; for all the eight elements. Since the sum of the eight-group is &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;m_{111}&lt;/math&gt; must also be &lt;math&gt;0&lt;/math&gt;. Therefore, all the elements are reduced to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> The key to the algorithm is to form a &lt;math&gt;2^k&lt;/math&gt; subset, which is guaranteed to be reducible to all the members of the same value, namely the mean. Then before that, if we could always choose &lt;math&gt;M\ge N-2^k&lt;/math&gt; members to form pairs, each yielding the average of the total group, then all the members are reduced to the average. Under the condition that two arbitrary elements are chosen first, we need only &lt;math&gt;N\ge4&lt;/math&gt; to guarantee this result. But for &lt;math&gt;N=2&lt;/math&gt; the first operation leads to equal elements, so &lt;math&gt;N=3&lt;/math&gt; is the only case when all the members may not be reduced to average.<br /> <br /> Sidenote: Actually, for &lt;math&gt;N=3&lt;/math&gt;, the members are all reduced to the average, as the sum of the terms is constant and does not change.<br /> <br /> == Solution 2 ==<br /> Consider any arithmetic sequence. WLOG, let it be &lt;math&gt;s = (1, 2, 3, \dots, 2015)&lt;/math&gt;, i.e. &lt;math&gt;s_i = i\ \forall\ 1\le i\le 2015&lt;/math&gt;. Define the move &lt;math&gt;(x, y)&lt;/math&gt; as replacing the numbers located at positions &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with their mean, assuming &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct. If they are the same integer, define it as not making a move. Now, suppose the initial move &lt;math&gt;m_0 = (a, b)&lt;/math&gt;. If &lt;math&gt;a+b=2016&lt;/math&gt;, then &lt;math&gt;s_a = s_b = \frac{a+b}{2} = 1008&lt;/math&gt;. Then, applying the moves<br /> &lt;cmath&gt;m_1 = (1, 2015), m_2 = (2, 2014), \dots, m_{1007} = (1007, 1009),&lt;/cmath&gt;<br /> we get &lt;math&gt;s_1 = s_2 = \cdots = s_{2015} = 1008&lt;/math&gt;.<br /> Otherwise, suppose &lt;math&gt;a+b\ne 2016&lt;/math&gt;. Then consider the following &lt;math&gt;3&lt;/math&gt; moves:<br /> &lt;cmath&gt;m_{-1} = (2016-a, 2016-b), m_{-2} = (a, 2016-a), m_{-3} = (b, 2016-b)&lt;/cmath&gt;<br /> We have &lt;math&gt;m_{-1}&lt;/math&gt; makes &lt;math&gt;s_{2016-a} = s_{2016-b} = 2016 - \frac{a+b}{2}&lt;/math&gt;. So, &lt;math&gt;m_{-2}&lt;/math&gt; makes<br /> &lt;cmath&gt;s_a = s_{2016-a} = \frac{2016 - \frac{a+b}{2} + \frac{a+b}{2}}{2} = 1008&lt;/cmath&gt;<br /> Similarly for &lt;math&gt;m_{-3}&lt;/math&gt; with &lt;math&gt;s_b&lt;/math&gt; and &lt;math&gt;s_{2016-b}&lt;/math&gt;. Then, finishing up with the moves<br /> &lt;cmath&gt;m_1 = (1, 2015), m_2 = (2, 2014), \dots, m_{1007} = (1007, 1009),&lt;/cmath&gt;<br /> we get &lt;math&gt;s_1 = s_2 = \cdots = s_{2015} = 1008&lt;/math&gt;.<br /> <br /> ==Solution 3 (INCORRECT)==<br /> Let the set be &lt;math&gt;{a_1,a_2,\dots ,a_{2015}}&lt;/math&gt;, where all the terms are nonnegative. Note that the sum of all the terms in this sequence will always be the same after any amount of moves. To prove this, let &lt;math&gt;i,j&lt;/math&gt; be integers with &lt;math&gt;1\le i&lt;j\le 2015&lt;/math&gt;, and we have &lt;math&gt;a_i+a_j = \frac{a_i+a_j}{2}+\frac{a_i+a_j}{2}&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;a_ia_j \le (\frac{a_i+a_j}{2})(\frac{a_i+a_j}{2})&lt;/math&gt; by AM-GM, so the product of all the terms will not decrease after any number of moves. However, the product will only stay the same when &lt;math&gt;a_i=a_j&lt;/math&gt;, so the product will always increase if &lt;math&gt;a_i\ne a_j&lt;/math&gt;.<br /> <br /> Finally, note that &lt;math&gt;a_1a_2\dots a_{2015}\le (\frac{a_1+a_2+\dots +a_{2015}}{2015})^{2015}&lt;/math&gt; by AM-GM, so because &lt;math&gt;a_1+a_2+\dots +a_{2015}&lt;/math&gt; is fixed, there is a maximum product that is reached after a finite number of moves as the product increases. This product is reached when &lt;math&gt;a_1=a_2=\dots =a_{2015}&lt;/math&gt;, so we are done.<br /> <br /> This solution is incorrect; the product may take an infinite number of moves to reach the maximum (for example, consider the sequence &lt;math&gt;1, 3, 4, 5... 2016&lt;/math&gt;)<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2015|beforetext=|before=First Problem|num-a=2}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2015_USAJMO_Problems/Problem_1&diff=119373 2015 USAJMO Problems/Problem 1 2020-03-14T14:34:22Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem==<br /> Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of &lt;math&gt;2015&lt;/math&gt; distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.<br /> <br /> ==Solution==<br /> Let the set be &lt;math&gt;{-1007, -1006, ...0,1,2...1006, 1007}&lt;/math&gt;, namely all the consecutive integers from &lt;math&gt;-1007&lt;/math&gt; to &lt;math&gt;1007&lt;/math&gt;. Notice that the operation we are applying in this problem does not change the sum or the mean of the set, which is &lt;math&gt;0&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;1007&lt;/math&gt; pairs of opposite integers &lt;math&gt;\{a,-a\}&lt;/math&gt;. After the first two elements are chosen, there are at least &lt;math&gt;1005&lt;/math&gt; such pairs. For each such pair we perform the operation of average, hence reducing these &lt;math&gt;2010&lt;/math&gt; elements to &lt;math&gt;0&lt;/math&gt;. Then use the other &lt;math&gt;5&lt;/math&gt; elements together with three &lt;math&gt;0&lt;/math&gt;'s produced to form the group of eight: &lt;math&gt;{a_1,a_2,a_3,a_4,a_5,a_6=0,a_7=0,a_8=0}&lt;/math&gt;, and perform the operation in the following order: &lt;cmath&gt;(a_1,a_2)\to(m_1,m_1), (a_3,a_4)\to(m_2,m_2), (a_5,a_6)\to(m_3,m_3), (a_7,a_8)\to(m_4,m_4),&lt;/cmath&gt; where &lt;math&gt;m_i=\frac{a_i+a_{i+1}}{2}&lt;/math&gt;. Then, &lt;math&gt;(m_1,m_2)\to(m_{11}, m_{11})&lt;/math&gt; for two groups, &lt;math&gt;(m_3,m_4)\to(m_{12}, m_{12})&lt;/math&gt; for the other two groups, and finally &lt;math&gt;(m_{11},m_{12})\to(m_{111}, m_{111})&lt;/math&gt; for all the eight elements. Since the sum of the eight-group is &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;m_{111}&lt;/math&gt; must also be &lt;math&gt;0&lt;/math&gt;. Therefore, all the elements are reduced to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> The key to the algorithm is to form a &lt;math&gt;2^k&lt;/math&gt; subset, which is guaranteed to be reducible to all the members of the same value, namely the mean. Then before that, if we could always choose &lt;math&gt;M\ge N-2^k&lt;/math&gt; members to form pairs, each yielding the average of the total group, then all the members are reduced to the average. Under the condition that two arbitrary elements are chosen first, we need only &lt;math&gt;N\ge4&lt;/math&gt; to guarantee this result. But for &lt;math&gt;N=2&lt;/math&gt; the first operation leads to equal elements, so &lt;math&gt;N=3&lt;/math&gt; is the only case when all the members may not be reduced to average.<br /> <br /> Sidenote: Actually, for &lt;math&gt;N=3&lt;/math&gt;, the members are all reduced to the average, as the sum of the terms is constant and does not change.<br /> <br /> == Solution 2 ==<br /> Consider any arithmetic sequence. WLOG, let it be &lt;math&gt;s = (1, 2, 3, \dots, 2015)&lt;/math&gt;, i.e. &lt;math&gt;s_i = i\ \forall\ 1\le i\le 2015&lt;/math&gt;. Define the move &lt;math&gt;(x, y)&lt;/math&gt; as replacing the numbers located at positions &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with their mean, assuming &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are distinct. If they are the same integer, define it as not making a move. Now, suppose the initial move &lt;math&gt;m_0 = (a, b)&lt;/math&gt;. If &lt;math&gt;a+b=2016&lt;/math&gt;, then &lt;math&gt;s_a = s_b = \frac{a+b}{2} = 1008&lt;/math&gt;. Then, applying the moves<br /> &lt;cmath&gt;m_1 = (1, 2015), m_2 = (2, 2014), \dots, m_{1007} = (1007, 1009),&lt;/cmath&gt;<br /> we get &lt;math&gt;s_1 = s_2 = \cdots = s_{2015} = 1008&lt;/math&gt;.<br /> Otherwise, suppose &lt;math&gt;a+b\ne 2016&lt;/math&gt;. Then consider the following &lt;math&gt;3&lt;/math&gt; moves:<br /> &lt;cmath&gt;m_{-1} = (2016-a, 2016-b), m_{-2} = (a, 2016-a), m_{-3} = (b, 2016-b)&lt;/cmath&gt;<br /> We have &lt;math&gt;m_{-1}&lt;/math&gt; makes &lt;math&gt;s_{2016-a} = s_{2016-b} = 2016 - \frac{a+b}{2}&lt;/math&gt;. So, &lt;math&gt;m_{-2}&lt;/math&gt; makes<br /> &lt;cmath&gt;s_a = s_{2016-a} = \frac{2016 - \frac{a+b}{2} + \frac{a+b}{2}}{2} = 1008&lt;/cmath&gt;<br /> Similarly for &lt;math&gt;m_{-3}&lt;/math&gt; with &lt;math&gt;s_b&lt;/math&gt; and &lt;math&gt;s_{2016-b}&lt;/math&gt;. Then, finishing up with the moves<br /> &lt;cmath&gt;m_1 = (1, 2015), m_2 = (2, 2014), \dots, m_{1007} = (1007, 1009),&lt;/cmath&gt;<br /> we get &lt;math&gt;s_1 = s_2 = \cdots = s_{2015} = 1008&lt;/math&gt;.<br /> <br /> ==Solution 2 (INCORRECT)==<br /> Let the set be &lt;math&gt;{a_1,a_2,\dots ,a_{2015}}&lt;/math&gt;, where all the terms are nonnegative. Note that the sum of all the terms in this sequence will always be the same after any amount of moves. To prove this, let &lt;math&gt;i,j&lt;/math&gt; be integers with &lt;math&gt;1\le i&lt;j\le 2015&lt;/math&gt;, and we have &lt;math&gt;a_i+a_j = \frac{a_i+a_j}{2}+\frac{a_i+a_j}{2}&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;a_ia_j \le (\frac{a_i+a_j}{2})(\frac{a_i+a_j}{2})&lt;/math&gt; by AM-GM, so the product of all the terms will not decrease after any number of moves. However, the product will only stay the same when &lt;math&gt;a_i=a_j&lt;/math&gt;, so the product will always increase if &lt;math&gt;a_i\ne a_j&lt;/math&gt;.<br /> <br /> Finally, note that &lt;math&gt;a_1a_2\dots a_{2015}\le (\frac{a_1+a_2+\dots +a_{2015}}{2015})^{2015}&lt;/math&gt; by AM-GM, so because &lt;math&gt;a_1+a_2+\dots +a_{2015}&lt;/math&gt; is fixed, there is a maximum product that is reached after a finite number of moves as the product increases. This product is reached when &lt;math&gt;a_1=a_2=\dots =a_{2015}&lt;/math&gt;, so we are done.<br /> <br /> This solution is incorrect; the product may take an infinite number of moves to reach the maximum (for example, consider the sequence &lt;math&gt;1, 3, 4, 5... 2016&lt;/math&gt;)<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2015|beforetext=|before=First Problem|num-a=2}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_7&diff=119314 2020 AIME I Problems/Problem 7 2020-03-13T17:55:55Z <p>Programjames1: /* Solution 3 (Vandermonde's identity) */</p> <hr /> <div><br /> == Problem ==<br /> A club consisting of &lt;math&gt;11&lt;/math&gt; men and &lt;math&gt;12&lt;/math&gt; women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as &lt;math&gt;1&lt;/math&gt; member or as many as &lt;math&gt;23&lt;/math&gt; members. Let &lt;math&gt;N&lt;/math&gt; be the number of such committees that can be formed. Find the sum of the prime numbers that divide &lt;math&gt;N.&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We will be selecting girls, but &lt;i&gt;not&lt;/i&gt; selecting boys. We claim that the amount of girls selected and the amount of guys not selected adds to &lt;math&gt;12&lt;/math&gt;. This is easy to see: if &lt;math&gt;k&lt;/math&gt; women were chosen, then &lt;math&gt;k + (11 - k + 1) = 12&lt;/math&gt;. Therefore, we simply take &lt;math&gt;\binom{23}{12} \implies \boxed{081}&lt;/math&gt;. ~awang11's sol<br /> <br /> == Solution 2 (Bash) ==<br /> <br /> We casework on the amount of men on the committee. <br /> <br /> If there are no men in the committee, there are &lt;math&gt;\dbinom{12}{1}&lt;/math&gt; ways to pick the women on the committee, for a total of &lt;math&gt;\dbinom{11}{0} \cdot \dbinom{12}{1}&lt;/math&gt;. Notice that &lt;math&gt;\dbinom{11}{0}&lt;/math&gt; is equal to &lt;math&gt;\dbinom{11}{11}&lt;/math&gt;, so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all females must also be picked, for a total of &lt;math&gt;\dbinom{12}{12}&lt;/math&gt;. Therefore, these cases can be combined to &lt;cmath&gt;\dbinom{11}{0} \cdot (\dbinom{12}{1} + \dbinom{12}{12})&lt;/cmath&gt; Since &lt;math&gt;\dbinom{12}{12} = \dbinom{12}{0}&lt;/math&gt;, and &lt;math&gt;\dbinom{12}{0} + \dbinom{12}{1} = \dbinom{13}{1}&lt;/math&gt;, we can further simplify this to &lt;cmath&gt;\dbinom{11}{0} \cdot \dbinom{13}{1}&lt;/cmath&gt;<br /> <br /> All other cases proceed similarly. For example, the case with one men or ten men is equal to &lt;math&gt;\dbinom{11}{1} \cdot \dbinom{13}{2}&lt;/math&gt;. Now, if we factor out a &lt;math&gt;13&lt;/math&gt;, then all cases except the first two have a factor of &lt;math&gt;121&lt;/math&gt;, so we can factor this out too to make our computation slightly easier. The first two cases (with &lt;math&gt;13&lt;/math&gt; factored out) give &lt;math&gt;1+66=67&lt;/math&gt;, and the rest gives &lt;math&gt;121(10+75+270+504) = 103,939&lt;/math&gt;. Adding the &lt;math&gt;67&lt;/math&gt; gives &lt;math&gt;104,006&lt;/math&gt;. Now, we can test for prime factors. We know there is a factor of &lt;math&gt;2&lt;/math&gt;, and the rest is &lt;math&gt;52,003&lt;/math&gt;. We can also factor out a &lt;math&gt;7&lt;/math&gt;, for &lt;math&gt;7,429&lt;/math&gt;, and the rest is &lt;math&gt;17 \cdot 19 \cdot 23&lt;/math&gt;. Adding up all the prime factors gives &lt;math&gt;2+7+13+17+19+23 = \boxed{081}&lt;/math&gt;.<br /> <br /> == Solution 3 (Vandermonde's identity) ==<br /> Applying Vandermonde's identity by setting &lt;math&gt;m=12&lt;/math&gt;, &lt;math&gt;n=11&lt;/math&gt;, and &lt;math&gt;r=11&lt;/math&gt;, we obtain &lt;math&gt;\binom{23}{11}\implies&lt;/math&gt; &lt;math&gt;\boxed{081}&lt;/math&gt;.<br /> ~Lcz<br /> <br /> &lt;h3&gt;Short Proof&lt;/h3&gt;<br /> Consider the following setup:<br /> &lt;asy&gt;<br /> size(1000, 100);<br /> for(int i=0; i&lt;23; ++i){<br /> dot((i, 0));<br /> }<br /> draw((10.5, -1.5)--(10.5, 1.5), dashed);<br /> &lt;/asy&gt;<br /> The dots to the left represent the men, and the dots to the right represent the women. Now, suppose we put a mark on &lt;math&gt;11&lt;/math&gt; people (the &lt;math&gt;*&lt;/math&gt;). Those to the left of the dashed line get to be &quot;in&quot; on the committee if they have a mark. Those on the right side of the dashed line are already on the committee, but if they're marked they get forcibly evicted from it. If there were &lt;math&gt;x&lt;/math&gt; people marked on the left, there ends up being &lt;math&gt;12-(11-x) = x+1&lt;/math&gt; people not marked on the right. Circles represent those in the committee.<br /> &lt;asy&gt;<br /> size(1000, 100);<br /> for(int i=0; i&lt;23; ++i){<br /> dot((i, 0));<br /> }<br /> for(int i=0; i&lt;23; ++i){<br /> if(i%2==0){<br /> if(i &gt;= 11){<br /> draw(circle((i, 0), 0.25));<br /> }<br /> continue;<br /> }<br /> label(&quot;$*$&quot;, (i,0.5), N);<br /> if(i &lt; 11){<br /> draw(circle((i, 0), 0.25));<br /> }<br /> }<br /> draw((10.5, -1.5)--(10.5, 1.5), dashed);<br /> &lt;/asy&gt;<br /> <br /> We have our bijection, so the number of ways will be &lt;math&gt;\binom{23}{11}&lt;/math&gt;.<br /> <br /> ~programjames1<br /> <br /> == Solution 4 ==<br /> Notice that the committee can consist of &lt;math&gt;k&lt;/math&gt; boys and &lt;math&gt;k+1&lt;/math&gt; girls. Summing over all possible &lt;math&gt;k&lt;/math&gt; gives <br /> &lt;cmath&gt;\sum_{k=0}^{11}\binom{11}{k}\binom{12}{k+1}=\binom{11}{0}\binom{12}{1}+\binom{11}{1}\binom{12}{2}+\cdots + \binom{11}{11}\binom{12}{12}&lt;/cmath&gt;<br /> Using the identity &lt;math&gt;\binom{n}{k}=\binom{n}{n-k}&lt;/math&gt;, and Pascal's Identity &lt;math&gt;\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}&lt;/math&gt;, we get<br /> &lt;cmath&gt;\sum_{k=0}^{11}\binom{11}{k}\binom{12}{k+1}=\binom{12}{12}+\binom{12}{1}\left(\binom{11}{0}+\binom{11}{1}\right)+\cdots &lt;/cmath&gt;<br /> &lt;cmath&gt;=\binom{12}{0}^2+\binom{12}{1}^2+\binom{12}{2}^2+\binom{12}{3}^2+\binom{12}{4}^2+\binom{12}{5}^2+\frac{\binom{12}{6}^2}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\frac{1}{2}\sum_{k=0}^{12}\binom{12}{k}^2&lt;/cmath&gt;<br /> Using the identity &lt;math&gt;\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}&lt;/math&gt;, this simplifies to <br /> &lt;cmath&gt;\frac{1}{2}\cdot \binom{24}{12}=\frac{24\cdot 23\cdot 22\cdot 21\cdot 20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13}{2\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}=2\cdot 7\cdot 13\cdot 17\cdot 19\cdot 23&lt;/cmath&gt;<br /> so the desired answer is &lt;math&gt;2+7+13+17+19+23=\boxed{081}&lt;/math&gt; ~ktong<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_6&diff=117720 2014 UMO Problems/Problem 6 2020-02-12T21:04:59Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Draw &lt;math&gt;n&lt;/math&gt; rows of &lt;math&gt;2n&lt;/math&gt; equilateral triangles each, stacked on top of each other in a diamond shape, as<br /> shown below when &lt;math&gt;n = 3&lt;/math&gt;. Set point &lt;math&gt;A&lt;/math&gt; as the southwest corner and point &lt;math&gt;B&lt;/math&gt; as the northeast corner.<br /> A step consists of moving from one point to an adjacent point along a drawn line segment, in one of<br /> the four legal directions indicated. A path is a series of steps, starting at &lt;math&gt;A&lt;/math&gt; and ending at &lt;math&gt;B&lt;/math&gt;, such<br /> that no line segment is used twice. One path is drawn below. Prove that for every positive integer &lt;math&gt;n&lt;/math&gt;,<br /> the number of distinct paths is a perfect square. (Note: A perfect square is a number of the form &lt;math&gt;k^2&lt;/math&gt;,<br /> where &lt;math&gt;k&lt;/math&gt; is an integer).<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> path p1=polygon(3),p2=shift(sqrt(3)/2,1/2)*rotate(180)*polygon(3);<br /> for(int k=0;k&lt;3;++k)<br /> {for (int j=0;j&lt;3;++j)<br /> {draw(shift(sqrt(3)*(k+j/2),1.5*j)*p1,black+linewidth(.4));<br /> draw(shift(sqrt(3)*(k+j/2),1.5*j)*p2,black+linewidth(.4));}<br /> }<br /> pair O=(-sqrt(3)/2,-1/2),E=(sqrt(3),0),NW=(-sqrt(3)/2,3/2),NE=(sqrt(3)/2,3/2),SE=(sqrt(3)/2,-3/2);<br /> D(O--(O+E)--(O+E+NE)--(O+2E+NE)--(O+2E+NE+SE)--(O+2E+2NE+SE)--(O+2E+2NE+SE+NW)--(O+3E+2NE+SE+NW)--(O+3E+3NE+SE+NW),black+linewidth(2));<br /> MP(&quot;A&quot;,O,SW);<br /> MP(&quot;B&quot;,(O+3E+3NE+SE+NW),NE);<br /> draw(shift(10,0)*p1,black+linewidth(.4));<br /> draw(shift(10,sqrt(3))*p1,black+linewidth(.4));<br /> draw(shift(12.5,0)*p1,black+linewidth(.4));<br /> draw(shift(12.5,sqrt(3))*p1,black+linewidth(.4));<br /> MP(&quot;Legal Steps&quot;,(11.4,4.2));<br /> draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow());<br /> draw((12.5-sqrt(3)/2,-1/2)--(12.5+sqrt(3)/2,-1/2),arrow=Arrow());<br /> draw((10,sqrt(3)+1)--(10+sqrt(3)/2,sqrt(3)-1/2),arrow=Arrow());<br /> draw((12.5+sqrt(3)/2,sqrt(3)-1/2)--(12.5,sqrt(3)+1),arrow=Arrow());<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> Consider just the equilateral triangle of length &lt;math&gt;n&lt;/math&gt; made by cutting the parallelogram in half. In how many ways can you get from the SW vertex to any point on the opposite edge? Call this value &lt;math&gt;T_n&lt;/math&gt; (in fact, it doesn't matter which edgepoint you go to - there are the same number of paths). Note that &lt;math&gt;T_{n} = 2n\cdot T_{n-1}&lt;/math&gt;. Why? Consider the equilateral triangle of length &lt;math&gt;n-1&lt;/math&gt;. Note that for each edgepoint in the &lt;math&gt;n&lt;/math&gt; triangle, we can get there in &lt;math&gt;2&lt;/math&gt; ways from each of the &lt;math&gt;n&lt;/math&gt; edgepoints in the &lt;math&gt;n-1&lt;/math&gt; triangle. Now it isn't hard to see that the general formula for &lt;math&gt;T&lt;/math&gt; is &lt;math&gt;T_n = 2^nn!&lt;/math&gt;.<br /> <br /> Now to find the number of ways to get from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;. It's the sum of the number of ways to get from &lt;math&gt;B&lt;/math&gt; to each point along the diagonal, times &lt;math&gt;T_n&lt;/math&gt;. But the former is just &lt;math&gt;2n\cdot T_{n-1} = T_n&lt;/math&gt;. Therefore, the total number of paths is &lt;math&gt;T_n^2&lt;/math&gt;, which is a square number.<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=5|after=Last Problem}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_5&diff=117719 2014 UMO Problems/Problem 5 2020-02-12T20:48:42Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Find all positive real numbers &lt;math&gt;x, y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; that satisfy both of the following equations.<br /> &lt;cmath&gt;<br /> \begin{align*} xyz &amp; = 1\\<br /> x^2 + y^2 + z^2 &amp; = 4x\sqrt{yz}- 2yz \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> == Solution ==<br /> By AM-GM<br /> &lt;cmath&gt;x^2+y^2+z^2 + 2yz\ge x^2 + 4yz\ge 4x\sqrt{yz}&lt;/cmath&gt;<br /> Hence, the second equation implies that &lt;math&gt;y=z&lt;/math&gt; and &lt;math&gt;x^2=4yz\implies x=2y=2z&lt;/math&gt;.<br /> <br /> Now we plug it into the first equation to get &lt;math&gt;(x,y,z) = \left(\sqrt{4}, \frac{\sqrt4}{2}, \frac{\sqrt4}{2}\right)&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_4&diff=117718 2014 UMO Problems/Problem 4 2020-02-12T20:45:27Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Joel is playing with ordered lists of integers in the following way. He starts out with an ordered list<br /> of nonnegative integers. Then, he counts the number of &lt;math&gt;0&lt;/math&gt;’s, &lt;math&gt;1&lt;/math&gt;’s, &lt;math&gt;2&lt;/math&gt;’s, and so on in the list, writing<br /> the counts out as a new list. He stops counting when he has counted everything in the previous list.<br /> Then he takes the second list and applies the same process to get a third list. He repeats this process<br /> indefinitely.<br /> <br /> For example, he could start out with the ordered list &lt;math&gt;(0, 0, 0, 2)&lt;/math&gt;. He counts three &lt;math&gt;0&lt;/math&gt;’s, zero &lt;math&gt;1&lt;/math&gt;’s, and one<br /> &lt;math&gt;2&lt;/math&gt;, and then stops counting, so the second list is &lt;math&gt;(3, 0, 1)&lt;/math&gt; In the second list, he counts one &lt;math&gt;0&lt;/math&gt;, one &lt;math&gt;1&lt;/math&gt;,<br /> zero &lt;math&gt;2&lt;/math&gt;’s, and one &lt;math&gt;3&lt;/math&gt;, so the third list is &lt;math&gt;(1, 1, 0, 1)&lt;/math&gt;. Then he counts one &lt;math&gt;0&lt;/math&gt; and three &lt;math&gt;1&lt;/math&gt;’s, so the fourth list<br /> is &lt;math&gt;(1, 3)&lt;/math&gt;. Here are the first few lists he writes down:<br /> &lt;math&gt;(0,0, 0, 2) \longrightarrow (3, 0,1) \longrightarrow (1, 1, 0, 1) \longrightarrow (1, 3) \longrightarrow \cdots &lt;/math&gt;<br /> If instead he started with &lt;math&gt;(0, 0)&lt;/math&gt;, he would write down:<br /> &lt;math&gt;(0, 0) \longrightarrow (2) \longrightarrow (0, 0, 1) \longrightarrow (2, 1) \longrightarrow \cdots&lt;/math&gt;<br /> If Joel starts out with an arbitrary list of nonnegative integers and then continues this process, there<br /> are certain lists &lt;math&gt;(m, n)&lt;/math&gt; of length two that he might end up writing an infinite number of times. Find<br /> all such pairs &lt;math&gt;(m, n)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;b&gt;Answer:&lt;/b&gt;<br /> &lt;i&gt;All&lt;/i&gt; pairs of &lt;math&gt;(m, n)&lt;/math&gt; work.<br /> <br /> &lt;b&gt;Proof:&lt;/b&gt;<br /> First, note that &lt;math&gt;(2, 2)&lt;/math&gt; repeats forever:<br /> &lt;cmath&gt;(2, 2)\to (0, 0, 2)\to (0, 0, 1)\to (2, 1)\to (0, 1, 1)\to (1,2)\to (0, 1, 1)\qquad (1)&lt;/cmath&gt;<br /> Now suppose we have &lt;math&gt;m\ne n&lt;/math&gt;. Let &lt;math&gt;M =\max(m, n)&lt;/math&gt; We get<br /> &lt;cmath&gt;(m, n)\to (0, 0, \dots, 0, 1, 0, \dots, 0, 1)\to (M-1, 2)\to \cdots \to (M-2, 2) \to \cdots \to (2, 2)\qquad (2)&lt;/cmath&gt;<br /> Otherwise, &lt;math&gt;m=n&lt;/math&gt;. If &lt;math&gt;n &gt; 1&lt;/math&gt; then<br /> &lt;cmath&gt;(n, n)\to (0, 0, \dots, 0, 2)\to (n, 0, 1)\to (1, 1, 0, \dots, 0, 1)\to (n-2, 3)&lt;/cmath&gt;<br /> If &lt;math&gt;n=5&lt;/math&gt; we get<br /> &lt;cmath&gt;(3,3)\to (1,3) \to \text{Apply (2)} \to (2, 2)&lt;/cmath&gt;<br /> Otherwise we get<br /> &lt;cmath&gt;(n-2, 3)\to\text{Apply (2)} \to (2, 2)&lt;/cmath&gt;<br /> Our final case is if &lt;math&gt;n = 1&lt;/math&gt;. Then we have<br /> &lt;cmath&gt;(1, 1)\to (0, 2)\to (0, 0, 1)&lt;/cmath&gt;<br /> which we know repeats from &lt;math&gt;(1)&lt;/math&gt;.<br /> <br /> Therefore, for all &lt;math&gt;(m, n)&lt;/math&gt; it will repeat forever.<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_1&diff=117717 2014 UMO Problems/Problem 1 2020-02-12T20:30:41Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Todd and Allison are playing a game on the grid shown below. At the beginning, an orange stone<br /> is placed in the center intersection on the grid. They take turns, with Todd going first. In each of<br /> Todd’s turns, he must move the orange stone from its current position to a horizontally or vertically<br /> adjacent intersection that is not occupied by a blue stone, and then he places a blue stone in the orange<br /> stone’s previous spot. In each of Allison’s turns, she places a blue stone on exactly one unoccupied<br /> intersection. Todd loses the game when he is forced to move into one of the corner intersections,<br /> labeled by &lt;math&gt;A, B, C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt; in the diagram below. Allison loses if Todd can’t move.<br /> Allison tries to force Todd to lose in as few as turns as possible, and Todd tries to survive as long as<br /> possible. If both of them play as best they can, how many blue stones will be on the board at the end<br /> of the game? (You may assume that Todd always loses.)<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> for(int k=0;k&lt;9;++k)<br /> {<br /> D((0,k)--(8,k),black);D((k,0)--(k,8),black);<br /> }<br /> MP(&quot;A&quot;,(0,8),NW);MP(&quot;B&quot;,(0,0),SW);MP(&quot;C&quot;,(8,0),SE);MP(&quot;D&quot;,(8,8),NE);<br /> D(CR((4,4),.2),black);<br /> fill(CR((4,4),.2),orange);<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> WLOG, assume Todd moves to the left. Note that you can always force Todd to move down or left. Then it's a matter of just counting how many steps it takes to get Todd to the bottom left corner, multiplying by 2, and subtracting one (because you don't play on the last turn). We get &lt;math&gt;15&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|before=First Problem|num-a=2}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_3&diff=117716 2014 UMO Problems/Problem 3 2020-02-12T20:27:47Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Completely describe the set of all right triangles with positive integer-valued legs such that when four<br /> copies of the triangle are arranged in square formation shown below, the incenters of the four triangles<br /> lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center<br /> of the circle inscribed in that triangle.)<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(T,black);<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> pair A, B, C, X, Y, Z, I;<br /> A = shift(10,0)*rotate(53)*(0,0);<br /> B = shift(10,0)*rotate(53)*(4, 0);<br /> C = shift(10,0)*rotate(53)*(4, 3);<br /> I = incenter(A, B, C);<br /> X = foot(I, B, C);<br /> Y = foot(I, A, C);<br /> Z = foot(I, A, B);<br /> D(incircle(A, B, C), black);<br /> D(I -- X, black);<br /> D(I -- Y, black);<br /> D(I -- Z, black);<br /> label(&quot;X&quot;, X, NE);<br /> label(&quot;Y&quot;, Y, W);<br /> label(&quot;Z&quot;, Z, S);<br /> label(&quot;A&quot;, A, S);<br /> label(&quot;B&quot;, B, NE);<br /> label(&quot;C&quot;, C, NW);<br /> label(&quot;I&quot;, I, N);<br /> D(A--Y, green);<br /> D(A--Z, green);<br /> D(B--Z, red);<br /> D(B--X, red);<br /> D(C--Y, blue);<br /> D(C--X, blue);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of a triangle. Drop &lt;math&gt;I&lt;/math&gt; onto the three sides of the triangle, and let the points be &lt;math&gt;X, Y, Z&lt;/math&gt;<br /> Finally, let &lt;math&gt;a = AB, b = BC&lt;/math&gt; so that &lt;math&gt;a &gt; b&lt;/math&gt; and let &lt;math&gt;s = BZ&lt;/math&gt;.<br /> Note that &lt;math&gt;Z&lt;/math&gt; is also a corner of the square, so &lt;math&gt;s = a - b&lt;/math&gt;. But then, &lt;math&gt;AC = CX + AZ = a + b - 2(a-b) = 3b-a&lt;/math&gt;. From the Pythogorean theorem, we also know that &lt;math&gt;AC^2 = a^2+b^2&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;8b^2 = 6ab&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;b = \frac34 a&lt;/cmath&gt;<br /> So, the only solutions are of the form &lt;math&gt;(3k, 4k, 5k).&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_6&diff=117715 2014 UMO Problems/Problem 6 2020-02-12T20:20:56Z <p>Programjames1: /* Problem */</p> <hr /> <div>==Problem ==<br /> <br /> Draw &lt;math&gt;n&lt;/math&gt; rows of &lt;math&gt;2n&lt;/math&gt; equilateral triangles each, stacked on top of each other in a diamond shape, as<br /> shown below when &lt;math&gt;n = 3&lt;/math&gt;. Set point &lt;math&gt;A&lt;/math&gt; as the southwest corner and point &lt;math&gt;B&lt;/math&gt; as the northeast corner.<br /> A step consists of moving from one point to an adjacent point along a drawn line segment, in one of<br /> the four legal directions indicated. A path is a series of steps, starting at &lt;math&gt;A&lt;/math&gt; and ending at &lt;math&gt;B&lt;/math&gt;, such<br /> that no line segment is used twice. One path is drawn below. Prove that for every positive integer &lt;math&gt;n&lt;/math&gt;,<br /> the number of distinct paths is a perfect square. (Note: A perfect square is a number of the form &lt;math&gt;k^2&lt;/math&gt;,<br /> where &lt;math&gt;k&lt;/math&gt; is an integer).<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> path p1=polygon(3),p2=shift(sqrt(3)/2,1/2)*rotate(180)*polygon(3);<br /> for(int k=0;k&lt;3;++k)<br /> {for (int j=0;j&lt;3;++j)<br /> {draw(shift(sqrt(3)*(k+j/2),1.5*j)*p1,black+linewidth(.4));<br /> draw(shift(sqrt(3)*(k+j/2),1.5*j)*p2,black+linewidth(.4));}<br /> }<br /> pair O=(-sqrt(3)/2,-1/2),E=(sqrt(3),0),NW=(-sqrt(3)/2,3/2),NE=(sqrt(3)/2,3/2),SE=(sqrt(3)/2,-3/2);<br /> D(O--(O+E)--(O+E+NE)--(O+2E+NE)--(O+2E+NE+SE)--(O+2E+2NE+SE)--(O+2E+2NE+SE+NW)--(O+3E+2NE+SE+NW)--(O+3E+3NE+SE+NW),black+linewidth(2));<br /> MP(&quot;A&quot;,O,SW);<br /> MP(&quot;B&quot;,(O+3E+3NE+SE+NW),NE);<br /> draw(shift(10,0)*p1,black+linewidth(.4));<br /> draw(shift(10,sqrt(3))*p1,black+linewidth(.4));<br /> draw(shift(12.5,0)*p1,black+linewidth(.4));<br /> draw(shift(12.5,sqrt(3))*p1,black+linewidth(.4));<br /> MP(&quot;Legal Steps&quot;,(11.4,4.2));<br /> draw((10-sqrt(3)/2,-1/2)--(10,1.5-1/2),arrow=Arrow());<br /> draw((12.5-sqrt(3)/2,-1/2)--(12.5+sqrt(3)/2,-1/2),arrow=Arrow());<br /> draw((10,sqrt(3)+1)--(10+sqrt(3)/2,sqrt(3)-1/2),arrow=Arrow());<br /> draw((12.5+sqrt(3)/2,sqrt(3)-1/2)--(12.5,sqrt(3)+1),arrow=Arrow());<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=5|after=Last Problem}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_3&diff=117702 2014 UMO Problems/Problem 3 2020-02-12T17:11:59Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Completely describe the set of all right triangles with positive integer-valued legs such that when four<br /> copies of the triangle are arranged in square formation shown below, the incenters of the four triangles<br /> lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center<br /> of the circle inscribed in that triangle.)<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(T,black);<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(T,black);<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> pair A, B, C, X, Y, Z, I;<br /> A = shift(10,0)*rotate(53)*(0,0);<br /> B = shift(10,0)*rotate(53)*(4, 0);<br /> C = shift(10,0)*rotate(53)*(4, 3);<br /> I = incenter(A, B, C);<br /> X = foot(I, B, C);<br /> Y = foot(I, A, C);<br /> Z = foot(I, A, B);<br /> D(incircle(A, B, C), black);<br /> D(I -- X, black);<br /> D(I -- Y, black);<br /> D(I -- Z, black);<br /> label(&quot;X&quot;, X, NE);<br /> label(&quot;Y&quot;, Y, W);<br /> label(&quot;Z&quot;, Z, S);<br /> label(&quot;A&quot;, A, S);<br /> label(&quot;B&quot;, B, NE);<br /> label(&quot;C&quot;, C, NW);<br /> label(&quot;I&quot;, I, N);<br /> D(A--Y, green);<br /> D(A--Z, green);<br /> D(B--Z, red);<br /> D(B--X, red);<br /> D(C--Y, blue);<br /> D(C--X, blue);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of a triangle. Drop &lt;math&gt;I&lt;/math&gt; onto the three sides of the triangle, and let the points be &lt;math&gt;X, Y, Z&lt;/math&gt;<br /> Finally, let &lt;math&gt;a = AB, b = BC&lt;/math&gt; so that &lt;math&gt;a &gt; b&lt;/math&gt; and let &lt;math&gt;s = BZ&lt;/math&gt;.<br /> Note that &lt;math&gt;Z&lt;/math&gt; is also a corner of the square, so &lt;math&gt;s = a - b&lt;/math&gt;. But then, &lt;math&gt;AC = CX + AZ = a + b - 2(a-b) = 3b-a&lt;/math&gt;. From the Pythogorean theorem, we also know that &lt;math&gt;AC^2 = a^2+b^2&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;8b^2 = 6ab&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;b = \frac34 a&lt;/cmath&gt;<br /> So, the only solutions are of the form &lt;math&gt;(3k, 4k, 5k).&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2014_UMO_Problems/Problem_3&diff=117701 2014 UMO Problems/Problem 3 2020-02-12T17:09:23Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem ==<br /> <br /> Completely describe the set of all right triangles with positive integer-valued legs such that when four<br /> copies of the triangle are arranged in square formation shown below, the incenters of the four triangles<br /> lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center<br /> of the circle inscribed in that triangle.)<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(T,black);<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> size(200);<br /> path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));<br /> D(T,black);<br /> D(shift(10,0)*rotate(53)*T,black);<br /> D(shift(15,5)*rotate(233)*T,black);<br /> D(shift(15,0)*rotate(143)*T,black);<br /> D(shift(10,5)*rotate(323)*T,black);<br /> pair A, B, C, X, Y, Z, I;<br /> A = shift(10,0)*rotate(53)*(0,0);<br /> B = shift(10,0)*rotate(53)*(4, 0);<br /> C = shift(10,0)*rotate(53)*(4, 3);<br /> I = incenter(A, B, C);<br /> X = foot(I, B, C);<br /> Y = foot(I, A, C);<br /> Z = foot(I, A, B);<br /> D(incircle(A, B, C));<br /> D(I -- X);<br /> D(I -- Y);<br /> D(I -- Z);<br /> label(&quot;X&quot;, X, NE,blue);<br /> label(&quot;Y&quot;, Y, W, blue);<br /> label(&quot;Z&quot;, Z, S, blue);<br /> label(&quot;A&quot;, A, S);<br /> label(&quot;B&quot;, B, NE);<br /> label(&quot;C&quot;, C, NW);<br /> label(&quot;I&quot;, I, N, blue);<br /> draw(A--Y, green);<br /> draw(A--Z, yellow);<br /> draw(B--Z, yellow);<br /> draw(B--X, red);<br /> draw(C--Y, green);<br /> draw(C--x, red);<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of a triangle. Drop &lt;math&gt;I&lt;/math&gt; onto the three sides of the triangle, and let the points be &lt;math&gt;X, Y, Z&lt;/math&gt;<br /> Finally, let &lt;math&gt;a = AB, b = BC&lt;/math&gt; so that &lt;math&gt;a &gt; b&lt;/math&gt; and let &lt;math&gt;s = BZ&lt;/math&gt;.<br /> Note that &lt;math&gt;Z&lt;/math&gt; is also a corner of the square, so &lt;math&gt;s = a - b&lt;/math&gt;. But then, &lt;math&gt;AC = CX + AZ = a + b - 2(a-b) = 3b-a&lt;/math&gt;. From the Pythogorean theorem, we also know that &lt;math&gt;AC^2 = a^2+b^2&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;8b^2 = 6ab&lt;/cmath&gt;<br /> &lt;cmath&gt;\Longleftrightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;b = \frac34 a&lt;/cmath&gt;<br /> So, the only solutions are of the form &lt;math&gt;(3k, 4k, 5k).&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{UMO box|year=2014|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=Utah_mathematics_competitions&diff=112851 Utah mathematics competitions 2019-12-14T17:07:58Z <p>Programjames1: /* Utah college mathematics competitions */</p> <hr /> <div>== Utah middle school mathematics competitions==<br /> [[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br /> <br /> * [[Utah MathCounts]] [http://utahmathcounts.org/ (website)] is part of the national [[MathCounts]] program.<br /> <br /> *Mission Math Utah [http://missionmathutah.com/ (website)] is a contest for elementary and middle school students in Utah, run by Utah high school students.<br /> <br /> == Utah high school mathematics competitions==<br /> [[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics <br /> <br /> * [[Utah ARML]] is the Utah team for the [[American Regions Mathematics League]]<br /> *[[Utah Math Olympiad]] [http://www.utahmath.org website]<br /> *[[Utah State Math Contest]] [http://www.math.utah.edu/smc/ website]<br /> <br /> == Utah college mathematics competitions==<br /> * Intermountain Math Contest[https://www.math.utah.edu/~bestvina/intermountain/index.html] is an college proof based math competition. It's format is similar to the Putnam, but the problems are much more approachable. High school students are welcome to compete, and have won the contest in the past.<br /> <br /> == National math contests ==<br /> * [[List of United States elementary school mathematics competitions]] for national contests.<br /> * [[List of United States middle school mathematics competitions]] for national contests.<br /> * [[List of United States high school mathematics competitions]] for national contests.<br /> <br /> <br /> == See also ==<br /> * [[Utah Math Circle]] [http://www.math.utah.edu/mathcircle/ website]<br /> * [[List of mathematics competitions]]<br /> * [[Mathematics competitions resources]]<br /> * [[Mathematics scholarships]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=Utah_mathematics_competitions&diff=112850 Utah mathematics competitions 2019-12-14T17:02:59Z <p>Programjames1: /* Utah high school mathematics competitions */</p> <hr /> <div>== Utah middle school mathematics competitions==<br /> [[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br /> <br /> * [[Utah MathCounts]] [http://utahmathcounts.org/ (website)] is part of the national [[MathCounts]] program.<br /> <br /> *Mission Math Utah [http://missionmathutah.com/ (website)] is a contest for elementary and middle school students in Utah, run by Utah high school students.<br /> <br /> == Utah high school mathematics competitions==<br /> [[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics <br /> <br /> * [[Utah ARML]] is the Utah team for the [[American Regions Mathematics League]]<br /> *[[Utah Math Olympiad]] [http://www.utahmath.org website]<br /> *[[Utah State Math Contest]] [http://www.math.utah.edu/smc/ website]<br /> <br /> == Utah college mathematics competitions==<br /> * [[Intermountain Math Contest]] [https://www.math.utah.edu/~bestvina/intermountain/index.html] is an college proof based math competition. It's format is similar to the Putnam, but the problems are much more approachable. High school students are welcome to compete, and have won the contest in the past.<br /> <br /> == National math contests ==<br /> * [[List of United States elementary school mathematics competitions]] for national contests.<br /> * [[List of United States middle school mathematics competitions]] for national contests.<br /> * [[List of United States high school mathematics competitions]] for national contests.<br /> <br /> <br /> == See also ==<br /> * [[Utah Math Circle]] [http://www.math.utah.edu/mathcircle/ website]<br /> * [[List of mathematics competitions]]<br /> * [[Mathematics competitions resources]]<br /> * [[Mathematics scholarships]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=103691 AMC historical results 2019-02-21T16:39:14Z <p>Programjames1: /* AMC 10B */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 111-114)<br /> *DHR: TBD<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 108 - 111)<br /> *DHR: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 75 -79.5)<br /> *DHR: TBD<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 82.5 - 88.5)<br /> *DHR: TBD<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score:5.70<br /> *Median score: 6<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 111<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 111<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: 120<br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=103690 AMC historical results 2019-02-21T16:39:01Z <p>Programjames1: /* AMC 10A */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 111-114)<br /> *DHR: TBD<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 100.5 - 105)<br /> *DHR: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 75 -79.5)<br /> *DHR: TBD<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 82.5 - 88.5)<br /> *DHR: TBD<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score:5.70<br /> *Median score: 6<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 111<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 111<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: 120<br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=103687 AMC historical results 2019-02-21T16:38:26Z <p>Programjames1: /* AMC 10A */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 115)<br /> *DHR: TBD<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 144 - 150)<br /> *DHR: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: TBD<br /> *AIME floor: TBD. (AoPSers predict 75 -79.5)<br /> *DHR: TBD<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD (AoPSers predict 82.5 - 88.5)<br /> *DHR: TBD<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score:5.70<br /> *Median score: 6<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 111<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 111<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: 120<br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_20&diff=102481 2019 AMC 10B Problems/Problem 20 2019-02-14T21:03:08Z <p>Programjames1: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #20]] and [[2019 AMC 12B Problems|2019 AMC 12B #15]]}}<br /> ==Problem==<br /> As shown in the figure, line segment &lt;math&gt;\overline{AD}&lt;/math&gt; is trisected by points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AB=BC=CD=2.&lt;/math&gt; Three semicircles of radius &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;\overarc{AEB},\overarc{BFC},&lt;/math&gt; and &lt;math&gt;\overarc{CGD},&lt;/math&gt; have their diameters on &lt;math&gt;\overline{AD},&lt;/math&gt; and are tangent to line &lt;math&gt;EG&lt;/math&gt; at &lt;math&gt;E,F,&lt;/math&gt; and &lt;math&gt;G,&lt;/math&gt; respectively. A circle of radius &lt;math&gt;2&lt;/math&gt; has its center on &lt;math&gt;F. &lt;/math&gt; The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form <br /> &lt;cmath&gt;\frac{a}{b}\cdot\pi-\sqrt{c}+d,&lt;/cmath&gt;<br /> where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17&lt;/math&gt;<br /> ==Solution==<br /> Divide the circle into four parts: The top semicircle (1), the bottom sector with arc length 120 degrees (2), the triangle formed by the radii of (2) and the chord (3), and the four parts which are the corners of a circle inscribed in a square. The area is just (1) + (2) - (3) + (4).<br /> <br /> Area of (1): &lt;math&gt;2\pi&lt;/math&gt;<br /> <br /> <br /> Area of (2): &lt;math&gt;\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> <br /> Area of (3): Radius of 2, distance of 1 to BC, creates 2 30-60-90 triangles, so area of it is &lt;math&gt;2\sqrt{3}*1/2=\sqrt{3}&lt;/math&gt;<br /> <br /> <br /> Area of (4): &lt;math&gt;4*1-1/4*\pi*4=4-\pi&lt;/math&gt;<br /> <br /> <br /> Total sum: &lt;math&gt;\frac{7\pi}{3}+\sqrt{3}+4&lt;/math&gt;<br /> <br /> &lt;math&gt;7+3+3+4=\boxed{17}&lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> filldraw(circle((0,0),2), grey);<br /> filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0));<br /> filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0));<br /> dot((-3,-1));<br /> label(&quot;$A$&quot;,(-3,-1),S);<br /> dot((-2,0));<br /> label(&quot;$E$&quot;,(-2,0),NW);<br /> dot((-1,-1));<br /> label(&quot;$B$&quot;,(-1,-1),S);<br /> dot((0,0));<br /> label(&quot;$F$&quot;,(0,0),N);<br /> dot((1,-1));<br /> label(&quot;$C$&quot;,(1,-1), S);<br /> dot((2,0));<br /> label(&quot;$G$&quot;, (2,0),NE);<br /> dot((3,-1));<br /> label(&quot;$D$&quot;, (3,-1), S);<br /> &lt;/asy&gt;<br /> -Arpitr20<br /> <br /> ==Solution 2==<br /> I don't have enough time to post it yet, but here is the outline:<br /> Divide it into the top semicircle, the bottom sector, and two rectangles. Using these rectangles we can find the area of those weird shapes that are shaded. Adding everything together, and we can finish.<br /> Sorry for not the full solution.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems&diff=102253 2019 AMC 12B Problems 2019-02-14T17:30:10Z <p>Programjames1: /* Problem 24 */</p> <hr /> <div>==Problem 1==<br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> <br /> [[2019 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Consider the statement, &quot;If &lt;math&gt;n&lt;/math&gt; is not prime, then &lt;math&gt;n-2&lt;/math&gt; is prime.&quot; Which of the following values of &lt;math&gt;n&lt;/math&gt; is a counterexample to this statement.<br /> <br /> &lt;math&gt;\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> [[2019 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A positive integer &lt;math&gt;n&lt;/math&gt; satisfies the equation &lt;math&gt;(n+1)!+(n+2)!=440\cdot n!&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or &lt;math&gt;n&lt;/math&gt; pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> In a given plane, points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;10&lt;/math&gt; units apart. How many points &lt;math&gt;C&lt;/math&gt; are there in the plane such that the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; units and the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which the median of the numbers &lt;math&gt;4,6,8,17,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is equal to the mean of those five numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> [[2019 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> [[2019 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> [[2019 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> How many unordered pairs of edges of a given cube determine a plane?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> [[2019 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> [[2019 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> [[2019 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> There are lily pads in a row numbered 0 to 11, in that order. There are predators on lily pads 3 and 6, and a morsel of food on lily pad 10. Fiona the frog starts on pad 0, and from any given lily pad, has a &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; chance to hop to the next pad, and an equal chance to jump 2 pads. What is the probability that Fiona reaches pad 10 without landing on either pad 3 or pad 6?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> How many nonzero complex numbers &lt;math&gt;z&lt;/math&gt; have the property that &lt;math&gt;0, z,&lt;/math&gt; and &lt;math&gt;z^3,&lt;/math&gt; when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Square pyramid &lt;math&gt;ABCDE&lt;/math&gt; has base &lt;math&gt;ABCD,&lt;/math&gt; which measures &lt;math&gt;3&lt;/math&gt; cm on a side, and altitude &lt;math&gt;\overline{AE}&lt;/math&gt; perpendicular to the base&lt;math&gt;,&lt;/math&gt; which measures &lt;math&gt;6&lt;/math&gt; cm. Point &lt;math&gt;P&lt;/math&gt; lies on &lt;math&gt;\overline{BE},&lt;/math&gt; one third of the way from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E;&lt;/math&gt; point &lt;math&gt;Q&lt;/math&gt; lies on &lt;math&gt;\overline{DE},&lt;/math&gt; one third of the way from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;E;&lt;/math&gt; and point &lt;math&gt;R&lt;/math&gt; lies on &lt;math&gt;\overline{CE},&lt;/math&gt; two thirds of the way from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E.&lt;/math&gt; What is the area, in square centimeters, of &lt;math&gt;\triangle PQR?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> [[2019 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is &lt;math&gt;ax^2+bx+c,a\neq 0,&lt;/math&gt; and the roots are &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;s,&lt;/math&gt; then the requirement is that &lt;math&gt;\{a,b,c\}=\{r,s\}&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and<br /> &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt;for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.&lt;/math&gt; Let &lt;math&gt;S&lt;/math&gt; denote all points in the complex plane of the form &lt;math&gt;a+b\omega+c\omega^2,&lt;/math&gt; where &lt;math&gt;0\leq a \leq 1,0\leq b\leq 1,&lt;/math&gt; and &lt;math&gt;0\leq c\leq 1.&lt;/math&gt; What is the area of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;CD=6.&lt;/math&gt; Suppose that the centroids of &lt;math&gt;\triangle ABC,\triangle BCD,&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt; form the vertices of an equilateral triangle. What is the maximum possible value of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 25|Solution]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems&diff=102252 2019 AMC 12B Problems 2019-02-14T17:29:53Z <p>Programjames1: /* Problem 24 */</p> <hr /> <div>==Problem 1==<br /> Alicia had two containers. The first was &lt;math&gt;\tfrac{5}{6}&lt;/math&gt; full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was &lt;math&gt;\tfrac{3}{4}&lt;/math&gt; full of water. What is the ratio of the volume of the first container to the volume of the second container?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{8} \qquad \textbf{(B) } \frac{4}{5} \qquad \textbf{(C) } \frac{7}{8} \qquad \textbf{(D) } \frac{9}{10} \qquad \textbf{(E) } \frac{11}{12}&lt;/math&gt;<br /> <br /> <br /> [[2019 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Consider the statement, &quot;If &lt;math&gt;n&lt;/math&gt; is not prime, then &lt;math&gt;n-2&lt;/math&gt; is prime.&quot; Which of the following values of &lt;math&gt;n&lt;/math&gt; is a counterexample to this statement.<br /> <br /> &lt;math&gt;\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> [[2019 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A positive integer &lt;math&gt;n&lt;/math&gt; satisfies the equation &lt;math&gt;(n+1)!+(n+2)!=440\cdot n!&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 10\qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or &lt;math&gt;n&lt;/math&gt; pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> In a given plane, points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;10&lt;/math&gt; units apart. How many points &lt;math&gt;C&lt;/math&gt; are there in the plane such that the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;50&lt;/math&gt; units and the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which the median of the numbers &lt;math&gt;4,6,8,17,&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; is equal to the mean of those five numbers?<br /> <br /> &lt;math&gt;\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> [[2019 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> [[2019 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> [[2019 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> How many unordered pairs of edges of a given cube determine a plane?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> [[2019 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> [[2019 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all positive integer divisors of &lt;math&gt;100,000.&lt;/math&gt; How many numbers are the product of two distinct elements of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> [[2019 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> There are lily pads in a row numbered 0 to 11, in that order. There are predators on lily pads 3 and 6, and a morsel of food on lily pad 10. Fiona the frog starts on pad 0, and from any given lily pad, has a &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; chance to hop to the next pad, and an equal chance to jump 2 pads. What is the probability that Fiona reaches pad 10 without landing on either pad 3 or pad 6?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> How many nonzero complex numbers &lt;math&gt;z&lt;/math&gt; have the property that &lt;math&gt;0, z,&lt;/math&gt; and &lt;math&gt;z^3,&lt;/math&gt; when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Square pyramid &lt;math&gt;ABCDE&lt;/math&gt; has base &lt;math&gt;ABCD,&lt;/math&gt; which measures &lt;math&gt;3&lt;/math&gt; cm on a side, and altitude &lt;math&gt;\overline{AE}&lt;/math&gt; perpendicular to the base&lt;math&gt;,&lt;/math&gt; which measures &lt;math&gt;6&lt;/math&gt; cm. Point &lt;math&gt;P&lt;/math&gt; lies on &lt;math&gt;\overline{BE},&lt;/math&gt; one third of the way from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E;&lt;/math&gt; point &lt;math&gt;Q&lt;/math&gt; lies on &lt;math&gt;\overline{DE},&lt;/math&gt; one third of the way from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;E;&lt;/math&gt; and point &lt;math&gt;R&lt;/math&gt; lies on &lt;math&gt;\overline{CE},&lt;/math&gt; two thirds of the way from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E.&lt;/math&gt; What is the area, in square centimeters, of &lt;math&gt;\triangle PQR?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> [[2019 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Points &lt;math&gt;A(6,13)&lt;/math&gt; and &lt;math&gt;B(12,11)&lt;/math&gt; lie on circle &lt;math&gt;\omega&lt;/math&gt; in the plane. Suppose that the tangent lines to &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; intersect at a point on the &lt;math&gt;x&lt;/math&gt;-axis. What is the area of &lt;math&gt;\omega&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br /> \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is &lt;math&gt;ax^2+bx+c,a\neq 0,&lt;/math&gt; and the roots are &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;s,&lt;/math&gt; then the requirement is that &lt;math&gt;\{a,b,c\}=\{r,s\}&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Define a sequence recursively by &lt;math&gt;x_0=5&lt;/math&gt; and<br /> &lt;cmath&gt;x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}&lt;/cmath&gt;for all nonnegative integers &lt;math&gt;n.&lt;/math&gt; Let &lt;math&gt;m&lt;/math&gt; be the least positive integer such that<br /> &lt;cmath&gt;x_m\leq 4+\frac{1}{2^{20}}.&lt;/cmath&gt;In which of the following intervals does &lt;math&gt;m&lt;/math&gt; lie?<br /> <br /> &lt;math&gt;\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.&lt;/math&gt; Let &lt;math&gt;S&lt;/math&gt; denote all points in the complex plane of the form &lt;math&gt;a+b\omega+c\omega^2,&lt;/math&gt; where &lt;math&gt;0\leq a \leq 1,0\leq b\leq 1,&lt;/math&gt; and &lt;math&gt;0\leq c\leq 1.&lt;/math&gt; What is the area of &lt;math&gt;S&lt;/math&gt;?<br /> &lt;cmath&gt;\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi&lt;/cmath&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;CD=6.&lt;/math&gt; Suppose that the centroids of &lt;math&gt;\triangle ABC,\triangle BCD,&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt; form the vertices of an equilateral triangle. What is the maximum possible value of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30&lt;/math&gt;<br /> <br /> [[2019 AMC 12B Problems/Problem 25|Solution]]</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_24&diff=102216 2019 AMC 12B Problems/Problem 24 2019-02-14T17:10:25Z <p>Programjames1: /* Problem */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.&lt;/math&gt; Let &lt;math&gt;S&lt;/math&gt; denote all points in the complex plane of the form &lt;math&gt;a+b\omega+c\omega^2,&lt;/math&gt; where &lt;math&gt;0\leq a \leq 1,0\leq b\leq 1,&lt;/math&gt; and &lt;math&gt;0\leq c\leq 1.&lt;/math&gt; What is the area of &lt;math&gt;S&lt;/math&gt;?<br /> &lt;cmath&gt;\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi&lt;/cmath&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\omega=e^{\frac{2i\pi}{3}}&lt;/math&gt; be the third root of unity. We wish to find the span of &lt;math&gt;a+b\omega+c\omega^2)&lt;/math&gt; for reals &lt;math&gt;0\le a,b,c\le 1&lt;/math&gt;.<br /> Note that if &lt;math&gt;a,b,c&gt;0&lt;/math&gt;, then &lt;math&gt;a-\min(a,b,c), b-\min(a,b,c), c-\min(a,b,c)&lt;/math&gt; forms the same point as &lt;math&gt;a,b,c&lt;/math&gt;. Therefore, assume that at least one of them is equal to &lt;math&gt;0&lt;/math&gt;. If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length &lt;math&gt;1&lt;/math&gt;. Similarly for if two are equal to zero. So the area of the six equilateral triangles is<br /> &lt;cmath&gt;\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}&lt;/cmath&gt;<br /> <br /> Here is a diagram:<br /> &lt;asy&gt;<br /> size(200,200);<br /> draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle);<br /> draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle);<br /> draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle);<br /> draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle);<br /> draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle);<br /> draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle);<br /> draw((-2,0)--(2,0));<br /> draw((0,-2)--(0,2));<br /> &lt;/asy&gt;<br /> <br /> -programjames1<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_24&diff=102214 2019 AMC 12B Problems/Problem 24 2019-02-14T17:09:27Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\omega=e^{\frac{2i\pi}{3}}&lt;/math&gt; be the third root of unity. We wish to find the span of &lt;math&gt;a+b\omega+c\omega^2)&lt;/math&gt; for reals &lt;math&gt;0\le a,b,c\le 1&lt;/math&gt;.<br /> Note that if &lt;math&gt;a,b,c&gt;0&lt;/math&gt;, then &lt;math&gt;a-\min(a,b,c), b-\min(a,b,c), c-\min(a,b,c)&lt;/math&gt; forms the same point as &lt;math&gt;a,b,c&lt;/math&gt;. Therefore, assume that at least one of them is equal to &lt;math&gt;0&lt;/math&gt;. If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length &lt;math&gt;1&lt;/math&gt;. Similarly for if two are equal to zero. So the area of the six equilateral triangles is<br /> &lt;cmath&gt;\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}&lt;/cmath&gt;<br /> <br /> Here is a diagram:<br /> &lt;asy&gt;<br /> size(200,200);<br /> draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle);<br /> draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle);<br /> draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle);<br /> draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle);<br /> draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle);<br /> draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle);<br /> draw((-2,0)--(2,0));<br /> draw((0,-2)--(0,2));<br /> &lt;/asy&gt;<br /> <br /> -programjames1<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_24&diff=102209 2019 AMC 12B Problems/Problem 24 2019-02-14T17:04:58Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\omega=e^{\frac{2i\pi}{3}}&lt;/math&gt; be the third root of unity. We wish to find the span of &lt;math&gt;a+b\omega+c\omega^2)&lt;/math&gt; for reals &lt;math&gt;0\le a,b,c\le 1&lt;/math&gt;.<br /> Note that if &lt;math&gt;a,b,c&gt;0&lt;/math&gt;, then &lt;math&gt;a-\min(a,b,c), b-\min(a,b,c), c-\min(a,b,c)&lt;/math&gt; forms the same point as &lt;math&gt;a,b,c&lt;/math&gt;. Therefore, assume that at least one of them is equal to &lt;math&gt;0&lt;/math&gt;. If only one of them is equal to zero, we can form an equilateral triangle with the remaining two, of side length &lt;math&gt;1&lt;/math&gt;. Similarly for if two are equal to zero. So the area of the six equilateral triangles is<br /> &lt;cmath&gt;\boxed{\text{(C) }\frac{3\sqrt{3}}{2}}&lt;/cmath&gt;<br /> -programjames1<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_25&diff=101922 2015 AMC 12B Problems/Problem 25 2019-02-12T19:40:36Z <p>Programjames1: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> A bee starts flying from point &lt;math&gt;P_0&lt;/math&gt;. She flies &lt;math&gt;1&lt;/math&gt; inch due east to point &lt;math&gt;P_1&lt;/math&gt;. For &lt;math&gt;j \ge 1&lt;/math&gt;, once the bee reaches point &lt;math&gt;P_j&lt;/math&gt;, she turns &lt;math&gt;30^{\circ}&lt;/math&gt; counterclockwise and then flies &lt;math&gt;j+1&lt;/math&gt; inches straight to point &lt;math&gt;P_{j+1}&lt;/math&gt;. When the bee reaches &lt;math&gt;P_{2015}&lt;/math&gt; she is exactly &lt;math&gt;a \sqrt{b} + c \sqrt{d}&lt;/math&gt; inches away from &lt;math&gt;P_0&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 positive integers and &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are not divisible by the square of any prime. What is &lt;math&gt;a+b+c+d&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;x = e^{i \pi / 6}&lt;/math&gt;, a &lt;math&gt;30^\circ&lt;/math&gt; counterclockwise rotation centered at the origin. Notice that &lt;math&gt;P_k&lt;/math&gt; on the complex plane is:<br /> <br /> &lt;cmath&gt;1 + 2x + 3x^2 + \cdots + (k+1)x^k&lt;/cmath&gt;<br /> <br /> We need to find the magnitude of &lt;math&gt;P_{2015}&lt;/math&gt; on the complex plane. This is an arithmetic/geometric series. <br /> <br /> &lt;cmath&gt;\begin{align*} S &amp;= 1 + 2x + 3x^2 + \cdots + 2015x^{2014} \\<br /> xS &amp;= x + 2x^2 + 3x^3 + \cdots + 2015x^{2015} \\<br /> (1-x)S &amp;= 1 + x + x^2 + \cdots + x^{2014} - 2015x^{2015} \\<br /> S &amp;= \frac{1 - x^{2015} }{(1-x)^2} - \frac{2015x^{2015}}{1-x} \end{align*} &lt;/cmath&gt;<br /> <br /> We want to find &lt;math&gt;|S|&lt;/math&gt;. First, note that &lt;math&gt;x^{2015} = x^{11} = x^{-1}&lt;/math&gt; because &lt;math&gt;x^{12} = 1&lt;/math&gt;. Therefore<br /> <br /> &lt;cmath&gt;S = \frac{1 - \frac{1}{x}}{(1-x)^2} - \frac{2015}{x(1-x)} = -\frac{1}{x(1-x)} - \frac{2015}{x(1-x)} = -\frac{2016}{x(1-x)}.&lt;/cmath&gt;<br /> <br /> Hence, since &lt;math&gt;|x|=1&lt;/math&gt;, we have &lt;math&gt;|S| = \frac{2016}{|1-x|}.&lt;/math&gt;<br /> <br /> Now we just have to find &lt;math&gt;|1-x|&lt;/math&gt;. This can just be computed directly:<br /> <br /> &lt;cmath&gt;1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;|1-x|^2 = \left(1 - \sqrt{3} + \frac{3}{4} \right) + \frac{1}{4} = 2 - \sqrt{3} = {\left( \frac{\sqrt{6}-\sqrt{2}}{2} \right)}^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.&lt;/cmath&gt;<br /> <br /> Therefore &lt;math&gt;|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.&lt;/math&gt;<br /> <br /> Thus the answer is &lt;math&gt;1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Here is an alternate solution that does not use complex numbers:<br /> <br /> We will calculate the distance from &lt;math&gt;P_{2015}&lt;/math&gt; to &lt;math&gt;P_0&lt;/math&gt; using the Pythagorean theorem. Assume &lt;math&gt;P_0&lt;/math&gt; lies at the origin, so we will calculate the distance to &lt;math&gt;P_{2015}&lt;/math&gt; by calculating the distance traveled in the x-direction and the distance traveled in the y-direction. We can calculate this by summing each movement:<br /> <br /> &lt;math&gt;x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+\cdot \cdot \cdot+2011\cos{180}+2012\cos {210}+2013\cos{240}+2014\cos{270}+2015\cos{300}&lt;/math&gt;<br /> <br /> A movement of &lt;math&gt;p&lt;/math&gt; units at &lt;math&gt;q&lt;/math&gt; degrees is the same thing as a movement of &lt;math&gt;-p&lt;/math&gt; units at &lt;math&gt;q-180&lt;/math&gt; degrees, so we can adjust all the cosines with arguments greater than 180 as follows:<br /> <br /> &lt;math&gt;x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdot \cdot \cdot -2015 \cos{120}&lt;/math&gt;<br /> <br /> Now we group terms with like-cosines and factor out the cosines:<br /> <br /> &lt;math&gt;x=(1-7+13-\cdot \cdot \cdot +2005-2011)\cos{0}+(2-8+14- \cdot \cdot \cdot +2006-2012)\cos{30}+\cdot \cdot \cdot +(5-11+17- \cdot \cdot \cdot +2008-2014)\cos{120}+(6-12+18- \cdot \cdot \cdot -2004+2010)\cos{150}&lt;/math&gt;<br /> <br /> Each sum in the parentheses has 336 terms (except the very last one, which has 335), so by pairing each term, we can see that there are &lt;math&gt;\frac {336}{2}&lt;/math&gt; pairs of &lt;math&gt;-6&lt;/math&gt;. So each sum evaluates to &lt;math&gt;168\cdot -6=-1008&lt;/math&gt;, except the very last sum, which has 167 pairs of &lt;math&gt;-6&lt;/math&gt; and an extra 2010, so it evaluates to &lt;math&gt;167\cdot -6+2010=1008&lt;/math&gt;. Plugging in these values:<br /> <br /> &lt;math&gt;x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}&lt;/math&gt;<br /> &lt;math&gt;x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})&lt;/math&gt;<br /> <br /> Now that we have how far was traveled in the x-direction, we need to find how far was traveled in the y-direction. Using the same logic as above, we arrive at the sum:<br /> <br /> &lt;math&gt;y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}&lt;/math&gt;<br /> <br /> &lt;math&gt;y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})&lt;/math&gt;<br /> <br /> The last step is to use the Pythagorean to find the distance from &lt;math&gt;P_0&lt;/math&gt;. This distance is given by:<br /> <br /> &lt;math&gt;\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot 1008^2 \cdot (1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}&lt;/math&gt;<br /> <br /> Multiplying out, we have &lt;math&gt;1008\sqrt{2}+1008\sqrt{6}&lt;/math&gt;, so the answer is &lt;math&gt;1008+2+1008+6= \boxed {\bold {(B)}\; 2024}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction? <br /> <br /> <br /> First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from &lt;math&gt;P_2&lt;/math&gt; to the line &lt;math&gt;\overline{P_0P_1}&lt;/math&gt; intersecting a point &lt;math&gt;C_0&lt;/math&gt;. We will also place the point &lt;math&gt;C_1&lt;/math&gt; at the intersection of &lt;math&gt;\overline{P_0P_1}&lt;/math&gt; and &lt;math&gt;\overline{P_3P_4}&lt;/math&gt;. In addition, the point &lt;math&gt;C_2&lt;/math&gt; is placed at the perpendicular dropped from &lt;math&gt;P_2&lt;/math&gt; to the line &lt;math&gt;\overline{P_3C_1}&lt;/math&gt;. We will also set the distance &lt;math&gt;\overline{P_0P_1} = n&lt;/math&gt; and thus &lt;math&gt;\overline{P_1P_2} = n+1&lt;/math&gt;. With this perpendicular we see that the triangle &lt;math&gt;\triangle{P_1P_2C_0}&lt;/math&gt; is a 30-60-90 triangle. This means that the length &lt;math&gt;\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}&lt;/math&gt; and the length &lt;math&gt;\overline{C_1C_2} = \frac{n+1}{2}&lt;/math&gt;. We can also see that the triangle &lt;math&gt;\triangle{P_2C_1P_3}&lt;/math&gt; is a 30-60-90 triangle and thus &lt;math&gt;\overline{C_0C_1} = \frac{n+2}{2}&lt;/math&gt; and &lt;math&gt;\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}&lt;/math&gt;. Now if we continue this across all &lt;math&gt;P_i&lt;/math&gt; and set the point &lt;math&gt;P_0&lt;/math&gt; to the coordinates &lt;math&gt;(0, 0)&lt;/math&gt;. As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length<br /> <br /> &lt;cmath&gt;\overline{P_0C_1} = n + \frac{(n+1)\sqrt{3}}{2} + \frac{n+2}{2}&lt;/cmath&gt;, &lt;cmath&gt;\overline{C_1C_4} = \frac{(n+1)}{2} + \frac{(n+2)\sqrt{3}}{2} + (n+3) + \frac{(n+4)\sqrt{3}}{2} + \frac{n+5}{2}&lt;/cmath&gt;, &lt;cmath&gt;\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}&lt;/cmath&gt;, &lt;cmath&gt;\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}&lt;/cmath&gt;, and finally &lt;cmath&gt;\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}&lt;/cmath&gt;. <br /> <br /> Here the point &lt;math&gt;C_4&lt;/math&gt; is defined as the intersection of lines &lt;math&gt;\overline{P_3P_4}&lt;/math&gt; and &lt;math&gt;\overline{P_6P_7}&lt;/math&gt;. The point &lt;math&gt;C_7&lt;/math&gt; is defined as the intersection of lines &lt;math&gt;\overline{P_6P_7}&lt;/math&gt; and &lt;math&gt;\overline{P_9P_{10}}&lt;/math&gt;. Finally, the point &lt;math&gt;C_10&lt;/math&gt; is defined as the intersection of lines &lt;math&gt;\overline{P_{9}P_{10}}&lt;/math&gt; and &lt;math&gt;\overline{P_{12}P_{13}}&lt;/math&gt;. Note that our spiral stops at &lt;math&gt;P_{12}&lt;/math&gt; before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point &lt;math&gt;P_{12}&lt;/math&gt;, is &lt;math&gt;({-6}, {-6}-12 \sqrt{3})&lt;/math&gt;. This is an interesting property that the points’ coordinate changes by a constant offset no matter what &lt;math&gt;n&lt;/math&gt; is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point &lt;math&gt;P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))&lt;/math&gt;. Using similar 30-60-90 triangle properties, we see that &lt;math&gt;P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)&lt;/math&gt;. Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is &lt;math&gt;(1008)(1+\sqrt{3})(\sqrt{2})&lt;/math&gt; which gives us a final answer of &lt;math&gt;\boxed{2024}&lt;/math&gt;.<br /> <br /> -bowmanrocks32<br /> <br /> <br /> ==Solution 4==<br /> Suppose that the bee makes a move of distance &lt;math&gt;i&lt;/math&gt;. After 6 turns it will be facing the opposite direction and move &lt;math&gt;i+6&lt;/math&gt; units. Combining these opposite movements gives a total movement of &lt;math&gt;-6&lt;/math&gt; units. The bee makes a total of &lt;math&gt;2016=12\cdot 168&lt;/math&gt; moves, so there are &lt;math&gt;168&lt;/math&gt; of these pairs, for each direction the bee faces (&lt;math&gt;0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ&lt;/math&gt;). To keep the numbers small, we will multiply by &lt;math&gt;6\cdot 168=1008&lt;/math&gt; at the end.<br /> <br /> We draw a quick diagram of one unit in each direction.<br /> &lt;asy&gt;<br /> draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)));<br /> draw((1,0)--(3/2+sqrt(3)/2, 0), dashed);<br /> draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed);<br /> draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed);<br /> draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed);<br /> draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed);<br /> &lt;/asy&gt;<br /> Using the 30-60-90 triangles, it is clear that the displacement vector is &lt;math&gt;\langle 1, 2+\sqrt{3}\rangle&lt;/math&gt;, or the displacement length is &lt;math&gt;\sqrt{1^2+(2+\sqrt{3})^2}=\sqrt{8+4\sqrt{3}}=\sqrt{2}+\sqrt{6}&lt;/math&gt;. Therefore the total displacement is &lt;math&gt;\left|-1008\left(\sqrt{2}+\sqrt{6}\right)\right|= 1008\sqrt{2}+1008\sqrt{6}\implies 1008+2+1008+6=\boxed{\text{(B) }2024}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|after=Last Problem|num-b=24}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_21&diff=101659 2019 AMC 10A Problems/Problem 21 2019-02-10T00:18:40Z <p>Programjames1: /* Diagram */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #21]] and [[2019 AMC 12A Problems|2019 AMC 12A #18]]}}<br /> <br /> ==Problem==<br /> <br /> A sphere with center &lt;math&gt;O&lt;/math&gt; has radius &lt;math&gt;6&lt;/math&gt;. A triangle with sides of length &lt;math&gt;15, 15,&lt;/math&gt; and &lt;math&gt;24&lt;/math&gt; is situated in space so that each of its sides is tangent to the sphere. What is the distance between &lt;math&gt;O&lt;/math&gt; and the plane determined by the triangle?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }2\sqrt{3}\qquad<br /> \textbf{(B) }4\qquad<br /> \textbf{(C) }3\sqrt{2}\qquad<br /> \textbf{(D) }2\sqrt{5}\qquad<br /> \textbf{(E) }5\qquad<br /> &lt;/math&gt;<br /> <br /> ==Diagram==<br /> 3D<br /> &lt;asy&gt;<br /> import graph3;<br /> import palette;<br /> size(200);<br /> currentprojection=orthographic(0,4,2);<br /> <br /> triple f(pair z) {return expi(z.x,z.y);}<br /> <br /> surface s=surface(f,(0,0),(pi,2pi),70,Spline);<br /> draw((0,-5/6,sqrt(5)/3)--(2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3)--cycle);<br /> draw(s,mean(palette(s.map(zpart),Grayscale())),nolight);<br /> draw((2,2/3,sqrt(5)/3)--(-2,2/3,sqrt(5)/3));<br /> &lt;/asy&gt;<br /> Plane through triangle.<br /> &lt;asy&gt;<br /> draw((0,0)--(12,9)--(24,0)--cycle);<br /> draw((12,9)--(12,0), dashed);<br /> draw((11.5,0)--(11.5,0.5)--(12,0.5));<br /> draw(circle((12,4),4));<br /> draw((12,4)--(48/5, 36/5));<br /> dot((12,4));<br /> label(&quot;$15$&quot;, (6,9/2),NW);<br /> label(&quot;$15$&quot;, (18,9/2),NE);<br /> label(&quot;$24$&quot;, (12,-1),S);<br /> label(&quot;$r$&quot;,(54/5, 28/5), SW);<br /> &lt;/asy&gt;<br /> -programjames1<br /> <br /> ==Solution==<br /> The triangle is placed on the sphere so that three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use &lt;math&gt;\text{area} = \text{inradius} \cdot \text{semiperimeter}&lt;/math&gt;. The area of the triangle can be found by drawing an altitude from the vertex between sides with length &lt;math&gt;15&lt;/math&gt; to the midpoint of the side with length &lt;math&gt;24&lt;/math&gt;. Pythagorean triples &lt;math&gt;9&lt;/math&gt; - &lt;math&gt;12&lt;/math&gt; - &lt;math&gt;15&lt;/math&gt; show the base is &lt;math&gt;24&lt;/math&gt; and height is 9. &lt;math&gt;\frac {\text{base} \cdot \text{height}} {2}&lt;/math&gt; can be used to find the area of the triangle, &lt;math&gt;108&lt;/math&gt;. The semiperimeter is &lt;math&gt;\frac {15 + 15 + 24} {2} = 27&lt;/math&gt; After plugging into the equation &lt;math&gt;108 = \text{inradius} \cdot 27&lt;/math&gt;, we get &lt;math&gt;\text{inradius} = 4&lt;/math&gt;. Let the distance between &lt;math&gt;O&lt;/math&gt; and the triangle be &lt;math&gt;x&lt;/math&gt;. Choose a point on the incircle and denote it &lt;math&gt;A&lt;/math&gt;. &lt;math&gt;\overline{OA}&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt; because it is the radius of the sphere. Point &lt;math&gt;A&lt;/math&gt; to the center of the incircle is length &lt;math&gt;4&lt;/math&gt; because it is the inradius of the incircle. By using Pythagorean Theorem, you will get that &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;\sqrt{6^2-4^2}=\sqrt{20}\implies\boxed{\textbf {(D)} 2 \sqrt {5}}&lt;/math&gt;.<br /> <br /> - ViolinGod(Argonauts16 latex)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_21&diff=101605 2019 AMC 10A Problems/Problem 21 2019-02-09T23:42:23Z <p>Programjames1: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #21]] and [[2019 AMC 12A Problems|2019 AMC 12A #18]]}}<br /> <br /> ==Problem==<br /> <br /> A sphere with center &lt;math&gt;O&lt;/math&gt; has radius &lt;math&gt;6&lt;/math&gt;. A triangle with sides of length &lt;math&gt;15, 15,&lt;/math&gt; and &lt;math&gt;24&lt;/math&gt; is situated in space so that each of its sides is tangent to the sphere. What is the distance between &lt;math&gt;O&lt;/math&gt; and the plane determined by the triangle?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }2\sqrt{3}\qquad<br /> \textbf{(B) }4\qquad<br /> \textbf{(C) }3\sqrt{2}\qquad<br /> \textbf{(D) }2\sqrt{5}\qquad<br /> \textbf{(E) }5\qquad<br /> &lt;/math&gt;<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> draw((0,0)--(12,9)--(24,0)--cycle);<br /> draw((12,9)--(12,0), dashed);<br /> draw((11.5,0)--(11.5,0.5)--(12,0.5));<br /> draw(circle((12,4),4));<br /> draw((12,4)--(48/5, 36/5));<br /> dot((12,4));<br /> label(&quot;$15$&quot;, (6,9/2),NW);<br /> label(&quot;$15$&quot;, (18,9/2),NE);<br /> label(&quot;$24$&quot;, (12,-1),S);<br /> label(&quot;$r$&quot;,(54/5, 28/5), SW);<br /> &lt;/asy&gt;<br /> -programjames1<br /> ==Solution==<br /> The triangle is placed on the sphere so that three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use &lt;math&gt;\text{area} = \text{inradius} \cdot \text{semiperimeter}&lt;/math&gt;. The area of the triangle can be found by drawing an altitude from the vertex between sides with length &lt;math&gt;15&lt;/math&gt; to the midpoint of the side with length &lt;math&gt;24&lt;/math&gt;. Pythagorean triples &lt;math&gt;9&lt;/math&gt; - &lt;math&gt;12&lt;/math&gt; - &lt;math&gt;15&lt;/math&gt; show the base is &lt;math&gt;24&lt;/math&gt; and height is 9. &lt;math&gt;\frac {\text{base} \cdot \text{height}} {2}&lt;/math&gt; can be used to find the area of the triangle, &lt;math&gt;108&lt;/math&gt;. The semiperimeter is &lt;math&gt;\frac {15 + 15 + 24} {2} = 27&lt;/math&gt; After plugging into the equation &lt;math&gt;108 = \text{inradius} \cdot 27&lt;/math&gt;, we get &lt;math&gt;\text{inradius} = 4&lt;/math&gt;. Let the distance between &lt;math&gt;O&lt;/math&gt; and the triangle be &lt;math&gt;x&lt;/math&gt;. Choose a point on the incircle and denote it &lt;math&gt;A&lt;/math&gt;. &lt;math&gt;\overline{OA}&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt; because it is the radius of the sphere. Point &lt;math&gt;A&lt;/math&gt; to the center of the incircle is length &lt;math&gt;4&lt;/math&gt; because it is the inradius of the incircle. By using Pythagorean Theorem, you will get that &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;\sqrt{6^2-4^2}=\sqrt{20}\implies\textbf {(D)} 2 \sqrt {5}&lt;/math&gt;.<br /> <br /> - ViolinGod(Argonauts16 latex)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_20&diff=101590 2019 AMC 10A Problems/Problem 20 2019-02-09T23:28:59Z <p>Programjames1: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #20]] and [[2019 AMC 12A Problems|2019 AMC 12A #16]]}}<br /> <br /> ==Problem==<br /> <br /> The numbers &lt;math&gt;1,2,\dots,9&lt;/math&gt; are randomly placed into the &lt;math&gt;9&lt;/math&gt; squares of a &lt;math&gt;3 \times 3&lt;/math&gt; grid. Each square gets one number, and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{21}\qquad\textbf{(B) }\frac{1}{14}\qquad\textbf{(C) }\frac{5}{63}\qquad\textbf{(D) }\frac{2}{21}\qquad\textbf{(E) }\frac{1}{7}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that odd sums can only be formed by &lt;math&gt;(e,e,o)&lt;/math&gt; or &lt;math&gt;(o,o,o),&lt;/math&gt; so we focus on placing the evens to get &lt;math&gt;\frac{5! \cdot 4! \cdot 9}{9!}=\boxed{B) \frac{1}{14}}&lt;/math&gt; (we need to have each even be with another even in each row/column)<br /> <br /> ==Solution 2==<br /> By the Pigeonhole Principle, there must be at least one row with 2 or more odd numbers in it. Therefore, that row must contain three odd numbers in order to sum to odd. The same thing can be done with the columns. Then we simply have to choose one row and one column to be filled with odd numbers - which can happen &lt;math&gt;3\cdot 3 = 9&lt;/math&gt; times. The answer is then<br /> &lt;cmath&gt;\frac{9}{\binom{9}{4}}=\boxed{\text{(B) }\tfrac{1}{14}}&lt;/cmath&gt;<br /> <br /> -programjames1<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2019|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_17&diff=101588 2019 AMC 10A Problems/Problem 17 2019-02-09T23:25:49Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A child builds towers using identically shaped cubes of different color. How many different towers with a height &lt;math&gt;8&lt;/math&gt; cubes can the child build with &lt;math&gt;2&lt;/math&gt; red cubes, &lt;math&gt;3&lt;/math&gt; blue cubes, and &lt;math&gt;4&lt;/math&gt; green cubes? (One cube will be left out.)<br /> <br /> &lt;math&gt;\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320&lt;/math&gt;<br /> <br /> ==Solution==<br /> Arranging eight cubes is the same as arranging the nine cubes first and then removing the last cube. Every arrangement of nine cubes corresponds to another arrangement that can be used (the first eight cubes in that order and the last cube is discarded). Thus, we get 9!. However, we overcounted because the red cubes can be permuted to have the same overall arrangement and the same with the blue and green cubes. Thus, we have to divide by the &lt;math&gt;2!&lt;/math&gt; ways to arrange the red cubes &lt;math&gt;3!&lt;/math&gt; ways to arrange the blue cubes, and &lt;math&gt;4!&lt;/math&gt; ways to arrange the green cubes. Thus we have &lt;math&gt;\frac {9!} {2! * 3! * 4!}&lt;/math&gt; = &lt;math&gt;\fbox {\textbf {(D)} 1,260}&lt;/math&gt; different arrangements of towers.<br /> <br /> - ViolinGod<br /> <br /> Note:<br /> This is written more compactly as<br /> &lt;cmath&gt;\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_2&diff=96319 1986 AIME Problems/Problem 2 2018-07-17T16:54:29Z <p>Programjames1: /* Solution */</p> <hr /> <div>== Problem ==<br /> Evaluate the product &lt;math&gt;(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Simplify by repeated application of the [[difference of squares]]. <br /> <br /> :&lt;math&gt;\left((\sqrt{6} + \sqrt{7})^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - (\sqrt{6} - \sqrt{7})^2\right)&lt;/math&gt;<br /> :&lt;math&gt;= (13 + 2\sqrt{42} - 5)(5 - (13 - 2\sqrt{42}))&lt;/math&gt;<br /> :&lt;math&gt;= (2\sqrt{42} + 8)(2\sqrt{42} - 8)&lt;/math&gt;<br /> :&lt;math&gt;= (2\sqrt{42})^2 - 8^2 =&lt;/math&gt; &lt;math&gt;\boxed{104}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Notice that in a triangle with side lengths &lt;math&gt;2\sqrt5,2\sqrt6,&lt;/math&gt; and &lt;math&gt;2\sqrt7&lt;/math&gt;, by Heron's formula, the area is the square root of what we are looking for. <br /> Let angle &lt;math&gt;\theta&lt;/math&gt; be opposite the &lt;math&gt;2\sqrt7&lt;/math&gt; side. By the Law of Cosines,<br /> &lt;cmath&gt;\cos\theta=\frac{(2\sqrt5)^2+(2\sqrt{6})^2-(2\sqrt7)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}&lt;/math&gt;. The area of the triangle is then<br /> &lt;cmath&gt;\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}&lt;/cmath&gt;<br /> <br /> So our answer is &lt;math&gt;\left(\sqrt{104}\right)^2=\boxed{104}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=1|num-a=3}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2000_USAMO_Problems/Problem_3&diff=91110 2000 USAMO Problems/Problem 3 2018-02-12T15:53:04Z <p>Programjames1: </p> <hr /> <div>== Problem ==<br /> <br /> A game of solitaire is played with &lt;math&gt;R&lt;/math&gt; red cards, &lt;math&gt;W&lt;/math&gt; white cards, and &lt;math&gt;B&lt;/math&gt; blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of &lt;math&gt;R, W,&lt;/math&gt; and &lt;math&gt;B,&lt;/math&gt; the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.<br /> <br /> == Solution ==<br /> We claim (inductively) that the minimum is just going to be &lt;math&gt;min(BW,2WR,3RB)&lt;/math&gt;. We'll start our induction with the case where one of the three quantities is zero, in which case we verify that we can indeed get away without any penalty by, for example, discarding blue if we are out of white.<br /> <br /> Now, for the inductive step, let &lt;math&gt;f(R,W,B)&lt;/math&gt; be the minimum we seek. Note that<br /> &lt;cmath&gt;f(B,W,R) = min(W+f(B-1,W,R),2R+f(B,W-1,R),3B+f(B,W,B-1)) &lt;/cmath&gt;<br /> By our inductive hypothesis, &lt;math&gt;f(B-1,W,R) = min((B-1)W,2WR,3R(B-1))&lt;/math&gt;. In order for this to cause our inductive step not to hold, we would require that &lt;math&gt;W+min((B-1)W,2WR,3R(B-1)) &lt; min(BW,2WR,3RB)&lt;/math&gt;. It is evident that the first two entries in the &lt;math&gt;min&lt;/math&gt; expression cannot cause this to happen, so that we need only consider &lt;math&gt;W+3R(B-1) &lt; min(BW,2WR,3RB)&lt;/math&gt;. So &lt;math&gt;W+3R(B-1) &lt; BW&lt;/math&gt;, whence &lt;math&gt;3R &lt; W&lt;/math&gt;. But &lt;math&gt;W+3R(B-1) &lt; 3RB&lt;/math&gt;, so that &lt;math&gt;W &lt; 3R&lt;/math&gt;, a contradiction.<br /> <br /> For the other two cases, we can get similar contradictions, so that our inductive step must hold, and so &lt;math&gt;f(B,W,R)&lt;/math&gt; is indeed &lt;math&gt;min(BW,2WR,3RB)&lt;/math&gt;.<br /> <br /> We now need only establish how many ways to do this. If one of these quantities is smaller, our induction and the fact that it is eventually zero guarantees that it will continue to be the smallest quantity as cards are discarded. (For example, if it is currently optimal to discard a blue card, it will continue to be so until we run out of blue cards.) Therefore, assuming that there is currently only one best choice of card to discard, this will continue to be so in the future, whence if &lt;math&gt;BW \neq 2WR \neq 3RB&lt;/math&gt;, there is only &lt;math&gt;1&lt;/math&gt; optimal strategy.<br /> <br /> Suppose, now, that &lt;math&gt;BW = 2WR&lt;/math&gt;. It is thus optimal to discard either a &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;W&lt;/math&gt; card. If we ever discard a blue card, then we will cause &lt;math&gt;BW &lt; 2WR&lt;/math&gt;, whence there is only one possible strategy from then on. However, if we discard a white card, then we will still have &lt;math&gt;BW = 2WR&lt;/math&gt;, meaning that we continue to have the choice of discarding a white or blue card. Since we can discard a white card at most &lt;math&gt;W&lt;/math&gt; times, there are &lt;math&gt;W+1&lt;/math&gt; choices for how many &lt;math&gt;W&lt;/math&gt; cards to discard (&lt;math&gt;0&lt;/math&gt; to &lt;math&gt;W&lt;/math&gt;), meaning that there are &lt;math&gt;W+1&lt;/math&gt; optimal strategies.<br /> <br /> By similar logic, we get &lt;math&gt;R+1&lt;/math&gt; optimal strategies if &lt;math&gt;2WR = 3RB&lt;/math&gt;, and &lt;math&gt;B+1&lt;/math&gt; optimal strategies if &lt;math&gt;3RB = BW&lt;/math&gt;.<br /> <br /> The final case, then, is if &lt;math&gt;BW = 2WR = 3RB&lt;/math&gt;. In this case, if we first discard a white card, we are left with the &lt;math&gt;BW = 2WR&lt;/math&gt; case, and similarly for a blue and white card. The total number of optimal strategies in this case is then the sum of the optimal strategies in those cases, or, in other words, &lt;math&gt;B+W+R&lt;/math&gt;.<br /> <br /> To summarize:<br /> The minimum penalty is &lt;math&gt;min(BW,2WR,3RB)&lt;/math&gt;.<br /> If &lt;math&gt;BW \neq 2WR \neq 3RB&lt;/math&gt;, there is &lt;math&gt;1&lt;/math&gt; optimal strategy.<br /> If &lt;math&gt;BW = 2WR &lt; 3RB&lt;/math&gt;, there are &lt;math&gt;W+1&lt;/math&gt; strategies.<br /> If &lt;math&gt;2WR = 3RB &lt; BW&lt;/math&gt;, there are &lt;math&gt;R+1&lt;/math&gt; strategies.<br /> If &lt;math&gt;3RB = BW &lt; 2WR&lt;/math&gt;, there are &lt;math&gt;B+1&lt;/math&gt; strategies.<br /> If &lt;math&gt;BW = 2WR = 3RB&lt;/math&gt;, there are &lt;math&gt;R+B+W&lt;/math&gt; strategies.<br /> <br /> By J Steinhardt, from AoPS Community<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=2000|num-b=2|num-a=4}}<br /> <br /> [[Category:Olympiad Combinatorics Problems]]<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_3&diff=89389 1996 AIME Problems/Problem 3 2018-01-03T21:12:42Z <p>Programjames1: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive [[integer]] &lt;math&gt;n&lt;/math&gt; for which the expansion of &lt;math&gt;(xy-3x+7y-21)^n&lt;/math&gt;, after like terms have been collected, has at least 1996 terms.<br /> <br /> == Solution ==<br /> Using [[Simon's Favorite Factoring Trick]], we rewrite as &lt;math&gt;[(x+7)(y-3)]^n = (x+7)^n(y-3)^n&lt;/math&gt;. Both [[binomial expansion]]s will contain &lt;math&gt;n+1&lt;/math&gt; non-like terms; their product will contain &lt;math&gt;(n+1)^2&lt;/math&gt; terms, as each term will have an unique power of &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;y&lt;/math&gt; and so none of the terms will need to be collected. Hence &lt;math&gt;(n+1)^2 \ge 1996&lt;/math&gt;, the smallest square after &lt;math&gt;1996&lt;/math&gt; is &lt;math&gt;2025 = 45^2&lt;/math&gt;, so our answer is &lt;math&gt;45 - 1 = \boxed{044}&lt;/math&gt;.<br /> <br /> Alternatively, when &lt;math&gt;n = k&lt;/math&gt;, the exponents of &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;y&lt;/math&gt; in &lt;math&gt;x^i y^i&lt;/math&gt; can be any integer between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; inclusive. Thus, when &lt;math&gt;n=1&lt;/math&gt;, there are &lt;math&gt;(2)(2)&lt;/math&gt; terms and, when &lt;math&gt;n = k&lt;/math&gt;, there are &lt;math&gt;(k+1)^2&lt;/math&gt; terms. Therefore, we need to find the smallest perfect square that is greater than &lt;math&gt;1996&lt;/math&gt;. From trial and error, we get &lt;math&gt;44^2 = 1936&lt;/math&gt; and &lt;math&gt;45^2 = 2025&lt;/math&gt;. Thus, &lt;math&gt;k = 45\rightarrow n = \boxed{044}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1996|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_24&diff=88924 2017 AMC 8 Problems/Problem 24 2017-12-15T23:33:59Z <p>Programjames1: /* Solution 1 */</p> <hr /> <div>==Problem 24==<br /> <br /> Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?<br /> <br /> &lt;math&gt;\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> In &lt;math&gt;360&lt;/math&gt; days, there are &lt;cmath&gt;360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144&lt;/cmath&gt; days without calls. Note that in the last five days of the year, day &lt;math&gt;361&lt;/math&gt; and &lt;math&gt;362&lt;/math&gt; also do not have any calls, as they are not multiples of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, or &lt;math&gt;5&lt;/math&gt;. Thus our answer is &lt;math&gt;144+2 = \boxed{\textbf{(D)}\ 146}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;&lt;/cmath&gt;<br /> Alternatively, there are &lt;math&gt;365\cdot \frac45 \cdot \frac 34 \cdot \frac23=\boxed{146}&lt;/math&gt; days without calls. Multiplying the fractions in this order prevents partial days, as &lt;math&gt;365&lt;/math&gt; is a multiple of &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;365\cdot \frac45&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;365\cdot \frac45 \cdot \frac 34&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We use Principle of Inclusion Exclusion. There are &lt;math&gt;365&lt;/math&gt; days in the year, and we subtract the days that she gets at least &lt;math&gt;1&lt;/math&gt; phone call, which is &lt;cmath&gt; \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor&lt;/cmath&gt;. <br /> <br /> To this result we add the number of days where she gets at least &lt;math&gt;2&lt;/math&gt; phone calls in a day because we double subtracted these days. This number is &lt;cmath&gt;\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor&lt;/cmath&gt;. <br /> <br /> We now subtract the number of days where she gets three phone calls, which is &lt;math&gt;\left \lfloor \frac{365}{60} \right \rfloor&lt;/math&gt;. Therefore, our answer is <br /> <br /> &lt;math&gt;365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor&lt;/math&gt; <br /> <br /> &lt;math&gt; = 365 - 285+72 - 6 = \boxed{\textbf{(D) } 146}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_4&diff=88730 1997 AIME Problems/Problem 4 2017-11-30T20:22:18Z <p>Programjames1: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Circle]]s of [[radii]] &lt;math&gt;5, 5, 8,&lt;/math&gt; and &lt;math&gt;\frac mn&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 /> == Solution ==<br /> [[Image:1997_AIME-4.png]]<br /> <br /> If (in the diagram above) we draw the line going through the centers of the circles with radii &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;\frac mn = r&lt;/math&gt;, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii &lt;math&gt;5&lt;/math&gt;. Then we form two [[right triangles]], of lengths &lt;math&gt;5, x, 5+r&lt;/math&gt; and &lt;math&gt;5, 8+r+x, 13&lt;/math&gt;, wher &lt;math&gt;x&lt;/math&gt; is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii &lt;math&gt;5&lt;/math&gt;. By the [[Pythagorean Theorem]], we now have two equations with two unknowns:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> 5^2 + x^2 &amp;=&amp; (5+r)^2 \\<br /> x &amp;=&amp; \sqrt{10r + r^2} \\<br /> &amp;&amp; \\<br /> (8 + r + \sqrt{10r+r^2})^2 + 5^2 &amp;=&amp; 13^2\\<br /> 8 + r + \sqrt{10r+r^2} &amp;=&amp; 12\\<br /> \sqrt{10r+r^2}&amp;=&amp; 4-r\\<br /> 10r+r^2 &amp;=&amp; 16 - 8r + r^2\\<br /> r &amp;=&amp; \frac{8}{9}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;m+n = \boxed{17}&lt;/math&gt;.<br /> <br /> <br /> NOTE: It can be seen that there is no apparent need to use the variable x as a 5,12,13 right triangle has been formed.<br /> == Solution 2 ==<br /> We may also use Descartes' theorem, &lt;math&gt;k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}&lt;/math&gt; where each of &lt;math&gt;k_i&lt;/math&gt; is the curvature of a circle with radius &lt;math&gt;r_i&lt;/math&gt;, and the curvature is defined as &lt;math&gt;k_i=\frac{1}{r_i}&lt;/math&gt;. The larger solution for &lt;math&gt;k_4&lt;/math&gt; will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get &lt;math&gt;k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}&lt;/math&gt;. Thus, &lt;math&gt;r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{017}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_7&diff=88248 1993 AIME Problems/Problem 7 2017-11-14T01:08:25Z <p>Programjames1: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Three numbers, &lt;math&gt;a_1\,&lt;/math&gt;, &lt;math&gt;a_2\,&lt;/math&gt;, &lt;math&gt;a_3\,&lt;/math&gt;, are drawn randomly and without replacement from the [[set]] &lt;math&gt;\{1, 2, 3, \dots, 1000\}\,&lt;/math&gt;. Three other numbers, &lt;math&gt;b_1\,&lt;/math&gt;, &lt;math&gt;b_2\,&lt;/math&gt;, &lt;math&gt;b_3\,&lt;/math&gt;, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let &lt;math&gt;p\,&lt;/math&gt; be the [[probability]] that, after a suitable rotation, a brick of dimensions &lt;math&gt;a_1 \times a_2 \times a_3\,&lt;/math&gt; can be enclosed in a box of dimensions &lt;math&gt;b_1 \times b_2 \times b_3\,&lt;/math&gt;, with the sides of the brick [[parallel]] to the sides of the box. If &lt;math&gt;p\,&lt;/math&gt; is written as a [[fraction]] in lowest terms, what is the sum of the numerator and denominator?<br /> <br /> == Solution 1 ==<br /> Call the six numbers selected &lt;math&gt;x_1 &gt; x_2 &gt; x_3 &gt; x_4 &gt; x_5 &gt; x_6&lt;/math&gt;. Clearly, &lt;math&gt;x_1&lt;/math&gt; must be a dimension of the box, and &lt;math&gt;x_6&lt;/math&gt; must be a dimension of the brick. <br /> <br /> *If &lt;math&gt;x_2&lt;/math&gt; is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us &lt;math&gt;3&lt;/math&gt; possibilities.<br /> *If &lt;math&gt;x_2&lt;/math&gt; is not a dimension of the box but &lt;math&gt;x_3&lt;/math&gt; is, then both remaining dimensions will work as a dimension of the box. That gives us &lt;math&gt;2&lt;/math&gt; possibilities.<br /> *If &lt;math&gt;x_4&lt;/math&gt; is a dimension of the box but &lt;math&gt;x_2,\ x_3&lt;/math&gt; aren’t, there are no possibilities (same for &lt;math&gt;x_5&lt;/math&gt;). <br /> <br /> The total number of arrangements is &lt;math&gt;{6\choose3} = 20&lt;/math&gt;; therefore, &lt;math&gt;p = \frac{3 + 2}{20} = \frac{1}{4}&lt;/math&gt;, and the answer is &lt;math&gt;1 + 4 = \boxed{005}&lt;/math&gt;.<br /> <br /> '''Note''' that the &lt;math&gt;1000&lt;/math&gt; in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers &lt;math&gt;x_1,x_2,x_3,x_4,x_5,x_6&lt;/math&gt; whether they may be &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; or &lt;math&gt;999,5,3,998,997,891&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Like solution &lt;math&gt;1&lt;/math&gt;, call the six numbers selected &lt;math&gt;x_1 &gt; x_2 &gt; x_3 &gt; x_4 &gt; x_5 &gt; x_6&lt;/math&gt;. Using the hook-length formula, the number of valid configuration is &lt;math&gt;\frac{6!}{4\cdot3\cdot2\cdot3\cdot2}&lt;/math&gt;. This gives us &lt;math&gt;5&lt;/math&gt;, and we proceed as &lt;math&gt;solution&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; did.<br /> == Solution 3==<br /> As in the preceding solutions, we let &lt;math&gt;x_1&gt;x_2&gt;x_3&gt;x_4&gt;x_5&gt;x_6&lt;/math&gt; where each &lt;math&gt;x_i&lt;/math&gt; is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the &lt;math&gt;n&lt;/math&gt;th Catalan number (where &lt;math&gt;n&lt;/math&gt; is the number of pairs of rising and falling sections) is &lt;cmath&gt;\frac{\binom{2n}{n}}{n+1}&lt;/cmath&gt;<br /> Thus, there are &lt;math&gt;\frac{\binom{6}{3}}{4}&lt;/math&gt; ways to pick which of &lt;math&gt;x_1,x_2,x_3,x_4,x_5,&lt;/math&gt; and &lt;math&gt;x_6&lt;/math&gt; are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are &lt;math&gt;\binom{6}{3}&lt;/math&gt; total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is &lt;math&gt;\frac{\left(\frac{\binom{6}{3}}{4}\right)}{\binom{6}{3}}=\frac14\Longrightarrow \boxed{005}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1993|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_2&diff=87683 2010 USAMO Problems/Problem 2 2017-10-03T23:20:08Z <p>Programjames1: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> There are &lt;math&gt;n&lt;/math&gt; students standing in a circle, one behind the<br /> other. The students have heights &lt;math&gt;h_1 &lt; h_2 &lt; \ldots &lt; h_n&lt;/math&gt;. If a<br /> student with height &lt;math&gt;h_k&lt;/math&gt; is standing directly behind a student<br /> with height &lt;math&gt;h_{k-2}&lt;/math&gt; or less, the two students are permitted to<br /> switch places. Prove that it is not possible to make more than<br /> &lt;math&gt;\binom{n}{3}&lt;/math&gt; such switches before reaching a position in which<br /> no further switches are possible.<br /> <br /> ==Solution 1==<br /> We adopt the usual convention that &lt;math&gt;\binom{i}{j} = 0&lt;/math&gt; unless &lt;math&gt;0 \le j \le i&lt;/math&gt;.<br /> With this, the binomial coefficients are defined for all integers via the<br /> recursion:<br /> &lt;center&gt;<br /> &lt;cmath&gt;<br /> \binom{0}{0} = 1, \quad \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}<br /> &lt;/cmath&gt;<br /> &lt;/center&gt;<br /> <br /> It is clear that the circle is oriented and all the students are facing in same direction (clockwise or counterclockwise). We'll call this direction &lt;i&gt;forward&lt;/i&gt;.<br /> <br /> In any switch consider the taller student to have moved &lt;i&gt;forward&lt;/i&gt; and the shorter student to have remained &lt;i&gt;stationary&lt;/i&gt;. No backward motion is allowed. With this definition of forward motion, the first two students with heights &lt;math&gt;h_1&lt;/math&gt; and &lt;math&gt;h_2&lt;/math&gt; are always stationary, while other students potentially move past them.<br /> <br /> For &lt;math&gt;k &gt; 2&lt;/math&gt;, the student with height &lt;math&gt;h_k&lt;/math&gt; can never switch places with the student with height &lt;math&gt;h_{k-1}&lt;/math&gt;, and the former can make at most &lt;math&gt;k-2&lt;/math&gt; more forward moves than the latter (when all the students of heights &lt;math&gt;h_1, \ldots h_{k-2}&lt;/math&gt; are between &lt;math&gt;h_k&lt;/math&gt; and &lt;math&gt;h_{k-1}&lt;/math&gt; in the forward direction).<br /> <br /> Therefore, if the &lt;math&gt;(k-1)^{\mathrm{st}}&lt;/math&gt; student can make &lt;math&gt;s_{k-1}&lt;/math&gt; forward steps, the &lt;math&gt;k^{\mathrm{th}}&lt;/math&gt; student can make at most &lt;math&gt;s_{k-1} + k - 2&lt;/math&gt; steps. With &lt;math&gt;s_1 = s_2 = 0&lt;/math&gt; and &lt;math&gt;s_3 = 0 + (3-2) = 1&lt;/math&gt;, and a constant second difference of &lt;math&gt;1&lt;/math&gt;, we quickly see that &lt;math&gt;s_k = \binom{k-1}{2}&lt;/math&gt;.<br /> <br /> With &lt;math&gt;n&lt;/math&gt; students in all, the total number of steps is therefore at most &lt;math&gt;\sum_{i=3}^{n}\binom{i-1}{2} = \binom{n}{3}&lt;/math&gt;. The sum is a telescoping sum since: &lt;math&gt;\binom{n}{3} - \binom{n-1}{3} = \binom{n-1}{2}.&lt;/math&gt;<br /> ==Solution 2==<br /> WLOG, let &lt;math&gt;h_k=k&lt;/math&gt;. Now, we find the end arrangement with the most switches possible. We claim that the arrangement will be &lt;math&gt;n,n-1,n-2,\dots ,1&lt;/math&gt;, where the left to right direction is the &quot;backwards&quot; direction. To prove this makes the most switches, we show that there is always at least one more switch that can be done for any other arrangement. This is elementary to show. There will always be one height &lt;math&gt;x&lt;/math&gt; such that the number to its right is &lt;math&gt;2&lt;/math&gt; less than &lt;math&gt;x&lt;/math&gt;, unless every number has &lt;math&gt;x-1&lt;/math&gt; to the right of &lt;math&gt;x&lt;/math&gt; (other than &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;). The exception occurs at our claim, so our claim is proven. Now we want to find the maximum ways we can &quot;undo&quot; our arrangement. But undoing a switch is just doing a switch from our arrangement in the opposite direction. So, the start arrangement with the most possible switches is the reverse of the end arrangement or &lt;math&gt;1,2,3,\dots ,n&lt;/math&gt;.<br /> We want to find how many switches must be done to get from the start arrangement to the end arrangement. We start by switching &lt;math&gt;n&lt;/math&gt; around until it cannot be switched anymore. We find that we can switch &lt;math&gt;n&lt;/math&gt; &lt;math&gt;n-2&lt;/math&gt; times. When we are switching (or, in other words, moving to the right) the number &lt;math&gt;k&lt;/math&gt;, we can switch it &lt;math&gt;k-2&lt;/math&gt; times before it is to the left of &lt;math&gt;k-1&lt;/math&gt;. But then we can also switch each of &lt;math&gt;k+1,k+2,\dots ,n&lt;/math&gt; (in that order) &lt;math&gt;k-2&lt;/math&gt; times before they get stopped again. Let &lt;math&gt;x=n-k+2&lt;/math&gt;. Then, the sum of the switches is &lt;math&gt;\sum\limits_{x=2}^{n-1} (n-x)(x-1)=\sum\limits_{x=1}^{n} (n-x)(x-1)&lt;/math&gt;.<br /> <br /> Rearranging and summing, we get &lt;cmath&gt;\sum\limits_{x=1}^{n} x(n+1)-\sum\limits_{x=1}^{n} n -\sum\limits_{x=1}^{n} x^2=\frac{n(n+1)^2}2-n^2-\frac{n(n+1)(2n+1)}6=\frac{3n(n^2+1)}6-\frac{n(n+1)(2n+1)}6=\frac{n(n^2-3n+2)}6=\frac{n(n-1)(n-2)}{3!}=\binom{n}{3}&lt;/cmath&gt;<br /> <br /> ==Comment==<br /> This process of switching is very similar to Bubble Sort, except that the terms must be in consecutive decreasing order, and wraps around &lt;math&gt;2,1,n,n-1\dots&lt;/math&gt;. This answer seems to disagree, though, because the worst case &lt;math&gt;O&lt;/math&gt; efficiency is &lt;math&gt;n^2&lt;/math&gt;, not &lt;math&gt;\frac{n(n-1)(n-2)}{6}&lt;/math&gt;.<br /> <br /> https://en.wikipedia.org/wiki/Bubble_sort<br /> <br /> ==See also==<br /> {{USAMO newbox|year=2010|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_2&diff=87682 2010 USAMO Problems/Problem 2 2017-10-03T23:19:42Z <p>Programjames1: /* Solution */</p> <hr /> <div>==Problem==<br /> There are &lt;math&gt;n&lt;/math&gt; students standing in a circle, one behind the<br /> other. The students have heights &lt;math&gt;h_1 &lt; h_2 &lt; \ldots &lt; h_n&lt;/math&gt;. If a<br /> student with height &lt;math&gt;h_k&lt;/math&gt; is standing directly behind a student<br /> with height &lt;math&gt;h_{k-2}&lt;/math&gt; or less, the two students are permitted to<br /> switch places. Prove that it is not possible to make more than<br /> &lt;math&gt;\binom{n}{3}&lt;/math&gt; such switches before reaching a position in which<br /> no further switches are possible.<br /> <br /> ==Solution 1==<br /> We adopt the usual convention that &lt;math&gt;\binom{i}{j} = 0&lt;/math&gt; unless &lt;math&gt;0 \le j \le i&lt;/math&gt;.<br /> With this, the binomial coefficients are defined for all integers via the<br /> recursion:<br /> &lt;center&gt;<br /> &lt;cmath&gt;<br /> \binom{0}{0} = 1, \quad \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}<br /> &lt;/cmath&gt;<br /> &lt;/center&gt;<br /> <br /> It is clear that the circle is oriented and all the students are facing in same direction (clockwise or counterclockwise). We'll call this direction &lt;i&gt;forward&lt;/i&gt;.<br /> <br /> In any switch consider the taller student to have moved &lt;i&gt;forward&lt;/i&gt; and the shorter student to have remained &lt;i&gt;stationary&lt;/i&gt;. No backward motion is allowed. With this definition of forward motion, the first two students with heights &lt;math&gt;h_1&lt;/math&gt; and &lt;math&gt;h_2&lt;/math&gt; are always stationary, while other students potentially move past them.<br /> <br /> For &lt;math&gt;k &gt; 2&lt;/math&gt;, the student with height &lt;math&gt;h_k&lt;/math&gt; can never switch places with the student with height &lt;math&gt;h_{k-1}&lt;/math&gt;, and the former can make at most &lt;math&gt;k-2&lt;/math&gt; more forward moves than the latter (when all the students of heights &lt;math&gt;h_1, \ldots h_{k-2}&lt;/math&gt; are between &lt;math&gt;h_k&lt;/math&gt; and &lt;math&gt;h_{k-1}&lt;/math&gt; in the forward direction).<br /> <br /> Therefore, if the &lt;math&gt;(k-1)^{\mathrm{st}}&lt;/math&gt; student can make &lt;math&gt;s_{k-1}&lt;/math&gt; forward steps, the &lt;math&gt;k^{\mathrm{th}}&lt;/math&gt; student can make at most &lt;math&gt;s_{k-1} + k - 2&lt;/math&gt; steps. With &lt;math&gt;s_1 = s_2 = 0&lt;/math&gt; and &lt;math&gt;s_3 = 0 + (3-2) = 1&lt;/math&gt;, and a constant second difference of &lt;math&gt;1&lt;/math&gt;, we quickly see that &lt;math&gt;s_k = \binom{k-1}{2}&lt;/math&gt;.<br /> <br /> With &lt;math&gt;n&lt;/math&gt; students in all, the total number of steps is therefore at most &lt;math&gt;\sum_{i=3}^{n}\binom{i-1}{2} = \binom{n}{3}&lt;/math&gt;. The sum is a telescoping sum since: &lt;math&gt;\binom{n}{3} - \binom{n-1}{3} = \binom{n-1}{2}.&lt;/math&gt;<br /> ==Solution 2==<br /> WLOG, let the &lt;math&gt;h_k=k&lt;/math&gt;. Now, we find the end arrangement with the most switches possible. We claim that the arrangement will be &lt;math&gt;n,n-1,n-2,\dots ,1&lt;/math&gt;, where the left to right direction is the &quot;backwards&quot; direction. To prove this makes the most switches, we show that there is always at least one more switch that can be done for any other arrangement. This is elementary to show. There will always be one height &lt;math&gt;x&lt;/math&gt; such that the number to its right is &lt;math&gt;2&lt;/math&gt; less than &lt;math&gt;x&lt;/math&gt;, unless every number has &lt;math&gt;x-1&lt;/math&gt; to the right of &lt;math&gt;x&lt;/math&gt; (other than &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;). The exception occurs at our claim, so our claim is proven. Now we want to find the maximum ways we can &quot;undo&quot; our arrangement. But undoing a switch is just doing a switch from our arrangement in the opposite direction. So, the start arrangement with the most possible switches is the reverse of the end arrangement or &lt;math&gt;1,2,3,\dots ,n&lt;/math&gt;.<br /> We want to find how many switches must be done to get from the start arrangement to the end arrangement. We start by switching &lt;math&gt;n&lt;/math&gt; around until it cannot be switched anymore. We find that we can switch &lt;math&gt;n&lt;/math&gt; &lt;math&gt;n-2&lt;/math&gt; times. When we are switching (or, in other words, moving to the right) the number &lt;math&gt;k&lt;/math&gt;, we can switch it &lt;math&gt;k-2&lt;/math&gt; times before it is to the left of &lt;math&gt;k-1&lt;/math&gt;. But then we can also switch each of &lt;math&gt;k+1,k+2,\dots ,n&lt;/math&gt; (in that order) &lt;math&gt;k-2&lt;/math&gt; times before they get stopped again. Let &lt;math&gt;x=n-k+2&lt;/math&gt;. Then, the sum of the switches is &lt;math&gt;\sum\limits_{x=2}^{n-1} (n-x)(x-1)=\sum\limits_{x=1}^{n} (n-x)(x-1)&lt;/math&gt;.<br /> <br /> Rearranging and summing, we get &lt;cmath&gt;\sum\limits_{x=1}^{n} x(n+1)-\sum\limits_{x=1}^{n} n -\sum\limits_{x=1}^{n} x^2=\frac{n(n+1)^2}2-n^2-\frac{n(n+1)(2n+1)}6=\frac{3n(n^2+1)}6-\frac{n(n+1)(2n+1)}6=\frac{n(n^2-3n+2)}6=\frac{n(n-1)(n-2)}{3!}=\binom{n}{3}&lt;/cmath&gt;<br /> <br /> ==Comment==<br /> This process of switching is very similar to Bubble Sort, except that the terms must be in consecutive decreasing order, and wraps around &lt;math&gt;2,1,n,n-1\dots&lt;/math&gt;. This answer seems to disagree, though, because the worst case &lt;math&gt;O&lt;/math&gt; efficiency is &lt;math&gt;n^2&lt;/math&gt;, not &lt;math&gt;\frac{n(n-1)(n-2)}{6}&lt;/math&gt;.<br /> <br /> https://en.wikipedia.org/wiki/Bubble_sort<br /> <br /> ==See also==<br /> {{USAMO newbox|year=2010|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_6&diff=81438 2016 AMC 8 Problems/Problem 6 2016-11-23T16:19:30Z <p>Programjames1: /* Solution */</p> <hr /> <div>The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br /> <br /> <br /> ==Solution==<br /> If you have a solution, then please help us out by posting it.<br /> <br /> {{AMC8 box|year=2016|num-b=5|num-a=7}}</div> Programjames1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_22&diff=81428 2016 AMC 8 Problems/Problem 22 2016-11-23T16:04:13Z <p>Programjames1: </p> <hr /> <div>Rectangle &lt;math&gt;DEFA&lt;/math&gt; below is a &lt;math&gt;3 \times 4&lt;/math&gt; rectangle with &lt;math&gt;DC=CB=BA&lt;/math&gt;. What is the area of the &quot;bat wings&quot; (shaded area)?<br /> &lt;asy&gt;<br /> draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br /> draw((3,0)--(1,4)--(0,0));<br /> fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br /> fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> {{solution}}<br /> The area of the trapezoid containing the shaded region and the two isosceles triangles is &lt;math&gt;\frac{1+3}2\cdot 4=8&lt;/math&gt;. Next we find the height of each triangle to calculate their area. The triangles are similar, and are in a &lt;math&gt;3:1&lt;/math&gt; ratio, so the height of the bigger one is 3, while the height of the smaller one is 1. Thus, their areas are &lt;math&gt;\frac12&lt;/math&gt; and &lt;math&gt;\frac92&lt;/math&gt;. Subtracting these areas from the trapezoid, we get &lt;math&gt;8-\frac12-\frac92 =\boxed3&lt;/math&gt;. The answer is (C).<br /> <br /> <br /> {{AMC8 box|year=2016|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Programjames1