https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Neutrinonerd3333&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-21T02:04:06Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1997_USAMO_Problems/Problem_1&diff=56794 1997 USAMO Problems/Problem 1 2013-08-03T10:12:36Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_1,p_2,p_3,...&lt;/math&gt; be the prime numbers listed in increasing order, and let &lt;math&gt;x_0&lt;/math&gt; be a real number between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. For positive integer &lt;math&gt;k&lt;/math&gt;, define<br /> <br /> &lt;math&gt; x_{k}=\begin{cases}0&amp;\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&amp;\text{ if }x_{k-1}\ne0\end{cases} &lt;/math&gt;<br /> <br /> where &lt;math&gt;\{x\}&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. (The fractional part of &lt;math&gt;x&lt;/math&gt; is given by &lt;math&gt;x-\lfloor{x}\rfloor&lt;/math&gt; where &lt;math&gt;\lfloor{x}\rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.) Find, with proof, all &lt;math&gt;x_0&lt;/math&gt; satisfying &lt;math&gt;0&lt;x_0&lt;1&lt;/math&gt; for which the sequence &lt;math&gt;x_0,x_1,x_2,...&lt;/math&gt; eventually becomes &lt;math&gt;0&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> All rational numbers between 0 and 1 will eventually become 0 under this iterative process. To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let &lt;math&gt;x_0 = \frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m,n&lt;/math&gt; are coprime positive integers. Since &lt;math&gt;0&lt;x_0&lt;1&lt;/math&gt;, &lt;math&gt;0&lt;m&lt;n&lt;/math&gt;. Now &lt;cmath&gt;x_1 = \left\{\frac{p_1}{\frac{m}{n}}\right\}=\left\{\frac{np_1}{m}\right\}.&lt;/cmath&gt; From this, we can see that applying the iterative process will decrease the value of the denominator, since &lt;math&gt;m&lt;n&lt;/math&gt;. Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.<br /> <br /> On the other hand, irrational &lt;math&gt;x_0&lt;/math&gt; will never become 0, because &lt;math&gt;x_k&lt;/math&gt; will always be irrational.<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=1997|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1997_USAMO_Problems/Problem_1&diff=56793 1997 USAMO Problems/Problem 1 2013-08-03T10:12:13Z <p>Neutrinonerd3333: added solution</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_1,p_2,p_3,...&lt;/math&gt; be the prime numbers listed in increasing order, and let &lt;math&gt;x_0&lt;/math&gt; be a real number between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. For positive integer &lt;math&gt;k&lt;/math&gt;, define<br /> <br /> &lt;math&gt; x_{k}=\begin{cases}0&amp;\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&amp;\text{ if }x_{k-1}\ne0\end{cases} &lt;/math&gt;<br /> <br /> where &lt;math&gt;\{x\}&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. (The fractional part of &lt;math&gt;x&lt;/math&gt; is given by &lt;math&gt;x-\lfloor{x}\rfloor&lt;/math&gt; where &lt;math&gt;\lfloor{x}\rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.) Find, with proof, all &lt;math&gt;x_0&lt;/math&gt; satisfying &lt;math&gt;0&lt;x_0&lt;1&lt;/math&gt; for which the sequence &lt;math&gt;x_0,x_1,x_2,...&lt;/math&gt; eventually becomes &lt;math&gt;0&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> All rational numbers between 0 and 1 will eventually become 0 under this iterative process. To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let &lt;math&gt;x_0 = \frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m,n&lt;/math&gt; are coprime positive integers. Since &lt;math&gt;0&lt;x_0&lt;1&lt;/math&gt;, &lt;math&gt;0&lt;m&lt;n&lt;/math&gt;. Now &lt;cmath&gt;x_1 = \left\{\frac{p_1}{\frac{m}{n}}\right\}=\left\{frac{np_1}{m}\right\}.&lt;/cmath&gt; From this, we can see that applying the iterative process will decrease the value of the denominator, since &lt;math&gt;m&lt;n&lt;/math&gt;. Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.<br /> <br /> On the other hand, irrational &lt;math&gt;x_0&lt;/math&gt; will never become 0, because &lt;math&gt;x_k&lt;/math&gt; will always be irrational.<br /> <br /> == See Also ==<br /> {{USAMO newbox|year=1997|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_4&diff=56789 1996 USAMO Problems/Problem 4 2013-08-02T07:33:18Z <p>Neutrinonerd3333: added solution</p> <hr /> <div>==Problem==<br /> <br /> An &lt;math&gt;n&lt;/math&gt;-term sequence &lt;math&gt;(x_1, x_2, \ldots, x_n)&lt;/math&gt; in which each term is either 0 or 1 is called a binary sequence of length &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;a_n&lt;/math&gt; be the number of binary sequences of length n containing no three consecutive terms equal to 0, 1, 0 in that order. Let &lt;math&gt;b_n&lt;/math&gt; be the number of binary sequences of length &lt;math&gt;n&lt;/math&gt; that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that &lt;math&gt;b_{n+1} = 2a_n&lt;/math&gt; for all positive integers &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> Given any binary sequence &lt;math&gt;B=(b_1,b_2,b_3,\dots,b_k)&lt;/math&gt;, define &lt;math&gt;f(B)=(|b_2-b_1|,|b_3-b_2|,\dots,|b_k-b_{k-1}|)&lt;/math&gt;. The operator &lt;math&gt;f&lt;/math&gt; basically takes pairs of consecutive terms and returns 0 if the terms are the same and 1 otherwise. Note that for every sequence &lt;math&gt;S&lt;/math&gt; of length &lt;math&gt;n&lt;/math&gt; there exist exactly two binary sequences &lt;math&gt;B&lt;/math&gt; of length &lt;math&gt;n+1&lt;/math&gt; such that &lt;math&gt;f(B)=S&lt;/math&gt;.<br /> <br /> If &lt;math&gt;f(B)&lt;/math&gt; does not contain the string 0, 1, 0, &lt;math&gt;B&lt;/math&gt; cannot contain either of the strings 0, 0, 1, 1 or 1, 1, 0, 0. Conversely, if &lt;math&gt;B&lt;/math&gt; does not contain the sequences 0, 0, 1, 1 or 1, 1, 0, 0, &lt;math&gt;f(B)&lt;/math&gt; cannot contain 0, 1, 0. There are &lt;math&gt;a_n&lt;/math&gt; such &lt;math&gt;f(B)&lt;/math&gt; and &lt;math&gt;b_{n+1}&lt;/math&gt; such &lt;math&gt;B&lt;/math&gt;. Since each &lt;math&gt;S&lt;/math&gt; corresponds with two &lt;math&gt;B&lt;/math&gt;, there are twice as many such &lt;math&gt;B&lt;/math&gt; as such &lt;math&gt;S&lt;/math&gt;; thus, &lt;math&gt;b_{n+1}=2a_n&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_5&diff=56782 1996 USAMO Problems/Problem 5 2013-08-01T14:25:34Z <p>Neutrinonerd3333: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle, and &lt;math&gt;M&lt;/math&gt; an interior point such that &lt;math&gt;\angle MAB=10^\circ &lt;/math&gt;, &lt;math&gt;\angle MBA=20^\circ&lt;/math&gt; , &lt;math&gt;\angle MAC= 40^\circ&lt;/math&gt; and &lt;math&gt;\angle MCA=30^\circ&lt;/math&gt;. Prove that the triangle is isosceles.<br /> <br /> ==Solution==<br /> <br /> Clearly, &lt;math&gt;\angle AMB = 150^\circ&lt;/math&gt; and &lt;math&gt;\angle AMC = 110^\circ&lt;/math&gt;. Now by the Law of Sines on triangles &lt;math&gt;ABM&lt;/math&gt; and &lt;math&gt;ACM&lt;/math&gt;, we have &lt;cmath&gt;\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}&lt;/cmath&gt; and &lt;cmath&gt;\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.&lt;/cmath&gt; Combining these equations gives us &lt;cmath&gt;\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.&lt;/cmath&gt; Without loss of generality, let &lt;math&gt;AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}&lt;/math&gt; and &lt;math&gt;AC = \sin 20^\circ \sin 110^\circ&lt;/math&gt;. Then by the Law of Cosines, we have<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> BC^2 &amp;= AB^2 + AC^2 - 2(AB)(BC)\cos\angle BAC\\<br /> &amp;= \frac{1}{16} + \sin^2 20^\circ\sin^2 110^\circ - 2\left(\frac{1}{4}\right)\sin 20^\circ\sin 110^\circ\cos 50^\circ \\<br /> &amp;= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \frac{1}{2}\sin 20^\circ\sin 110^\circ\sin 40^\circ \\<br /> &amp;= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \sin 20^\circ\sin 110^\circ\sin 20^\circ\cos 20^\circ \\<br /> &amp;= \frac{1}{16}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;AB = BC&lt;/math&gt;, our desired conclusion.<br /> <br /> {{MAA Notice}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_2&diff=56768 1996 USAMO Problems/Problem 2 2013-07-30T06:56:53Z <p>Neutrinonerd3333: clarified solution</p> <hr /> <div>==Problem==<br /> For any nonempty set &lt;math&gt;S&lt;/math&gt; of real numbers, let &lt;math&gt;\sigma(S)&lt;/math&gt; denote the sum of the elements of &lt;math&gt;S&lt;/math&gt;. Given a set &lt;math&gt;A&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; positive integers, consider the collection of all distinct sums &lt;math&gt;\sigma(S)&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt; ranges over the nonempty subsets of &lt;math&gt;A&lt;/math&gt;. Prove that this collection of sums can be partitioned into &lt;math&gt;n&lt;/math&gt; classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.<br /> <br /> ==Solution==<br /> Let set &lt;math&gt;A&lt;/math&gt; consist of the integers &lt;math&gt;a_1\le a_2\le a_3\le\dots\le a_n&lt;/math&gt;. For &lt;math&gt;k\ge 2&lt;/math&gt;, call &lt;math&gt;a_k&lt;/math&gt; greedy if &lt;math&gt;\sum_{i=1}^{k-1}a_i &lt; a_k&lt;/math&gt;. Also call &lt;math&gt;a_1&lt;/math&gt; greedy. Now put all elements of &lt;math&gt;A&lt;/math&gt; into groups of consecutive terms in such a way that each group &lt;math&gt;G&lt;/math&gt; begins with a greedy term, call it &lt;math&gt;a_p&lt;/math&gt;, and ends on the term &lt;math&gt;a_{q-1}&lt;/math&gt; just before the next greedy term after &lt;math&gt;a_p&lt;/math&gt;. (If &lt;math&gt;a_p&lt;/math&gt; is the last greedy term, let &lt;math&gt;q-1=n&lt;/math&gt;.) We introduce some more terminology. A sum &lt;math&gt;\sigma(S)&lt;/math&gt; is said to &quot;belong to&quot; a group &lt;math&gt;G&lt;/math&gt; if &lt;math&gt;\max(S)\in G&lt;/math&gt;. Denote by &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; the set of all sums belonging to &lt;math&gt;G&lt;/math&gt;.<br /> <br /> '''We now show that we can divide &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; into &lt;math&gt;|G|&lt;/math&gt; (the cardinality of &lt;math&gt;G&lt;/math&gt;) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2.''' Using the previous notation, we first prove that &lt;math&gt;\frac{\max{\mathcal{S}(G)}}{\min{\mathcal{S}(G)}}&lt; 2^{|G|}=2^{q-p}&lt;/math&gt;. Note that &lt;cmath&gt;\max{\mathcal{S}(G)}=\sum_{i=1}^{q-1}a_i&lt;/cmath&gt; and &lt;cmath&gt;\min{\mathcal{S}(G)}=a_p.&lt;/cmath&gt; Taking note of that fact that &lt;math&gt;a_{p+1},a_{p+2},\dots,a_{q-1}&lt;/math&gt; are not greedy numbers, we write:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \max{\mathcal{S}}&amp;=\sum_{i=1}^{q-1}a_i \\<br /> &amp;=a_{q-1}+\sum_{i=1}^{q-2}a_i \\<br /> &amp;\le 2\sum_{i=1}^{q-2}a_i\\<br /> &amp;\le 2^2\sum_{i=1}^{q-3}a_i\\<br /> &amp;\vdots\\<br /> &amp;\le 2^{q-p-1}\sum_{i=1}^{p}a_i\\<br /> &amp;= 2^{q-p-1}(\sum_{i=1}^{p-1}a_i + a_p)\\<br /> &amp;&lt;2^{q-p-1}(2a_p)\\<br /> &amp;=2^{q-p}\min\mathcal{S}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> where the inequalities after the ellipses result from the fact that &lt;math&gt;a_p&lt;/math&gt; is a greedy number (which implies by definition that &lt;math&gt;\sum_{i=1}^{p-1}a_i&lt;a_p&lt;/math&gt;). This proves the desired inequality. Now we can prove the result in bold above. Divide &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; into &lt;math&gt;|G|=q-p&lt;/math&gt; classes by taking those terms in &lt;math&gt;[a_p,2a_p)&lt;/math&gt; and placing them in the first class, taking those terms in &lt;math&gt;[2a_p,2^2a_p)&lt;/math&gt; and placing them in another, and so on, until we reach &lt;math&gt;[2^{q-p-1}a_p,2^{q-p}a_p)&lt;/math&gt;. The inequality we proved above shows that all of the sums in &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by &lt;math&gt;\min\mathcal{S}(G)=a_p&lt;/math&gt; on the bottom and &lt;math&gt;2^{q-p}a_p&gt;\max\mathcal{S}(G)&lt;/math&gt; on the top. This proves the result in bold.<br /> <br /> However, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely &lt;math&gt;n&lt;/math&gt; classes, because every term &lt;math&gt;a_i&lt;/math&gt; belongs to a group and for each group &lt;math&gt;G&lt;/math&gt;, there are as many classes for &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; as there are terms in &lt;math&gt;G&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_2&diff=56767 1996 USAMO Problems/Problem 2 2013-07-30T06:53:11Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>==Problem==<br /> For any nonempty set &lt;math&gt;S&lt;/math&gt; of real numbers, let &lt;math&gt;\sigma(S)&lt;/math&gt; denote the sum of the elements of &lt;math&gt;S&lt;/math&gt;. Given a set &lt;math&gt;A&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; positive integers, consider the collection of all distinct sums &lt;math&gt;\sigma(S)&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt; ranges over the nonempty subsets of &lt;math&gt;A&lt;/math&gt;. Prove that this collection of sums can be partitioned into &lt;math&gt;n&lt;/math&gt; classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.<br /> <br /> ==Solution==<br /> Let set &lt;math&gt;A&lt;/math&gt; consist of the integers &lt;math&gt;a_1\le a_2\le a_3\le\dots\le a_n&lt;/math&gt;. For &lt;math&gt;k\ge 2&lt;/math&gt;, call &lt;math&gt;a_k&lt;/math&gt; greedy if &lt;math&gt;\sum_{i=1}^{k-1}a_i &lt; a_k&lt;/math&gt;. Also call &lt;math&gt;a_1&lt;/math&gt; greedy. Now put all elements of &lt;math&gt;A&lt;/math&gt; into groups of consecutive terms in such a way that each group &lt;math&gt;G&lt;/math&gt; begins with a greedy term, call it &lt;math&gt;a_p&lt;/math&gt;, and ends on the term &lt;math&gt;a_{q-1}&lt;/math&gt; just before the next greedy term after &lt;math&gt;a_p&lt;/math&gt;. (If &lt;math&gt;a_p&lt;/math&gt; is the last greedy term, let &lt;math&gt;q-1=n&lt;/math&gt;.) We introduce some more terminology. A sum &lt;math&gt;\sigma(S)&lt;/math&gt; is said to &quot;belong to&quot; a group &lt;math&gt;G&lt;/math&gt; if &lt;math&gt;\max(S)\in G&lt;/math&gt;. Denote by &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; the set of all sums belonging to &lt;math&gt;G&lt;/math&gt;.<br /> <br /> '''We now show that we can divide &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; into &lt;math&gt;|G|&lt;/math&gt; (the cardinality of &lt;math&gt;G&lt;/math&gt;) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2.''' Using the previous notation, we first prove that &lt;math&gt;\frac{\max{\mathcal{S}(G)}}{\min{\mathcal{S}(G)}}&lt; 2^{|G|}=2^{q-p}&lt;/math&gt;. Note that &lt;cmath&gt;\max{\mathcal{S}(G)}=\sum_{i=1}^{q-1}a_i&lt;/cmath&gt; and &lt;cmath&gt;\min{\mathcal{S}(G)}=a_p.&lt;/cmath&gt; Taking note of that fact that &lt;math&gt;a_{p+1},a_{p+2},\dots,a_{q-1}&lt;/math&gt; are not greedy numbers, we write:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \max{\mathcal{S}}&amp;=\sum_{i=1}^{q-1}a_i \\<br /> &amp;=a_{q-1}+\sum_{i=1}^{q-2}a_i \\<br /> &amp;\le 2\sum_{i=1}^{q-2}a_i\\<br /> &amp;\le 2^2\sum_{i=1}^{q-3}a_i\\<br /> &amp;\vdots\\<br /> &amp;\le 2^{q-p-1}\sum_{i=1}^{p}a_i\\<br /> &amp;= 2^{q-p-1}(\sum_{i=1}^{p-1}a_i + a_p)\\<br /> &amp;&lt;2^{q-p-1}(2a_p)\\<br /> &amp;=2^{q-p}\min\mathcal{S}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> where the inequalities after the ellipses result from the fact that &lt;math&gt;a_p&lt;/math&gt; is a greedy number (which implies by definition that &lt;math&gt;\sum_{i=1}^{p-1}a_i&lt;a_p&lt;/math&gt;). This proves the desired inequality. Now we can prove the result in bold above. Divide &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; into &lt;math&gt;|G|=q-p&lt;/math&gt; classes by taking those terms in &lt;math&gt;[a_p,2a_p)&lt;/math&gt; and placing them in the first class, taking those terms in &lt;math&gt;[2a_p,2^2a_p)&lt;/math&gt; and placing them in another, and so on, until we reach &lt;math&gt;[2^{q-p-1}a_p,2^{q-p}a_p)&lt;/math&gt;. The inequality we proved above shows that all of the sums in &lt;math&gt;\mathcal{S}(G)&lt;/math&gt; will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by &lt;math&gt;\min\mathcal{S}(G)=a_p&lt;/math&gt; on the bottom and &lt;math&gt;2^{q-p}a_p&gt;\max\mathcal{S}(G)&lt;/math&gt; on the top. This proves the result in bold.<br /> <br /> However, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely &lt;math&gt;n&lt;/math&gt; classes, because every term &lt;math&gt;a_i&lt;/math&gt; belongs to a group and in each group, there are as many classes as there are terms.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_2&diff=56766 1996 USAMO Problems/Problem 2 2013-07-30T06:42:23Z <p>Neutrinonerd3333: Added solution</p> <hr /> <div>==Problem==<br /> For any nonempty set &lt;math&gt;S&lt;/math&gt; of real numbers, let &lt;math&gt;\sigma(S)&lt;/math&gt; denote the sum of the elements of &lt;math&gt;S&lt;/math&gt;. Given a set &lt;math&gt;A&lt;/math&gt; of &lt;math&gt;n&lt;/math&gt; positive integers, consider the collection of all distinct sums &lt;math&gt;\sigma(S)&lt;/math&gt; as &lt;math&gt;S&lt;/math&gt; ranges over the nonempty subsets of &lt;math&gt;A&lt;/math&gt;. Prove that this collection of sums can be partitioned into &lt;math&gt;n&lt;/math&gt; classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.<br /> <br /> ==Solution==<br /> Let set &lt;math&gt;A&lt;/math&gt; consist of the integers &lt;math&gt;a_1\le a_2\le a_3\le\dots\le a_n&lt;/math&gt;. For &lt;math&gt;k\ge 2&lt;/math&gt;, call &lt;math&gt;a_k&lt;/math&gt; greedy if &lt;math&gt;\sum_{i=1}^{k-1}a_i &lt; a_k&lt;/math&gt;. Also call &lt;math&gt;a_1&lt;/math&gt; greedy. Now put all elements of &lt;math&gt;A&lt;/math&gt; into groups of consecutive terms in such a way that each group &lt;math&gt;G&lt;/math&gt; begins with a greedy term, call it &lt;math&gt;a_p&lt;/math&gt;, and ends on the term &lt;math&gt;a_{q-1}&lt;/math&gt; just before the next greedy term after &lt;math&gt;a_p&lt;/math&gt;. (If &lt;math&gt;a_p&lt;/math&gt; is the last greedy term, let &lt;math&gt;q-1=n&lt;/math&gt;.) We introduce some more terminology. A sum &lt;math&gt;\sigma(S)&lt;/math&gt; is said to &quot;belong to&quot; a group &lt;math&gt;G&lt;/math&gt; if &lt;math&gt;\max(S)\in G&lt;/math&gt;. Denote by &lt;math&gt;\mathcal{S}&lt;/math&gt; the set of all sums belonging to &lt;math&gt;G&lt;/math&gt;.<br /> <br /> '''We now show that we can divide &lt;math&gt;\mathcal{S}&lt;/math&gt; into &lt;math&gt;|G|&lt;/math&gt; (the cardinality of &lt;math&gt;G&lt;/math&gt;) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2.''' Using the previous notation, we first prove that &lt;math&gt;\frac{\max{\mathcal{S}}}{\min{\mathcal{S}}}&lt; 2^{|G|}=2^{q-p}&lt;/math&gt;. Note that &lt;cmath&gt;\max{\mathcal{S}}=\sum_{i=1}^{q-1}a_i&lt;/cmath&gt; and &lt;cmath&gt;\min{\mathcal{S}}=a_p.&lt;/cmath&gt; Taking note of that fact that &lt;math&gt;a_{p+1},a_{p+2},\dots,a_{q-1}&lt;/math&gt; are not greedy numbers, we write:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \max{\mathcal{S}}&amp;=\sum_{i=1}^{q-1}a_i \\<br /> &amp;=a_{q-1}+\sum_{i=1}^{q-2}a_i \\<br /> &amp;\le 2\sum_{i=1}^{q-2}a_i\\<br /> &amp;\le 2^2\sum_{i=1}^{q-3}a_i\\<br /> &amp;\vdots\\<br /> &amp;\le 2^{q-p-1}\sum_{i=1}^{p}a_i\\<br /> &amp;= 2^{q-p-1}(\sum_{i=1}^{p-1}a_i + a_p)\\<br /> &amp;&lt;2^{q-p-1}(2a_p)\\<br /> &amp;=2^{q-p}\min\mathcal{S}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> where the inequalities after the ellipses result from the fact that &lt;math&gt;a_p&lt;/math&gt; is a greedy number (which implies by definition that &lt;math&gt;\sum_{i=1}^{p-1}a_i&lt;a_p&lt;/math&gt;). This proves the desired inequality. Now we can prove the result in bold above. Divide &lt;math&gt;\mathcal{S}&lt;/math&gt; into classes by taking those terms in &lt;math&gt;[a_p,2a_p)&lt;/math&gt; and placing them in the first class, taking those terms in &lt;math&gt;[2a_p,2^2a_p)&lt;/math&gt; and placing them in another, and so on, until we reach &lt;math&gt;[2^{q-p-1}a_p,2^{q-p}a_p)&lt;/math&gt;. The inequality we proved above shows that all of the sums in &lt;math&gt;\mathcal{S}&lt;/math&gt; will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by &lt;math&gt;\min\mathcal{S}=a_p&lt;/math&gt; on the bottom and &lt;math&gt;2^{q-p}a_p&gt;\max\mathcal{S}&lt;/math&gt; on the top. This proves the result in bold.<br /> <br /> However, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely &lt;math&gt;n&lt;/math&gt; classes, because every term &lt;math&gt;a_i&lt;/math&gt; belongs to a group and in each group, there are as many classes as there are terms.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=51744 2013 AIME I Problems/Problem 15 2013-03-16T02:21:43Z <p>Neutrinonerd3333: /* Problem 15 */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of ordered triples &lt;math&gt;(A,B,C)&lt;/math&gt; of integers satisfying the conditions (a) &lt;math&gt;0\le A&lt;B&lt;C\le99&lt;/math&gt;, (b) there exist integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, and prime &lt;math&gt;p&lt;/math&gt; where &lt;math&gt;0\le b&lt;a&lt;c&lt;p&lt;/math&gt;, (c) &lt;math&gt;p&lt;/math&gt; divides &lt;math&gt;A-a&lt;/math&gt;, &lt;math&gt;B-b&lt;/math&gt;, and &lt;math&gt;C-c&lt;/math&gt;, and (d) each ordered triple &lt;math&gt;(A,B,C)&lt;/math&gt; and each ordered triple &lt;math&gt;(b,a,c)&lt;/math&gt; form arithmetic sequences. Find &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==Solution==<br /> From condition (d), we have &lt;math&gt;(A,B,C)=(B-\Delta,B,B+\Delta)&lt;/math&gt; and &lt;math&gt;(b,a,c)=(a-\delta,a,a+\delta)&lt;/math&gt;. Condition (c) states that &lt;math&gt;p|B-\Delta-a&lt;/math&gt;, &lt;math&gt;p|B-a+\delta&lt;/math&gt;, and &lt;math&gt;p|B+\Delta-a-\delta&lt;/math&gt;. We subtract the first two to get &lt;math&gt;p|-\delta-\Delta&lt;/math&gt;, and we do the same for the last two to get &lt;math&gt;p|2\delta-\Delta&lt;/math&gt;. We subtract these two to get &lt;math&gt;p|3\delta&lt;/math&gt;. So &lt;math&gt;p|3&lt;/math&gt; or &lt;math&gt;p|\delta&lt;/math&gt;. The second case is clearly impossible, because that would make &lt;math&gt;c=a+\delta&gt;p&lt;/math&gt;, violating condition (b). So we have &lt;math&gt;p|3&lt;/math&gt;, meaning &lt;math&gt;p=3&lt;/math&gt;. Condition (b) implies that &lt;math&gt;(b,a,c)=(0,1,2)&lt;/math&gt;. Now we return to condition (c), which now implies that &lt;math&gt;(A,B,C)\equiv(-2,0,2)\pmod{3}&lt;/math&gt;. Now, we set &lt;math&gt;B=3k&lt;/math&gt; for increasing integer values of &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;B=0&lt;/math&gt; yields no solutions. &lt;math&gt;B=3&lt;/math&gt; gives &lt;math&gt;(A,B,C)=(1,3,5)&lt;/math&gt;, giving us one solution. If &lt;math&gt;B=6&lt;/math&gt;, we get two solutions. Proceeding in the manner, we see that if &lt;math&gt;B=48&lt;/math&gt;, we get 16 solutions. However, &lt;math&gt;B=51&lt;/math&gt; still gives 16 solutions. &lt;math&gt;B=54&lt;/math&gt; gives 15 solutions. This continues until &lt;math&gt;B=96&lt;/math&gt; gives one solution. &lt;math&gt;B=99&lt;/math&gt; gives no solution. Thus, &lt;math&gt;N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=51743 2013 AIME I Problems/Problem 15 2013-03-16T02:20:46Z <p>Neutrinonerd3333: Created page, added problem and solution</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;N&lt;/math&gt; be the number of ordered triple &lt;math&gt;(A,B,C)&lt;/math&gt; of integers satisfying the conditions (a) &lt;math&gt;0\le A&lt;B&lt;C\le99&lt;/math&gt;, (b) there exist integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, and prime &lt;math&gt;p&lt;/math&gt; where &lt;math&gt;0\le b&lt;a&lt;c&lt;p&lt;/math&gt;, (c) &lt;math&gt;p&lt;/math&gt; divides &lt;math&gt;A-a&lt;/math&gt;, &lt;math&gt;B-b&lt;/math&gt;, and &lt;math&gt;C-c&lt;/math&gt;, and (d) each ordered triple &lt;math&gt;(A,B,C)&lt;/math&gt; and each ordered triple &lt;math&gt;(b,a,c)&lt;/math&gt; form arithmetic sequences. Find &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==Solution==<br /> From condition (d), we have &lt;math&gt;(A,B,C)=(B-\Delta,B,B+\Delta)&lt;/math&gt; and &lt;math&gt;(b,a,c)=(a-\delta,a,a+\delta)&lt;/math&gt;. Condition (c) states that &lt;math&gt;p|B-\Delta-a&lt;/math&gt;, &lt;math&gt;p|B-a+\delta&lt;/math&gt;, and &lt;math&gt;p|B+\Delta-a-\delta&lt;/math&gt;. We subtract the first two to get &lt;math&gt;p|-\delta-\Delta&lt;/math&gt;, and we do the same for the last two to get &lt;math&gt;p|2\delta-\Delta&lt;/math&gt;. We subtract these two to get &lt;math&gt;p|3\delta&lt;/math&gt;. So &lt;math&gt;p|3&lt;/math&gt; or &lt;math&gt;p|\delta&lt;/math&gt;. The second case is clearly impossible, because that would make &lt;math&gt;c=a+\delta&gt;p&lt;/math&gt;, violating condition (b). So we have &lt;math&gt;p|3&lt;/math&gt;, meaning &lt;math&gt;p=3&lt;/math&gt;. Condition (b) implies that &lt;math&gt;(b,a,c)=(0,1,2)&lt;/math&gt;. Now we return to condition (c), which now implies that &lt;math&gt;(A,B,C)\equiv(-2,0,2)\pmod{3}&lt;/math&gt;. Now, we set &lt;math&gt;B=3k&lt;/math&gt; for increasing integer values of &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;B=0&lt;/math&gt; yields no solutions. &lt;math&gt;B=3&lt;/math&gt; gives &lt;math&gt;(A,B,C)=(1,3,5)&lt;/math&gt;, giving us one solution. If &lt;math&gt;B=6&lt;/math&gt;, we get two solutions. Proceeding in the manner, we see that if &lt;math&gt;B=48&lt;/math&gt;, we get 16 solutions. However, &lt;math&gt;B=51&lt;/math&gt; still gives 16 solutions. &lt;math&gt;B=54&lt;/math&gt; gives 15 solutions. This continues until &lt;math&gt;B=96&lt;/math&gt; gives one solution. &lt;math&gt;B=99&lt;/math&gt; gives no solution. Thus, &lt;math&gt;N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=\boxed{272}&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_23&diff=50582 2006 AMC 12B Problems/Problem 23 2013-01-22T02:19:36Z <p>Neutrinonerd3333: Solution 2 -- more of an outline</p> <hr /> <div>== Problem ==<br /> Isosceles &lt;math&gt;\triangle ABC&lt;/math&gt; has a right angle at &lt;math&gt;C&lt;/math&gt;. Point &lt;math&gt;P&lt;/math&gt; is inside &lt;math&gt;\triangle ABC&lt;/math&gt;, such that &lt;math&gt;PA=11&lt;/math&gt;, &lt;math&gt;PB=7&lt;/math&gt;, and &lt;math&gt;PC=6&lt;/math&gt;. Legs &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; have length &lt;math&gt;s=\sqrt{a+b\sqrt{2}{&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> pathpen = linewidth(0.7);<br /> pen f = fontsize(10);<br /> size(5cm);<br /> pair B = (0,sqrt(85+42*sqrt(2)));<br /> pair A = (B.y,0);<br /> pair C = (0,0);<br /> pair P = IP(arc(B,7,180,360),arc(C,6,0,90));<br /> D(A--B--C--cycle);<br /> D(P--A);<br /> D(P--B);<br /> D(P--C);<br /> MP(&quot;A&quot;,D(A),plain.E,f);<br /> MP(&quot;B&quot;,D(B),plain.N,f);<br /> MP(&quot;C&quot;,D(C),plain.SW,f);<br /> MP(&quot;P&quot;,D(P),plain.NE,f);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 85<br /> \qquad<br /> \mathrm{(B)}\ 91<br /> \qquad<br /> \mathrm{(C)}\ 108<br /> \qquad<br /> \mathrm{(D)}\ 121<br /> \qquad<br /> \mathrm{(E)}\ 127<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;asy&gt;<br /> pathpen = linewidth(0.7);<br /> pen f = fontsize(10);<br /> size(5cm);<br /> pair B = (0,sqrt(85+42*sqrt(2)));<br /> pair A = (B.y,0);<br /> pair C = (0,0);<br /> pair P = IP(arc(B,7,180,360),arc(C,6,0,90));<br /> D(A--B--C--cycle);<br /> D(P--A);<br /> D(P--B);<br /> D(P--C);<br /> MP(&quot;A&quot;,D(A),plain.E,f);<br /> MP(&quot;B&quot;,D(B),plain.N,f);<br /> MP(&quot;C&quot;,D(C),plain.SW,f);<br /> MP(&quot;P&quot;,D(P),plain.NE,f);<br /> MP(&quot;\alpha&quot;,C,5*dir(80),f);<br /> MP(&quot;90^\circ-\alpha&quot;,C,3*dir(30),f);<br /> MP(&quot;s&quot;,(A+C)/2,plain.S,f);<br /> MP(&quot;s&quot;,(B+C)/2,plain.W,f);<br /> &lt;/asy&gt;<br /> Using the Law of Cosines on &lt;math&gt;\triangle PBC&lt;/math&gt;, we have:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> PB^2&amp;=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Using the Law of Cosines on &lt;math&gt;\triangle PAC&lt;/math&gt;, we have:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> PA^2&amp;=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now we use &lt;math&gt;\sin^2(\alpha) + \cos^2(\alpha) = 1&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \sin^2(\alpha)+\cos^2(\alpha) = 1 &amp;\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\<br /> &amp;\Rightarrow 2s^4-340s^2+7394 = 0 \\<br /> &amp;\Rightarrow s^4-170s^2+3697 = 0 \\<br /> &amp;\Rightarrow s^2 = \dfrac{170 \pm \sqrt{170^2 - 4\cdot3697}}{2}\\<br /> &amp;\Rightarrow s^2 = \dfrac{170 \pm \sqrt{28900 - 14788}}{2}\\<br /> &amp;\Rightarrow s^2 = \dfrac{170 \pm \sqrt{14112}}{2}\\<br /> &amp;\Rightarrow s^2 = \dfrac{170 \pm \sqrt{2^5\cdot3^2\cdot7^2}}{2}\\<br /> &amp;\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Note that we know that we want the solution with &lt;math&gt;s^2 &gt; 85&lt;/math&gt; since we know that &lt;math&gt;\sin(\alpha) &gt; 0&lt;/math&gt;. Thus, &lt;math&gt;a+b=85+42=\boxed{127}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Rotate triangle &lt;math&gt;PAC&lt;/math&gt; 90 degrees counterclockwise about &lt;math&gt;C&lt;/math&gt; so that the image of &lt;math&gt;A&lt;/math&gt; rests on &lt;math&gt;B&lt;/math&gt;. Now let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P'&lt;/math&gt;. Note that &lt;math&gt;P'C=6&lt;/math&gt;, meaning triangle &lt;math&gt;PCP'&lt;/math&gt; is right isosceles, and &lt;math&gt;\angle PP'C=45^\circ&lt;/math&gt;. Then &lt;math&gt;PP'=6\sqrt{2}&lt;/math&gt;. Now because &lt;math&gt;PB=7&lt;/math&gt; and &lt;math&gt;P'B=11&lt;/math&gt;, we observe that &lt;math&gt;\angle P'PB=90^\circ&lt;/math&gt;, by the Pythagorean Theorem on &lt;math&gt;P'PB&lt;/math&gt;. Now we have that &lt;math&gt;\angle APC=\angle BP'C=\angle BP'P + \angle PP'C&lt;/math&gt;. So we take the cosine of the second equality, using that fact that &lt;math&gt;\angle PP'C=45^\circ&lt;/math&gt;, to get &lt;math&gt;\cos(BP'C)=\frac{6\sqrt{2}-7}{11\sqrt{2}}&lt;/math&gt;. Finally, we use the fact that &lt;math&gt;\cos(BP'C)=\cos(CPA)&lt;/math&gt; and use the Law of Cosines on triangle &lt;math&gt;CPA&lt;/math&gt; to arrive at the value of &lt;math&gt;s^2&lt;/math&gt;.<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=22|num-a=24}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1993_USAMO_Problems/Problem_1&diff=47590 1993 USAMO Problems/Problem 1 2012-07-07T22:44:06Z <p>Neutrinonerd3333: Solution 2</p> <hr /> <div>==Problem==<br /> For each integer &lt;math&gt;n\ge 2&lt;/math&gt;, determine, with proof, which of the two positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; satisfying<br /> &lt;cmath&gt;a^n=a+1,\qquad b^{2n}=b+3a&lt;/cmath&gt;<br /> is larger.<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> Square and rearrange the first equation and also rearrange the second.<br /> &lt;cmath&gt;\begin{align}<br /> a^{2n}-a&amp;=a^2+a+1\\<br /> b^{2n}-b&amp;=3a<br /> \end{align}&lt;/cmath&gt;<br /> It is [[Trivial Inequality|trivial]] that<br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)^2 &gt; 0 \tag{3}<br /> \end{align*}&lt;/cmath&gt;<br /> since &lt;math&gt;a-1&lt;/math&gt; clearly cannot equal &lt;math&gt;0&lt;/math&gt; (Otherwise &lt;math&gt;a^n=1\neq 1+1&lt;/math&gt;). Thus<br /> &lt;cmath&gt;\begin{align*}<br /> a^2+a+1&amp;&gt;3a \tag{4}\\<br /> a^{2n}-a&amp;&gt;b^{2n}-b \tag{5}<br /> \end{align*}&lt;/cmath&gt;<br /> where we substituted in equations (1) and (2) to achieve (5). Notice that from &lt;math&gt;a^{n}=a+1&lt;/math&gt; we have &lt;math&gt;a&gt;1&lt;/math&gt;. Thus, if &lt;math&gt;b&gt;a&lt;/math&gt;, then &lt;math&gt;b^{2n-1}-1&gt;a^{2n-1}-1&lt;/math&gt;. Since &lt;math&gt;a&gt;1\Rightarrow a^{2n-1}-1&gt;0&lt;/math&gt;, multiplying the two inequalities yields &lt;math&gt;b^{2n}-b&gt;a^{2n}-a&lt;/math&gt;, a contradiction, so &lt;math&gt;a&gt; b&lt;/math&gt;. However, when &lt;math&gt;n&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;, the first equation becomes meaningless, so we conclude that for each integer &lt;math&gt;n\ge 2&lt;/math&gt;, we always have &lt;math&gt;a&gt;b&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Define &lt;math&gt;f(x)=x^n-x-1&lt;/math&gt; and &lt;math&gt;g(x)=x^{2n}-x-3a&lt;/math&gt;. By Descarte's Rule of Signs, both polynomials' only positive roots are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, respectively. With the Intermediate Value Theorem and the fact that &lt;math&gt;f(1)=-1&lt;/math&gt; and &lt;math&gt;f(2)=2^n-3&gt;0&lt;/math&gt;, we have &lt;math&gt;a\in(1,2)&lt;/math&gt;.<br /> Thus, &lt;math&gt;-3a\in(-6,-3)&lt;/math&gt;, which means that &lt;math&gt;g(1)=-3a&lt;0&lt;/math&gt;. Also, we find that &lt;math&gt;g(a)=a^{2n}-4a&lt;/math&gt;. All that remains to prove is that &lt;math&gt;g(a)&gt;0&lt;/math&gt;, or &lt;math&gt;a^{2n}-4a&gt;0&lt;/math&gt;. We can then conclude that &lt;math&gt;b&lt;/math&gt; is between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; from the Intermediate Value Theorem. From the first equation given, &lt;math&gt;a^{2n}=(a+1)^2=a^2+2a+1&lt;/math&gt;. Subtracting &lt;math&gt;4a&lt;/math&gt; gives us &lt;math&gt;a^2-2a+1&gt;0&lt;/math&gt;, which is clearly true, as &lt;math&gt;a\neq1&lt;/math&gt;. Therefore, we conclude that &lt;math&gt;1&lt;b&lt;a&lt;2&lt;/math&gt;.<br /> <br /> <br /> <br /> == See also ==<br /> {{USAMO box|year=1993|before=First Problem|num-a=2}}<br /> <br /> [[Category:Olympiad Algebra Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=47501 2003 AMC 12A Problems/Problem 17 2012-06-30T22:04:24Z <p>Neutrinonerd3333: solution 2</p> <hr /> <div>== Problem ==<br /> Square &lt;math&gt;ABCD&lt;/math&gt; has sides of length &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{CD}&lt;/math&gt;. A circle with radius &lt;math&gt;2&lt;/math&gt; and center &lt;math&gt;M&lt;/math&gt; intersects a circle with radius &lt;math&gt;4&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. What is the distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AD}&lt;/math&gt;?<br /> <br /> [[Image:5d50417537c6cddfb70810403c62787b889cdcb1.png]]<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the origin. &lt;math&gt;A&lt;/math&gt; is the point &lt;math&gt;(0,4)&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; is the point &lt;math&gt;(2,0)&lt;/math&gt;. We are given the radius of the quarter circle and semicircle as &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, respectively, so their equations, respectively, are:<br /> <br /> &lt;math&gt;x^2 + (y-4)^2 = 4^2&lt;/math&gt;<br /> <br /> &lt;math&gt;(x-2)^2 + y^2 = 2^2&lt;/math&gt;<br /> <br /> <br /> Algebraically manipulating the second equation gives:<br /> <br /> &lt;math&gt;y^2 = 2^2 - (x-2)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 = (2-(x-2)(2+(x-2))&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 = (4-x)(x)&lt;/math&gt;<br /> <br /> &lt;math&gt;y = \sqrt{4x - x^2}&lt;/math&gt;<br /> <br /> <br /> Substituting this back into the first equation:<br /> <br /> &lt;math&gt;x^2 + (\sqrt{4x - x^2} - 4)^2 = 4^2&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + 4x - x^2 - 8\sqrt{4x - x^2} + 16 = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;4x - 8\sqrt{4x - x^2} = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;4x = 8\sqrt{4x - x^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;16x^2 = 64(4x - x^2)&lt;/math&gt;<br /> <br /> &lt;math&gt;16x^2 = 256x - 64x^2&lt;/math&gt;<br /> <br /> &lt;math&gt;80x^2 - 256x = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x(80x - 256) = 0&lt;/math&gt;<br /> <br /> <br /> Solving each factor for 0 yields &lt;math&gt;x = 0 , \frac{16}{5}&lt;/math&gt;. The first value of &lt;math&gt;0&lt;/math&gt; is obviously referring to the x-coordinate of the point where the circles intersect at the origin, &lt;math&gt;D&lt;/math&gt;, so the second value must be referring to the x coordinate of &lt;math&gt;P&lt;/math&gt;. Since &lt;math&gt;\overline{AD}&lt;/math&gt; is the y-axis, the distance to it from &lt;math&gt;P&lt;/math&gt; is the same as the x-value of the coordinate of &lt;math&gt;P&lt;/math&gt;, so the distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AD}&lt;/math&gt; is &lt;math&gt;\frac{16}{5} \Rightarrow B&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;P&lt;/math&gt; is merely a reflection of &lt;math&gt;D&lt;/math&gt; over &lt;math&gt;AM&lt;/math&gt;. Call the intersection of &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;DP&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;. Drop perpendiculars from &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt;, and denote their respective points of intersection by &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;. We then have &lt;math&gt;\triangle DXJ\sim\triangle DPK&lt;/math&gt;, with a scale factor of 2. Thus, we can find &lt;math&gt;XJ&lt;/math&gt; and double it to get our answer. With some analytical geometry, we find that &lt;math&gt;XJ=\frac{8}{5}&lt;/math&gt;, implying that &lt;math&gt;PK=\frac{16}{5}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> *[[2003 AMC 12A Problems]]<br /> *[[2003 AMC 12A Problems/Problem 16|Previous Problem]]<br /> *[[2003 AMC 12A Problems/Problem 18|Next Problem]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_14&diff=44585 2009 AMC 12A Problems/Problem 14 2012-02-09T03:24:27Z <p>Neutrinonerd3333: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A triangle has vertices &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(1,1)&lt;/math&gt;, and &lt;math&gt;(6m,0)&lt;/math&gt;, and the line &lt;math&gt;y = mx&lt;/math&gt; divides the triangle into two triangles of equal area. What is the sum of all possible values of &lt;math&gt;m&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let's label the three points as &lt;math&gt;A=(0,0)&lt;/math&gt;, &lt;math&gt;B=(1,1)&lt;/math&gt;, and &lt;math&gt;C=(6m,0)&lt;/math&gt;.<br /> <br /> Clearly, whenever the line &lt;math&gt;y=mx&lt;/math&gt; intersects the inside of the triangle, it will intersect the side &lt;math&gt;BC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the point of intersection. <br /> <br /> The triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; have the same height, which is the distance between the point &lt;math&gt;A&lt;/math&gt; and the line &lt;math&gt;BC&lt;/math&gt;.<br /> Hence they have equal areas if and only if &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> The midpoint of the segment &lt;math&gt;BC&lt;/math&gt; has coordinates &lt;math&gt;\left( \frac{6m+1}2, \frac 12 \right)&lt;/math&gt;. This point lies on the line &lt;math&gt;y=mx&lt;/math&gt; if and only if &lt;math&gt;\frac 12 = m \cdot \frac{6m+1}2&lt;/math&gt;. This simplifies to &lt;math&gt;6m^2 + m - 1 = 0&lt;/math&gt;. This is a quadratic equation with roots <br /> &lt;math&gt;m=\frac 13&lt;/math&gt; and &lt;math&gt;m=-\frac 12&lt;/math&gt;. Both roots represent valid solutions, and their sum is &lt;math&gt;\frac 13 - \frac 12 = \boxed{-\frac 16}&lt;/math&gt;.<br /> <br /> For illustration, below are pictures of the situation for &lt;math&gt;m=1.5&lt;/math&gt;, &lt;math&gt;m=0.5&lt;/math&gt;, &lt;math&gt;m=1/3&lt;/math&gt;, and &lt;math&gt;m=-1/2&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> real m=1.5;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),NE);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=0.5;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),N);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=1/3;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),N);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=-1/2;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-2)*(1,m)) -- (1*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),S);<br /> label(&quot;$B$&quot;,(1,1),NE);<br /> label(&quot;$C$&quot;,(6*m,0),W);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 2==<br /> <br /> The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus &lt;math&gt;\frac{\frac{1}{3}}{\frac{1+6m}{3}}&lt;/math&gt;, which is equal to &lt;math&gt;m&lt;/math&gt;. This becomes a quadratic, and using Viete's Formulas, we get our answer, &lt;math&gt;\boxed{-\frac 1 6}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=13|num-a=15}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_14&diff=44584 2009 AMC 12A Problems/Problem 14 2012-02-09T03:23:19Z <p>Neutrinonerd3333: /* Solution 2*/</p> <hr /> <div>== Problem ==<br /> <br /> A triangle has vertices &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(1,1)&lt;/math&gt;, and &lt;math&gt;(6m,0)&lt;/math&gt;, and the line &lt;math&gt;y = mx&lt;/math&gt; divides the triangle into two triangles of equal area. What is the sum of all possible values of &lt;math&gt;m&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let's label the three points as &lt;math&gt;A=(0,0)&lt;/math&gt;, &lt;math&gt;B=(1,1)&lt;/math&gt;, and &lt;math&gt;C=(6m,0)&lt;/math&gt;.<br /> <br /> Clearly, whenever the line &lt;math&gt;y=mx&lt;/math&gt; intersects the inside of the triangle, it will intersect the side &lt;math&gt;BC&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the point of intersection. <br /> <br /> The triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; have the same height, which is the distance between the point &lt;math&gt;A&lt;/math&gt; and the line &lt;math&gt;BC&lt;/math&gt;.<br /> Hence they have equal areas if and only if &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> The midpoint of the segment &lt;math&gt;BC&lt;/math&gt; has coordinates &lt;math&gt;\left( \frac{6m+1}2, \frac 12 \right)&lt;/math&gt;. This point lies on the line &lt;math&gt;y=mx&lt;/math&gt; if and only if &lt;math&gt;\frac 12 = m \cdot \frac{6m+1}2&lt;/math&gt;. This simplifies to &lt;math&gt;6m^2 + m - 1 = 0&lt;/math&gt;. This is a quadratic equation with roots <br /> &lt;math&gt;m=\frac 13&lt;/math&gt; and &lt;math&gt;m=-\frac 12&lt;/math&gt;. Both roots represent valid solutions, and their sum is &lt;math&gt;\frac 13 - \frac 12 = \boxed{-\frac 16}&lt;/math&gt;.<br /> <br /> For illustration, below are pictures of the situation for &lt;math&gt;m=1.5&lt;/math&gt;, &lt;math&gt;m=0.5&lt;/math&gt;, &lt;math&gt;m=1/3&lt;/math&gt;, and &lt;math&gt;m=-1/2&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> real m=1.5;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),NE);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=0.5;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),N);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=1/3;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-1)*(1,m)) -- (3*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),NW);<br /> label(&quot;$B$&quot;,(1,1),N);<br /> label(&quot;$C$&quot;,(6*m,0),E);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(0.8);<br /> real m=-1/2;<br /> draw( (0,0)--(1,1)--(6*m,0)--cycle );<br /> draw( ((-2)*(1,m)) -- (1*(1,m)), dashed );<br /> label(&quot;$A$&quot;,(0,0),S);<br /> label(&quot;$B$&quot;,(1,1),NE);<br /> label(&quot;$C$&quot;,(6*m,0),W);<br /> pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) );<br /> dot(D); label(&quot;$D$&quot;,D,S);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 2==<br /> <br /> The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus &lt;math&gt;\frac{\frac{1}{3}}{\frac{1+6m}{3}}&lt;/math&gt;, which is equal to &lt;math&gt;m&lt;/math&gt;. Solving this equation, we arrive at the answer: &lt;math&gt;\boxed{-\frac{1}{6}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=13|num-a=15}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems&diff=42453 1995 AHSME Problems 2011-10-04T04:09:42Z <p>Neutrinonerd3333: /* Problem 17 */</p> <hr /> <div>== Problem 1 ==<br /> Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will <br /> <br /> &lt;math&gt; \mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> If &lt;math&gt;\sqrt {2 + \sqrt {x}} = 3&lt;/math&gt;, then &lt;math&gt;x =&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \sqrt{7} } \qquad \mathrm{(C) \ 7 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 121 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> The total in-store price for an appliance is &lt;math&gt;\textdollar 99.99&lt;/math&gt;. A television commercial advertises the same product for three easy payments of &lt;math&gt;\textdollar 29.98&lt;/math&gt; and a one-time shipping and handling charge of &lt;math&gt;\textdollar 9.98&lt;/math&gt;. How much is saved by buying the appliance from the television advertiser?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> If &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;30 \%&lt;/math&gt; of &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;20 \%&lt;/math&gt; of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;50 \%&lt;/math&gt; of &lt;math&gt;P&lt;/math&gt;, then &lt;math&gt;\frac {M}{N} =&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ \frac {3}{250} } \qquad \mathrm{(B) \ \frac {3}{25} } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is <br /> <br /> &lt;math&gt; \mathrm{(A) \ \text{500 thousand} } \qquad \mathrm{(B) \ \text{5 million} } \qquad \mathrm{(C) \ \text{50 million} } \qquad \mathrm{(D) \ \text{500 million} } \qquad \mathrm{(E) \ \text{5 billion} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked ? <br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7));<br /> path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3);<br /> draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin);<br /> draw(shift(1,0)*p, dashed);<br /> label(&quot;$x$&quot;, (0.3,0.5), E);<br /> label(&quot;$A$&quot;, (1.3,0.5), E);<br /> label(&quot;$B$&quot;, (1.3,1.5), E);<br /> label(&quot;$C$&quot;, (2.3,1.5), E);<br /> label(&quot;$D$&quot;, (2.3,2.5), E);<br /> label(&quot;$E$&quot;, (3.3,2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ A } \qquad \mathrm{(B) \ B } \qquad \mathrm{(C) \ C } \qquad \mathrm{(D) \ D } \qquad \mathrm{(E) \ E } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is: <br /> <br /> &lt;math&gt; \mathrm{(A) \ 8 } \qquad \mathrm{(B) \ 25 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 75 } \qquad \mathrm{(E) \ 100 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\angle C = 90^\circ, AC = 6&lt;/math&gt; and &lt;math&gt;BC = 8&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, respectively, and &lt;math&gt;\angle BED = 90^\circ&lt;/math&gt;. If &lt;math&gt;DE = 4&lt;/math&gt;, then &lt;math&gt;BD =&lt;/math&gt;<br /> &lt;asy&gt; <br /> size(100); pathpen = linewidth(0.7); pointpen = black+linewidth(3);<br /> pair A = (0,0), C = (6,0), B = (6,8), D = (2*A+B)/3, E = (2*C+B)/3; D(D(&quot;A&quot;,A,SW)--D(&quot;B&quot;,B,NW)--D(&quot;C&quot;,C,SE)--cycle); D(D(&quot;D&quot;,D,NW)--D(&quot;E&quot;,E,plain.E)); D(rightanglemark(D,E,B,16)); D(rightanglemark(A,C,B,16)); <br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is <br /> &lt;asy&gt;<br /> size(100); defaultpen(linewidth(0.7)); draw(unitsquare^^(0,0)--(1,1)^^(0,1)--(1,0)^^(.5,0)--(.5,1)^^(0,.5)--(1,.5));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The area of the triangle bounded by the lines &lt;math&gt;y = x, y = - x&lt;/math&gt; and &lt;math&gt;y = 6&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 12\sqrt{2} } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 24\sqrt{2} } \qquad \mathrm{(E) \ 36 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> How many base 10 four-digit numbers, &lt;math&gt;N = \underline{a} \underline{b} \underline{c} \underline{d}&lt;/math&gt;, satisfy all three of the following conditions?<br /> <br /> (i) &lt;math&gt;4,000 \leq N &lt; 6,000;&lt;/math&gt; (ii) &lt;math&gt;N&lt;/math&gt; is a multiple of 5; (iii) &lt;math&gt;3 \leq b &lt; c \leq 6&lt;/math&gt;.<br /> <br /> &lt;math&gt; \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \ 48 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Let &lt;math&gt;f&lt;/math&gt; be a linear function with the properties that &lt;math&gt;f(1) \leq f(2), f(3) \geq f(4),&lt;/math&gt; and &lt;math&gt;f(5) = 5&lt;/math&gt;. Which of the following is true?<br /> <br /> &lt;math&gt; \mathrm{(A) \ f(0) &lt; 0 } \qquad \mathrm{(B) \ f(0) = 0 } \qquad \mathrm{(C) \ f(1) &lt; f(0) &lt; f( - 1) } \qquad \mathrm{(D) \ f(0) = 5 } \qquad \mathrm{(E) \ f(0) &gt; 5 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> The addition below is incorrect. The display can be made correct by changing one digit &lt;math&gt;d&lt;/math&gt;, wherever it occurs, to another digit &lt;math&gt;e&lt;/math&gt;. Find the sum of &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt;.<br /> <br /> &lt;math&gt;\begin{tabular}{ccccccc} &amp; 7 &amp; 4 &amp; 2 &amp; 5 &amp; 8 &amp; 6 \\<br /> + &amp; 8 &amp; 2 &amp; 9 &amp; 4 &amp; 3 &amp; 0 \\<br /> \hline 1 &amp; 2 &amp; 1 &amp; 2 &amp; 0 &amp; 1 &amp; 6 \end{tabular}&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> If &lt;math&gt;f(x) = ax^4 - bx^2 + x + 5&lt;/math&gt; and &lt;math&gt;f( - 3) = 2&lt;/math&gt;, then &lt;math&gt;f(3) =&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point <br /> &lt;asy&gt;<br /> size(80); defaultpen(linewidth(0.7)+fontsize(10)); draw(unitcircle);<br /> for(int i = 0; i &lt; 5; ++i) { pair P = dir(90-i*72); dot(P); label(&quot;$&quot;+string(i+1)+&quot;$&quot;,P,1.4*P); }<br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:<br /> <br /> i. The actual attendance in Atlanta is within &lt;math&gt;10 \%&lt;/math&gt; of Anita's estimate.<br /> ii. Bob's estimate is within &lt;math&gt;10 \%&lt;/math&gt; of the actual attendance in Boston.<br /> <br /> To the nearest 1,000, the largest possible difference between the numbers attending the two games is<br /> <br /> &lt;math&gt; \mathrm{(A) \ 10000 } \qquad \mathrm{(B) \ 11000 } \qquad \mathrm{(C) \ 20000 } \qquad \mathrm{(D) \ 21000 } \qquad \mathrm{(E) \ 22000 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> Given regular pentagon &lt;math&gt;ABCDE&lt;/math&gt;, a circle can be drawn that is tangent to &lt;math&gt;\overline{DC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and to &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;A&lt;/math&gt;. The number of degrees in minor arc &lt;math&gt;AD&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7));<br /> draw(rotate(18)*polygon(5));<br /> real x=0.6180339887;<br /> draw(Circle((-x,0), 1));<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> dot(origin+1*dir(36+72*i));<br /> }<br /> label(&quot;$B$&quot;, origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0)));<br /> label(&quot;$A$&quot;, origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72)));<br /> label(&quot;$E$&quot;, origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144)));<br /> label(&quot;$D$&quot;, origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3)));<br /> label(&quot;$C$&quot;, origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4)));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> Two rays with common endpoint &lt;math&gt;O&lt;/math&gt; forms a &lt;math&gt;30^\circ&lt;/math&gt; angle. Point &lt;math&gt;A&lt;/math&gt; lies on one ray, point &lt;math&gt;B&lt;/math&gt; on the other ray, and &lt;math&gt;AB = 1&lt;/math&gt;. The maximum possible length of &lt;math&gt;OB&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ \frac{4}{\sqrt{3}} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> Equilateral triangle &lt;math&gt;DEF&lt;/math&gt; is inscribed in equilateral triangle &lt;math&gt;ABC&lt;/math&gt; such that &lt;math&gt;\overline{DE} \perp \overline{BC}&lt;/math&gt;. The reatio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is<br /> &lt;asy&gt;<br /> pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);<br /> pair B = (0,0), C = (1,0), A = dir(60), D = C*2/3, E = (2*A+C)/3, F = (2*B+A)/3;<br /> D(D(&quot;A&quot;,A,N)--D(&quot;B&quot;,B,SW)--D(&quot;C&quot;,C,SE)--cycle); D(D(&quot;D&quot;,D)--D(&quot;E&quot;,E,NE)--D(&quot;F&quot;,F,NW)--cycle); D(rightanglemark(C,D,E,1.5));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> If &lt;math&gt;a,b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are three (not necessarily different) numbers chosen randomly and with replacement from the set &lt;math&gt;\{1,2,3,4,5 \}&lt;/math&gt;, the probability that &lt;math&gt;ab + c&lt;/math&gt; is even is<br /> <br /> &lt;math&gt; \mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } \qquad \mathrm{(E) \ \frac {3}{5} } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> Two nonadjacent vertices of a rectangle are &lt;math&gt;(4,3)&lt;/math&gt; and &lt;math&gt;(-4,-3)&lt;/math&gt;, and the coordinates of the other two vertices are integers. The number of such rectangles is<br /> <br /> &lt;math&gt; \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths 13,19,20,25 and 31, although this is not necessarily their order around the pentagon. The area of the pentagon is<br /> <br /> &lt;math&gt; \mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> The sides of a triangle have lengths 11,15, and &lt;math&gt;k&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is an integer. For how many values of &lt;math&gt;k&lt;/math&gt; is the triangle obtuse?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> There exist positive integers &lt;math&gt;A,B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, with no common factor greater than 1, such that<br /> <br /> &lt;cmath&gt;A \log_{200} 5 + B \log_{200} 2 = C&lt;/cmath&gt;<br /> <br /> What is &lt;math&gt;A + B + C&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> A list of five positive integers has mean 12 and range 18. The mode and median are both 8. How many different values are possible for the second largest element of the list? <br /> <br /> &lt;math&gt; \mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> In the figure, &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; are diameters of the circle with center &lt;math&gt;O&lt;/math&gt;, &lt;math&gt;\overline{AB} \perp \overline{CD}&lt;/math&gt;, and chord &lt;math&gt;\overline{DF}&lt;/math&gt; intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. If &lt;math&gt;DE = 6&lt;/math&gt; and &lt;math&gt;EF = 2&lt;/math&gt;, then the area of the circle is <br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7));<br /> draw(Circle(origin, 5));<br /> pair O=origin, A=(-5,0), B=(5,0), C=(0,5), D=(0,-5), F=5*dir(40), E=intersectionpoint(A--B, F--D);<br /> draw(A--B^^C--D--F);<br /> dot(O^^A^^B^^C^^D^^E^^F);<br /> markscalefactor=0.05;<br /> draw(rightanglemark(B, O, D));<br /> label(&quot;$A$&quot;, A, dir(O--A));<br /> label(&quot;$B$&quot;, B, dir(O--B));<br /> label(&quot;$C$&quot;, C, dir(O--C));<br /> label(&quot;$D$&quot;, D, dir(O--D));<br /> label(&quot;$F$&quot;, F, dir(O--F));<br /> label(&quot;$O$&quot;, O, NW);<br /> label(&quot;$E$&quot;, E, SE);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.<br /> <br /> &lt;cmath&gt;\begin{tabular}{ccccccccccc} &amp; &amp; &amp; &amp; &amp; 0 &amp; &amp; &amp; &amp; &amp; \\<br /> &amp; &amp; &amp; &amp; 1 &amp; &amp; 1 &amp; &amp; &amp; &amp; \\<br /> &amp; &amp; &amp; 2 &amp; &amp; 2 &amp; &amp; 2 &amp; &amp; &amp; \\<br /> &amp; &amp; 3 &amp; &amp; 4 &amp; &amp; 4 &amp; &amp; 3 &amp; &amp; \\<br /> &amp; 4 &amp; &amp; 7 &amp; &amp; 8 &amp; &amp; 7 &amp; &amp; 4 &amp; \\<br /> 5 &amp; &amp; 11 &amp; &amp; 15 &amp; &amp; 15 &amp; &amp; 11 &amp; &amp; 5 &amp; \end{tabular}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;f(n)&lt;/math&gt; denote the sum of the numbers in row &lt;math&gt;n&lt;/math&gt;. What is the remainder when &lt;math&gt;f(100)&lt;/math&gt; is divided by 100?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> Two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6. The chord parallel to these chords and midway between them is of length &lt;math&gt;\sqrt {a}&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is<br /> &lt;asy&gt;<br /> // note: diagram deliberately not to scale -- azjps<br /> void htick(pair A, pair B, real r){ D(A--B); D(A-(r,0)--A+(r,0)); D(B-(r,0)--B+(r,0)); }<br /> size(120); pathpen = linewidth(0.7); pointpen = black+linewidth(3);<br /> real min = -0.6, step = 0.5;<br /> pair[] A, B; D(unitcircle);<br /> for(int i = 0; i &lt; 3; ++i) {<br /> A.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle)); B.push(intersectionpoints((-9,min+i*step)--(9,min+i*step),unitcircle));<br /> D(D(A[i])--D(B[i]));<br /> }<br /> MP(&quot;10&quot;,(A+B)/2,N);<br /> MP(&quot;\sqrt{a}&quot;,(A+B)/2,N);<br /> MP(&quot;14&quot;,(A+B)/2,N);<br /> htick((B.x+0.1,B.y),(B.x+0.1,B.y),0.06); MP(&quot;6&quot;,(B.x+0.1,B.y/2+B.y/2),E);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ 144 } \qquad \mathrm{(B) \ 156 } \qquad \mathrm{(C) \ 168 } \qquad \mathrm{(D) \ 176 } \qquad \mathrm{(E) \ 184 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> For how many three-element sets of positive integers &lt;math&gt;\{a,b,c\}&lt;/math&gt; is it true that &lt;math&gt;a \times b \times c = 2310&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is<br /> &lt;asy&gt;<br /> size(150); defaultpen(linewidth(0.7)); pair slant = (2,1); <br /> for(int i = 0; i &lt; 4; ++i) <br /> draw((0,i)--(3,i)^^(i,0)--(i,3)^^(3,i)--(3,i)+slant^^(i,3)--(i,3)+slant); <br /> for(int i = 1; i &lt; 4; ++i)<br /> draw((0,3)+slant*i/3--(3,3)+slant*i/3^^(3,0)+slant*i/3--(3,3)+slant*i/3);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 } &lt;/math&gt;<br /> <br /> [[1995 AHSME Problems/Problem 30|Solution]]<br /> <br /> == See also ==<br /> {{AHSME box|year=1995|before=[[1994 AHSME Problems|1994 AHSME]]|after=[[1996 AHSME Problems|1996 AHSME]]}}<br /> * [[AHSME]]<br /> * [[AHSME Problems and Solutions]]<br /> * [[Mathematics competition resources]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_15,_2011&diff=41574 AoPS Wiki talk:Problem of the Day/July 15, 2011 2011-08-18T08:52:24Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/July 15, 2011}}<br /> ==Solution==<br /> There are &lt;math&gt;\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}&lt;/math&gt; ways to satisfy the baby storks, out of a total of &lt;math&gt;2^{10}&lt;/math&gt; ways to catch the fish. Using the fact that &lt;math&gt;\binom{n}{k}=\binom{n}{n-k}&lt;/math&gt;, we can quickly evaluate the probability as &lt;math&gt;\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_15,_2011&diff=41573 AoPS Wiki talk:Problem of the Day/July 15, 2011 2011-08-18T08:51:57Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/July 15, 2011}}<br /> ==Solution==<br /> There are &lt;math&gt;\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}&lt;/math&gt; ways to satisfy the baby storks, out of a total of &lt;math&gt;2^10&lt;/math&gt; ways to catch the fish. Using the fact that &lt;math&gt;\binom{n}{k}=\binom{n}{n-k}&lt;/math&gt;, we can quickly evaluate the probability as &lt;math&gt;\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}&lt;/math&gt;.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_15,_2011&diff=41572 AoPS Wiki talk:Problem of the Day/July 15, 2011 2011-08-18T08:51:31Z <p>Neutrinonerd3333: solution</p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/July 15, 2011}}<br /> ==Solution==<br /> There are &lt;math&gt;\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}&lt;/math&gt; ways to satisfy the baby storks, out of a total of &lt;math&gt;2^10&lt;/math&gt; ways to catch the fish. Using the fact that &lt;math&gt;\binom{n}{k}=\binom{n}{n-k}&lt;/math&gt;, we can quickly evaluate the probability as \$\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Problem_of_the_Day/August_18,_2011&diff=41571 AoPS Wiki:Problem of the Day/August 18, 2011 2011-08-18T08:43:59Z <p>Neutrinonerd3333: Created page with &quot;{{:AoPSWiki:Problem of the Day/August 18, 2011}}&quot;</p> <hr /> <div>{{:AoPSWiki:Problem of the Day/August 18, 2011}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_3,_2011&diff=41570 AoPS Wiki talk:Problem of the Day/August 3, 2011 2011-08-18T08:39:10Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/August 3, 2011}}<br /> ==Solution==<br /> Since &lt;math&gt;\sum_{n=1}^{\infty}\dfrac{5^n+n}{n} = \sum_{n=1}^{\infty}\left(1+\frac{5^n}{n}\right)&lt;/math&gt;, the sum diverges, since the sum is always greater than the index, due to constant 1.</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_3,_2011&diff=41569 AoPS Wiki talk:Problem of the Day/August 3, 2011 2011-08-18T08:38:42Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/August 3, 2011}}<br /> ==Solution==<br /> Since &lt;math&gt;\sum_{n=1}^{\infty}\dfrac{5^n+n}{n} = \sum_{n=1}^{\infty}\left(1+\frac{5^n}{n}\right)&lt;/math&gt;, the sum diverges, since the sum is always greater than the index, due to constant 1.<br /> <br /> {{potd_solution}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_3,_2011&diff=41568 AoPS Wiki talk:Problem of the Day/August 3, 2011 2011-08-18T08:37:57Z <p>Neutrinonerd3333: </p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/August 3, 2011}}<br /> ==Solution==<br /> Since &lt;math&gt;\sum_{n=1}{\infty}\dfrac{5^n+n}{n} = \sum_{n=1}{\infty}\left(1+\frac{5^n}{n}\right)&lt;/math&gt;, the sum diverges, since the sum is always greater than the index, due to constant 1.<br /> <br /> {{potd_solution}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/August_3,_2011&diff=41567 AoPS Wiki talk:Problem of the Day/August 3, 2011 2011-08-18T08:37:08Z <p>Neutrinonerd3333: </p> <hr /> <div>==Problem==<br /> {{:AoPSWiki:Problem of the Day/August 3, 2011}}<br /> ==Solution==<br /> Since &lt;math&gt;\sum_{1 \to \infty}\dfrac{5^n+n}{n} = \sum_{1 \to \infty}\left(1+\frac{5^n}{n}\right)&lt;/math&gt;, the sum diverges, since the sum is always greater than the index, due to constant 1.<br /> <br /> {{potd_solution}}</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=Pythagorean_Theorem&diff=41566 Pythagorean Theorem 2011-08-18T07:00:30Z <p>Neutrinonerd3333: 20-21-29</p> <hr /> <div>The '''Pythagorean Theorem''' states that for a [[right triangle]] with legs of length &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; and [[hypotenuse]] of length &lt;math&gt;c&lt;/math&gt; we have the relationship &lt;math&gt;{a}^{2}+{b}^{2}={c}^{2}&lt;/math&gt;. This theorem has been know since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually. The Pythagorean Theorem is one of the most frequently used theorems in [[geometry]], and is one of the many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theorem. <br /> <br /> This is generalized by the [[Geometric inequality#Pythagorean_Inequality | Pythagorean Inequality]] and the [[Law of Cosines]].<br /> <br /> == Proofs ==<br /> <br /> In these proofs, we will let &lt;math&gt;ABC &lt;/math&gt; be any right triangle with a right angle at &lt;math&gt;{} C &lt;/math&gt;.<br /> <br /> === Proof 1 ===<br /> <br /> We use &lt;math&gt;[ABC] &lt;/math&gt; to denote the area of triangle &lt;math&gt;ABC &lt;/math&gt;.<br /> <br /> Let &lt;math&gt;H &lt;/math&gt; be the perpendicular to side &lt;math&gt;AB &lt;/math&gt; from &lt;math&gt;{} C &lt;/math&gt;.<br /> <br /> &lt;center&gt;[[Image:Pyth1.png]]&lt;/center&gt;<br /> <br /> Since &lt;math&gt;ABC, CBH, ACH &lt;/math&gt; are similar right triangles, and the areas of similar triangles are proportionate to the squares of corresponding side lengths,<br /> &lt;center&gt;<br /> &lt;math&gt; \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} &lt;/math&gt;.<br /> &lt;/center&gt;<br /> But since triangle &lt;math&gt;ABC &lt;/math&gt; is composed of triangles &lt;math&gt;CBH &lt;/math&gt; and &lt;math&gt;ACH &lt;/math&gt;, &lt;math&gt;[ABC] = [CBH] + [ACH] &lt;/math&gt;, so &lt;math&gt;AB^2 = CB^2 + AC^2 &lt;/math&gt;. {{Halmos}}<br /> <br /> === Proof 2 ===<br /> <br /> Consider a circle &lt;math&gt;\omega &lt;/math&gt; with center &lt;math&gt;B &lt;/math&gt; and radius &lt;math&gt;BC &lt;/math&gt;. Since &lt;math&gt;BC &lt;/math&gt; and &lt;math&gt;AC &lt;/math&gt; are perpendicular, &lt;math&gt;AC &lt;/math&gt; is tangent to &lt;math&gt;\omega &lt;/math&gt;. Let the line &lt;math&gt;AB &lt;/math&gt; meet &lt;math&gt;\omega &lt;/math&gt; at &lt;math&gt;Y &lt;/math&gt; and &lt;math&gt;X &lt;/math&gt;, as shown in the diagram:<br /> <br /> &lt;center&gt;[[Image:Pyth2.png]]&lt;/center&gt;<br /> <br /> Evidently, &lt;math&gt;AY = AB - BC &lt;/math&gt; and &lt;math&gt;AX = AB + BC &lt;/math&gt;. By considering the [[power of a point | power]] of point &lt;math&gt;A &lt;/math&gt; with respect to &lt;math&gt;\omega &lt;/math&gt;, we see<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 &lt;/math&gt;. {{Halmos}}<br /> &lt;/center&gt;<br /> <br /> == Common Pythagorean Triples ==<br /> A [[Pythagorean Triple]] is a [[set]] of 3 [[positive integer]]s such that &lt;math&gt;a^{2}+b^{2}=c^{2}&lt;/math&gt;, i.e. the 3 numbers can be the lengths of the sides of a right triangle. Among these, the [[Primitive Pythagorean Triple]]s, those in which the three numbers have no common [[divisor]], are most interesting. A few of them are:<br /> <br /> &lt;cmath&gt;3-4-5&lt;/cmath&gt;<br /> &lt;cmath&gt;5-12-13&lt;/cmath&gt;<br /> &lt;cmath&gt;7-24-25&lt;/cmath&gt;<br /> &lt;cmath&gt;8-15-17&lt;/cmath&gt;<br /> &lt;cmath&gt;9-40-41&lt;/cmath&gt;<br /> &lt;cmath&gt;20-21-29&lt;/cmath&gt;<br /> <br /> == Problems ==<br /> === Introductory ===<br /> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]<br /> * [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]<br /> <br /> == External links ==<br /> *[http://www.cut-the-knot.org/pythagoras/index.shtml 75 proofs of the Pythagorean Theorem]<br /> <br /> [[Category:Geometry]]<br /> <br /> [[Category:Theorems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_3&diff=41482 2006 AIME I Problems/Problem 3 2011-08-15T10:44:38Z <p>Neutrinonerd3333: # style</p> <hr /> <div>== Problem ==<br /> Find the least [[positive]] [[integer]] such that when its leftmost [[digit]] is deleted, the resulting integer is &lt;math&gt;\frac{1}{29}&lt;/math&gt; of the original integer.<br /> <br /> == Solution ==<br /> The number can be represented as &lt;math&gt;10^na+b&lt;/math&gt;, where &lt;math&gt; a &lt;/math&gt; is the leftmost digit, and &lt;math&gt; b &lt;/math&gt; is the rest of the number.* We know that &lt;math&gt;b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na&lt;/math&gt;. Thus &lt;math&gt; a &lt;/math&gt; has to be 7 since &lt;math&gt; 10^n &lt;/math&gt; can not have 7 as a factor, and the smallest &lt;math&gt; 10^n &lt;/math&gt; can be and have a factor of &lt;math&gt; 2^2 &lt;/math&gt; is &lt;math&gt; 10^2=100. &lt;/math&gt; We find that &lt;math&gt;b=25&lt;/math&gt;, so the number is &lt;math&gt;725&lt;/math&gt;.<br /> <br /> *It is quite obvious that &lt;math&gt;n=2&lt;/math&gt;, since the desired number can't be single or double digit, and cannot exceed &lt;math&gt;999&lt;/math&gt;. From &lt;math&gt;100a+b=29b&lt;/math&gt;, proceed as above.<br /> <br /> == See also ==<br /> * [[Number Theory]]<br /> {{AIME box|year=2006|n=I|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&diff=41481 2006 AIME I Problems/Problem 2 2011-08-15T10:43:55Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let [[set]] &lt;math&gt; \mathcal{A} &lt;/math&gt; be a 90-[[element]] [[subset]] of &lt;math&gt; \{1,2,3,\ldots,100\}, &lt;/math&gt; and let &lt;math&gt; S &lt;/math&gt; be the sum of the elements of &lt;math&gt; \mathcal{A}. &lt;/math&gt; Find the number of possible values of &lt;math&gt; S. &lt;/math&gt;<br /> <br /> == Solution ==<br /> The smallest &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;1+2+ \ldots +90 = 91 \cdot 45 = 4095&lt;/math&gt;. The largest &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;11+12+ \ldots +100=111\cdot 45=4995&lt;/math&gt;. All numbers between &lt;math&gt;4095&lt;/math&gt; and &lt;math&gt;4995&lt;/math&gt; are possible values of S, so the number of possible values of S is &lt;math&gt;4995-4095+1=901&lt;/math&gt;.<br /> <br /> Alternatively, for ease of calculation, let set &lt;math&gt;\mathcal{B}&lt;/math&gt; be a 10-element subset of &lt;math&gt;\{1,2,3,\ldots,100\}&lt;/math&gt;, and let &lt;math&gt;T&lt;/math&gt; be the sum of the elements of &lt;math&gt;\mathcal{B}&lt;/math&gt;. Note that the number of possible &lt;math&gt;S&lt;/math&gt; is the number of possible &lt;math&gt;T=5050-S&lt;/math&gt;. The smallest possible &lt;math&gt;T&lt;/math&gt; is &lt;math&gt;1+2+ \ldots +10 = 55&lt;/math&gt; and the largest is &lt;math&gt;91+92+ \ldots + 100 = 955&lt;/math&gt;, so the number of possible values of T, and therefore S, is &lt;math&gt;955-55+1=\boxed{901}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_3&diff=41480 2006 AIME I Problems/Problem 3 2011-08-15T10:43:11Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the least [[positive]] [[integer]] such that when its leftmost [[digit]] is deleted, the resulting integer is &lt;math&gt;\frac{1}{29}&lt;/math&gt; of the original integer.<br /> <br /> == Solution ==<br /> The number can be represented as &lt;math&gt;10^na+b&lt;/math&gt;, where &lt;math&gt; a &lt;/math&gt; is the leftmost digit, and &lt;math&gt; b &lt;/math&gt; is the rest of the number.* We know that &lt;math&gt;b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na&lt;/math&gt;. Thus &lt;math&gt; a &lt;/math&gt; has to be 7 since &lt;math&gt; 10^n &lt;/math&gt; can not have 7 as a factor, and the smallest &lt;math&gt; 10^n &lt;/math&gt; can be and have a factor of &lt;math&gt; 2^2 &lt;/math&gt; is &lt;math&gt; 10^2=100. &lt;/math&gt; We find that &lt;math&gt;b=25&lt;/math&gt;, so the number is &lt;math&gt;725&lt;/math&gt;.<br /> <br /> *It is quite obvious that &lt;math&gt;n=2&lt;/math&gt;, since the desired number can't be single or double digit, and cannot exceed 999. From &lt;math&gt;100a+b=29b&lt;/math&gt;, proceed as above.<br /> <br /> == See also ==<br /> * [[Number Theory]]<br /> {{AIME box|year=2006|n=I|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&diff=41479 2006 AIME I Problems/Problem 2 2011-08-15T10:39:23Z <p>Neutrinonerd3333: </p> <hr /> <div>== Problem ==<br /> Let [[set]] &lt;math&gt; \mathcal{A} &lt;/math&gt; be a 90-[[element]] [[subset]] of &lt;math&gt; \{1,2,3,\ldots,100\}, &lt;/math&gt; and let &lt;math&gt; S &lt;/math&gt; be the sum of the elements of &lt;math&gt; \mathcal{A}. &lt;/math&gt; Find the number of possible values of &lt;math&gt; S. &lt;/math&gt;<br /> <br /> == Solution ==<br /> The smallest &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;1+2+ \ldots +90 = 91 \cdot 45 = 4095&lt;/math&gt;. The largest &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;11+12+ \ldots +100=111\cdot 45=4995&lt;/math&gt;. All numbers between &lt;math&gt;4095&lt;/math&gt; and &lt;math&gt;4995&lt;/math&gt; are possible values of S, so the number of possible values of S is &lt;math&gt;4995-4095+1=901&lt;/math&gt;.<br /> <br /> Alternatively, for ease of calculation, let set &lt;math&gt;\mathcal{B}&lt;/math&gt; be a 10-element subset of &lt;math&gt;\{1,2,3,\ldots,100\}, and let &lt;/math&gt;T&lt;math&gt; be the sum of the elements of &lt;/math&gt;\mathcal{B}&lt;math&gt;. Note that the number of possible &lt;/math&gt;S&lt;math&gt; is the number of possible &lt;/math&gt;T=5050-S&lt;math&gt;. The smallest possible &lt;/math&gt;T&lt;math&gt; is &lt;/math&gt;1+2+ \ldots +10 = 55&lt;math&gt; and the largest is &lt;/math&gt;91+92+ \ldots + 100 = 955&lt;math&gt;, so the number of possible values of T, and therefore S, is &lt;/math&gt;955-55+1=\boxed{901}.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_8&diff=41475 2005 AIME II Problems/Problem 8 2011-08-15T02:51:43Z <p>Neutrinonerd3333: Solution 2</p> <hr /> <div>== Problem ==<br /> [[Circle]]s &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2 &lt;/math&gt; are externally [[tangent]], and they are both internally tangent to circle &lt;math&gt; C_3. &lt;/math&gt; The radii of &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2 &lt;/math&gt; are 4 and 10, respectively, and the [[center]]s of the three circles are all [[collinear]]. A [[chord]] of &lt;math&gt; C_3 &lt;/math&gt; is also a common external tangent of &lt;math&gt; C_1 &lt;/math&gt; and &lt;math&gt; C_2. &lt;/math&gt; Given that the length of the chord is &lt;math&gt; \frac{m\sqrt{n}}p &lt;/math&gt; where &lt;math&gt; m,n, &lt;/math&gt; and &lt;math&gt; p &lt;/math&gt; are positive integers, &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; p &lt;/math&gt; are [[relatively prime]], and &lt;math&gt; n &lt;/math&gt; is not divisible by the square of any [[prime]], find &lt;math&gt; m+n+p. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> pointpen = black; <br /> pathpen = black + linewidth(0.7); <br /> size(200);<br /> <br /> pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); <br /> path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H);<br /> pair A = OP(cir3, t), B = IP(cir3, t), T1 = IP(cir1, t), T2 = IP(cir2, t);<br /> <br /> draw(cir1); draw(cir2); draw(cir3); <br /> draw((14,0)--(-14,0));<br /> draw(A--B); <br /> MP(&quot;H&quot;,H,W);<br /> draw((-14,0)--H--A, linewidth(0.7) + linetype(&quot;4 4&quot;));<br /> <br /> draw(MP(&quot;O_1&quot;,C1)); <br /> draw(MP(&quot;O_2&quot;,C2)); <br /> draw(MP(&quot;O_3&quot;,C3)); <br /> <br /> draw(MP(&quot;T&quot;,T,N)); <br /> draw(MP(&quot;A&quot;,A,NW)); <br /> draw(MP(&quot;B&quot;,B,NE)); <br /> draw(C1--MP(&quot;T_1&quot;,T1,N)); <br /> <br /> draw(C2--MP(&quot;T_2&quot;,T2,N)); <br /> draw(C3--T); <br /> draw(rightanglemark(C3,T,H));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;O_1, O_2, O_3&lt;/math&gt; be the centers and &lt;math&gt;r_1 = 4, r_2 = 10,r_3 = 14&lt;/math&gt; the radii of the circles &lt;math&gt;C_1, C_2, C_3&lt;/math&gt;. Let &lt;math&gt;T_1, T_2&lt;/math&gt; be the points of tangency from the common external tangent of &lt;math&gt;C_1, C_2&lt;/math&gt;, respectively, and let the extension of &lt;math&gt;\overline{T_1T_2}&lt;/math&gt; intersect the extension of &lt;math&gt;\overline{O_1O_2}&lt;/math&gt; at a point &lt;math&gt;H&lt;/math&gt;. Let the endpoints of the chord/tangent be &lt;math&gt;A,B&lt;/math&gt;, and the foot of the perpendicular from &lt;math&gt;O_3&lt;/math&gt; to &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;T&lt;/math&gt;. From the similar [[right triangle]]s &lt;math&gt;\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T &lt;/math&gt;,<br /> <br /> &lt;cmath&gt; \frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}. &lt;/cmath&gt;<br /> <br /> It follows that &lt;math&gt;HO_1 = \frac{28}{3}&lt;/math&gt;, and that &lt;math&gt;O_3T = \frac{58}{7}&lt;/math&gt;. By the [[Pythagorean Theorem]] on &lt;math&gt;\triangle ATO_3&lt;/math&gt;, we find that <br /> <br /> &lt;cmath&gt;AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}&lt;/cmath&gt;<br /> <br /> and the answer is &lt;math&gt;m+n+p=\boxed{405}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Call our desired length &lt;math&gt;x&lt;/math&gt;. Note for any &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;\overline{O_1O_2}&lt;/math&gt; such that &lt;math&gt;\overline{XY}\perp\overline{AB}&lt;/math&gt; that the function &lt;math&gt;f&lt;/math&gt; such that &lt;math&gt;f(\overline{O_1Y})=\overline{XY}&lt;/math&gt; is linear. Since &lt;math&gt;(0,4)&lt;/math&gt; and &lt;math&gt;(14,10)&lt;/math&gt;, we can quickly interpolate that &lt;math&gt;f(10)=\overline{O_3T}=\frac{58}{7}&lt;/math&gt;. Then, extend &lt;math&gt;\overline{O_3T}&lt;/math&gt; until it reaches the circle on both sides; call them &lt;math&gt;P,Q&lt;/math&gt;. By Power of a Point,<br /> &lt;math&gt;\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}&lt;/math&gt;.<br /> Since &lt;math&gt;\overline{AT}=\overline{TB}=\frac{1}{2}x&lt;/math&gt;,<br /> &lt;cmath&gt;(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2&lt;/cmath&gt;<br /> After solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x=\frac{8\sqrt{390}}{7}&lt;/math&gt;, so our answer is &lt;math&gt;8+390+7=\boxed{405}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&diff=41407 2005 AIME II Problems/Problem 5 2011-08-14T10:04:53Z <p>Neutrinonerd3333: Variable change</p> <hr /> <div>== Problem ==<br /> Determine the number of [[ordered pair]]s &lt;math&gt; (a,b) &lt;/math&gt; of [[integer]]s such that &lt;math&gt; \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, &lt;/math&gt; and &lt;math&gt; 2 \leq b \leq 2005. &lt;/math&gt;<br /> <br /> == Solution ==<br /> The equation can be rewritten as &lt;math&gt; \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 &lt;/math&gt; Multiplying through by &lt;math&gt;\log a \log b &lt;/math&gt; and factoring yields &lt;math&gt;(\log b - 3\log a)(\log b - 2\log a)=0 &lt;/math&gt;. Therefore, &lt;math&gt;\log b=3\log a &lt;/math&gt; or &lt;math&gt;\log b=2\log a &lt;/math&gt;, so either &lt;math&gt; b=a^3 &lt;/math&gt; or &lt;math&gt; b=a^2 &lt;/math&gt;. <br /> *For the case &lt;math&gt; b=a^2 &lt;/math&gt;, note that &lt;math&gt; 44^2=1936 &lt;/math&gt; and &lt;math&gt; 45^2=2025 &lt;/math&gt;. Thus, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; will work. <br /> *For the case &lt;math&gt; b=a^3 &lt;/math&gt;, note that &lt;math&gt; 12^3=1728 &lt;/math&gt; while &lt;math&gt; 13^3=2197 &lt;/math&gt;. Therefore, for this case, all values of &lt;math&gt;a&lt;/math&gt; from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt; work. <br /> There are &lt;math&gt; 44-2+1=43 &lt;/math&gt; possibilities for the square case and &lt;math&gt; 12-2+1=11 &lt;/math&gt; possibilities for the cube case. Thus, the answer is &lt;math&gt; 43+11= \boxed{054}&lt;/math&gt;.<br /> <br /> Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs &lt;math&gt;(a,b)&lt;/math&gt;, and not for the number of possible values of &lt;math&gt;b&lt;/math&gt;. Were the problem to ask for the number of possible values of &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;b^6&lt;/math&gt; under &lt;math&gt;2005&lt;/math&gt; would have to be subtracted, which would just be &lt;math&gt;2&lt;/math&gt; values: &lt;math&gt;2^6&lt;/math&gt; and &lt;math&gt;3^6&lt;/math&gt;.<br /> <br /> <br /> <br /> ==Solution II ==<br /> Let &lt;math&gt;k=\log_a b&lt;/math&gt;. Then our equation becomes &lt;math&gt;k+\frac{6}{k}=5&lt;/math&gt;. Multiplying through by &lt;math&gt;k&lt;/math&gt; and solving the quadratic gives us &lt;math&gt;k=2&lt;/math&gt; or &lt;math&gt;k=3&lt;/math&gt;. Hence &lt;math&gt;a^2=b&lt;/math&gt; or &lt;math&gt;a^3=b&lt;/math&gt;. <br /> <br /> For the first case &lt;math&gt;a^2=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 44, a total of 43 values.<br /> For the second case &lt;math&gt;a^3=b&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; can range from 2 to 12, a total of 11 values.<br /> <br /> <br /> Thus the total number of possible values is &lt;math&gt;43+11=\boxed{54}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_1&diff=41406 2005 AIME II Problems/Problem 1 2011-08-14T09:42:37Z <p>Neutrinonerd3333: </p> <hr /> <div>== Problem ==<br /> <br /> A game uses a deck of &lt;math&gt; n &lt;/math&gt; different cards, where &lt;math&gt; n &lt;/math&gt; is an integer and &lt;math&gt; n \geq 6. &lt;/math&gt; The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find &lt;math&gt; n. &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The number of ways to draw six cards from &lt;math&gt;n&lt;/math&gt; is given by the [[binomial coefficient]] &lt;math&gt;{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}&lt;/math&gt;. <br /> <br /> The number of ways to choose three cards from &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;{n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}&lt;/math&gt;.<br /> <br /> We are given that &lt;math&gt;{n\choose 6} = 6 {n \choose 3}&lt;/math&gt;, so &lt;math&gt;\frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 6 \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}&lt;/math&gt;.<br /> <br /> Cancelling like terms, we get &lt;math&gt;(n - 3)(n - 4)(n - 5) = 720&lt;/math&gt;.<br /> <br /> We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecutive [[integer]]s. Since 720 is close to &lt;math&gt;9^3=729&lt;/math&gt;, we try 8, 9, and 10, which works, so &lt;math&gt;n - 3 = 10&lt;/math&gt; and &lt;math&gt;n = \boxed{013}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2005|n=II|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_1&diff=40049 1998 AIME Problems/Problem 1 2011-06-29T10:17:57Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> For how many values of &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;12^{12}&lt;/math&gt; the [[least common multiple]] of the positive integers &lt;math&gt;6^6&lt;/math&gt;, &lt;math&gt;8^8&lt;/math&gt;, and &lt;math&gt;k&lt;/math&gt;? <br /> == Solution ==<br /> It is evident that &lt;math&gt;k&lt;/math&gt; has only 2s and 3s in its prime factorization, or &lt;math&gt;k = 2^a3^b&lt;/math&gt;. <br /> <br /> *&lt;math&gt;6^6 = 2^6\cdot3^6&lt;/math&gt;<br /> *&lt;math&gt;8^8 = 2^{24}&lt;/math&gt;<br /> *&lt;math&gt;12^{12} = 2^{24}\cdot3^{12}&lt;/math&gt;<br /> <br /> The [[LCM]] of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. &lt;math&gt;[6^6,8^8] = 2^{24}3^6&lt;/math&gt;. Therefore &lt;math&gt;12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}&lt;/math&gt;, and &lt;math&gt;b = 12&lt;/math&gt;. Since &lt;math&gt;0 \le a \le 24&lt;/math&gt;, there are &lt;math&gt;\boxed{025}&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|before=First question|num-a=2}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=40048 1997 AIME Problems/Problem 12 2011-06-29T09:44:23Z <p>Neutrinonerd3333: change of variables</p> <hr /> <div>== Problem ==<br /> The [[function]] &lt;math&gt;f&lt;/math&gt; defined by &lt;math&gt;f(x)= \frac{ax+b}{cx+d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;,&lt;math&gt;b&lt;/math&gt;,&lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are nonzero real numbers, has the properties &lt;math&gt;f(19)=19&lt;/math&gt;, &lt;math&gt;f(97)=97&lt;/math&gt; and &lt;math&gt;f(f(x))=x&lt;/math&gt; for all values except &lt;math&gt;\frac{-d}{c}&lt;/math&gt;. Find the unique number that is not in the range of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> First, we use the fact that &lt;math&gt;f(f(x)) = x&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in the domain. Substituting the function definition, we have &lt;math&gt;{\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x}&lt;/math&gt;, which reduces to<br /> &lt;cmath&gt;{\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =<br /> \frac {px + q}{rx + s} = x}. &lt;/cmath&gt;<br /> In order for this fraction to reduce to &lt;math&gt;x&lt;/math&gt;, we must have &lt;math&gt;q = r = 0&lt;/math&gt; and &lt;math&gt;p = s\not = 0&lt;/math&gt;. From &lt;math&gt;c(a + d) = b(a + d) = 0&lt;/math&gt;, we get &lt;math&gt;a = - d&lt;/math&gt; or &lt;math&gt;b = c = 0&lt;/math&gt;. The second cannot be true, since we are given that &lt;math&gt;a,b,c,d&lt;/math&gt; are nonzero. This means &lt;math&gt;a = - d&lt;/math&gt;, so &lt;math&gt;f(x) = \frac {ax + b}{cx - a}&lt;/math&gt;.<br /> <br /> The only value that is not in the range of this function is &lt;math&gt;\frac {a}{c}&lt;/math&gt;. To find &lt;math&gt;\frac {a}{c}&lt;/math&gt;, we use the two values of the function given to us. We get &lt;math&gt;2(97)a + b = 97^2c&lt;/math&gt; and &lt;math&gt;2(19)a + b = 19^2c&lt;/math&gt;. Subtracting the second equation from the first will eliminate &lt;math&gt;b&lt;/math&gt;, and this results in &lt;math&gt;2(97 - 19)a = (97^2 - 19^2)c&lt;/math&gt;, so<br /> &lt;cmath&gt;{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . &lt;/cmath&gt;<br /> <br /> Alternatively, we could have found out that &lt;math&gt;a = -d&lt;/math&gt; by using the fact that &lt;math&gt;f(f(-b/a))=-b/a&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First, we note that &lt;math&gt;e = \frac ac&lt;/math&gt; is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of &lt;math&gt;f&lt;/math&gt; will be &lt;math&gt;e&lt;/math&gt;. &lt;math&gt;\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}&lt;/math&gt;. [[Without loss of generality]], let &lt;math&gt;c=1&lt;/math&gt;, so the function becomes &lt;math&gt;\frac{b- \frac{d}{a}}{x+d} + e&lt;/math&gt;. <br /> <br /> (Considering &lt;math&gt;\infty&lt;/math&gt; as a limit) By the given, &lt;math&gt;f(f(\infty)) = \infty&lt;/math&gt;. &lt;math&gt;\lim_{x \rightarrow \infty} f(x) = e&lt;/math&gt;, so &lt;math&gt;f(e) = \infty&lt;/math&gt;. &lt;math&gt;f(x) \rightarrow \infty&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt; reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at &lt;math&gt;-\frac{d}{c} = -d&lt;/math&gt;. Hence &lt;math&gt;e = -d&lt;/math&gt;. Substituting the givens, we get <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 17 &amp;= \frac{b - \frac da}{17 - e} + e\\<br /> 97 &amp;= \frac{b - \frac da}{97 - e} + e\\<br /> b - \frac da &amp;= (17 - e)^2 = (97 - e)^2\\<br /> 17 - e &amp;= \pm (97 - e)<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Clearly we can discard the positive root, so &lt;math&gt;e = 58&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> &lt;!-- some linear algebra --&gt;<br /> We first note (as before) that the number not in the range of<br /> &lt;cmath&gt; f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} &lt;/cmath&gt;<br /> is &lt;math&gt;a/c&lt;/math&gt;, as &lt;math&gt;\frac{b-ad/c}{cx+d}&lt;/math&gt; is evidently never 0 (otherwise, &lt;math&gt;f&lt;/math&gt;<br /> would be a constant function, violating the condition &lt;math&gt;f(19) \neq f(97)&lt;/math&gt;).<br /> <br /> We may represent the real number &lt;math&gt;x/y&lt;/math&gt; as<br /> &lt;math&gt;\begin{pmatrix}x \\ y\end{pmatrix}&lt;/math&gt;, with two such [[column vectors]]<br /> considered equivalent if they are scalar multiples of each other. Similarly,<br /> we can represent a function &lt;math&gt;F(x) = \frac{Ax + B}{Cx + D}&lt;/math&gt; as a matrix<br /> &lt;math&gt;\begin{pmatrix} A &amp; B\\ C&amp; D \end{pmatrix}&lt;/math&gt;. Function composition and<br /> evaluation then become matrix multiplication.<br /> <br /> Now in general,<br /> &lt;cmath&gt; f^{-1} = \begin{pmatrix} a &amp; b\\ c&amp;d \end{pmatrix}^{-1} =<br /> \frac{1}{\det(f)} \begin{pmatrix} d &amp; -b \\ -c &amp; a \end{pmatrix} .&lt;/cmath&gt;<br /> In our problem &lt;math&gt;f^2(x) = x&lt;/math&gt;. It follows that<br /> &lt;cmath&gt; \begin{pmatrix} a &amp; b \\ c&amp; d \end{pmatrix} = K<br /> \begin{pmatrix} d &amp; -b \\ -c &amp; a \end{pmatrix} , &lt;/cmath&gt;<br /> for some nonzero real &lt;math&gt;K&lt;/math&gt;. Since<br /> &lt;cmath&gt; \frac{a}{d} = \frac{b}{-b} = K, &lt;/cmath&gt;<br /> it follows that &lt;math&gt;a = -d&lt;/math&gt;. (In fact, this condition condition is equivalent<br /> to the condition that &lt;math&gt;f(f(x)) = x&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in the domain of &lt;math&gt;f&lt;/math&gt;.)<br /> <br /> We next note that the function<br /> &lt;cmath&gt; g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} &lt;/cmath&gt;<br /> evaluates to 0 when &lt;math&gt;x&lt;/math&gt; equals 19 and 97. Therefore<br /> &lt;cmath&gt; \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. &lt;/cmath&gt;<br /> Thus &lt;math&gt;-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}&lt;/math&gt;, so &lt;math&gt;a/c = (19+97)/2 = 58&lt;/math&gt;,<br /> our answer. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> Any number that is not in the domain of the inverse of &lt;math&gt;f(x)&lt;/math&gt; cannot be in the range of &lt;math&gt;f(x)&lt;/math&gt;. Starting with &lt;math&gt;f(x) = \frac{ax+b}{cx+d}&lt;/math&gt;, we rearrange some things to get &lt;math&gt;x = \frac{b-f(x)d}{f(x)c-a}&lt;/math&gt;. Clearly, &lt;math&gt;\frac{a}{c}&lt;/math&gt; is the number that is outside the range of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> <br /> Since we are given &lt;math&gt;f(f(x))=x&lt;/math&gt;, we have that &lt;cmath&gt;x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)&lt;/cmath&gt;<br /> All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that &lt;math&gt;a = -d&lt;/math&gt;.<br /> <br /> This solution follows in the same manner as the last paragraph of the first solution.<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_7&diff=40047 1997 AIME Problems/Problem 7 2011-06-29T07:46:06Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> A car travels due east at &lt;math&gt;\frac 23&lt;/math&gt; mile per minute on a long, straight road. At the same time, a circular storm, whose radius is &lt;math&gt;51&lt;/math&gt; miles, moves southeast at &lt;math&gt;\frac 12\sqrt{2}&lt;/math&gt; mile per minute. At time &lt;math&gt;t=0&lt;/math&gt;, the center of the storm is &lt;math&gt;110&lt;/math&gt; miles due north of the car. At time &lt;math&gt;t=t_1&lt;/math&gt; minutes, the car enters the storm circle, and at time &lt;math&gt;t=t_2&lt;/math&gt; minutes, the car leaves the storm circle. Find &lt;math&gt;\frac 12(t_1+t_2)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We set up a coordinate system, with the starting point of the car at the [[origin]]. At time &lt;math&gt;t&lt;/math&gt;, the car is at &lt;math&gt;\left(\frac 23t,0\right)&lt;/math&gt; and the center of the storm is at &lt;math&gt;\left(\frac{t}{2}, 110 - \frac{t}{2}\right)&lt;/math&gt;. Using the distance formula, <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \sqrt{\left(\frac{2}{3}t - \frac 12t\right)^2 + \left(110-\frac{t}{2}\right)^2} &amp;\le&amp; 51\\<br /> \frac{t^2}{36} + \frac{t^2}{4} - 110t + 110^2 &amp;\le&amp; 51^2\\<br /> \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &amp;\le&amp; 0\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Noting that &lt;math&gt;\frac 12(t_1+t_2)&lt;/math&gt; is at the maximum point of the parabola, we can use &lt;math&gt;-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_6&diff=40046 1997 AIME Problems/Problem 6 2011-06-29T07:43:13Z <p>Neutrinonerd3333: a third solution</p> <hr /> <div>== Problem ==<br /> [[Point]] &lt;math&gt;B&lt;/math&gt; is in the exterior of the [[regular polygon|regular]] &lt;math&gt;n&lt;/math&gt;-sided polygon &lt;math&gt;A_1A_2\cdots A_n&lt;/math&gt;, and &lt;math&gt;A_1A_2B&lt;/math&gt; is an [[equilateral triangle]]. What is the largest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;A_n&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; are consecutive vertices of a regular polygon?<br /> <br /> == Solution 1==<br /> [[Image:1997_AIME-6.png]]<br /> <br /> Let the other regular polygon have &lt;math&gt;m&lt;/math&gt; sides. Using the interior angle of a regular polygon formula, we have &lt;math&gt;\angle A_2A_1A_n = \frac{(n-2)180}{n}&lt;/math&gt;, &lt;math&gt;\angle A_nA_1B = \frac{(m-2)180}{m}&lt;/math&gt;, and &lt;math&gt;\angle A_2A_1B = 60^{\circ}&lt;/math&gt;. Since those three angles add up to &lt;math&gt;360^{\circ}&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{(n-2)180}{n} + \frac{(m-2)180}{m} &amp;=&amp; 300\\<br /> m(n-2)180 + n(m-2)180 &amp;=&amp; 300mn\\<br /> 360mn - 360m - 360n &amp;=&amp; 300mn\\<br /> mn - 6m - 6n &amp;=&amp; 0<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> Using [[Simon's Favorite Factoring Trick|SFFT]], <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> (m-6)(n-6) &amp;=&amp; 36<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> Clearly &lt;math&gt;n&lt;/math&gt; is maximized when &lt;math&gt;m = 7, n = \boxed{42}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> As above, find that &lt;math&gt;mn - 6m - 6n = 0&lt;/math&gt; using the formula for the interior angle of a polygon. <br /> <br /> Solve for &lt;math&gt;n&lt;/math&gt; to find that &lt;math&gt;n = \frac{6m}{m-6}&lt;/math&gt;. Clearly, &lt;math&gt;m&gt;6&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; to be positive.<br /> <br /> With this restriction of &lt;math&gt;m&gt;6&lt;/math&gt;, the larger &lt;math&gt;m&lt;/math&gt; gets, the smaller the fraction &lt;math&gt;\frac{6m}{m-6}&lt;/math&gt; becomes. This can be proven either by calculus, by noting that &lt;math&gt;n = \frac{6m}{m-6}&lt;/math&gt; is a transformed hyperbola, or by dividing out the rational function to get &lt;math&gt;n = 6 + \frac{36}{m - 6}.&lt;/math&gt;<br /> <br /> Either way, minimizng &lt;math&gt;m&lt;/math&gt; will maximize &lt;math&gt;n&lt;/math&gt;, and the smallest integer &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;n&lt;/math&gt; is positive is &lt;math&gt;m=7&lt;/math&gt;, giving &lt;math&gt;n = \boxed{42}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> From the formula for the measure for an individual angle of a regular n-gon, &lt;math&gt;180 - \frac{360}{n}&lt;/math&gt;, the measure of &lt;math&gt;\angle A_2A_1A_n = 180 - \frac{360}{n}&lt;/math&gt;. Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of &lt;math&gt;\angle A_nA_1B = 120 + \frac{360}{n}&lt;/math&gt; (Notice that this value decreases as &lt;math&gt;n&lt;/math&gt; increases; hence, we are looking for the least possible value of &lt;math&gt;\angle A_nA_1B&lt;/math&gt;). For &lt;math&gt;A_n, A_1, B&lt;/math&gt; to be vertices of a regular polygon, &lt;math&gt;\angle A_nA_1B&lt;/math&gt; must be of the form &lt;math&gt;180 - \frac{360}{n}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is a natural number greater than or equal to 3. It is obvious that &lt;math&gt;\angle A_nA_1B &gt; 120&lt;/math&gt;. The least angle satisfying this condition is &lt;math&gt;180 - \frac{360}{7}&lt;/math&gt;. Equating this with &lt;math&gt;120 + \frac{360}{n}&lt;/math&gt; and solving yields &lt;math&gt;n = \boxed{42}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&diff=38133 1989 AIME Problems/Problem 9 2011-04-17T19:09:51Z <p>Neutrinonerd3333: </p> <hr /> <div>== Problem ==<br /> One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^{5}&lt;/math&gt;. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Note that &lt;math&gt;n&lt;/math&gt; is even, since the &lt;math&gt;LHS&lt;/math&gt; consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 7 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;-1 + 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt;, so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n \geq 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large, so &lt;math&gt;n&lt;/math&gt; must be &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_6&diff=37649 2011 AIME I Problems/Problem 6 2011-03-19T04:35:36Z <p>Neutrinonerd3333: </p> <hr /> <div>==Problem==<br /> Suppose that a parabola has vertex &lt;math&gt;\left(\frac{1}{4},-\frac{9}{8}\right)&lt;/math&gt; and equation &lt;math&gt;y = ax^2 + bx + c&lt;/math&gt;, where &lt;math&gt;a &gt; 0&lt;/math&gt; and &lt;math&gt;a + b + c&lt;/math&gt; is an integer. The minimum possible value of &lt;math&gt;a&lt;/math&gt; can be written in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> ==Solution==<br /> If the vertex is at &lt;math&gt;\left(\frac{1}{4}, -\frac{9}{8}\right)&lt;/math&gt;, the equation of the parabola can be expressed in the form &lt;math&gt;y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}&lt;/math&gt;.<br /> Expanding, we find that &lt;math&gt;y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8}&lt;/math&gt; , and &lt;math&gt;y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}&lt;/math&gt;. From the problem, we know that the parabola can be expressed in the form &lt;math&gt;y=ax^2+bx+c&lt;/math&gt;, where &lt;math&gt;a+b+c&lt;/math&gt; is an integer. From the above equation, we can conclude that &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;-\frac{a}{2}=b&lt;/math&gt;, and &lt;math&gt;\frac{a}{16}-\frac{9}{8}=c&lt;/math&gt;. Adding up all of these gives us &lt;math&gt;\frac{9a-18}{16}=a+b+c&lt;/math&gt;. We know that &lt;math&gt;a+b+c&lt;/math&gt; is an integer, so 9a-18 must be divisible by 16. Let &lt;math&gt;9a=z&lt;/math&gt;. If &lt;math&gt;{z-18}\equiv {0} \pmod{16}&lt;/math&gt;, then &lt;math&gt;{z}\equiv {2} \pmod{16}&lt;/math&gt;. Therefore, if &lt;math&gt;9a=2&lt;/math&gt;, &lt;math&gt;a=\frac{2}{9}&lt;/math&gt;. Adding up gives us &lt;math&gt;2+9=\boxed{011}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2011|before=Problem 5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_6&diff=37648 2011 AIME I Problems/Problem 6 2011-03-19T04:28:09Z <p>Neutrinonerd3333: </p> <hr /> <div>If the vertex is at &lt;math&gt;(\frac{1}{4}, -\frac{9}{8})&lt;/math&gt;, the equation of the parabola can be expressed in the form &lt;math&gt;y=a(x-\frac{1}{4})^2-\frac{9}{8}&lt;/math&gt;.<br /> Expanding, we find that &lt;math&gt;y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}&lt;/math&gt; , and &lt;math&gt;y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}&lt;/math&gt;. From the problem, we know that the parabola can be expressed in the form &lt;math&gt;y=ax^2+bx+c&lt;/math&gt;, where &lt;math&gt;a+b+c&lt;/math&gt; is an integer. From the above equation, we can conclude that &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;-\frac{a}{2}=b&lt;/math&gt;, and &lt;math&gt;\frac{a}{16}-\frac{9}{8}=c&lt;/math&gt;. Adding up all of these gives us &lt;math&gt;\frac{9a-18}{16}=a+b+c&lt;/math&gt;. We know that &lt;math&gt;a+b+c&lt;/math&gt; is an integer, so 9a-18 must be divisible by 16. Let &lt;math&gt;9a=z&lt;/math&gt;. If &lt;math&gt;{z-18}\equiv {0} \pmod{16}&lt;/math&gt;, then &lt;math&gt;{z}\equiv {2} \pmod{16}&lt;/math&gt;. Therefore, if &lt;math&gt;9a=2&lt;/math&gt;, &lt;math&gt;a=\frac{2}{9}&lt;/math&gt;. Adding up gives us &lt;math&gt;2+9=\boxed{011}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2011|before=Problem 5|num-a=7}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_6&diff=37641 2011 AIME I Problems/Problem 6 2011-03-18T23:39:26Z <p>Neutrinonerd3333: </p> <hr /> <div>If the vertex is at &lt;math&gt;(\frac{1}{4}, -\frac{9}{8})&lt;/math&gt;, the equation of the parabola can be expressed in the form &lt;math&gt;y=a(x-\frac{1}{4})^2-\frac{9}{8}&lt;/math&gt;.<br /> Expanding, we find that &lt;math&gt;y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}&lt;/math&gt; , and &lt;math&gt;y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}&lt;/math&gt;. From the problem, we know that the parabola can be expressed in the form &lt;math&gt;y=ax^2+bx+c&lt;/math&gt;, where &lt;math&gt;a+b+c&lt;/math&gt; is an integer. From the above equation, we can conclude that &lt;math&gt;a=a&lt;/math&gt;, &lt;math&gt;-\frac{a}{2}=b&lt;/math&gt;, and &lt;math&gt;\frac{a}{16}-\frac{9}{8}=c&lt;/math&gt;. Adding up all of these gives us &lt;math&gt;\frac{9a-18}{16}=a+b+c&lt;/math&gt;. We know that &lt;math&gt;a+b+c&lt;/math&gt; is an integer, so 9a-18 must be divisible by 16. Let &lt;math&gt;9a=z&lt;/math&gt;. If &lt;math&gt;{z-18}\equiv {0} \pmod{16}&lt;/math&gt;, then &lt;math&gt;{z}\equiv {2} \pmod{16}&lt;/math&gt;. Therefore, if &lt;math&gt;9a=2&lt;/math&gt;, &lt;math&gt;a=\frac{2}{9}&lt;/math&gt;. Adding up gives us &lt;math&gt;2+9=\boxed{011}&lt;/math&gt;</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=37221 1984 AIME Problems/Problem 14 2011-03-03T05:35:17Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_10&diff=37220 1984 AIME Problems/Problem 10 2011-03-03T05:33:36Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> Mary told John her score on the American High School Mathematics Examination (AHSME), which was over &lt;math&gt;80&lt;/math&gt;. From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over &lt;math&gt;80&lt;/math&gt;, John could not have determined this. What was Mary's score? (Recall that the AHSME consists of &lt;math&gt;30&lt;/math&gt; multiple choice problems and that one's score, &lt;math&gt;s&lt;/math&gt;, is computed by the formula &lt;math&gt;s=30+4c-w&lt;/math&gt;, where &lt;math&gt;c&lt;/math&gt; is the number of correct answers and &lt;math&gt;w&lt;/math&gt; is the number of wrong answers. (Students are not penalized for problems left unanswered.)<br /> <br /> == Solution ==<br /> Let Mary's score, number correct, and number wrong be &lt;math&gt;s,c,w&lt;/math&gt; respectively. Then<br /> <br /> &lt;math&gt;s=30+4c-w=30+4(c-1)-(w-4)=30+4(c+1)-(w+4)&lt;/math&gt;.<br /> <br /> Therefore, Mary could not have left at least five blank; otherwise, 1 more correct and 4 more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have score above 80, or even 30.)<br /> <br /> It follows that &lt;math&gt;c+w\geq 26&lt;/math&gt; and &lt;math&gt;w\leq 3&lt;/math&gt;, so &lt;math&gt;c\geq 23&lt;/math&gt; and &lt;math&gt;s=30+4c-w\geq 30+4(23)-3=119&lt;/math&gt;. So Mary scored at least 119. To see that no result other than 23 right/3 wrong produces 119, note that &lt;math&gt;s=119\Rightarrow 4c-w=89&lt;/math&gt; so &lt;math&gt;w\equiv 3\pmod{4}&lt;/math&gt;. But if &lt;math&gt;w=3&lt;/math&gt;, then &lt;math&gt;c=23&lt;/math&gt;, which was the result given; otherwise &lt;math&gt;w\geq 7&lt;/math&gt; and &lt;math&gt;c\geq 24&lt;/math&gt;, but this implies at least 31 questions, a contradiction. This makes the minimum score &lt;math&gt;\boxed{119}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=9|num-a=11}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Neutrinonerd3333 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_5&diff=37213 1984 AIME Problems/Problem 5 2011-03-02T05:27:50Z <p>Neutrinonerd3333: /* Solution */</p> <hr /> <div>== Problem ==<br /> Determine the value of &lt;math&gt;ab&lt;/math&gt; if &lt;math&gt;\log_8a+\log_4b^2=5&lt;/math&gt; and &lt;math&gt;\log_8b+\log_4a^2=7&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Use the [[change of base formula]] to see that &lt;math&gt;\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5&lt;/math&gt;; combine [[denominator]]s to find that &lt;math&gt;\frac{\log a^3b}{3\log 2} = 5&lt;/math&gt;. Doing the same thing with the second equation yields that &lt;math&gt;\frac{\log b^3a}{3\log 2} = 7&lt;/math&gt;. This means that &lt;math&gt;\log a^3b = 15\log 2 \Longrightarrow a^3b = 2^{15}&lt;/math&gt; and that &lt;math&gt;\log ab^3 = 21\log 2 \Longrightarrow ab^3 = 2^{21}&lt;/math&gt;. If we multiply the two equations together, we get that &lt;math&gt;a^4b^4 = 2^{36}&lt;/math&gt;, so taking the fourth root of that, &lt;math&gt;ab = 2^9 = \boxed{512}&lt;/math&gt;.<br /> <br /> == See also ==<br /> *[[Logarithm]]<br /> <br /> {{AIME box|year=1984|num-b=4|num-a=6}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Neutrinonerd3333