https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Rahulkk&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-25T22:28:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_9&diff=74884 2015 AIME I Problems/Problem 9 2016-01-29T02:49:23Z <p>Rahulkk: /* Solution */ Boxed answer.</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all ordered triple of integers &lt;math&gt;(a_1,a_2,a_3)&lt;/math&gt; with &lt;math&gt;1 \le a_1,a_2,a_3 \le 10&lt;/math&gt;. Each ordered triple in &lt;math&gt;S&lt;/math&gt; generates a sequence according to the rule &lt;math&gt;a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |&lt;/math&gt; for all &lt;math&gt;n\ge 4&lt;/math&gt;. Find the number of such sequences for which &lt;math&gt;a_n=0&lt;/math&gt; for some &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;a_1=x, a_2=y, a_3=z&lt;/math&gt;. First note that if any absolute value equals 0, then &lt;math&gt;a_n=0&lt;/math&gt;.<br /> Also note that if at any position, &lt;math&gt;a_n=a_{n-1}&lt;/math&gt;, then &lt;math&gt;a_{n+2}=0&lt;/math&gt;.<br /> Then, if any absolute value equals 1, then &lt;math&gt;a_n=0&lt;/math&gt;.<br /> Therefore, if either &lt;math&gt;|y-x|&lt;/math&gt; or &lt;math&gt;|z-y|&lt;/math&gt; is less than or equal to 1, then that ordered triple meets the criteria.<br /> Assume that to be the only way the criteria is met.<br /> To prove, let &lt;math&gt;|y-x|&gt;1&lt;/math&gt;, and &lt;math&gt;|z-y|&gt;1&lt;/math&gt;. Then, &lt;math&gt;a_4 \ge 2z&lt;/math&gt;, &lt;math&gt;a_5 \ge 4z&lt;/math&gt;, and &lt;math&gt;a_6 \ge 4z&lt;/math&gt;.<br /> However, since the minimum values of &lt;math&gt;a_5&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt; are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be &lt;math&gt;z=1&lt;/math&gt;, &lt;math&gt;|y-x|=2&lt;/math&gt;. Again assume that any other scenario will not meet criteria.<br /> To prove, divide the other scenarios into two cases: &lt;math&gt;z&gt;1&lt;/math&gt;, &lt;math&gt;|y-x|&gt;1&lt;/math&gt;, and &lt;math&gt;|z-y|&gt;1&lt;/math&gt;; and &lt;math&gt;z=1&lt;/math&gt;, &lt;math&gt;|y-x|&gt;2&lt;/math&gt;, and &lt;math&gt;|z-y|&gt;1&lt;/math&gt;.<br /> For the first one, &lt;math&gt;a_4 \ge 2z&lt;/math&gt;, &lt;math&gt;a_5 \ge 4z&lt;/math&gt;, &lt;math&gt;a_6 \ge 8z&lt;/math&gt;, and &lt;math&gt;a_7 \ge 16z&lt;/math&gt;, by which point we see that this function diverges.<br /> For the second one, &lt;math&gt;a_4 \ge 3&lt;/math&gt;, &lt;math&gt;a_5 \ge 6&lt;/math&gt;, &lt;math&gt;a_6 \ge 18&lt;/math&gt;, and &lt;math&gt;a_7 \ge 54&lt;/math&gt;, by which point we see that this function diverges.<br /> Therefore, the only scenarios where &lt;math&gt;a_n=0&lt;/math&gt; is when any of the following are met:<br /> &lt;math&gt;|y-x|&lt;2&lt;/math&gt; (280 options)<br /> &lt;math&gt;|z-y|&lt;2&lt;/math&gt; (280 options, 80 of which coincide with option 1)<br /> &lt;math&gt;z=1&lt;/math&gt;, &lt;math&gt;|y-x|=2&lt;/math&gt;. (16 options, 2 of which coincide with either option 1 or option 2)<br /> Adding the total number of such ordered triples yields &lt;math&gt;280+280-80+16-2=\boxed{494}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2015|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Intermediate Combinatorics Problems]]</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_7&diff=71473 1998 AIME Problems/Problem 7 2015-08-06T17:32:17Z <p>Rahulkk: Corrected minor typo. (&quot;Out&quot; -&gt; &quot;Our&quot;)</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the number of ordered quadruples &lt;math&gt;(x_1,x_2,x_3,x_4)&lt;/math&gt; of positive odd [[integer]]s that satisfy &lt;math&gt;\sum_{i = 1}^4 x_i = 98.&lt;/math&gt; Find &lt;math&gt;\frac n{100}.&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define &lt;math&gt;x_i = 2y_i - 1&lt;/math&gt;. Then &lt;math&gt;2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98&lt;/math&gt;, so &lt;math&gt;\sum_{i = 1}^4 y_i = 51&lt;/math&gt;. <br /> <br /> So we want to find four integers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is &lt;math&gt;n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600&lt;/math&gt;, and &lt;math&gt;\frac n{100} = \boxed{196}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> <br /> Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into &lt;math&gt;4&lt;/math&gt; boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have &lt;math&gt;94&lt;/math&gt; stones left. Because we want an odd number in each box, we pair the stones, creating &lt;math&gt;47&lt;/math&gt; sets of &lt;math&gt;2&lt;/math&gt;. Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).<br /> <br /> Our problem can now be restated: how many ways are there to partition a line of &lt;math&gt;47&lt;/math&gt; stones? We can easily solve this by using &lt;math&gt;3&lt;/math&gt; sticks to separate the stones into &lt;math&gt;4&lt;/math&gt; groups, and this is the same as arranging a line of &lt;math&gt;3&lt;/math&gt; sticks and &lt;math&gt;47&lt;/math&gt; stones. &lt;cmath&gt;\frac{50!}{47! \cdot 3!} = 19600&lt;/cmath&gt; &lt;cmath&gt;\frac{50 * 49 * 48}{3 * 2} = 19600&lt;/cmath&gt; Our answer is therefore &lt;math&gt;\frac{19600}{100} = \boxed{196}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_6&diff=71438 2009 AIME I Problems/Problem 6 2015-08-03T01:18:00Z <p>Rahulkk: Corrected minor typo. (&quot;Theerefore&quot; -&gt; &quot;Therefore&quot;)</p> <hr /> <div>== Problem ==<br /> <br /> How many positive integers &lt;math&gt;N&lt;/math&gt; less than &lt;math&gt;1000&lt;/math&gt; are there such that the equation &lt;math&gt;x^{\lfloor x\rfloor} = N&lt;/math&gt; has a solution for &lt;math&gt;x&lt;/math&gt;? (The notation &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denotes the greatest integer that is less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> First, &lt;math&gt;x&lt;/math&gt; must be less than &lt;math&gt;5&lt;/math&gt;, since otherwise &lt;math&gt;x^{\lfloor x\rfloor}&lt;/math&gt; would be at least &lt;math&gt;3125&lt;/math&gt; which is greater than &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> Because &lt;math&gt;{\lfloor x\rfloor}&lt;/math&gt; must be an integer, we can do some simple case work:<br /> <br /> For &lt;math&gt;{\lfloor x\rfloor}=0&lt;/math&gt;, &lt;math&gt;N=1&lt;/math&gt; as long as &lt;math&gt;x \neq 0&lt;/math&gt;. This gives us &lt;math&gt;1&lt;/math&gt; value of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> For &lt;math&gt;{\lfloor x\rfloor}=1&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; can be anything between &lt;math&gt;1^1&lt;/math&gt; to &lt;math&gt;2^1&lt;/math&gt; excluding &lt;math&gt;2^1&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;N=1&lt;/math&gt;. However, we got N=1 in case 1 so it got counted twice.<br /> <br /> For &lt;math&gt;{\lfloor x\rfloor}=2&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; can be anything between &lt;math&gt;2^2&lt;/math&gt; to &lt;math&gt;3^2&lt;/math&gt; excluding &lt;math&gt;3^2&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;3^2-2^2=5&lt;/math&gt; &lt;math&gt;N&lt;/math&gt;'s<br /> <br /> For &lt;math&gt;{\lfloor x\rfloor}=3&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; can be anything between &lt;math&gt;3^3&lt;/math&gt; to &lt;math&gt;4^3&lt;/math&gt; excluding &lt;math&gt;4^3&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;4^3-3^3=37&lt;/math&gt; &lt;math&gt;N&lt;/math&gt;'s<br /> <br /> For &lt;math&gt;{\lfloor x\rfloor}=4&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; can be anything between &lt;math&gt;4^4&lt;/math&gt; to &lt;math&gt;5^4&lt;/math&gt; excluding &lt;math&gt;5^4&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;5^4-4^4=369&lt;/math&gt; &lt;math&gt;N&lt;/math&gt;'s<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; must be less than &lt;math&gt;5&lt;/math&gt;, we can stop here and the answer is &lt;math&gt;1+5+37+369= \boxed {412}&lt;/math&gt; possible values for &lt;math&gt;N&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems&diff=71430 2011 AIME II Problems 2015-08-01T23:27:08Z <p>Rahulkk: Corrected minor typo. (&quot;expresses&quot; -&gt; &quot;expressed&quot;)</p> <hr /> <div>{{AIME Problems|year=2011|n=II}}<br /> <br /> == Problem 1 ==<br /> Gary purchased a large beverage, but only drank &lt;math&gt;m/n&lt;/math&gt; of it, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only &lt;math&gt;2/9&lt;/math&gt; as much beverage. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> On square &lt;math&gt;ABCD&lt;/math&gt;, point &lt;math&gt;E&lt;/math&gt; lies on side &lt;math&gt;AD&lt;/math&gt; and point &lt;math&gt;F&lt;/math&gt; lies on side &lt;math&gt;BC&lt;/math&gt;, so that &lt;math&gt;BE=EF=FD=30&lt;/math&gt;. Find the area of the square &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.<br /> <br /> [[2011 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=\frac{20}{11}AC&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;D&lt;/math&gt;, and point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of intersection of &lt;math&gt;AC&lt;/math&gt; and the line &lt;math&gt;BM&lt;/math&gt;. The ratio of &lt;math&gt;CP&lt;/math&gt; to &lt;math&gt;PA&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;. <br /> <br /> [[2011 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms. <br /> <br /> [[2011 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Define an ordered quadruple of integers (a, b, c, d) as ''interesting'' if &lt;math&gt;1 \le a&lt;b&lt;c&lt;d \le 10&lt;/math&gt;, and &lt;math&gt; a+d&gt;b+c &lt;/math&gt;. How many interesting ordered quadruples are there?<br /> <br /> [[2011 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let &lt;math&gt;m&lt;/math&gt; be the maximum number of red marbles for which such an arrangement is possible, and let &lt;math&gt;N&lt;/math&gt; be the number of ways he can arrange the &lt;math&gt;m+5&lt;/math&gt; marbles to satisfy the requirement. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Let &lt;math&gt;z_1&lt;/math&gt;, &lt;math&gt;z_2&lt;/math&gt;, &lt;math&gt;z_3&lt;/math&gt;, &lt;math&gt;\dots&lt;/math&gt;, &lt;math&gt;z_{12}&lt;/math&gt; be the 12 zeroes of the polynomial &lt;math&gt;z^{12} - 2^{36}&lt;/math&gt;. For each &lt;math&gt;j&lt;/math&gt;, let &lt;math&gt;w_j&lt;/math&gt; be one of &lt;math&gt;z_j&lt;/math&gt; or &lt;math&gt;iz_j&lt;/math&gt;. Then the maximum possible value of the real part of &lt;math&gt;\sum_{j = 1}^{12} w_j&lt;/math&gt; can be written as &lt;math&gt;m + \sqrt{n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Let &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;\dots&lt;/math&gt;, &lt;math&gt;x_6&lt;/math&gt; be nonnegative real numbers such that &lt;math&gt;x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1&lt;/math&gt;, and &lt;math&gt;x_1x_3x_5 + x_2x_4x_6 \ge {\scriptstyle\frac{1}{540}}&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive relatively prime integers such that &lt;math&gt;\frac{p}{q}&lt;/math&gt; is the maximum possible value of &lt;math&gt;x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2&lt;/math&gt;. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> A circle with center &lt;math&gt;O&lt;/math&gt; has radius 25. Chord &lt;math&gt;\overline{AB}&lt;/math&gt; of length 30 and chord &lt;math&gt;\overline{CD}&lt;/math&gt; of length 14 intersect at point &lt;math&gt;P&lt;/math&gt;. The distance between the midpoints of the two chords is 12. The quantity &lt;math&gt;OP^2&lt;/math&gt; can be represented as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find the remainder when &lt;math&gt;m + n&lt;/math&gt; is divided by 1000.<br /> <br /> [[2011 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt;M_n&lt;/math&gt; be the &lt;math&gt;n \times n&lt;/math&gt; matrix with entries as follows: for &lt;math&gt;1 \le i \le n&lt;/math&gt;, &lt;math&gt;m_{i,i} = 10&lt;/math&gt;; for &lt;math&gt;1 \le i \le n - 1&lt;/math&gt;, &lt;math&gt;m_{i+1,i} = m_{i,i+1} = 3&lt;/math&gt;; all other entries in &lt;math&gt;M_n&lt;/math&gt; are zero. Let &lt;math&gt;D_n&lt;/math&gt; be the determinant of matrix &lt;math&gt;M_n&lt;/math&gt;. Then &lt;math&gt;\sum_{n=1}^{\infty} \frac{1}{8D_n+1}&lt;/math&gt; can be represented as &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 /> Note: The determinant of the &lt;math&gt;1 \times 1&lt;/math&gt; matrix &lt;math&gt;[a]&lt;/math&gt; is &lt;math&gt;a&lt;/math&gt;, and the determinant of the &lt;math&gt;2 \times 2&lt;/math&gt; matrix &lt;math&gt;\left[ {\begin{array}{cc}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{array} } \right] = ad - bc&lt;/math&gt;; for &lt;math&gt;n \ge 2&lt;/math&gt;, the determinant of an &lt;math&gt;n \times n&lt;/math&gt; matrix with first row or first column &lt;math&gt;a_1&lt;/math&gt; &lt;math&gt;a_2&lt;/math&gt; &lt;math&gt;a_3&lt;/math&gt; &lt;math&gt;\dots&lt;/math&gt; &lt;math&gt;a_n&lt;/math&gt; is equal to &lt;math&gt;a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n&lt;/math&gt;, where &lt;math&gt;C_i&lt;/math&gt; is the determinant of the &lt;math&gt;(n - 1) \times (n - 1)&lt;/math&gt; matrix formed by eliminating the row and column containing &lt;math&gt;a_i&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; be the circumcenters of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt;, respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_1PO_2 = 120\textdegree&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{a} + \sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> There are &lt;math&gt;N&lt;/math&gt; permutations &lt;math&gt;(a_1, a_2, \dots, a_{30})&lt;/math&gt; of &lt;math&gt;1, 2, \dots, 30&lt;/math&gt; such that for &lt;math&gt;m \in \{2,3,5\}&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;a_{n+m} - a_n&lt;/math&gt; for all integers &lt;math&gt;n&lt;/math&gt; with &lt;math&gt;1 \le n &lt; n+m \le 30&lt;/math&gt;. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000.<br /> <br /> [[2011 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let &lt;math&gt;P(x) = x^2 - 3x - 9&lt;/math&gt;. A real number &lt;math&gt;x&lt;/math&gt; is chosen at random from the interval &lt;math&gt;5 \le x \le 15&lt;/math&gt;. The probability that &lt;math&gt;\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; is equal to &lt;math&gt;\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}&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;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b + c + d + e&lt;/math&gt;.<br /> <br /> [[2011 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems&diff=71397 2003 AIME II Problems 2015-07-30T19:07:00Z <p>Rahulkk: Added a question mark at the end of the question.</p> <hr /> <div>{{AIME Problems|year=2003|n=II}}<br /> <br /> == Problem 1 ==<br /> The product &lt;math&gt;N&lt;/math&gt; of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest integer multiple of 8, whose digits are all different. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000?<br /> <br /> [[2003 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Define a &lt;math&gt;good~word&lt;/math&gt; as a sequence of letters that consists only of the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; - some of these letters may not appear in the sequence - and in which &lt;math&gt;A&lt;/math&gt; is never immediately followed by &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; is never immediately followed by &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; is never immediately followed by &lt;math&gt;A&lt;/math&gt;. How many seven-letter good words are there?<br /> <br /> [[2003 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is &lt;math&gt;m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A cylindrical log has diameter &lt;math&gt;12&lt;/math&gt; inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a &lt;math&gt;45^\circ&lt;/math&gt; angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as &lt;math&gt;n\pi&lt;/math&gt;, where n is a positive integer. Find &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;AB = 13,&lt;/math&gt; &lt;math&gt;BC = 14,&lt;/math&gt; &lt;math&gt;AC = 15,&lt;/math&gt; and point &lt;math&gt;G&lt;/math&gt; is the intersection of the medians. Points &lt;math&gt;A',&lt;/math&gt; &lt;math&gt;B',&lt;/math&gt; and &lt;math&gt;C',&lt;/math&gt; are the images of &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; respectively, after a &lt;math&gt;180^\circ&lt;/math&gt; rotation about &lt;math&gt;G.&lt;/math&gt; What is the area of the union of the two regions enclosed by the triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'?&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Find the area of rhombus &lt;math&gt;ABCD&lt;/math&gt; given that the radii of the circles circumscribed around triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; are &lt;math&gt;12.5&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt;, respectively.<br /> <br /> [[2003 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Find the eighth term of the sequence &lt;math&gt;1440,&lt;/math&gt; &lt;math&gt;1716,&lt;/math&gt; &lt;math&gt;1848,\ldots,&lt;/math&gt; whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br /> <br /> [[2003 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Consider the polynomials &lt;math&gt;P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x&lt;/math&gt; and &lt;math&gt;Q(x) = x^{4} - x^{3} - x^{2} - 1.&lt;/math&gt; Given that &lt;math&gt;z_{1},z_{2},z_{3},&lt;/math&gt; and &lt;math&gt;z_{4}&lt;/math&gt; are the roots of &lt;math&gt;Q(x) = 0,&lt;/math&gt; find &lt;math&gt;P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Two positive integers differ by &lt;math&gt;60.&lt;/math&gt; The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?<br /> <br /> [[2003 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is a right triangle with &lt;math&gt;AC = 7,&lt;/math&gt; &lt;math&gt;BC = 24,&lt;/math&gt; and right angle at &lt;math&gt;C.&lt;/math&gt; Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AB,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; is on the same side of line &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AD = BD = 15.&lt;/math&gt; Given that the area of triangle &lt;math&gt;CDM&lt;/math&gt; may be expressed as &lt;math&gt;\frac {m\sqrt {n}}{p},&lt;/math&gt; where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;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 /> [[2003 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The members of a distinguished committee were choosing a president, and each member gave one vote to one of the &lt;math&gt;27&lt;/math&gt; candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least &lt;math&gt;1&lt;/math&gt; than the number of votes for that candidate. What is the smallest possible number of members of the committee?<br /> <br /> [[2003 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;A = (0,0)&lt;/math&gt; and &lt;math&gt;B = (b,2)&lt;/math&gt; be points on the coordinate plane. Let &lt;math&gt;ABCDEF&lt;/math&gt; be a convex equilateral hexagon such that &lt;math&gt;\angle FAB = 120^\circ,&lt;/math&gt; &lt;math&gt;\overline{AB}\parallel \overline{DE},&lt;/math&gt; &lt;math&gt;\overline{BC}\parallel \overline{EF,}&lt;/math&gt; &lt;math&gt;\overline{CD}\parallel \overline{FA},&lt;/math&gt; and the y-coordinates of its vertices are distinct elements of the set &lt;math&gt;\{0,2,4,6,8,10\}.&lt;/math&gt; The area of the hexagon can be written in the form &lt;math&gt;m\sqrt {n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and n is not divisible by the square of any prime. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let<br /> &lt;center&gt;&lt;math&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/math&gt;&lt;/center&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_4&diff=71388 2005 AIME II Problems/Problem 4 2015-07-30T05:17:17Z <p>Rahulkk: Boxed the final answer.</p> <hr /> <div>== Problem ==<br /> Find the number of [[positive integer]]s that are divisors of at least one of &lt;math&gt; 10^{10},15^7,18^{11}. &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;10^{10} = 2^{10}\cdot 5^{10}&lt;/math&gt; so &lt;math&gt;10^{10}&lt;/math&gt; has &lt;math&gt;11\cdot11 = 121&lt;/math&gt; [[divisor]]s.<br /> <br /> &lt;math&gt;15^7 = 3^7\cdot5^7&lt;/math&gt; so &lt;math&gt;15^7&lt;/math&gt; has &lt;math&gt;8\cdot8 = 64&lt;/math&gt; divisors.<br /> <br /> &lt;math&gt;18^{11} = 2^{11}\cdot3^{22}&lt;/math&gt; so &lt;math&gt;18^{11}&lt;/math&gt; has &lt;math&gt;12\cdot23 = 276&lt;/math&gt; divisors.<br /> <br /> Now, we use the [[Principle of Inclusion-Exclusion]]. We have &lt;math&gt;121 + 64 + 276&lt;/math&gt; total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise [[greatest common divisor]]s.<br /> <br /> &lt;math&gt;\gcd(10^{10},15^7) = 5^7 &lt;/math&gt; which has 8 divisors.<br /> <br /> &lt;math&gt;\gcd(15^7, 18^{11}) = 3^7 &lt;/math&gt; which has 8 divisors.<br /> <br /> &lt;math&gt;\gcd(18^{11}, 10^{10}) = 2^{10} &lt;/math&gt; which has 11 divisors.<br /> <br /> So now we have &lt;math&gt;121 + 64 + 276 - 8 -8 -11&lt;/math&gt; potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of &lt;math&gt;121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems&diff=71043 2014 AIME I Problems 2015-07-08T19:35:13Z <p>Rahulkk: /* Problem 4 */</p> <hr /> <div>{{AIME Problems|year=2014|n=I}}<br /> <br /> ==Problem 1==<br /> The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.7));<br /> path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin;<br /> path laceR=reflect((75,0),(75,-240))*laceL;<br /> draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5);<br /> unfill(circ1); draw(circ1);<br /> unfill(circ2); draw(circ2);<br /> }<br /> draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));&lt;/asy&gt;<br /> [[2014 AIME I Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2== <br /> <br /> An urn contains &lt;math&gt;4&lt;/math&gt; green balls and &lt;math&gt;6&lt;/math&gt; blue balls. A second urn contains &lt;math&gt;16&lt;/math&gt; green balls and &lt;math&gt;N&lt;/math&gt; blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is &lt;math&gt;0.58&lt;/math&gt;. Find &lt;math&gt;N&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Find the number of rational numbers &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;0&lt;r&lt;1,&lt;/math&gt; such that when &lt;math&gt;r&lt;/math&gt; is written as a fraction in lowest terms, the numerator and the denominator have a sum of &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at &lt;math&gt;20&lt;/math&gt; miles per hour, and Steve rides west at &lt;math&gt;20&lt;/math&gt; miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly &lt;math&gt;1&lt;/math&gt; minute to go past Jon. The westbound train takes &lt;math&gt;10&lt;/math&gt; times as long as the eastbound train to go past Steve. The length of each train is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Let the set &lt;math&gt;S = \{P_1, P_2, \dots, P_{12}\}&lt;/math&gt; consist of the twelve vertices of a regular &lt;math&gt;12&lt;/math&gt;-gon. A subset &lt;math&gt;Q&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt; is called communal if there is a circle such that all points of &lt;math&gt;Q&lt;/math&gt; are inside the circle, and all points of &lt;math&gt;S&lt;/math&gt; not in &lt;math&gt;Q&lt;/math&gt; are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)<br /> <br /> [[2014 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The graphs &lt;math&gt;y = 3(x-h)^2 + j&lt;/math&gt; and &lt;math&gt;y = 2(x-h)^2 + k&lt;/math&gt; have y-intercepts of &lt;math&gt;2013&lt;/math&gt; and &lt;math&gt;2014&lt;/math&gt;, respectively, and each graph has two positive integer x-intercepts. Find &lt;math&gt;h&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Let &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be complex numbers such that &lt;math&gt;|w| = 1&lt;/math&gt; and &lt;math&gt;|z| = 10&lt;/math&gt;. Let &lt;math&gt;\theta = \arg \left(\tfrac{w-z}{z}\right) &lt;/math&gt;. The maximum possible value of &lt;math&gt;\tan^2 \theta&lt;/math&gt; can be written as &lt;math&gt;\tfrac{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;. (Note that &lt;math&gt;\arg(w)&lt;/math&gt;, for w &lt;math&gt;\neq 0&lt;/math&gt;, denotes the measure of the angle that the ray from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;w&lt;/math&gt; makes with the positive real axis in the complex plane.<br /> <br /> [[2014 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> The positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;N^2&lt;/math&gt; both end in the same sequence of four digits &lt;math&gt;abcd&lt;/math&gt; when written in base 10, where digit &lt;math&gt;a&lt;/math&gt; is not zero. Find the three-digit number &lt;math&gt;abc&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;x_1&lt; x_2 &lt; x_3&lt;/math&gt; be the three real roots of the equation &lt;math&gt;\sqrt{2014} x^3 - 4029x^2 + 2 = 0&lt;/math&gt;. Find &lt;math&gt;x_2(x_1+x_3)&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A disk with radius &lt;math&gt;1&lt;/math&gt; is externally tangent to a disk with radius &lt;math&gt;5&lt;/math&gt;. Let &lt;math&gt;A&lt;/math&gt; be the point where the disks are tangent, &lt;math&gt;C&lt;/math&gt; be the center of the smaller disk, and &lt;math&gt;E&lt;/math&gt; be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of &lt;math&gt;360^\circ&lt;/math&gt;. That is, if the center of the smaller disk has moved to the point &lt;math&gt;D&lt;/math&gt;, and the point on the smaller disk that began at &lt;math&gt;A&lt;/math&gt; has now moved to point &lt;math&gt;B&lt;/math&gt;, then &lt;math&gt;\overline{AC}&lt;/math&gt; is parallel to &lt;math&gt;\overline{BD}&lt;/math&gt;. Then &lt;math&gt;\sin^2(\angle BEA)=\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> A token starts at the point &lt;math&gt;(0,0)&lt;/math&gt; of an &lt;math&gt;xy&lt;/math&gt;-coordinate grid and them makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of &lt;math&gt;|y|=|x|&lt;/math&gt; is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A=\{1,2,3,4\}&lt;/math&gt;, and &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; be randomly chosen (not necessarily distinct) functions from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;. The probability that the range of &lt;math&gt;f&lt;/math&gt; and the range of &lt;math&gt;g&lt;/math&gt; are disjoint is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> On square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E,F,G&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; lie on sides &lt;math&gt;\overline{AB},\overline{BC},\overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; respectively, so that &lt;math&gt;\overline{EG} \perp \overline{FH}&lt;/math&gt; and &lt;math&gt;EG=FH = 34&lt;/math&gt;. Segments &lt;math&gt;\overline{EG}&lt;/math&gt; and &lt;math&gt;\overline{FH}&lt;/math&gt; intersect at a point &lt;math&gt;P&lt;/math&gt;, and the areas of the quadrilaterals &lt;math&gt;AEPH, BFPE, CGPF,&lt;/math&gt; and &lt;math&gt;DHPG&lt;/math&gt; are in the ratio &lt;math&gt;269:275:405:411.&lt;/math&gt; Find the area of square &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> pair A = (0,sqrt(850));<br /> pair B = (0,0);<br /> pair C = (sqrt(850),0);<br /> pair D = (sqrt(850),sqrt(850));<br /> draw(A--B--C--D--cycle);<br /> dotfactor = 3;<br /> dot(&quot;$A$&quot;,A,dir(135));<br /> dot(&quot;$B$&quot;,B,dir(215));<br /> dot(&quot;$C$&quot;,C,dir(305));<br /> dot(&quot;$D$&quot;,D,dir(45));<br /> pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));<br /> pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);<br /> dot(&quot;$H$&quot;,H,dir(90));<br /> dot(&quot;$F$&quot;,F,dir(270));<br /> draw(H--F);<br /> pair E = (0,(sqrt(850)-6)/2);<br /> pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);<br /> dot(&quot;$E$&quot;,E,dir(180));<br /> dot(&quot;$G$&quot;,G,dir(0));<br /> draw(E--G);<br /> pair P = extension(H,F,E,G);<br /> dot(&quot;$P$&quot;,P,dir(60));<br /> label(&quot;$w$&quot;, intersectionpoint( A--P, E--H ));<br /> label(&quot;$x$&quot;, intersectionpoint( B--P, E--F ));<br /> label(&quot;$y$&quot;, intersectionpoint( C--P, G--F ));<br /> label(&quot;$z$&quot;, intersectionpoint( D--P, G--H ));&lt;/asy&gt;<br /> <br /> [[2014 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;m&lt;/math&gt; be the largest real solution to the equation<br /> <br /> &lt;cmath&gt; \dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4&lt;/cmath&gt;<br /> <br /> There are positive integers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; such that &lt;math&gt;m = a + \sqrt{b + \sqrt{c}}&lt;/math&gt;. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> In &lt;math&gt;\triangle ABC, AB = 3, BC = 4,&lt;/math&gt; and &lt;math&gt;CA = 5&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B, \overline{BC}&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;. Given that &lt;math&gt;EF=DF&lt;/math&gt; and &lt;math&gt;\frac{DG}{EG} = \frac{3}{4},&lt;/math&gt; length &lt;math&gt;DE=\frac{a\sqrt{b}}{c},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 15|Solution]]<br /> <br /> {{MAA Notice}}</div> Rahulkk https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_8&diff=71035 2015 AIME II Problems/Problem 8 2015-07-08T02:45:18Z <p>Rahulkk: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be positive integers satisfying &lt;math&gt;\frac{ab+1}{a+b} &lt; \frac{3}{2}&lt;/math&gt;. The maximum possible value of &lt;math&gt;\frac{a^3b^3+1}{a^3+b^3}&lt;/math&gt; is &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 /> Let us call the quantity &lt;math&gt;\frac{a^3b^3+1}{a^3+b^3}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; for convenience. Knowing that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers, we can legitimately rearrange the given inequality so that &lt;math&gt;a&lt;/math&gt; is by itself, which makes it easier to determine the pairs of &lt;math&gt;(a, b)&lt;/math&gt; that work. Doing so, we have &lt;cmath&gt;\frac{ab+1}{a+b} &lt; \frac{3}{2}&lt;/cmath&gt; &lt;cmath&gt;\implies 2ab + 2 &lt; 3a + 3b \implies 2ab - 3a &lt; 3b - 2&lt;/cmath&gt; &lt;cmath&gt;\implies a &lt; \frac{3b - 2}{2b - 3}.&lt;/cmath&gt; Now, observe that if &lt;math&gt;b = 1&lt;/math&gt; we have that &lt;math&gt;N = \frac{a + 1}{a + 1} = 1&lt;/math&gt;, regardless of the value of &lt;math&gt;a&lt;/math&gt;. If &lt;math&gt;a = 1&lt;/math&gt;, we have the same result: that &lt;math&gt;N = \frac{b + 1}{b + 1} = 1&lt;/math&gt;, regardless of the value of &lt;math&gt;b&lt;/math&gt;. Hence, we want to find pairs of positive integers &lt;math&gt;(a, b)&lt;/math&gt; existing such that neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; is equal to &lt;math&gt;1&lt;/math&gt;, and that the conditions given in the problem are satisfied in order to check that the maximum value for &lt;math&gt;N&lt;/math&gt; is not &lt;math&gt;1&lt;/math&gt;.<br /> <br /> <br /> To avoid the possibility that &lt;math&gt;a = 1&lt;/math&gt;, we want to find values of &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\frac{3b - 2}{2b - 3} &gt; 2&lt;/math&gt;. If we do this, we will have that &lt;math&gt;a &lt; \frac{3b - 2}{2b - 3} = k&lt;/math&gt;, where &lt;math&gt;k&lt;/math&gt; is greater than &lt;math&gt;2&lt;/math&gt;, and this allows us to choose values of &lt;math&gt;a&lt;/math&gt; greater than &lt;math&gt;1&lt;/math&gt;. Again, since &lt;math&gt;b&lt;/math&gt; is a positive integer, and we want &lt;math&gt;b &gt; 1&lt;/math&gt;, we can legitimately multiply both sides of &lt;math&gt;\frac{3b - 2}{2b - 3} &gt; 2&lt;/math&gt; by &lt;math&gt;2b - 3&lt;/math&gt; to get &lt;math&gt;3b - 2 &gt; 4b - 6 \implies b &lt; 4&lt;/math&gt;. For &lt;math&gt;b = 3&lt;/math&gt;, we have that &lt;math&gt;a &lt; \frac{7}{3}&lt;/math&gt;, so the only possibility for &lt;math&gt;a&lt;/math&gt; greater than &lt;math&gt;1&lt;/math&gt; is obviously &lt;math&gt;2&lt;/math&gt;. Plugging these values into &lt;math&gt;N&lt;/math&gt;, we have that &lt;math&gt;N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}&lt;/math&gt;. For &lt;math&gt;b = 2&lt;/math&gt;, we have that &lt;math&gt;a &lt; \frac{4}{1} = 4&lt;/math&gt;. Plugging &lt;math&gt;a = 3&lt;/math&gt; and &lt;math&gt;b = 2&lt;/math&gt; in for &lt;math&gt;N&lt;/math&gt; yields the same result of &lt;math&gt;\frac{31}{5}&lt;/math&gt;, but plugging &lt;math&gt;a = 2&lt;/math&gt; and &lt;math&gt;b = 2&lt;/math&gt; into &lt;math&gt;N&lt;/math&gt; yields that &lt;math&gt;N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}&lt;/math&gt;. Clearly, &lt;math&gt;\frac{31}{5}&lt;/math&gt; is the largest value we can have for &lt;math&gt;N&lt;/math&gt;, so our answer is &lt;math&gt;31 + 5 = \boxed{036}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Proof without words)==<br /> &lt;cmath&gt;\frac{ab + 1}{a + b} &lt; \frac{3}{2} \rightarrow 2ab + 2 &lt; 3a + 3b,&lt;/cmath&gt;<br /> &lt;cmath&gt;\rightarrow 4ab - 6a - 6b + 4 &lt; 0 \rightarrow (2a - 3)(2b - 3) &lt; 5.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow&lt;/cmath&gt;<br /> &lt;cmath&gt;(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).&lt;/cmath&gt;<br /> &lt;cmath&gt;(a, b) = (2, 2), (2, 3), (3, 2).&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{31}{5} \rightarrow \boxed{036}.&lt;/cmath&gt;<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Rahulkk