https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pleaseletmewin&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-04T01:43:28Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_2&diff=134549 2013 AIME I Problems/Problem 2 2020-10-03T06:30:40Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>== Problem 2 ==<br /> Find the number of five-digit positive integers, &lt;math&gt;n&lt;/math&gt;, that satisfy the following conditions:<br /> <br /> &lt;UL&gt;<br /> (a) the number &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;5,&lt;/math&gt;<br /> &lt;/UL&gt;<br /> <br /> &lt;UL&gt;<br /> (b) the first and last digits of &lt;math&gt;n&lt;/math&gt; are equal, and<br /> &lt;/UL&gt;<br /> <br /> &lt;UL&gt;<br /> (c) the sum of the digits of &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;5.&lt;/math&gt;<br /> &lt;/UL&gt;<br /> <br /> <br /> == Solution==<br /> The number takes a form of &lt;math&gt;5\text{x,y,z}5&lt;/math&gt;, in which &lt;math&gt;5|x+y+z&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be arbitrary digits. For each pair of &lt;math&gt;x,y&lt;/math&gt;, there are exactly two values of &lt;math&gt;z&lt;/math&gt; that satisfy the condition of &lt;math&gt;5|x+y+z&lt;/math&gt;. Therefore, the answer is &lt;math&gt;10\times10\times2=\boxed{200}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=kz3ZX4PT-_0<br /> ~Shreyas S<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_2&diff=134548 2013 AIME I Problems/Problem 2 2020-10-03T06:30:20Z <p>Pleaseletmewin: </p> <hr /> <div>== Problem 2 ==<br /> Find the number of five-digit positive integers, &lt;math&gt;n&lt;/math&gt;, that satisfy the following conditions:<br /> <br /> &lt;UL&gt;<br /> (a) the number &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;5,&lt;/math&gt;<br /> &lt;/UL&gt;<br /> <br /> &lt;UL&gt;<br /> (b) the first and last digits of &lt;math&gt;n&lt;/math&gt; are equal, and<br /> &lt;/UL&gt;<br /> <br /> &lt;UL&gt;<br /> (c) the sum of the digits of &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;5.&lt;/math&gt;<br /> &lt;/UL&gt;<br /> <br /> <br /> == Solution==<br /> The number takes a form of &lt;math&gt;5\text{x,y,z}5&lt;/math&gt;, in which &lt;math&gt;5|x+y+z&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be arbitrary digits. For each pair of &lt;math&gt;x,y&lt;/math&gt;, there are exactly two values of &lt;math&gt;z&lt;/math&gt; that satisfy the condition of &lt;math&gt;5|x+y+z&lt;/math&gt;. Therefore, the answer is &lt;math&gt;10\times10\times2=\boxed{200}&lt;/math&gt;<br /> <br /> This 9-line code in Python also gives the answer too.<br /> import math<br /> counter=0<br /> for integer in range(10000,99999):<br /> if str(integer) != '5' or str(integer) != '5':<br /> counter=counter<br /> else:<br /> if math.remainder(int(str(integer)+str(integer)+str(integer)),5)==0:<br /> counter+=1<br /> print(counter)<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=kz3ZX4PT-_0<br /> ~Shreyas S<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems&diff=133704 2011 AIME II Problems 2020-09-16T06:41:26Z <p>Pleaseletmewin: /* Problem 15 */</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=20&lt;/math&gt; and &lt;math&gt;AC=11&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 &lt;math&gt;2011&lt;/math&gt; terms of a geometric sequence is &lt;math&gt;200&lt;/math&gt;. The sum of the first &lt;math&gt;4022&lt;/math&gt; terms is &lt;math&gt;380&lt;/math&gt;. Find the sum of the first &lt;math&gt;6033&lt;/math&gt; terms. <br /> <br /> [[2011 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Define an ordered quadruple of integers &lt;math&gt;(a, b, c, d)&lt;/math&gt; 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,z_2,z_3,\dots,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;i z_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 relatively prime positive 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 /> <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 ^{\circ}&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;\left\lfloor\sqrt{P(x)}\right\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 /> <br /> {{AIME box|year=2011|n=II|before=[[2011 AIME I Problems]]|after=[[2012 AIME I Problems]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_9&diff=132695 2005 AIME II Problems/Problem 9 2020-08-29T23:08:27Z <p>Pleaseletmewin: </p> <hr /> <div>== Problem ==<br /> For how many positive integers &lt;math&gt; n &lt;/math&gt; less than or equal to &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt; (\sin t + i \cos t)^n = \sin nt + i \cos nt &lt;/math&gt; true for all real &lt;math&gt; t &lt;/math&gt;?<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> We know by [[De Moivre's Theorem]] that &lt;math&gt;(\cos t + i \sin t)^n = \cos nt + i \sin nt&lt;/math&gt; for all [[real number]]s &lt;math&gt;t&lt;/math&gt; and all [[integer]]s &lt;math&gt;n&lt;/math&gt;. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. <br /> <br /> Recall the [[trigonometric identities]] &lt;math&gt;\cos \left(\frac{\pi}2 - u\right) = \sin u&lt;/math&gt; and &lt;math&gt;\sin \left(\frac{\pi}2 - u\right) = \cos u&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;. If our original equation holds for all &lt;math&gt;t&lt;/math&gt;, it must certainly hold for &lt;math&gt;t = \frac{\pi}2 - u&lt;/math&gt;. Thus, the question is equivalent to asking for how many [[positive integer]]s &lt;math&gt;n \leq 1000&lt;/math&gt; we have that &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)&lt;/math&gt; holds for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu&lt;/math&gt;. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\cos n u = \sin n\left(\frac\pi2 - u\right)&lt;/math&gt; and &lt;math&gt;\sin nu = \cos n\left(\frac\pi2 - u\right)&lt;/math&gt; hold for all real &lt;math&gt;u&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sin x = \cos y&lt;/math&gt; if and only if either &lt;math&gt;x + y = \frac \pi 2 + 2\pi \cdot k&lt;/math&gt; or &lt;math&gt;x - y = \frac\pi2 + 2\pi\cdot k&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt;. So from the equality of the real parts we need either &lt;math&gt;nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n = 1 + 4k&lt;/math&gt;, or we need &lt;math&gt;-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k&lt;/math&gt;, in which case &lt;math&gt;n&lt;/math&gt; will depend on &lt;math&gt;u&lt;/math&gt; and so the equation will not hold for all real values of &lt;math&gt;u&lt;/math&gt;. Checking &lt;math&gt;n = 1 + 4k&lt;/math&gt; in the equation for the imaginary parts, we see that it works there as well, so exactly those values of &lt;math&gt;n&lt;/math&gt; congruent to &lt;math&gt;1 \pmod 4&lt;/math&gt; work. There are &lt;math&gt;\boxed{250}&lt;/math&gt; of them in the given range.<br /> <br /> === Solution 2 ===<br /> This problem begs us to use the familiar identity &lt;math&gt;e^{it} = \cos(t) + i \sin(t)&lt;/math&gt;. Notice, &lt;math&gt;\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}&lt;/math&gt; since &lt;math&gt;\sin(-t) = -\sin(t)&lt;/math&gt;. Using this, &lt;math&gt;(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)&lt;/math&gt; is recast as &lt;math&gt;(i e^{-it})^n = i e^{-itn}&lt;/math&gt;. Hence we must have &lt;math&gt;i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}&lt;/math&gt;. Thus since &lt;math&gt;1000&lt;/math&gt; is a multiple of &lt;math&gt;4&lt;/math&gt; exactly one quarter of the residues are congruent to &lt;math&gt;1&lt;/math&gt; hence we have &lt;math&gt;\boxed{250}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can rewrite &lt;math&gt;\sin(t)&lt;/math&gt; as &lt;math&gt;\cos\left(\frac{\pi}{2}-t\right)&lt;/math&gt; and &lt;math&gt;\cos(t)&lt;/math&gt; as &lt;math&gt;\sin\left(\frac{\pi}{2}-t\right)&lt;/math&gt;. This means that &lt;math&gt;\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}&lt;/math&gt;. This theorem also tells us that &lt;math&gt;e^{\frac{\pi i}{2}}=i&lt;/math&gt;, so &lt;math&gt;\sin t + i\cos t = \frac{i}{e^{it}}&lt;/math&gt;. By the same line of reasoning, we have &lt;math&gt;\sin nt + i\cos nt = \frac{i}{e^{int}}&lt;/math&gt;.<br /> <br /> For the statement in the question to be true, we must have &lt;math&gt;\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}&lt;/math&gt;. The left hand side simplifies to &lt;math&gt;\frac{i^n}{e^{int}}&lt;/math&gt;. We cancel the denominators and find that the only thing that needs to be true is that &lt;math&gt;i^n=i&lt;/math&gt;. This is true if &lt;math&gt;n\equiv1\pmod{4}&lt;/math&gt;, and there are &lt;math&gt;\boxed{250}&lt;/math&gt; such numbers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1000&lt;/math&gt;. Solution by Zeroman<br /> <br /> === Solution 4 ===<br /> We are using degrees in this solution instead of radians. I just process stuff better that way.<br /> <br /> We can see that the LHS is &lt;math&gt;cis(n(90^{\circ}-t))&lt;/math&gt;, and the RHS is &lt;math&gt;cis(90^{\circ}-nt)&lt;/math&gt; So, &lt;math&gt;n(90-t) \equiv 90-nt \mod 360&lt;/math&gt; Expanding and canceling the nt terms, we will get &lt;math&gt;90n \equiv 90 \mod 360&lt;/math&gt;. Canceling gets &lt;math&gt;n \equiv 1 \mod 4&lt;/math&gt;, and thus there are &lt;math&gt;\boxed{250}&lt;/math&gt; values of n.<br /> <br /> -AlexLikeMath<br /> <br /> === Solution 5(CHEAP) ===<br /> Let &lt;math&gt;t=0&lt;/math&gt;. Then, we have &lt;math&gt;i^n=i&lt;/math&gt; which means &lt;math&gt;n\equiv 1\pmod{4}&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{250}&lt;/math&gt;.<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> <br /> {{MAA Notice}}<br /> <br /> ==See Also==<br /> {{AIME box|year=2005|n=II|num-b=8|num-a=10}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_23&diff=132688 1993 AHSME Problems/Problem 23 2020-08-29T21:00:09Z <p>Pleaseletmewin: /* Problem */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> draw(circle((0,0),10),black+linewidth(.75));<br /> draw((-10,0)--(10,0),black+linewidth(.75));<br /> draw((-10,0)--(9,sqrt(19)),black+linewidth(.75));<br /> draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75));<br /> draw((2,0)--(9,sqrt(19)),black+linewidth(.75));<br /> draw((2,0)--(9,-sqrt(19)),black+linewidth(.75));<br /> MP(&quot;X&quot;,(2,0),N);MP(&quot;A&quot;,(-10,0),W);MP(&quot;D&quot;,(10,0),E);MP(&quot;B&quot;,(9,sqrt(19)),E);MP(&quot;C&quot;,(9,-sqrt(19)),E);<br /> &lt;/asy&gt;<br /> <br /> Points &lt;math&gt;A,B,C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are on a circle of diameter &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is on diameter &lt;math&gt;\overline{AD}.&lt;/math&gt;<br /> <br /> If &lt;math&gt;BX=CX&lt;/math&gt; and &lt;math&gt;3\angle{BAC}=\angle{BXC}=36^\circ&lt;/math&gt;, then &lt;math&gt;AX=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A) } \cos(6^\circ)\cos(12^\circ)\sec(18^\circ)\quad\\<br /> \text{(B) } \cos(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\<br /> \text{(C) } \cos(6^\circ)\sin(12^\circ)\sec(18^\circ)\quad\\<br /> \text{(D) } \sin(6^\circ)\sin(12^\circ)\csc(18^\circ)\quad\\<br /> \text{(E) } \sin(6^\circ)\sin(12^\circ)\sec(18^\circ)&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We have all the angles we need, but most obviously, we see that right angle in triangle &lt;math&gt;ABD&lt;/math&gt;.<br /> <br /> Note also that angle &lt;math&gt;BAD&lt;/math&gt; is 6 degrees, so length &lt;math&gt;AB = cos(6)&lt;/math&gt; because the diameter, &lt;math&gt;AD&lt;/math&gt;, is 1.<br /> <br /> Now, we can concentrate on triangle &lt;math&gt;ABX&lt;/math&gt; (after all, now we can decipher all angles easily and use Law of Sines).<br /> <br /> We get:<br /> <br /> &lt;math&gt;\frac{AB}{\sin(\angle{AXB})} =\frac{AX}{\sin(\angle{ABX})}&lt;/math&gt;<br /> <br /> That's equal to<br /> <br /> &lt;math&gt;\frac{\cos(6)}{\sin(180-18)} =\frac{AX}{\sin(12)}&lt;/math&gt;<br /> <br /> Therefore, our answer is equal to:<br /> &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\sin(162) = \sin(18)&lt;/math&gt;, and don't accidentally put &lt;math&gt;\fbox{C}&lt;/math&gt; because you thought &lt;math&gt;\frac{1}{\sin}&lt;/math&gt; was &lt;math&gt;\sec&lt;/math&gt;!<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=22|num-a=24}} <br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131177 1989 AIME Problems/Problem 8 2020-08-09T00:01:50Z <p>Pleaseletmewin: /* Solution 5(Very Cheap)(Not advised) */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have <br /> <br /> &lt;math&gt;\begin{cases} x_1+4x_2+9x_3=1\\<br /> 4x_1+9x_2+16x_3=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{cases}&lt;/math&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer. <br /> <br /> -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131176 1989 AIME Problems/Problem 8 2020-08-09T00:01:35Z <p>Pleaseletmewin: /* Solution 5(Very Cheap)(Not advised) */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have <br /> <br /> &lt;math&gt;\begin{cases} x_1+4x_2+9x_3=1\\<br /> 4x_1+9x_2+16x_3=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{cases}&lt;/math&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer. \\ -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131175 1989 AIME Problems/Problem 8 2020-08-09T00:01:21Z <p>Pleaseletmewin: /* Solution 5(Very Cheap)(Not advised) */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have &lt;math&gt;\\ \begin{cases} x_1+4x_2+9x_3=1\\<br /> 4x_1+9x_2+16x_3=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{cases}&lt;/math&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer. \\ -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131174 1989 AIME Problems/Problem 8 2020-08-09T00:01:04Z <p>Pleaseletmewin: /* Solution 5(Very Cheap)(Not advised) */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have \\ &lt;math&gt;\begin{cases} x_1+4x_2+9x_3=1\\<br /> 4x_1+9x_2+16x_3=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{cases}&lt;/math&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer. \\ -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131173 1989 AIME Problems/Problem 8 2020-08-09T00:00:47Z <p>Pleaseletmewin: /* Solution 5(Very Cheap)(Not advised) */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have &lt;math&gt;\begin{cases} x_1+4x_2+9x_3=1\\<br /> 4x_1+9x_2+16x_3=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{cases}&lt;/math&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer.<br /> -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=131172 1989 AIME Problems/Problem 8 2020-08-09T00:00:22Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>== Problem ==<br /> Assume that &lt;math&gt;x_1,x_2,\ldots,x_7&lt;/math&gt; are real numbers such that<br /> &lt;cmath&gt;\begin{align*}x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123.\end{align*}&lt;/cmath&gt;<br /> <br /> Find the value of &lt;math&gt;16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1===<br /> Notice that because we are given a system of &lt;math&gt;3&lt;/math&gt; equations with &lt;math&gt;7&lt;/math&gt; unknowns, the values &lt;math&gt;(x_1, x_2, \ldots, x_7)&lt;/math&gt; are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three. <br /> <br /> <br /> Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of &lt;math&gt;x_i&lt;/math&gt; in the first equation be &lt;math&gt;y_i^2&lt;/math&gt;; then its coefficients in the second equation is &lt;math&gt;(y_i+1)^{2}&lt;/math&gt; and the third as &lt;math&gt;(y_i+2)^2&lt;/math&gt;. We need to find a way to sum these to make &lt;math&gt;(y_i+3)^2&lt;/math&gt; [this is in fact a specific approach generalized by the next solution below]. <br /> <br /> Thus, we hope to find constants &lt;math&gt;a,b,c&lt;/math&gt; satisfying &lt;math&gt;ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2&lt;/math&gt;. [[FOIL]]ing out all of the terms, we get <br /> <br /> &lt;center&gt;&lt;math&gt;[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.&lt;/math&gt;&lt;/center&gt; <br /> <br /> Comparing coefficents gives us the three equation system:<br /> <br /> &lt;center&gt;<br /> &lt;cmath&gt;\begin{align*}a + b + c &amp;= 1 \\ 2b + 4c &amp;= 6 \\ b + 4c &amp;= 9 \end{align*}&lt;/cmath&gt;<br /> &lt;/center&gt; <br /> <br /> Subtracting the second and third equations yields that &lt;math&gt;b = -3&lt;/math&gt;, so &lt;math&gt;c = 3&lt;/math&gt; and &lt;math&gt;a = 1&lt;/math&gt;. It follows that the desired expression is &lt;math&gt;a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> Notice that we may rewrite the equations in the more compact form as:<br /> <br /> &lt;math&gt;\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,&lt;/math&gt; and &lt;math&gt;\sum_{i=1}^{7}(i+3)^2x_i=c_4,&lt;/math&gt;<br /> <br /> where &lt;math&gt;c_1=1, c_2=12, c_3=123,&lt;/math&gt; and &lt;math&gt;c_4&lt;/math&gt; is what we're trying to find.<br /> <br /> Now consider the polynomial given by &lt;math&gt; f(z) := \sum_{i=1}^7 (z+i)^2x_i &lt;/math&gt; (we are only treating the &lt;math&gt;x_i&lt;/math&gt; as coefficients).<br /> <br /> Notice that &lt;math&gt;f&lt;/math&gt; is in fact a quadratic. We are given &lt;math&gt;f(0), \ f(1), \ f(2)&lt;/math&gt; as &lt;math&gt;c_1, \ c_2, \ c_3&lt;/math&gt; and are asked to find &lt;math&gt;c_4&lt;/math&gt;. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find &lt;math&gt;c_4=334&lt;/math&gt;. <br /> <br /> <br /> Alternatively, applying finite differences, one obtains &lt;math&gt;c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Notice that &lt;math&gt;3(n+2)^2-3(n+1)^2+n^2=(n+3)^2&lt;/math&gt;<br /> <br /> I'll number the equations for convenience <br /> <br /> &lt;cmath&gt;<br /> \begin{align}<br /> x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&amp;=1\\<br /> 4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&amp;=12\\<br /> 9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&amp;=123\\ <br /> 16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7&amp;=\end{align}&lt;/cmath&gt;<br /> <br /> Let the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(1)&lt;/math&gt; be &lt;math&gt;n^2&lt;/math&gt;. Then the coefficient of &lt;math&gt;x_i&lt;/math&gt; in &lt;math&gt;(2)&lt;/math&gt; is &lt;math&gt;(n+1)^2&lt;/math&gt; etc.<br /> <br /> Therefore, &lt;math&gt;3*(3)-3*(2)+(1)=(4)&lt;/math&gt;<br /> <br /> So &lt;math&gt;(4)=3*123-3*12+1=\boxed{334}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Notice subtracting the first equation from the second yields &lt;math&gt;3x_1 + 5x_2 + ... + 15x_7 = 11&lt;/math&gt;. Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get &lt;math&gt;2x_1 + 2x_2 + ... +2x_7 = 100&lt;/math&gt;. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get &lt;math&gt;\boxed{334}.&lt;/math&gt;<br /> <br /> ==Solution 5(Very Cheap)(Not advised)==<br /> We let &lt;math&gt;(x_4,x_5,x_6,x_7)=(0,0,0,0)&lt;/math&gt;. Thus, we have &lt;cmath&gt;\begin{align} x_1+4x_2+9x_3&amp;=1\\<br /> 4x_1+9x_2+16x_3&amp;=12\\<br /> 9x_1+16x_2+25x_3=123\\ \end{align}&lt;/cmath&gt;<br /> <br /> Grinding this out, we have &lt;math&gt;(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)&lt;/math&gt; which gives &lt;math&gt;\boxed{334}&lt;/math&gt; as our final answer.<br /> -Pleaseletmewin<br /> <br /> ===Video Solution===<br /> https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=7|num-a=9}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_11&diff=130943 2014 AIME II Problems/Problem 11 2020-08-07T05:38:03Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>==Problem 11==<br /> In &lt;math&gt;\triangle RED&lt;/math&gt;, &lt;math&gt;\measuredangle DRE=75^{\circ}&lt;/math&gt; and &lt;math&gt;\measuredangle RED=45^{\circ}&lt;/math&gt;. &lt;math&gt;RD=1&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;\overline{RD}&lt;/math&gt;. Point &lt;math&gt;C&lt;/math&gt; lies on side &lt;math&gt;\overline{ED}&lt;/math&gt; such that &lt;math&gt;\overline{RC}\perp\overline{EM}&lt;/math&gt;. Extend segment &lt;math&gt;\overline{DE}&lt;/math&gt; through &lt;math&gt;E&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt; such that &lt;math&gt;CA=AR&lt;/math&gt;. Then &lt;math&gt;AE=\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. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;P&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;\overline{CR}&lt;/math&gt;, so &lt;math&gt;\overline{AP}\parallel\overline{EM}&lt;/math&gt;. Since triangle &lt;math&gt;ARC&lt;/math&gt; is isosceles, &lt;math&gt;P&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{CR}&lt;/math&gt;, and &lt;math&gt;\overline{PM}\parallel\overline{CD}&lt;/math&gt;. Thus, &lt;math&gt;APME&lt;/math&gt; is a parallelogram and &lt;math&gt;AE = PM = \frac{CD}{2}&lt;/math&gt;. We can then use coordinates. Let &lt;math&gt;O&lt;/math&gt; be the foot of altitude &lt;math&gt;RO&lt;/math&gt; and set &lt;math&gt;O&lt;/math&gt; as the origin. Now we notice special right triangles! In particular, &lt;math&gt;DO = \frac{1}{2}&lt;/math&gt; and &lt;math&gt;EO = RO = \frac{\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;D\left(\frac{1}{2}, 0\right)&lt;/math&gt;, &lt;math&gt;E\left(-\frac{\sqrt{3}}{2}, 0\right)&lt;/math&gt;, and &lt;math&gt;R\left(0, \frac{\sqrt{3}}{2}\right).&lt;/math&gt; &lt;math&gt;M =&lt;/math&gt; midpoint&lt;math&gt;(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)&lt;/math&gt; and the slope of &lt;math&gt;ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}&lt;/math&gt;, so the slope of &lt;math&gt;RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.&lt;/math&gt; Instead of finding the equation of the line, we use the definition of slope: for every &lt;math&gt;CO = x&lt;/math&gt; to the left, we go &lt;math&gt;\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}&lt;/math&gt; up. Thus, &lt;math&gt;x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.&lt;/math&gt; &lt;math&gt;DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}&lt;/math&gt;, and &lt;math&gt;AE = \frac{7 - \sqrt{27}}{22}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(8cm);<br /> pair a, o, d, r, e, m, cm, c,p;<br /> o =(0,0);<br /> d = (0.5, 0);<br /> r = (0,sqrt(3)/2);<br /> e = (-sqrt(3)/2,0);<br /> <br /> m = midpoint(d--r);<br /> draw(e--m);<br /> cm = foot(r, e, m);<br /> draw(L(r, cm,1, 1));<br /> c = IP(L(r, cm, 1, 1), e--d);<br /> clip(r--d--e--cycle);<br /> draw(r--d--e--cycle);<br /> draw(rightanglemark(e, cm, c, 1.5));<br /> a = -(4sqrt(3)+9)/11+0.5;<br /> dot(a);<br /> draw(a--r, dashed);<br /> draw(a--c, dashed);<br /> pair[] PPAP = {a, o, d, r, e, m, c};<br /> for(int i = 0; i&lt;7; ++i) {<br /> dot(PPAP[i]);<br /> }<br /> label(&quot;$A$&quot;, a, W);<br /> label(&quot;$E$&quot;, e, SW);<br /> label(&quot;$C$&quot;, c, S);<br /> label(&quot;$O$&quot;, o, S);<br /> label(&quot;$D$&quot;, d, SE);<br /> label(&quot;$M$&quot;, m, NE);<br /> label(&quot;$R$&quot;, r, N);<br /> p = foot(a, r, c);<br /> label(&quot;$P$&quot;, p, NE);<br /> draw(p--m, dashed);<br /> draw(a--p, dashed);<br /> dot(p);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;MP = x.&lt;/math&gt; Meanwhile, because &lt;math&gt;\triangle RPM&lt;/math&gt; is similar to &lt;math&gt;\triangle RCD&lt;/math&gt; (angle, side, and side- &lt;math&gt;RP&lt;/math&gt; and &lt;math&gt;RC&lt;/math&gt; ratio), &lt;math&gt;CD&lt;/math&gt; must be 2&lt;math&gt;x&lt;/math&gt;. Now, notice that &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt;, because of the parallel segments &lt;math&gt;\overline A\overline E&lt;/math&gt; and &lt;math&gt;\overline P\overline M&lt;/math&gt;.<br /> <br /> Now we just have to calculate &lt;math&gt;ED&lt;/math&gt;. Using the Law of Sines, or perhaps using altitude &lt;math&gt;\overline R\overline O&lt;/math&gt;, we get &lt;math&gt;ED = \frac{\sqrt{3}+1}{2}&lt;/math&gt;. &lt;math&gt;CA=RA&lt;/math&gt;, which equals &lt;math&gt;ED - x&lt;/math&gt;<br /> <br /> Finally, what is &lt;math&gt;RE&lt;/math&gt;? It comes out to &lt;math&gt;\frac{\sqrt{6}}{2}&lt;/math&gt;. <br /> <br /> We got the three sides. Now all that is left is using the Law of Cosines. There we can equate &lt;math&gt;x&lt;/math&gt; and solve for it.<br /> <br /> Taking &lt;math&gt;\triangle AER&lt;/math&gt; and using &lt;math&gt;\angle AER&lt;/math&gt;, of course, we find out (after some calculation) that &lt;math&gt;AE = \frac{7 - \sqrt{27}}{22}&lt;/math&gt;. The step before? &lt;math&gt;x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems&diff=130942 2016 AIME II Problems 2020-08-07T05:33:35Z <p>Pleaseletmewin: /* Problem 13 */</p> <hr /> <div>{{AIME Problems|year=2016|n=II}}<br /> ==Problem 1==<br /> Initially Alex, Betty, and Charlie had a total of &lt;math&gt;444&lt;/math&gt; peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats &lt;math&gt;5&lt;/math&gt; of his peanuts, Betty eats &lt;math&gt;9&lt;/math&gt; of her peanuts, and Charlie eats &lt;math&gt;25&lt;/math&gt; of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.<br /> <br /> [[2016 AIME II Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> There is a &lt;math&gt;40\%&lt;/math&gt; chance of rain on Saturday and a &lt;math&gt;30\%&lt;/math&gt; chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;x,y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be real numbers satisfying the system<br /> &lt;cmath&gt;\log_2(xyz-3+\log_5 x)=5,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_3(xyz-3+\log_5 y)=4,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_4(xyz-3+\log_5 z)=4.&lt;/cmath&gt;<br /> Find the value of &lt;math&gt;|\log_5 x|+|\log_5 y|+|\log_5 z|&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> An &lt;math&gt;a \times b \times c&lt;/math&gt; rectangular box is built from &lt;math&gt;a \cdot b \cdot c&lt;/math&gt; unit cubes. Each unit cube is colored red, green, or yellow. Each of the &lt;math&gt;a&lt;/math&gt; layers of size &lt;math&gt;1 \times b \times c&lt;/math&gt; parallel to the &lt;math&gt;(b \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;9&lt;/math&gt; red cubes, exactly &lt;math&gt;12&lt;/math&gt; green cubes, and some yellow cubes. Each of the &lt;math&gt;b&lt;/math&gt; layers of size &lt;math&gt;a \times 1 \times c&lt;/math&gt; parallel to the &lt;math&gt;(a \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;20&lt;/math&gt; green cubes, exactly &lt;math&gt;25&lt;/math&gt; yellow cubes, and some red cubes. Find the smallest possible volume of the box.<br /> <br /> [[2016 AIME II Problems/Problem 4 | Solution]]<br /> ==Problem 5==<br /> Triangle &lt;math&gt;ABC_0&lt;/math&gt; has a right angle at &lt;math&gt;C_0&lt;/math&gt;. Its side lengths are pairwise relatively prime positive integers, and its perimeter is &lt;math&gt;p&lt;/math&gt;. Let &lt;math&gt;C_1&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;, and for &lt;math&gt;n \geq 2&lt;/math&gt;, let &lt;math&gt;C_n&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{C_{n-2}B}&lt;/math&gt; in &lt;math&gt;\triangle C_{n-2}C_{n-1}B&lt;/math&gt;. The sum &lt;math&gt;\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p&lt;/math&gt;. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> For polynomial &lt;math&gt;P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}&lt;/math&gt;, define<br /> &lt;math&gt;Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}&lt;/math&gt;.<br /> Then &lt;math&gt;\sum_{i=0}^{50} |a_i|=\dfrac{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 /> [[2016 AIME II Problems/Problem 6 | Solution]]<br /> ==Problem 7==<br /> Squares &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;EFGH&lt;/math&gt; have a common center and &lt;math&gt;\overline{AB} || \overline{EF}&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is 2016, and the area of &lt;math&gt;EFGH&lt;/math&gt; is a smaller positive integer. Square &lt;math&gt;IJKL&lt;/math&gt; is constructed so that each of its vertices lies on a side of &lt;math&gt;ABCD&lt;/math&gt; and each vertex of &lt;math&gt;EFGH&lt;/math&gt; lies on a side of &lt;math&gt;IJKL&lt;/math&gt;. Find the difference between the largest and smallest positive integer values for the area of &lt;math&gt;IJKL&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of sets &lt;math&gt;\{a,b,c\}&lt;/math&gt; of three distinct positive integers with the property that the product of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is equal to the product of &lt;math&gt;11,21,31,41,51,&lt;/math&gt; and &lt;math&gt;61&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> The sequences of positive integers &lt;math&gt;1,a_2, a_3,...&lt;/math&gt; and &lt;math&gt;1,b_2, b_3,...&lt;/math&gt; are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let &lt;math&gt;c_n=a_n+b_n&lt;/math&gt;. There is an integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;c_{k-1}=100&lt;/math&gt; and &lt;math&gt;c_{k+1}=1000&lt;/math&gt;. Find &lt;math&gt;c_k&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 9 | Solution]]<br /> ==Problem 10==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are on side &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AP&lt;AQ&lt;/math&gt;. Rays &lt;math&gt;CP&lt;/math&gt; and &lt;math&gt;CQ&lt;/math&gt; meet &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (other than &lt;math&gt;C&lt;/math&gt;), respectively. If &lt;math&gt;AP=4,PQ=3,QB=6,BT=5,&lt;/math&gt; and &lt;math&gt;AS=7&lt;/math&gt;, then &lt;math&gt;ST=\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 /> [[2016 AIME II Problems/Problem 10 | Solution]]<br /> ==Problem 11==<br /> For positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, define &lt;math&gt;N&lt;/math&gt; to be &lt;math&gt;k&lt;/math&gt;-nice if there exists a positive integer &lt;math&gt;a&lt;/math&gt; such that &lt;math&gt;a^{k}&lt;/math&gt; has exactly &lt;math&gt;N&lt;/math&gt; positive divisors. Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that are neither &lt;math&gt;7&lt;/math&gt;-nice nor &lt;math&gt;8&lt;/math&gt;-nice.<br /> <br /> [[2016 AIME II Problems/Problem 11 | Solution]]<br /> ==Problem 12==<br /> The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0), 4));<br /> draw(Circle((0,0), 3));<br /> draw((0,4)--(0,3));<br /> draw((0,-4)--(0,-3));<br /> draw((-2.598, 1.5)--(-3.4641, 2));<br /> draw((-2.598, -1.5)--(-3.4641, -2));<br /> draw((2.598, -1.5)--(3.4641, -2));<br /> draw((2.598, 1.5)--(3.4641, 2));<br /> &lt;/asy&gt;<br /> <br /> [[2016 AIME II Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Beatrix is going to place six rooks on a &lt;math&gt;6 \times 6&lt;/math&gt; chessboard where both the rows and columns are labeled &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;; the rooks are placed so that no two rooks are in the same row or the same column. The &lt;math&gt;value&lt;/math&gt; of a square is the sum of its row number and column number. The &lt;math&gt;score&lt;/math&gt; of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations 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 /> [[2016 AIME II Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;600&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie outside the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; and are on opposite sides of the plane. Furthermore, &lt;math&gt;PA=PB=PC&lt;/math&gt;, and &lt;math&gt;QA=QB=QC&lt;/math&gt;, and the planes of &lt;math&gt;\triangle PAB&lt;/math&gt; and &lt;math&gt;\triangle QAB&lt;/math&gt; form a &lt;math&gt;120^{\circ}&lt;/math&gt; dihedral angle (the angle between the two planes). There is a point &lt;math&gt;O&lt;/math&gt; whose distance from each of &lt;math&gt;A,B,C,P,&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt;. Find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 14 | Solution]]<br /> ==Problem 15==<br /> For &lt;math&gt;1 \leq i \leq 215&lt;/math&gt; let &lt;math&gt;a_i = \dfrac{1}{2^{i}}&lt;/math&gt; and &lt;math&gt;a_{216} = \dfrac{1}{2^{215}}&lt;/math&gt;. Let &lt;math&gt;x_1, x_2, ..., x_{216}&lt;/math&gt; be positive real numbers such that &lt;math&gt;\sum_{i=1}^{216} x_i=1&lt;/math&gt; and &lt;math&gt;\sum_{1 \leq i &lt; j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}&lt;/math&gt;. The maximum possible value of &lt;math&gt;x_2=\dfrac{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 /> [[2016 AIME II Problems/Problem 15 | Solution]]<br /> <br /> <br /> {{AIME box|year=2016|n=II|before=[[2016 AIME I Problems]]|after=[[2017 AIME I Problems]]}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems&diff=130941 2016 AIME II Problems 2020-08-07T05:32:58Z <p>Pleaseletmewin: /* Problem 13 */</p> <hr /> <div>{{AIME Problems|year=2016|n=II}}<br /> ==Problem 1==<br /> Initially Alex, Betty, and Charlie had a total of &lt;math&gt;444&lt;/math&gt; peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats &lt;math&gt;5&lt;/math&gt; of his peanuts, Betty eats &lt;math&gt;9&lt;/math&gt; of her peanuts, and Charlie eats &lt;math&gt;25&lt;/math&gt; of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.<br /> <br /> [[2016 AIME II Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> There is a &lt;math&gt;40\%&lt;/math&gt; chance of rain on Saturday and a &lt;math&gt;30\%&lt;/math&gt; chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;x,y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be real numbers satisfying the system<br /> &lt;cmath&gt;\log_2(xyz-3+\log_5 x)=5,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_3(xyz-3+\log_5 y)=4,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_4(xyz-3+\log_5 z)=4.&lt;/cmath&gt;<br /> Find the value of &lt;math&gt;|\log_5 x|+|\log_5 y|+|\log_5 z|&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> An &lt;math&gt;a \times b \times c&lt;/math&gt; rectangular box is built from &lt;math&gt;a \cdot b \cdot c&lt;/math&gt; unit cubes. Each unit cube is colored red, green, or yellow. Each of the &lt;math&gt;a&lt;/math&gt; layers of size &lt;math&gt;1 \times b \times c&lt;/math&gt; parallel to the &lt;math&gt;(b \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;9&lt;/math&gt; red cubes, exactly &lt;math&gt;12&lt;/math&gt; green cubes, and some yellow cubes. Each of the &lt;math&gt;b&lt;/math&gt; layers of size &lt;math&gt;a \times 1 \times c&lt;/math&gt; parallel to the &lt;math&gt;(a \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;20&lt;/math&gt; green cubes, exactly &lt;math&gt;25&lt;/math&gt; yellow cubes, and some red cubes. Find the smallest possible volume of the box.<br /> <br /> [[2016 AIME II Problems/Problem 4 | Solution]]<br /> ==Problem 5==<br /> Triangle &lt;math&gt;ABC_0&lt;/math&gt; has a right angle at &lt;math&gt;C_0&lt;/math&gt;. Its side lengths are pairwise relatively prime positive integers, and its perimeter is &lt;math&gt;p&lt;/math&gt;. Let &lt;math&gt;C_1&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;, and for &lt;math&gt;n \geq 2&lt;/math&gt;, let &lt;math&gt;C_n&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{C_{n-2}B}&lt;/math&gt; in &lt;math&gt;\triangle C_{n-2}C_{n-1}B&lt;/math&gt;. The sum &lt;math&gt;\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p&lt;/math&gt;. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> For polynomial &lt;math&gt;P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}&lt;/math&gt;, define<br /> &lt;math&gt;Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}&lt;/math&gt;.<br /> Then &lt;math&gt;\sum_{i=0}^{50} |a_i|=\dfrac{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 /> [[2016 AIME II Problems/Problem 6 | Solution]]<br /> ==Problem 7==<br /> Squares &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;EFGH&lt;/math&gt; have a common center and &lt;math&gt;\overline{AB} || \overline{EF}&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is 2016, and the area of &lt;math&gt;EFGH&lt;/math&gt; is a smaller positive integer. Square &lt;math&gt;IJKL&lt;/math&gt; is constructed so that each of its vertices lies on a side of &lt;math&gt;ABCD&lt;/math&gt; and each vertex of &lt;math&gt;EFGH&lt;/math&gt; lies on a side of &lt;math&gt;IJKL&lt;/math&gt;. Find the difference between the largest and smallest positive integer values for the area of &lt;math&gt;IJKL&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of sets &lt;math&gt;\{a,b,c\}&lt;/math&gt; of three distinct positive integers with the property that the product of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is equal to the product of &lt;math&gt;11,21,31,41,51,&lt;/math&gt; and &lt;math&gt;61&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> The sequences of positive integers &lt;math&gt;1,a_2, a_3,...&lt;/math&gt; and &lt;math&gt;1,b_2, b_3,...&lt;/math&gt; are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let &lt;math&gt;c_n=a_n+b_n&lt;/math&gt;. There is an integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;c_{k-1}=100&lt;/math&gt; and &lt;math&gt;c_{k+1}=1000&lt;/math&gt;. Find &lt;math&gt;c_k&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 9 | Solution]]<br /> ==Problem 10==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are on side &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AP&lt;AQ&lt;/math&gt;. Rays &lt;math&gt;CP&lt;/math&gt; and &lt;math&gt;CQ&lt;/math&gt; meet &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (other than &lt;math&gt;C&lt;/math&gt;), respectively. If &lt;math&gt;AP=4,PQ=3,QB=6,BT=5,&lt;/math&gt; and &lt;math&gt;AS=7&lt;/math&gt;, then &lt;math&gt;ST=\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 /> [[2016 AIME II Problems/Problem 10 | Solution]]<br /> ==Problem 11==<br /> For positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, define &lt;math&gt;N&lt;/math&gt; to be &lt;math&gt;k&lt;/math&gt;-nice if there exists a positive integer &lt;math&gt;a&lt;/math&gt; such that &lt;math&gt;a^{k}&lt;/math&gt; has exactly &lt;math&gt;N&lt;/math&gt; positive divisors. Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that are neither &lt;math&gt;7&lt;/math&gt;-nice nor &lt;math&gt;8&lt;/math&gt;-nice.<br /> <br /> [[2016 AIME II Problems/Problem 11 | Solution]]<br /> ==Problem 12==<br /> The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0), 4));<br /> draw(Circle((0,0), 3));<br /> draw((0,4)--(0,3));<br /> draw((0,-4)--(0,-3));<br /> draw((-2.598, 1.5)--(-3.4641, 2));<br /> draw((-2.598, -1.5)--(-3.4641, -2));<br /> draw((2.598, -1.5)--(3.4641, -2));<br /> draw((2.598, 1.5)--(3.4641, 2));<br /> &lt;/asy&gt;<br /> <br /> [[2016 AIME II Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Beatrix is going to place six rooks on a &lt;math&gt;6 \times 6&lt;/math&gt; chessboard where both the rows and columns are labeled &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;; the rooks are placed so that no two rooks are in the same row or the same column. The [/i]value[i] of a square is the sum of its row number and column number. The [/i]score[i] of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations 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 /> [[2016 AIME II Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;600&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie outside the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; and are on opposite sides of the plane. Furthermore, &lt;math&gt;PA=PB=PC&lt;/math&gt;, and &lt;math&gt;QA=QB=QC&lt;/math&gt;, and the planes of &lt;math&gt;\triangle PAB&lt;/math&gt; and &lt;math&gt;\triangle QAB&lt;/math&gt; form a &lt;math&gt;120^{\circ}&lt;/math&gt; dihedral angle (the angle between the two planes). There is a point &lt;math&gt;O&lt;/math&gt; whose distance from each of &lt;math&gt;A,B,C,P,&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt;. Find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 14 | Solution]]<br /> ==Problem 15==<br /> For &lt;math&gt;1 \leq i \leq 215&lt;/math&gt; let &lt;math&gt;a_i = \dfrac{1}{2^{i}}&lt;/math&gt; and &lt;math&gt;a_{216} = \dfrac{1}{2^{215}}&lt;/math&gt;. Let &lt;math&gt;x_1, x_2, ..., x_{216}&lt;/math&gt; be positive real numbers such that &lt;math&gt;\sum_{i=1}^{216} x_i=1&lt;/math&gt; and &lt;math&gt;\sum_{1 \leq i &lt; j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}&lt;/math&gt;. The maximum possible value of &lt;math&gt;x_2=\dfrac{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 /> [[2016 AIME II Problems/Problem 15 | Solution]]<br /> <br /> <br /> {{AIME box|year=2016|n=II|before=[[2016 AIME I Problems]]|after=[[2017 AIME I Problems]]}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems&diff=130940 2016 AIME II Problems 2020-08-07T05:32:37Z <p>Pleaseletmewin: /* Problem 13 */</p> <hr /> <div>{{AIME Problems|year=2016|n=II}}<br /> ==Problem 1==<br /> Initially Alex, Betty, and Charlie had a total of &lt;math&gt;444&lt;/math&gt; peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats &lt;math&gt;5&lt;/math&gt; of his peanuts, Betty eats &lt;math&gt;9&lt;/math&gt; of her peanuts, and Charlie eats &lt;math&gt;25&lt;/math&gt; of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.<br /> <br /> [[2016 AIME II Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> There is a &lt;math&gt;40\%&lt;/math&gt; chance of rain on Saturday and a &lt;math&gt;30\%&lt;/math&gt; chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;x,y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be real numbers satisfying the system<br /> &lt;cmath&gt;\log_2(xyz-3+\log_5 x)=5,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_3(xyz-3+\log_5 y)=4,&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_4(xyz-3+\log_5 z)=4.&lt;/cmath&gt;<br /> Find the value of &lt;math&gt;|\log_5 x|+|\log_5 y|+|\log_5 z|&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> An &lt;math&gt;a \times b \times c&lt;/math&gt; rectangular box is built from &lt;math&gt;a \cdot b \cdot c&lt;/math&gt; unit cubes. Each unit cube is colored red, green, or yellow. Each of the &lt;math&gt;a&lt;/math&gt; layers of size &lt;math&gt;1 \times b \times c&lt;/math&gt; parallel to the &lt;math&gt;(b \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;9&lt;/math&gt; red cubes, exactly &lt;math&gt;12&lt;/math&gt; green cubes, and some yellow cubes. Each of the &lt;math&gt;b&lt;/math&gt; layers of size &lt;math&gt;a \times 1 \times c&lt;/math&gt; parallel to the &lt;math&gt;(a \times c)&lt;/math&gt; faces of the box contains exactly &lt;math&gt;20&lt;/math&gt; green cubes, exactly &lt;math&gt;25&lt;/math&gt; yellow cubes, and some red cubes. Find the smallest possible volume of the box.<br /> <br /> [[2016 AIME II Problems/Problem 4 | Solution]]<br /> ==Problem 5==<br /> Triangle &lt;math&gt;ABC_0&lt;/math&gt; has a right angle at &lt;math&gt;C_0&lt;/math&gt;. Its side lengths are pairwise relatively prime positive integers, and its perimeter is &lt;math&gt;p&lt;/math&gt;. Let &lt;math&gt;C_1&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{AB}&lt;/math&gt;, and for &lt;math&gt;n \geq 2&lt;/math&gt;, let &lt;math&gt;C_n&lt;/math&gt; be the foot of the altitude to &lt;math&gt;\overline{C_{n-2}B}&lt;/math&gt; in &lt;math&gt;\triangle C_{n-2}C_{n-1}B&lt;/math&gt;. The sum &lt;math&gt;\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p&lt;/math&gt;. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> For polynomial &lt;math&gt;P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}&lt;/math&gt;, define<br /> &lt;math&gt;Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}&lt;/math&gt;.<br /> Then &lt;math&gt;\sum_{i=0}^{50} |a_i|=\dfrac{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 /> [[2016 AIME II Problems/Problem 6 | Solution]]<br /> ==Problem 7==<br /> Squares &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;EFGH&lt;/math&gt; have a common center and &lt;math&gt;\overline{AB} || \overline{EF}&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is 2016, and the area of &lt;math&gt;EFGH&lt;/math&gt; is a smaller positive integer. Square &lt;math&gt;IJKL&lt;/math&gt; is constructed so that each of its vertices lies on a side of &lt;math&gt;ABCD&lt;/math&gt; and each vertex of &lt;math&gt;EFGH&lt;/math&gt; lies on a side of &lt;math&gt;IJKL&lt;/math&gt;. Find the difference between the largest and smallest positive integer values for the area of &lt;math&gt;IJKL&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of sets &lt;math&gt;\{a,b,c\}&lt;/math&gt; of three distinct positive integers with the property that the product of &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is equal to the product of &lt;math&gt;11,21,31,41,51,&lt;/math&gt; and &lt;math&gt;61&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> The sequences of positive integers &lt;math&gt;1,a_2, a_3,...&lt;/math&gt; and &lt;math&gt;1,b_2, b_3,...&lt;/math&gt; are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let &lt;math&gt;c_n=a_n+b_n&lt;/math&gt;. There is an integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;c_{k-1}=100&lt;/math&gt; and &lt;math&gt;c_{k+1}=1000&lt;/math&gt;. Find &lt;math&gt;c_k&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 9 | Solution]]<br /> ==Problem 10==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are on side &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AP&lt;AQ&lt;/math&gt;. Rays &lt;math&gt;CP&lt;/math&gt; and &lt;math&gt;CQ&lt;/math&gt; meet &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (other than &lt;math&gt;C&lt;/math&gt;), respectively. If &lt;math&gt;AP=4,PQ=3,QB=6,BT=5,&lt;/math&gt; and &lt;math&gt;AS=7&lt;/math&gt;, then &lt;math&gt;ST=\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 /> [[2016 AIME II Problems/Problem 10 | Solution]]<br /> ==Problem 11==<br /> For positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, define &lt;math&gt;N&lt;/math&gt; to be &lt;math&gt;k&lt;/math&gt;-nice if there exists a positive integer &lt;math&gt;a&lt;/math&gt; such that &lt;math&gt;a^{k}&lt;/math&gt; has exactly &lt;math&gt;N&lt;/math&gt; positive divisors. Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that are neither &lt;math&gt;7&lt;/math&gt;-nice nor &lt;math&gt;8&lt;/math&gt;-nice.<br /> <br /> [[2016 AIME II Problems/Problem 11 | Solution]]<br /> ==Problem 12==<br /> The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.<br /> <br /> &lt;asy&gt;<br /> draw(Circle((0,0), 4));<br /> draw(Circle((0,0), 3));<br /> draw((0,4)--(0,3));<br /> draw((0,-4)--(0,-3));<br /> draw((-2.598, 1.5)--(-3.4641, 2));<br /> draw((-2.598, -1.5)--(-3.4641, -2));<br /> draw((2.598, -1.5)--(3.4641, -2));<br /> draw((2.598, 1.5)--(3.4641, 2));<br /> &lt;/asy&gt;<br /> <br /> [[2016 AIME II Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> Beatrix is going to place six rooks on a &lt;math&gt;6 \times 6&lt;/math&gt; chessboard where both the rows and columns are labeled &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;; the rooks are placed so that no two rooks are in the same row or the same column. The [i]value[/i] of a square is the sum of its row number and column number. The [i]score[/i] of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations 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 /> [[2016 AIME II Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;600&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie outside the plane of &lt;math&gt;\triangle ABC&lt;/math&gt; and are on opposite sides of the plane. Furthermore, &lt;math&gt;PA=PB=PC&lt;/math&gt;, and &lt;math&gt;QA=QB=QC&lt;/math&gt;, and the planes of &lt;math&gt;\triangle PAB&lt;/math&gt; and &lt;math&gt;\triangle QAB&lt;/math&gt; form a &lt;math&gt;120^{\circ}&lt;/math&gt; dihedral angle (the angle between the two planes). There is a point &lt;math&gt;O&lt;/math&gt; whose distance from each of &lt;math&gt;A,B,C,P,&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt;. Find &lt;math&gt;d&lt;/math&gt;.<br /> <br /> [[2016 AIME II Problems/Problem 14 | Solution]]<br /> ==Problem 15==<br /> For &lt;math&gt;1 \leq i \leq 215&lt;/math&gt; let &lt;math&gt;a_i = \dfrac{1}{2^{i}}&lt;/math&gt; and &lt;math&gt;a_{216} = \dfrac{1}{2^{215}}&lt;/math&gt;. Let &lt;math&gt;x_1, x_2, ..., x_{216}&lt;/math&gt; be positive real numbers such that &lt;math&gt;\sum_{i=1}^{216} x_i=1&lt;/math&gt; and &lt;math&gt;\sum_{1 \leq i &lt; j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}&lt;/math&gt;. The maximum possible value of &lt;math&gt;x_2=\dfrac{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 /> [[2016 AIME II Problems/Problem 15 | Solution]]<br /> <br /> <br /> {{AIME box|year=2016|n=II|before=[[2016 AIME I Problems]]|after=[[2017 AIME I Problems]]}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_11&diff=130939 2018 AIME I Problems/Problem 11 2020-08-07T05:21:25Z <p>Pleaseletmewin: /* Modular Arithmetic Solution- Strange (MASS) */</p> <hr /> <div><br /> ==Problem==<br /> Find the least positive integer &lt;math&gt;n&lt;/math&gt; such that when &lt;math&gt;3^n&lt;/math&gt; is written in base &lt;math&gt;143&lt;/math&gt;, its two right-most digits in base &lt;math&gt;143&lt;/math&gt; are &lt;math&gt;01&lt;/math&gt;.<br /> <br /> ==Solutions==<br /> <br /> ==Modular Arithmetic Solution- Strange (MASS)==<br /> Note that the given condition is equivalent to &lt;math&gt;3^n \equiv 1 \pmod{143^2}&lt;/math&gt; and &lt;math&gt;143=11\cdot 13&lt;/math&gt;. Because &lt;math&gt;\gcd(11^2, 13^2) = 1&lt;/math&gt;, the desired condition is equivalent to &lt;math&gt;3^n \equiv 1 \pmod{121}&lt;/math&gt; and &lt;math&gt;3^n \equiv 1 \pmod{169}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;3^n \equiv 1 \pmod{121}&lt;/math&gt;, one can see the sequence &lt;math&gt;1, 3, 9, 27, 81, 1, 3, 9...&lt;/math&gt; so &lt;math&gt;5|n&lt;/math&gt;.<br /> <br /> Now if &lt;math&gt;3^n \equiv 1 \pmod{169}&lt;/math&gt;, it is harder. But we do observe that &lt;math&gt;3^3 \equiv 1 \pmod{13}&lt;/math&gt;, therefore &lt;math&gt;3^3 = 13a + 1&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. So our goal is to find the first number &lt;math&gt;p_1&lt;/math&gt; such that &lt;math&gt;(13a+1)^ {p_1} \equiv 1 \pmod{169}&lt;/math&gt;. In other words, the &lt;math&gt;p_1 \equiv 0 \pmod{13}&lt;/math&gt;. It is not difficult to see that the smallest &lt;math&gt;p_1=13&lt;/math&gt;, so ultimately &lt;math&gt;3^{39} \equiv 1 \pmod{169}&lt;/math&gt;. Therefore, &lt;math&gt;39|n&lt;/math&gt;.<br /> <br /> The first &lt;math&gt;n&lt;/math&gt; satisfying both criteria is thus &lt;math&gt;5\cdot 39=\boxed{195}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==Solution 2==<br /> Note that Euler's Totient Theorem would not necessarily lead to the smallest &lt;math&gt;n&lt;/math&gt; and that in this case that &lt;math&gt;n&lt;/math&gt; is greater than &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> We wish to find the least &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;3^n \equiv 1 \pmod{143^2}&lt;/math&gt;. This factors as &lt;math&gt;143^2=11^{2}*13^{2}&lt;/math&gt;. Because &lt;math&gt;gcd(121, 169) = 1&lt;/math&gt;, we can simply find the least &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;3^n \equiv 1 \pmod{121}&lt;/math&gt; and &lt;math&gt;3^n \equiv 1 \pmod{169}&lt;/math&gt;.<br /> <br /> Quick inspection yields &lt;math&gt;3^5 \equiv 1 \pmod{121}&lt;/math&gt; and &lt;math&gt;3^3 \equiv 1 \pmod{13}&lt;/math&gt;. Now we must find the smallest &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;3^{3k} \equiv 1 \pmod{13}&lt;/math&gt;. Euler's gives &lt;math&gt;3^{156} \equiv 1 \pmod{169}&lt;/math&gt;. So &lt;math&gt;3k&lt;/math&gt; is a factor of &lt;math&gt;156&lt;/math&gt;. This gives &lt;math&gt;k=1,2, 4, 13, 26, 52&lt;/math&gt;. Some more inspection yields &lt;math&gt;k=13&lt;/math&gt; is the smallest valid &lt;math&gt;k&lt;/math&gt;. So &lt;math&gt;3^5 \equiv 1 \pmod{121}&lt;/math&gt; and &lt;math&gt;3^{39} \equiv 1 \pmod{169}&lt;/math&gt;. The least &lt;math&gt;n&lt;/math&gt; satisfying both is &lt;math&gt;lcm(5, 39)=\boxed{195}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3 (Big Bash)==<br /> Listing out the powers of &lt;math&gt;3&lt;/math&gt;, modulo &lt;math&gt;169&lt;/math&gt; and modulo &lt;math&gt;121&lt;/math&gt;, we have:<br /> &lt;cmath&gt;\begin{array}{c|c|c}<br /> n &amp; 3^n\mod{169} &amp; 3^n\mod{121}\\ \hline<br /> 0 &amp; 1 &amp; 1\\<br /> 1 &amp; 3 &amp; 3\\<br /> 2 &amp; 9 &amp; 9\\<br /> 3 &amp; 27 &amp; 27\\<br /> 4 &amp; 81 &amp; 81\\<br /> 5 &amp; 74 &amp; 1\\<br /> 6 &amp; 53\\<br /> 7 &amp; 159\\<br /> 8 &amp; 139\\<br /> 9 &amp; 79\\<br /> 10 &amp; 68\\<br /> 11 &amp; 35\\<br /> 12 &amp; 105\\<br /> 13 &amp; 146\\<br /> 14 &amp; 100\\<br /> 15 &amp; 131\\<br /> 16 &amp; 55\\<br /> 17 &amp; 165\\<br /> 18 &amp; 157\\<br /> 19 &amp; 133\\<br /> 20 &amp; 61\\<br /> 21 &amp; 14\\<br /> 22 &amp; 42\\<br /> 23 &amp; 126\\<br /> 24 &amp; 40\\<br /> 25 &amp; 120\\<br /> 26 &amp; 22\\<br /> 27 &amp; 66\\<br /> 28 &amp; 29\\<br /> 29 &amp; 87\\<br /> 30 &amp; 92\\<br /> 31 &amp; 107\\<br /> 32 &amp; 152\\<br /> 33 &amp; 118\\<br /> 34 &amp; 16\\<br /> 35 &amp; 48\\<br /> 36 &amp; 144\\<br /> 37 &amp; 94\\<br /> 38 &amp; 113\\<br /> 39 &amp; 1\\<br /> \end{array}&lt;/cmath&gt;<br /> <br /> The powers of &lt;math&gt;3&lt;/math&gt; repeat in cycles of &lt;math&gt;5&lt;/math&gt; an &lt;math&gt;39&lt;/math&gt; in modulo &lt;math&gt;121&lt;/math&gt; and modulo &lt;math&gt;169&lt;/math&gt;, respectively. The answer is &lt;math&gt;\text{lcm}(5, 39) = \boxed{195}&lt;/math&gt;.<br /> <br /> ==Solution 4(Order+Bash)==<br /> We have that &lt;cmath&gt;3^n \equiv 1 \pmod{143^2}.&lt;/cmath&gt;Now, &lt;math&gt;3^{110} \equiv 1 \pmod{11^2}&lt;/math&gt; so by the Fundamental Theorem of Orders, &lt;math&gt;\text{ord}_{11^2}(3)|110&lt;/math&gt; and with some bashing, we get that it is &lt;math&gt;5&lt;/math&gt;. Similarly, we get that &lt;math&gt;\text{ord}_{13^2}(3)=39&lt;/math&gt;. Now, &lt;math&gt;\text{lcm}(39,5)=\boxed{195}&lt;/math&gt; which is our desired solution.<br /> <br /> ==Solution 5 (Easy Binomial Theorem)==<br /> We wish to find the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;3^n\equiv 1\pmod{143^2}&lt;/math&gt;, so we want &lt;math&gt;n\equiv 1\pmod{121}&lt;/math&gt; and &lt;math&gt;n\equiv 1\pmod{169}&lt;/math&gt;. Note that &lt;math&gt;3^5\equiv 1\pmod{121}&lt;/math&gt;, so &lt;math&gt;3^n&lt;/math&gt; repeats &lt;math&gt;121&lt;/math&gt; with a period of &lt;math&gt;5&lt;/math&gt;, so &lt;math&gt;5|n&lt;/math&gt;. Now, in order for &lt;math&gt;n\equiv 1\pmod{169}&lt;/math&gt;, then &lt;math&gt;n\equiv 1\pmod{13}&lt;/math&gt;. Because &lt;math&gt;3^3\equiv 1\pmod{13}&lt;/math&gt;, &lt;math&gt;3^n&lt;/math&gt; repeats with a period of &lt;math&gt;3&lt;/math&gt;, so &lt;math&gt;3|n&lt;/math&gt;. <br /> Hence, we have that for some positive integer &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}&lt;/math&gt;, so &lt;math&gt;26p\equiv 0\pmod{169}&lt;/math&gt; and &lt;math&gt;p\equiv 0\pmod{13}&lt;/math&gt;. Thus, we have that &lt;math&gt;5|n&lt;/math&gt;, &lt;math&gt;3|n&lt;/math&gt;, and &lt;math&gt;13|n&lt;/math&gt;, so the smallest possible value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;3\times5\times13=\boxed{195}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==Solution 6(LTE)==<br /> We can see that &lt;math&gt;3^n-1 = 143^2*x&lt;/math&gt;, which means that &lt;math&gt;v_{11}(3^n-1) \geq 2&lt;/math&gt;, &lt;math&gt;v_{13}(3^n-1) \geq 2&lt;/math&gt;. &lt;math&gt;v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})&lt;/math&gt;, &lt;math&gt;v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})&lt;/math&gt; by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as &lt;math&gt;v_{13}(3^n-1) \geq 2&lt;/math&gt;. Therefore the minimum possible value of n is &lt;math&gt;3\times5\times13=\boxed{195}&lt;/math&gt;.<br /> <br /> -bradleyguo<br /> <br /> ==Solution 7==<br /> Note that the problem is basically asking for the least positive integer &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;11^2 \cdot 13^2 | 3^n - 1.&lt;/math&gt; It is easy to see that &lt;math&gt;n = \text{lcm } (a, b),&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is the least positive integer satisfying &lt;math&gt;11^2 | 3^a - 1&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; the least positive integer satisfying &lt;math&gt;13^2 | 3^b - 1&lt;/math&gt;. Luckily, finding &lt;math&gt;a&lt;/math&gt; is a relatively trivial task, as one can simply notice that &lt;math&gt;3^5 = 243 \equiv 1 \mod 121&lt;/math&gt;. However, finding &lt;math&gt;b&lt;/math&gt; is slightly more nontrivial. The order of &lt;math&gt;3^k&lt;/math&gt; modulo &lt;math&gt;13&lt;/math&gt; (which is &lt;math&gt;3&lt;/math&gt;) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of &lt;math&gt;3&lt;/math&gt; by &lt;math&gt;13&lt;/math&gt;, or one can notice that &lt;math&gt;3^3 = 27 \equiv 1 \mod 13&lt;/math&gt; (the latter which is the definition of period/orders by FLT). We can thus rewrite &lt;math&gt;3^3&lt;/math&gt; as &lt;math&gt;(2 \cdot 13 + 1) \mod 13^2&lt;/math&gt;. Now suppose that &lt;cmath&gt;3^{3k} \equiv (13n + 1) \mod 13^2. &lt;/cmath&gt; I claim that &lt;math&gt;3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2. &lt;/math&gt;<br /> <br /> Proof:<br /> To find &lt;math&gt;3^{3(k+1)}, &lt;/math&gt; we can simply multiply &lt;math&gt;3^{3k}&lt;/math&gt; by &lt;math&gt;3^3, &lt;/math&gt; which is congruent to &lt;math&gt;2 \cdot 13 + 1&lt;/math&gt; modulo &lt;math&gt;13^2&lt;/math&gt;.<br /> By expanding the product out, we obtain &lt;cmath&gt;3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2, &lt;/cmath&gt; and since the &lt;math&gt;13^2&lt;/math&gt; on the LHS cancels out, we're left with &lt;cmath&gt;13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2&lt;/cmath&gt;. Thus, our claim is proven.<br /> Let &lt;math&gt;f(n)&lt;/math&gt; be the second to last digit when &lt;math&gt;3^{3k}&lt;/math&gt; is written in base &lt;math&gt;13^2&lt;/math&gt;. <br /> Using our proof, it is easy to see that &lt;math&gt;f(n)&lt;/math&gt; satisfies the recurrence &lt;math&gt;f(1) = 2&lt;/math&gt; and &lt;math&gt;f(n+1) = f(n) + 2&lt;/math&gt;. Since this implies &lt;math&gt;f(n) = 2n, &lt;/math&gt; we just have to find the least positive integer &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;2n&lt;/math&gt; is a multiple of &lt;math&gt;13&lt;/math&gt;, which is trivially obtained as &lt;math&gt;13&lt;/math&gt;.<br /> The least integer &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;3^n - 1&lt;/math&gt; is divisible by &lt;math&gt;13^2&lt;/math&gt; is &lt;math&gt;3 \cdot 13 = 39, &lt;/math&gt; so our final answer is &lt;math&gt;\text{lcm } (5, 39) = \boxed{195}.&lt;/math&gt;<br /> <br /> -fidgetboss_4000<br /> -minor edits made by srisainandan6<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_10&diff=130756 2016 AIME I Problems/Problem 10 2020-08-06T06:18:11Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem==<br /> <br /> A strictly increasing sequence of positive integers &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;\cdots&lt;/math&gt; has the property that for every positive integer &lt;math&gt;k&lt;/math&gt;, the subsequence &lt;math&gt;a_{2k-1}&lt;/math&gt;, &lt;math&gt;a_{2k}&lt;/math&gt;, &lt;math&gt;a_{2k+1}&lt;/math&gt; is geometric and the subsequence &lt;math&gt;a_{2k}&lt;/math&gt;, &lt;math&gt;a_{2k+1}&lt;/math&gt;, &lt;math&gt;a_{2k+2}&lt;/math&gt; is arithmetic. Suppose that &lt;math&gt;a_{13} = 2016&lt;/math&gt;. Find &lt;math&gt;a_1&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> We first create a similar sequence where &lt;math&gt;a_1=1&lt;/math&gt; and &lt;math&gt;a_2=2&lt;/math&gt;. Continuing the sequence,<br /> <br /> &lt;cmath&gt;1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots&lt;/cmath&gt;<br /> <br /> Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. This can be proven by induction. Similarly, &lt;math&gt;a_{13}&lt;/math&gt; would also need to be the end of a geometric sequence (divisible by a square). We see that &lt;math&gt;2016&lt;/math&gt; is &lt;math&gt;2^5 \cdot 3^2 \cdot 7&lt;/math&gt;, so the squares that would fit in &lt;math&gt;2016&lt;/math&gt; are &lt;math&gt;1^2=1&lt;/math&gt;, &lt;math&gt;2^2=4&lt;/math&gt;, &lt;math&gt;3^2=9&lt;/math&gt;, &lt;math&gt;2^4=16&lt;/math&gt;, &lt;math&gt;2^2 \cdot 3^2 = 36&lt;/math&gt;, and &lt;math&gt;2^4 \cdot 3^2 = 144&lt;/math&gt;. By simple inspection &lt;math&gt;144&lt;/math&gt; is the only plausible square, since the other squares in the sequence don't have enough elements before them to go all the way back to &lt;math&gt;a_1&lt;/math&gt; while still staying as positive integers. &lt;math&gt;a_{13}=2016=14\cdot 144&lt;/math&gt;, so &lt;math&gt;a_1=14\cdot 36=\fbox{504}&lt;/math&gt;.<br /> <br /> ~IYN~<br /> <br /> ==Solution 2==<br /> <br /> Setting &lt;math&gt;a_1 = a&lt;/math&gt; and &lt;math&gt;a_2 = ka&lt;/math&gt;, the sequence becomes:<br /> <br /> &lt;cmath&gt;a, ka, k^2a, k(2k-1)a, (2k-1)^2a, (2k-1)(3k-2)a, (3k-2)^2a, \cdots&lt;/cmath&gt;<br /> and so forth, with &lt;math&gt;a_{2n+1} = (nk-(n-1))^2a&lt;/math&gt;. Then, &lt;math&gt;a_{13} = (6k-5)^2a = 2016&lt;/math&gt;. Keep in mind, &lt;math&gt;k&lt;/math&gt; need not be an integer, only &lt;math&gt;k^2a, (k+1)^2a,&lt;/math&gt; etc. does. &lt;math&gt;2016 = 2^5*3^2*7&lt;/math&gt;, so only the squares &lt;math&gt;1, 4, 9, 16, 36,&lt;/math&gt; and &lt;math&gt;144&lt;/math&gt; are plausible for &lt;math&gt;(6k-5)^2&lt;/math&gt;. But when that is anything other than &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;k^2a&lt;/math&gt; is not an integer. Therefore, &lt;math&gt;a = 2016/2^2 = 504&lt;/math&gt;.<br /> <br /> Thanks for reading, Rowechen Zhong.<br /> <br /> <br /> ==Solution 3==<br /> <br /> This is not a hard bash. You can try the ratios &lt;math&gt;\frac 2 3&lt;/math&gt;, &lt;math&gt;\frac 3 4&lt;/math&gt;, and &lt;math&gt;\frac {11} {12}.&lt;/math&gt; Working backwards from &lt;math&gt;\frac {11} {12},&lt;/math&gt; we get &lt;math&gt;504.&lt;/math&gt;<br /> <br /> ==Solution 4(Very Risky and Very Stupid)==<br /> The thirteenth term of the sequence is &lt;math&gt;2016&lt;/math&gt;, which makes that fourteenth term of the sequence &lt;math&gt;2016+r&lt;/math&gt; and the &lt;math&gt;15^{\text{th}}&lt;/math&gt; term &lt;math&gt;\displaystyle \frac{(2016+r)^2}{2016}&lt;/math&gt;. We note that &lt;math&gt;r&lt;/math&gt; is an integer so that means &lt;math&gt;\displaystyle \frac{r^2}{2016}&lt;/math&gt; is an integer. Thus, we assume the smallest value of &lt;math&gt;r&lt;/math&gt;, which is &lt;math&gt;168&lt;/math&gt;. We bash all the way back to the first term and get our answer of &lt;math&gt;\boxed{504}&lt;/math&gt;.<br /> <br /> -Pleaseletmewin<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_25&diff=130276 1972 AHSME Problems/Problem 25 2020-08-02T21:02:23Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>Inscribed in a circle is a quadrilateral having sides of lengths &lt;math&gt;25,~39,~52&lt;/math&gt;, and &lt;math&gt;60&lt;/math&gt; taken consecutively. The diameter of this circle has length<br /> <br /> &lt;math&gt;\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We note that &lt;math&gt;25^2+60^2=65^2&lt;/math&gt; and &lt;math&gt;39^2+52^2=65^2&lt;/math&gt; so our answer is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> -Pleaseletmewin</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_25&diff=130275 1972 AHSME Problems/Problem 25 2020-08-02T21:01:41Z <p>Pleaseletmewin: Created page with &quot;==Solution== We note that &lt;math&gt;25^2+60^2=65^2&lt;/math&gt; and &lt;math&gt;39^2+52^2=65^2&lt;/math&gt; so our answer is &lt;math&gt;\boxed{65}&lt;/math&gt;. -Pleaseletmewin&quot;</p> <hr /> <div>==Solution==<br /> <br /> We note that &lt;math&gt;25^2+60^2=65^2&lt;/math&gt; and &lt;math&gt;39^2+52^2=65^2&lt;/math&gt; so our answer is &lt;math&gt;\boxed{65}&lt;/math&gt;.<br /> <br /> -Pleaseletmewin</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_23&diff=130221 1993 AHSME Problems/Problem 23 2020-08-02T05:26:55Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> draw(circle((0,0),10),black+linewidth(.75));<br /> draw((-10,0)--(10,0),black+linewidth(.75));<br /> draw((-10,0)--(9,sqrt(19)),black+linewidth(.75));<br /> draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75));<br /> draw((2,0)--(9,sqrt(19)),black+linewidth(.75));<br /> draw((2,0)--(9,-sqrt(19)),black+linewidth(.75));<br /> MP(&quot;X&quot;,(2,0),N);MP(&quot;A&quot;,(-10,0),W);MP(&quot;D&quot;,(10,0),E);MP(&quot;B&quot;,(9,sqrt(19)),E);MP(&quot;C&quot;,(9,-sqrt(19)),E);<br /> &lt;/asy&gt;<br /> <br /> Points &lt;math&gt;A,B,C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are on a circle of diameter &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is on diameter &lt;math&gt;\overline{AD}.&lt;/math&gt;<br /> <br /> If &lt;math&gt;BX=CX&lt;/math&gt; and &lt;math&gt;3\angle{BAC}=\angle{BXC}=36^\circ&lt;/math&gt;, then &lt;math&gt;AX=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A) } cos(6^\circ)cos(12^\circ)sec(18^\circ)\quad\\<br /> \text{(B) } cos(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\<br /> \text{(C) } cos(6^\circ)sin(12^\circ)sec(18^\circ)\quad\\<br /> \text{(D) } sin(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\<br /> \text{(E) } sin(6^\circ)sin(12^\circ)sec(18^\circ)&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We have all the angles we need, but most obviously, we see that right angle in triangle &lt;math&gt;ABD&lt;/math&gt;.<br /> <br /> Note also that angle &lt;math&gt;BAD&lt;/math&gt; is 6 degrees, so length &lt;math&gt;AB = cos(6)&lt;/math&gt; because the diameter, &lt;math&gt;AD&lt;/math&gt;, is 1.<br /> <br /> Now, we can concentrate on triangle &lt;math&gt;ABX&lt;/math&gt; (after all, now we can decipher all angles easily and use Law of Sines).<br /> <br /> We get:<br /> <br /> &lt;math&gt;\frac{AB}{\sin(\angle{AXB})} =\frac{AX}{\sin(\angle{ABX})}&lt;/math&gt;<br /> <br /> That's equal to<br /> <br /> &lt;math&gt;\frac{\cos(6)}{\sin(180-18)} =\frac{AX}{\sin(12)}&lt;/math&gt;<br /> <br /> Therefore, our answer is equal to:<br /> &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\sin(162) = \sin(18)&lt;/math&gt;, and don't accidentally put &lt;math&gt;\fbox{C}&lt;/math&gt; because you thought &lt;math&gt;\frac{1}{\sin}&lt;/math&gt; was &lt;math&gt;\sec&lt;/math&gt;!<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=22|num-a=24}} <br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&diff=130084 1985 AHSME Problems/Problem 16 2020-08-01T01:49:35Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem==<br /> If &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, then the value of &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> First, let's leave everything in variables and see if we can simplify &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt;.<br /> <br /> <br /> We can write everything in terms of sine and cosine to get &lt;math&gt; \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> <br /> We can multiply out the numerator to get &lt;math&gt; \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:<br /> <br /> <br /> &lt;math&gt; \cos(A-B)=\sin A\sin B+\cos A\cos B &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin(A+B)=\sin A\cos B+\sin B\cos A &lt;/math&gt;<br /> <br /> <br /> Therefore, our fraction is equal to &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> We can also use the product-to-sum formula<br /> <br /> &lt;math&gt; \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) &lt;/math&gt; to simplify the denominator:<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} &lt;/math&gt;.<br /> <br /> <br /> But now we seem stuck. However, we can note that since &lt;math&gt; A+B=45^\circ &lt;/math&gt;, we have &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we get<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} &lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; \frac{1}{\frac{1}{2}} &lt;/math&gt;<br /> <br /> &lt;math&gt; 2, \boxed{\text{B}} &lt;/math&gt;<br /> <br /> Note that we only used the fact that &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we have in fact not just shown that &lt;math&gt; (1+\tan A)(1+\tan B)=2 &lt;/math&gt; for &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, but for all &lt;math&gt; A, B &lt;/math&gt; such that &lt;math&gt; A+B=45^\circ+n180^\circ &lt;/math&gt;, for integer &lt;math&gt; n &lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We can see that &lt;math&gt;25^o+20^o=45^o&lt;/math&gt;. We also know that &lt;math&gt;\tan 45=1&lt;/math&gt;. First, let us expand &lt;math&gt;(1+\tan A)(1+\tan B)&lt;/math&gt;.<br /> <br /> We get &lt;math&gt;1+\tan A+\tan B+\tan A\tan B&lt;/math&gt;. <br /> <br /> Now, let us look at &lt;math&gt;\tan45=\tan(20+25)&lt;/math&gt;.<br /> <br /> By the &lt;math&gt;\tan&lt;/math&gt; sum formula, we know that &lt;math&gt;\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}&lt;/math&gt;<br /> <br /> Then, since &lt;math&gt;\tan 45=1&lt;/math&gt;, we can see that &lt;math&gt;\tan A+\tan B=1-\tan A\tan B&lt;/math&gt;<br /> <br /> Then &lt;math&gt;1=\tan A+\tan B+\tan A\tan B&lt;/math&gt;<br /> <br /> Thus, the sum become &lt;math&gt;1+1=2&lt;/math&gt; and the answer is &lt;math&gt;\fbox{\text{(B)}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to &lt;cmath&gt;\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = &lt;/cmath&gt; &lt;cmath&gt;\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Clearly, that is equal to &lt;cmath&gt;1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Now, we note that &lt;cmath&gt;\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ&lt;/cmath&gt; is equal to &lt;math&gt;\sin 45^\circ&lt;/math&gt;. Now, we would like to get &lt;math&gt;\sin 20^\circ \sin 25^\circ&lt;/math&gt; in the denominator. What springs to mind is the fact that &lt;cmath&gt;\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.&lt;/cmath&gt; Therefore, we can express the desired value as &lt;cmath&gt;1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.&lt;/cmath&gt; Because &lt;math&gt;\cos 45^\circ = \sin 45^\circ&lt;/math&gt;, we see that the fractional part is &lt;math&gt;1&lt;/math&gt;, and so the sum is &lt;math&gt;1 + 1 = 2&lt;/math&gt;, which brings us to the answer &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AHSME box|year=1985|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&diff=130083 1985 AHSME Problems/Problem 16 2020-08-01T01:47:37Z <p>Pleaseletmewin: /* Solution 4(Easiest) */</p> <hr /> <div>==Problem==<br /> If &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, then the value of &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> First, let's leave everything in variables and see if we can simplify &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt;.<br /> <br /> <br /> We can write everything in terms of sine and cosine to get &lt;math&gt; \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> <br /> We can multiply out the numerator to get &lt;math&gt; \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:<br /> <br /> <br /> &lt;math&gt; \cos(A-B)=\sin A\sin B+\cos A\cos B &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin(A+B)=\sin A\cos B+\sin B\cos A &lt;/math&gt;<br /> <br /> <br /> Therefore, our fraction is equal to &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> We can also use the product-to-sum formula<br /> <br /> &lt;math&gt; \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) &lt;/math&gt; to simplify the denominator:<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} &lt;/math&gt;.<br /> <br /> <br /> But now we seem stuck. However, we can note that since &lt;math&gt; A+B=45^\circ &lt;/math&gt;, we have &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we get<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} &lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; \frac{1}{\frac{1}{2}} &lt;/math&gt;<br /> <br /> &lt;math&gt; 2, \boxed{\text{B}} &lt;/math&gt;<br /> <br /> Note that we only used the fact that &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we have in fact not just shown that &lt;math&gt; (1+\tan A)(1+\tan B)=2 &lt;/math&gt; for &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, but for all &lt;math&gt; A, B &lt;/math&gt; such that &lt;math&gt; A+B=45^\circ+n180^\circ &lt;/math&gt;, for integer &lt;math&gt; n &lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We can see that &lt;math&gt;25^o+20^o=45^o&lt;/math&gt;. We also know that &lt;math&gt;\tan 45=1&lt;/math&gt;. First, let us expand &lt;math&gt;(1+\tan A)(1+\tan B)&lt;/math&gt;.<br /> <br /> We get &lt;math&gt;1+\tan A+\tan B+\tan A\tan B&lt;/math&gt;. <br /> <br /> Now, let us look at &lt;math&gt;\tan45=\tan(20+25)&lt;/math&gt;.<br /> <br /> By the &lt;math&gt;\tan&lt;/math&gt; sum formula, we know that &lt;math&gt;\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}&lt;/math&gt;<br /> <br /> Then, since &lt;math&gt;\tan 45=1&lt;/math&gt;, we can see that &lt;math&gt;\tan A+\tan B=1-\tan A\tan B&lt;/math&gt;<br /> <br /> Then &lt;math&gt;1=\tan A+\tan B+\tan A\tan B&lt;/math&gt;<br /> <br /> Thus, the sum become &lt;math&gt;1+1=2&lt;/math&gt; and the answer is &lt;math&gt;\fbox{\text{(B)}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to &lt;cmath&gt;\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = &lt;/cmath&gt; &lt;cmath&gt;\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Clearly, that is equal to &lt;cmath&gt;1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Now, we note that &lt;cmath&gt;\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ&lt;/cmath&gt; is equal to &lt;math&gt;\sin 45^\circ&lt;/math&gt;. Now, we would like to get &lt;math&gt;\sin 20^\circ \sin 25^\circ&lt;/math&gt; in the denominator. What springs to mind is the fact that &lt;cmath&gt;\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.&lt;/cmath&gt; Therefore, we can express the desired value as &lt;cmath&gt;1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.&lt;/cmath&gt; Because &lt;math&gt;\cos 45^\circ = \sin 45^\circ&lt;/math&gt;, we see that the fractional part is &lt;math&gt;1&lt;/math&gt;, and so the sum is &lt;math&gt;1 + 1 = 2&lt;/math&gt;, which brings us to the answer &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ===Solution 4(Easiest)===<br /> <br /> Notice that &lt;math&gt;\tan(20^{\circ}+25^{\circ})=\frac{\tan 20^{\circ} + \tan 25^{\circ}}{1-\tan 20^{\circ}\tan 25^{\circ}}&lt;/math&gt;. Because &lt;math&gt;\tan(20^{\circ}+25^{\circ})=1,&lt;/math&gt; we have &lt;math&gt;1-\tan 20^\circ \tan 25^\circ=\tan 20^{\circ} + \tan 25^{\circ}&lt;/math&gt; which means that &lt;math&gt;\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ}\tan25^{\circ}+1=2&lt;/math&gt; which gives us &lt;math&gt;\boxed{B}&lt;/math&gt;. \\ -Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AHSME box|year=1985|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&diff=130082 1985 AHSME Problems/Problem 16 2020-08-01T01:46:40Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem==<br /> If &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, then the value of &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> First, let's leave everything in variables and see if we can simplify &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt;.<br /> <br /> <br /> We can write everything in terms of sine and cosine to get &lt;math&gt; \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> <br /> We can multiply out the numerator to get &lt;math&gt; \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:<br /> <br /> <br /> &lt;math&gt; \cos(A-B)=\sin A\sin B+\cos A\cos B &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin(A+B)=\sin A\cos B+\sin B\cos A &lt;/math&gt;<br /> <br /> <br /> Therefore, our fraction is equal to &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> We can also use the product-to-sum formula<br /> <br /> &lt;math&gt; \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) &lt;/math&gt; to simplify the denominator:<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} &lt;/math&gt;.<br /> <br /> <br /> But now we seem stuck. However, we can note that since &lt;math&gt; A+B=45^\circ &lt;/math&gt;, we have &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we get<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} &lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; \frac{1}{\frac{1}{2}} &lt;/math&gt;<br /> <br /> &lt;math&gt; 2, \boxed{\text{B}} &lt;/math&gt;<br /> <br /> Note that we only used the fact that &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we have in fact not just shown that &lt;math&gt; (1+\tan A)(1+\tan B)=2 &lt;/math&gt; for &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, but for all &lt;math&gt; A, B &lt;/math&gt; such that &lt;math&gt; A+B=45^\circ+n180^\circ &lt;/math&gt;, for integer &lt;math&gt; n &lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We can see that &lt;math&gt;25^o+20^o=45^o&lt;/math&gt;. We also know that &lt;math&gt;\tan 45=1&lt;/math&gt;. First, let us expand &lt;math&gt;(1+\tan A)(1+\tan B)&lt;/math&gt;.<br /> <br /> We get &lt;math&gt;1+\tan A+\tan B+\tan A\tan B&lt;/math&gt;. <br /> <br /> Now, let us look at &lt;math&gt;\tan45=\tan(20+25)&lt;/math&gt;.<br /> <br /> By the &lt;math&gt;\tan&lt;/math&gt; sum formula, we know that &lt;math&gt;\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}&lt;/math&gt;<br /> <br /> Then, since &lt;math&gt;\tan 45=1&lt;/math&gt;, we can see that &lt;math&gt;\tan A+\tan B=1-\tan A\tan B&lt;/math&gt;<br /> <br /> Then &lt;math&gt;1=\tan A+\tan B+\tan A\tan B&lt;/math&gt;<br /> <br /> Thus, the sum become &lt;math&gt;1+1=2&lt;/math&gt; and the answer is &lt;math&gt;\fbox{\text{(B)}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to &lt;cmath&gt;\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = &lt;/cmath&gt; &lt;cmath&gt;\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Clearly, that is equal to &lt;cmath&gt;1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Now, we note that &lt;cmath&gt;\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ&lt;/cmath&gt; is equal to &lt;math&gt;\sin 45^\circ&lt;/math&gt;. Now, we would like to get &lt;math&gt;\sin 20^\circ \sin 25^\circ&lt;/math&gt; in the denominator. What springs to mind is the fact that &lt;cmath&gt;\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.&lt;/cmath&gt; Therefore, we can express the desired value as &lt;cmath&gt;1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.&lt;/cmath&gt; Because &lt;math&gt;\cos 45^\circ = \sin 45^\circ&lt;/math&gt;, we see that the fractional part is &lt;math&gt;1&lt;/math&gt;, and so the sum is &lt;math&gt;1 + 1 = 2&lt;/math&gt;, which brings us to the answer &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ===Solution 4(Easiest)===<br /> <br /> Notice that &lt;math&gt;\tan(20^{\circ}+25^{\circ})=\frac{\tan 20^{\circ} + \tan 25^{\circ}}{1-\tan 20^{\circ}\tan 25^{\circ}}&lt;/math&gt;. Because &lt;math&gt;\tan(20^{\circ}+25^{\circ})=1,&lt;/math&gt; we have &lt;math&gt;1-\tan 20^\circ \tan 25^\circ=\tan 20^{\circ} + \tan 25^{\circ}&lt;/math&gt; which means that &lt;math&gt;\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ}\tan25^{\circ}+1=2&lt;/math&gt; which gives us &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> -Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AHSME box|year=1985|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_16&diff=130081 1985 AHSME Problems/Problem 16 2020-08-01T01:46:04Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem==<br /> If &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, then the value of &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> First, let's leave everything in variables and see if we can simplify &lt;math&gt; (1+\tan A)(1+\tan B) &lt;/math&gt;.<br /> <br /> <br /> We can write everything in terms of sine and cosine to get &lt;math&gt; \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> <br /> We can multiply out the numerator to get &lt;math&gt; \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:<br /> <br /> <br /> &lt;math&gt; \cos(A-B)=\sin A\sin B+\cos A\cos B &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin(A+B)=\sin A\cos B+\sin B\cos A &lt;/math&gt;<br /> <br /> <br /> Therefore, our fraction is equal to &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B} &lt;/math&gt;.<br /> <br /> <br /> We can also use the product-to-sum formula<br /> <br /> &lt;math&gt; \cos A\cos B=\frac{1}{2}(\cos(A-B)+\cos(A+B)) &lt;/math&gt; to simplify the denominator:<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\cos(A+B))} &lt;/math&gt;.<br /> <br /> <br /> But now we seem stuck. However, we can note that since &lt;math&gt; A+B=45^\circ &lt;/math&gt;, we have &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we get<br /> <br /> <br /> &lt;math&gt; \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}(\cos(A-B)+\sin(A+B))} &lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; \frac{1}{\frac{1}{2}} &lt;/math&gt;<br /> <br /> &lt;math&gt; 2, \boxed{\text{B}} &lt;/math&gt;<br /> <br /> Note that we only used the fact that &lt;math&gt; \sin(A+B)=\cos(A+B) &lt;/math&gt;, so we have in fact not just shown that &lt;math&gt; (1+\tan A)(1+\tan B)=2 &lt;/math&gt; for &lt;math&gt; A=20^\circ &lt;/math&gt; and &lt;math&gt; B=25^\circ &lt;/math&gt;, but for all &lt;math&gt; A, B &lt;/math&gt; such that &lt;math&gt; A+B=45^\circ+n180^\circ &lt;/math&gt;, for integer &lt;math&gt; n &lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We can see that &lt;math&gt;25^o+20^o=45^o&lt;/math&gt;. We also know that &lt;math&gt;\tan 45=1&lt;/math&gt;. First, let us expand &lt;math&gt;(1+\tan A)(1+\tan B)&lt;/math&gt;.<br /> <br /> We get &lt;math&gt;1+\tan A+\tan B+\tan A\tan B&lt;/math&gt;. <br /> <br /> Now, let us look at &lt;math&gt;\tan45=\tan(20+25)&lt;/math&gt;.<br /> <br /> By the &lt;math&gt;\tan&lt;/math&gt; sum formula, we know that &lt;math&gt;\tan45=\dfrac{\text{tan A}+\text{tan B}}{1- \text{tan A} \text{tan B}}&lt;/math&gt;<br /> <br /> Then, since &lt;math&gt;\tan 45=1&lt;/math&gt;, we can see that &lt;math&gt;\tan A+\tan B=1-\tan A\tan B&lt;/math&gt;<br /> <br /> Then &lt;math&gt;1=\tan A+\tan B+\tan A\tan B&lt;/math&gt;<br /> <br /> Thus, the sum become &lt;math&gt;1+1=2&lt;/math&gt; and the answer is &lt;math&gt;\fbox{\text{(B)}}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to &lt;cmath&gt;\left(1+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(1+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = \left(\frac{\cos 20^\circ}{\cos 20^\circ}+\frac{\sin 20^\circ}{\cos 20^\circ}\right)\left(\frac{\cos 25^\circ}{\cos 25^\circ}+\frac{\sin 25^\circ}{\cos 25^\circ}\right) = &lt;/cmath&gt; &lt;cmath&gt;\frac{\cos 20^\circ \cos 25^\circ + \cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Clearly, that is equal to &lt;cmath&gt;1 + \frac{\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ + \sin 20^\circ \sin 25^\circ}{\cos 20^\circ \cos 25^\circ}.&lt;/cmath&gt; Now, we note that &lt;cmath&gt;\cos 20^\circ \sin 25^\circ + \sin 20^\circ \cos 25^\circ&lt;/cmath&gt; is equal to &lt;math&gt;\sin 45^\circ&lt;/math&gt;. Now, we would like to get &lt;math&gt;\sin 20^\circ \sin 25^\circ&lt;/math&gt; in the denominator. What springs to mind is the fact that &lt;cmath&gt;\cos 20^\circ \cos 25^\circ - \sin 20^\circ \sin 25^\circ = \cos 45^\circ.&lt;/cmath&gt; Therefore, we can express the desired value as &lt;cmath&gt;1 + \frac{\sin 45^\circ + \sin 20^\circ \sin 25^\circ}{\cos 45^\circ + \sin 20^\circ \sin 25^\circ}.&lt;/cmath&gt; Because &lt;math&gt;\cos 45^\circ = \sin 45^\circ&lt;/math&gt;, we see that the fractional part is &lt;math&gt;1&lt;/math&gt;, and so the sum is &lt;math&gt;1 + 1 = 2&lt;/math&gt;, which brings us to the answer &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ===Solution 4(Easiest)===<br /> <br /> Notice that &lt;math&gt;\tan(20^{\circ}+25^{\circ})=\frac{\tan 20^{\circ} + \tan 25^{\circ}}{1-\tan 20^{\circ}\tan 25^{\circ}}&lt;/math&gt;. Because &lt;math&gt;\tan(20^{\circ}+25^{\circ})=1,&lt;/math&gt; we have &lt;math&gt;1-\tan 20^\circ \tan 25^\circ=\tan 20^{\circ} + \tan 25^{\circ}&lt;/math&gt; which means that &lt;math&gt;\tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ}\tan25^{\circ}+1=2&lt;/math&gt; which gives us &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AHSME box|year=1985|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=129615 2020 AMC 12B Problems/Problem 22 2020-07-28T06:40:04Z <p>Pleaseletmewin: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2}&lt;/math&gt; &lt;math&gt;\ge \sqrt{(2^t-3t)(3t)}&lt;/math&gt;. Thus, squaring gives us that &lt;math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;/math&gt;. Rembering what we want to find(divide by &lt;math&gt;4^t&lt;/math&gt;), we get the maximal values as &lt;math&gt;\frac{1}{12}&lt;/math&gt;, and we are done.<br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let the maximum value of the function be &lt;math&gt;m&lt;/math&gt;. Then we have &lt;cmath&gt;\frac{(2^t-3t)t}{4^t} = m \implies m2^{2t} - t2^t + 3t^2 = 0.&lt;/cmath&gt;<br /> Solving for &lt;math&gt;2^{t}&lt;/math&gt;, we see &lt;cmath&gt;2^{t} = \frac{t}{m} \pm \frac{\sqrt{t^2 - 12mt^2}}{m} = \frac{t}{m} \pm \frac{t\sqrt{1 - 12m}}{m}.&lt;/cmath&gt; We see that &lt;math&gt;1 - 12m \geq 0 \implies m \leq \frac{1}{12}.&lt;/math&gt; Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=129614 2020 AMC 12B Problems/Problem 22 2020-07-28T06:39:50Z <p>Pleaseletmewin: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2}&lt;/math&gt; &lt;math&gt;\ge \sqrt{((2^t-3t)(3t))}&lt;/math&gt;. Thus, squaring gives us that &lt;math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;/math&gt;. Rembering what we want to find(divide by &lt;math&gt;4^t&lt;/math&gt;), we get the maximal values as &lt;math&gt;\frac{1}{12}&lt;/math&gt;, and we are done.<br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Let the maximum value of the function be &lt;math&gt;m&lt;/math&gt;. Then we have &lt;cmath&gt;\frac{(2^t-3t)t}{4^t} = m \implies m2^{2t} - t2^t + 3t^2 = 0.&lt;/cmath&gt;<br /> Solving for &lt;math&gt;2^{t}&lt;/math&gt;, we see &lt;cmath&gt;2^{t} = \frac{t}{m} \pm \frac{\sqrt{t^2 - 12mt^2}}{m} = \frac{t}{m} \pm \frac{t\sqrt{1 - 12m}}{m}.&lt;/cmath&gt; We see that &lt;math&gt;1 - 12m \geq 0 \implies m \leq \frac{1}{12}.&lt;/math&gt; Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=128363 2019 AMC 8 Problems/Problem 3 2020-07-15T23:03:57Z <p>Pleaseletmewin: /* Solution 1 */</p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> ==Solution 1==<br /> Each one is in the form &lt;math&gt;\frac{x+4}{x}&lt;/math&gt; so we are really comparing &lt;math&gt;\frac{4}{11}, \frac{4}{15},&lt;/math&gt; and &lt;math&gt;\frac{4}{13}&lt;/math&gt; where you can see &lt;math&gt;\frac{4}{11}&gt;\frac{4}{13}&gt;\frac{4}{15}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We take a common denominator:<br /> &lt;cmath&gt;\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> ~ dolphin7 - I took your idea and made it an explanation.<br /> <br /> ==Solution 3==<br /> When &lt;math&gt;\frac{x}{y}&gt;1&lt;/math&gt; and &lt;math&gt;z&gt;0&lt;/math&gt;, &lt;math&gt;\frac{x+z}{y+z}&lt;\frac{x}{y}&lt;/math&gt;. Hence, the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> ~ ryjs<br /> <br /> This is also similar to Problem 20 on the AMC 2012.<br /> <br /> ==Solution 4(probably won't use this solution)==<br /> We use our insane mental calculator to find out that &lt;math&gt;\frac{15}{11} \approx 1.36&lt;/math&gt;, &lt;math&gt;\frac{19}{15} \approx 1.27&lt;/math&gt;, and &lt;math&gt;\frac{17}{13} \approx 1.31&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ~~ by an insane math guy<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}<br /> Wot blitz</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_8&diff=127869 1994 AIME Problems/Problem 8 2020-07-09T04:34:11Z <p>Pleaseletmewin: /* Solution */</p> <hr /> <div>== Problem ==<br /> The points &lt;math&gt;(0,0)\,&lt;/math&gt;, &lt;math&gt;(a,11)\,&lt;/math&gt;, and &lt;math&gt;(b,37)\,&lt;/math&gt; are the vertices of an equilateral triangle. Find the value of &lt;math&gt;ab\,&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Consider the points on the [[complex plane]]. The point &lt;math&gt;b+37i&lt;/math&gt; is then a rotation of &lt;math&gt;60&lt;/math&gt; degrees of &lt;math&gt;a+11i&lt;/math&gt; about the origin, so:<br /> <br /> &lt;cmath&gt;(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.&lt;/cmath&gt;<br /> <br /> Equating the real and imaginary parts, we have:<br /> <br /> &lt;cmath&gt;\begin{align*}b&amp;=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&amp;=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}&lt;/cmath&gt;<br /> <br /> Solving this system, we find that &lt;math&gt;a=21\sqrt{3}, b=5\sqrt{3}&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> '''Note''': There is another solution where the point &lt;math&gt;b+37i&lt;/math&gt; is a rotation of &lt;math&gt;-60&lt;/math&gt; degrees of &lt;math&gt;a+11i&lt;/math&gt;; however, this triangle is just a reflection of the first triangle by the &lt;math&gt;y&lt;/math&gt;-axis, and the signs of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are flipped. However, the product &lt;math&gt;ab&lt;/math&gt; is unchanged.<br /> <br /> == Solution Two ==<br /> Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? &lt;math&gt;\sqrt{3}&lt;/math&gt; and perpendiculars inspires this solution:<br /> <br /> First, drop a perpendicular from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. Call this midpoint of &lt;math&gt;AB M&lt;/math&gt;. Thus, &lt;math&gt;M=(\frac{a+b}{2}, 24)&lt;/math&gt;. The vector from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;[\frac{a+b}{2}, 24]&lt;/math&gt;. Meanwhile from point &lt;math&gt;M&lt;/math&gt; we can use a vector with &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt; the distance; we have to switch the &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; and our displacement is &lt;math&gt;[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}]&lt;/math&gt;. (Do you see why we switched &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; due to the rotation of 90 degrees?)<br /> <br /> <br /> We see this displacement from &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;[\frac{a-b}{2}, 13]&lt;/math&gt; as well. Equating the two vectors, we get &lt;math&gt;a+b=26\sqrt{3}&lt;/math&gt; and &lt;math&gt;a-b=16\sqrt{3}&lt;/math&gt;. Therefore, &lt;math&gt;a=21\sqrt{3}&lt;/math&gt; and &lt;math&gt;b=5\sqrt{3}&lt;/math&gt;. And the answer is &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> '''Note''': This solution was also present in Titu Andreescu and Zuming Feng's &quot;103 Trigonometry Problems&quot;.<br /> <br /> == Solution Three ==<br /> Plot this equilateral triangle on the complex plane.<br /> Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives &lt;math&gt;(\frac{a+b}{3}, 16i)&lt;/math&gt;. The new coordinates of the equilateral triangle are &lt;math&gt;(-\frac{a+b}{3}-16i), (a-\frac{a+b}{3}-5i), (b-\frac{a+b}{3}+21i)&lt;/math&gt;. These three vertices are solutions of a cubic polynomial of form &lt;math&gt;x^3 + C&lt;/math&gt;. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots r1, r2, and r3, r1r2 + r2r3 + r3r1 = 0.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation &lt;math&gt;5a = 21b.&lt;/math&gt;<br /> Now use the equation with only real parts. This should give you a quadratic &lt;math&gt;a^2 - ab + b^2 = 1083&lt;/math&gt;. Use your previously obtained equation to plug in for a and solve for b, which should yield &lt;math&gt;5\sqrt{3}&lt;/math&gt;. a is then &lt;math&gt;21/5\sqrt{3}&lt;/math&gt;. Multiplying a and b yields &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> {{AIME box|year=1994|num-b=7|num-a=9}}<br /> <br /> <br /> <br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&diff=126112 2017 AMC 12B Problems/Problem 15 2020-06-21T04:54:24Z <p>Pleaseletmewin: /* Solution 4: Computing the Areas */</p> <hr /> <div>==Problem 15==<br /> Let &lt;math&gt;ABC&lt;/math&gt; be an equilateral triangle. Extend side &lt;math&gt;\overline{AB}&lt;/math&gt; beyond &lt;math&gt;B&lt;/math&gt; to a point &lt;math&gt;B'&lt;/math&gt; so that &lt;math&gt;BB'=3AB&lt;/math&gt;. Similarly, extend side &lt;math&gt;\overline{BC}&lt;/math&gt; beyond &lt;math&gt;C&lt;/math&gt; to a point &lt;math&gt;C'&lt;/math&gt; so that &lt;math&gt;CC'=3BC&lt;/math&gt;, and extend side &lt;math&gt;\overline{CA}&lt;/math&gt; beyond &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;A'&lt;/math&gt; so that &lt;math&gt;AA'=3CA&lt;/math&gt;. What is the ratio of the area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; to the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1&lt;/math&gt;<br /> <br /> <br /> ==Solution 1: Law of Cosines==<br /> Solution by HydroQuantum<br /> <br /> <br /> Let &lt;math&gt;AB=BC=CA=x&lt;/math&gt;.<br /> <br /> <br /> Recall The Law of Cosines. Letting &lt;math&gt;A'B'=B'C'=C'A'=y&lt;/math&gt;, &lt;cmath&gt;y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = &lt;/cmath&gt; &lt;cmath&gt;(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24x(cos120)=25x^2+12x^2=37x^2.&lt;/cmath&gt; Since both &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle A'B'C'&lt;/math&gt; are both equilateral triangles, they must be similar due to &lt;math&gt;AA&lt;/math&gt; similarity. This means that &lt;math&gt;\frac{A'B'}{AB}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{B'C'}{BC}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{C'A'}{CA}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{[\triangle A'B'C']}{[\triangle ABC]}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{37}{1}&lt;/math&gt;.<br /> <br /> <br /> Therefore, our answer is &lt;math&gt;\boxed{\textbf{(E) }37:1}&lt;/math&gt;.<br /> <br /> ==Solution 2: Inspection(easiest solution)==<br /> Note that the height and base of &lt;math&gt;\triangle A'CC'&lt;/math&gt; are respectively 4 times and 3 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;. Therefore the area of &lt;math&gt;\triangle A'CC'&lt;/math&gt; is 12 times that of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> By symmetry, &lt;math&gt;\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'&lt;/math&gt;. Adding the areas of these three triangles and &lt;math&gt;\triangle ABC&lt;/math&gt; for the total area of &lt;math&gt;\triangle A'B'C'&lt;/math&gt; gives a ratio of &lt;math&gt;(12 + 12 + 12 + 1) : 1&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;.<br /> <br /> ==Solution 3: Coordinates==<br /> <br /> First we note that &lt;math&gt;A'B'C'\sim ABC&lt;/math&gt; due to symmetry. WLOG, let &lt;math&gt;B = (0, 0)&lt;/math&gt; and &lt;math&gt;AB = 1&lt;/math&gt; Therefore, &lt;math&gt;C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)&lt;/math&gt;. Using the condition that &lt;math&gt;CC' = 3&lt;/math&gt;, we get &lt;math&gt;C' = (4, 0)&lt;/math&gt; and &lt;math&gt;B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)&lt;/math&gt;. It is easy to check that &lt;math&gt;B'C' = \sqrt{37}&lt;/math&gt;. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is &lt;math&gt;\boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;<br /> <br /> Solution by &lt;i&gt;mathwiz0803&lt;/i&gt;<br /> <br /> <br /> ==Solution 4: Computing the Areas==<br /> <br /> Note that angle &lt;math&gt;C'BB'&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;°, as it is supplementary to the equilateral triangle. Then, using area &lt;math&gt;= \frac{1}{2}ab\sin\theta&lt;/math&gt; and letting side &lt;math&gt;AB = 1&lt;/math&gt; for ease, we get: &lt;math&gt;4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}&lt;/math&gt; as the area of &lt;math&gt;C'BB'&lt;/math&gt;. Then, the area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt;, so the ratio is &lt;math&gt;\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}&lt;/math&gt;<br /> <br /> Solution by Aadileo<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems&diff=125943 1985 AIME Problems 2020-06-19T18:50:08Z <p>Pleaseletmewin: /* Problem 9 */</p> <hr /> <div>{{AIME Problems|year=1985}}<br /> <br /> ==Problem 1==<br /> Let &lt;math&gt;x_1=97&lt;/math&gt;, and for &lt;math&gt;n&gt;1&lt;/math&gt; let &lt;math&gt;x_n=\frac{n}{x_{n-1}}&lt;/math&gt;. Calculate the product &lt;math&gt;x_1x_2 \ldots x_8&lt;/math&gt;.<br /> <br /> [[1985 AIME Problems/Problem 1 | Solution]]<br /> ==Problem 2==<br /> When a right triangle is rotated about one leg, the volume of the cone produced is &lt;math&gt;800\pi \;\textrm{cm}^3&lt;/math&gt;. When the triangle is rotated about the other leg, the volume of the cone produced is &lt;math&gt;1920\pi \;\textrm{cm}^3&lt;/math&gt;. What is the length (in cm) of the hypotenuse of the triangle?<br /> <br /> [[1985 AIME Problems/Problem 2 | Solution]]<br /> ==Problem 3==<br /> Find &lt;math&gt;c&lt;/math&gt; if &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers which satisfy &lt;math&gt;c=(a + bi)^3 - 107i&lt;/math&gt;, where &lt;math&gt;i^2 = -1&lt;/math&gt;.<br /> <br /> [[1985 AIME Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> A small square is constructed inside a square of area &lt;math&gt;1&lt;/math&gt; by dividing each side of the unit square into &lt;math&gt;n&lt;/math&gt; equal parts, and then connecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of &lt;math&gt;n&lt;/math&gt; if the the area of the small square is exactly &lt;math&gt;\frac1{1985}&lt;/math&gt;.<br /> <br /> [[Image:AIME 1985 Problem 4.png]]<br /> <br /> [[1985 AIME Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> A sequence of integers &lt;math&gt;a_1, a_2, a_3, \ldots&lt;/math&gt; is chosen so that &lt;math&gt;a_n = a_{n - 1} - a_{n - 2}&lt;/math&gt; for each &lt;math&gt;n \ge 3&lt;/math&gt;. What is the sum of the first &lt;math&gt;2001&lt;/math&gt; terms of this sequence if the sum of the first &lt;math&gt;1492&lt;/math&gt; terms is &lt;math&gt;1985&lt;/math&gt;, and the sum of the first &lt;math&gt;1985&lt;/math&gt; terms is &lt;math&gt;1492&lt;/math&gt;?<br /> <br /> [[1985 AIME Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> As shown in the figure, &lt;math&gt;\triangle ABC&lt;/math&gt; is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> [[Image:AIME 1985 Problem 6.png]]<br /> <br /> [[1985 AIME Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> Assume that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive integers such that &lt;math&gt;a^5 = b^4&lt;/math&gt;, &lt;math&gt;c^3 = d^2&lt;/math&gt; and &lt;math&gt;c - a = 19&lt;/math&gt;. Determine &lt;math&gt;d - b&lt;/math&gt;.<br /> <br /> [[1985 AIME Problems/Problem 7 | Solution]]<br /> <br /> ==Problem 8==<br /> The sum of the following seven numbers is exactly 19: &lt;math&gt;a_1 = 2.56&lt;/math&gt;, &lt;math&gt;a_2 = 2.61&lt;/math&gt;, &lt;math&gt;a_3 = 2.65&lt;/math&gt;, &lt;math&gt;a_4 = 2.71&lt;/math&gt;, &lt;math&gt;a_5 = 2.79&lt;/math&gt;, &lt;math&gt;a_6 = 2.82&lt;/math&gt;, &lt;math&gt;a_7 = 2.86&lt;/math&gt;. It is desired to replace each &lt;math&gt;a_i&lt;/math&gt; by an integer approximation &lt;math&gt;A_i&lt;/math&gt;, &lt;math&gt;1\le i \le 7&lt;/math&gt;, so that the sum of the &lt;math&gt;A_i&lt;/math&gt;'s is also &lt;math&gt;19&lt;/math&gt; and so that &lt;math&gt;M&lt;/math&gt;, the maximum of the &quot;errors&quot; &lt;math&gt;\lvert A_i-a_i \rvert&lt;/math&gt;, is as small as possible. For this minimum &lt;math&gt;M&lt;/math&gt;, what is &lt;math&gt;100M&lt;/math&gt;?<br /> <br /> [[1985 AIME Problems/Problem 8 | Solution]]<br /> <br /> ==Problem 9==<br /> In a circle, parallel chords of lengths &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt; determine central angles of &lt;math&gt;\alpha&lt;/math&gt;, &lt;math&gt;\beta&lt;/math&gt;, and &lt;math&gt;\alpha + \beta &lt;/math&gt; radians, respectively, where &lt;math&gt;\alpha + \beta &lt; \pi&lt;/math&gt;. If &lt;math&gt;\cos \alpha&lt;/math&gt;, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?<br /> <br /> [[1985 AIME Problems/Problem 9 | Solution]]<br /> <br /> ==Problem 10==<br /> How many of the first &lt;math&gt;1000&lt;/math&gt; positive integers can be expressed in the form<br /> <br /> &lt;math&gt;\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor&lt;/math&gt;,<br /> <br /> where &lt;math&gt;x&lt;/math&gt; is a real number, and &lt;math&gt;\lfloor z \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;z&lt;/math&gt;? <br /> <br /> [[1985 AIME Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> An ellipse has foci at &lt;math&gt;(9, 20)&lt;/math&gt; and &lt;math&gt;(49, 55)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane and is tangent to the &lt;math&gt;x&lt;/math&gt;-axis. What is the length of its major axis?<br /> <br /> [[1985 AIME Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> Let &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; be the [[vertex | vertices]] of a regular [[tetrahedron]], each of whose [[edge]]s measures &lt;math&gt;1&lt;/math&gt; meter. A bug, starting from vertex &lt;math&gt;A&lt;/math&gt;, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let &lt;math&gt;p = \frac{n}{729}&lt;/math&gt; be the [[probability]] that the bug is at vertex &lt;math&gt;A&lt;/math&gt; when it has crawled exactly &lt;math&gt;7&lt;/math&gt; meters. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[1985 AIME Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> The numbers in the [[sequence]] &lt;math&gt;101&lt;/math&gt;, &lt;math&gt;104&lt;/math&gt;, &lt;math&gt;109&lt;/math&gt;, &lt;math&gt;116&lt;/math&gt;,&lt;math&gt;\ldots&lt;/math&gt; are of the form &lt;math&gt;a_n=100+n^2&lt;/math&gt;, where &lt;math&gt;n=1,2,3,\ldots&lt;/math&gt;. For each &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;d_n&lt;/math&gt; be the greatest common divisor of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;a_{n+1}&lt;/math&gt;. Find the maximum value of &lt;math&gt;d_n&lt;/math&gt; as &lt;math&gt;n&lt;/math&gt; ranges through the [[positive integer]]s.<br /> <br /> [[1985 AIME Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned &lt;math&gt;\frac{1}{2}&lt;/math&gt; point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?<br /> <br /> [[1985 AIME Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> Three &lt;math&gt;12&lt;/math&gt; cm &lt;math&gt;\times 12&lt;/math&gt; cm [[square (geometry) | squares]] are each cut into two pieces &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, as shown in the first figure below, by joining the [[midpoint]]s of two adjacent sides. These six pieces are then attached to a [[regular polygon | regular]] [[hexagon]], as shown in the second figure, so as to fold into a [[polyhedron]]. What is the [[volume]] (in &lt;math&gt;\mathrm{cm}^3&lt;/math&gt;) of this polyhedron?<br /> <br /> [[Image:AIME 1985 Problem 15.png]]<br /> <br /> [[1985 AIME Problems/Problem 15 | Solution]]<br /> <br /> ==See also==<br /> <br /> {{AIME box|year=1985|before=[[1984 AIME Problems]]|after=[[1986 AIME Problems]]}}<br /> <br /> * [[1985 AIME]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{MAA Notice}}<br /> [[Category:AIME Problems]]</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_13&diff=125730 2018 AIME II Problems/Problem 13 2020-06-17T19:34:17Z <p>Pleaseletmewin: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Misha rolls a standard, fair six-sided die until she rolls &lt;math&gt;1-2-3&lt;/math&gt; in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is &lt;math&gt;\dfrac{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 /> ==Solution 1==<br /> Let &lt;math&gt;P_{odd}=\frac{m}{n}&lt;/math&gt;, with the subscript indicating an odd number of rolls. Then &lt;math&gt;P_{even}=1-\frac{m}{n}&lt;/math&gt;.<br /> <br /> The ratio of &lt;math&gt;\frac{P_{odd}}{P_{even}}&lt;/math&gt; is just &lt;math&gt;\frac{P_{odd}}{1-P_{odd}}&lt;/math&gt;.<br /> <br /> We see that &lt;math&gt;P_{odd}&lt;/math&gt; is the sum of &lt;math&gt;P_3&lt;/math&gt;,&lt;math&gt;P_5&lt;/math&gt;,&lt;math&gt;P_7&lt;/math&gt;,... , while &lt;math&gt;P_{even}&lt;/math&gt; is the sum of &lt;math&gt;P_4&lt;/math&gt;, &lt;math&gt;P_6&lt;/math&gt;, &lt;math&gt;P_8&lt;/math&gt;,... .<br /> <br /> &lt;math&gt;P_3&lt;/math&gt;, the probability of getting rolls of 1-2-3 in exactly 3 rolls, is obviously &lt;math&gt;\frac{1}{216}&lt;/math&gt;.<br /> <br /> We set this probability of &lt;math&gt;P_3&lt;/math&gt; aside, meaning we totally remove the chance of getting 1-2-3 in 3 rolls. Now the ratio of &lt;math&gt;P_4+P_6+P_8+...&lt;/math&gt; to &lt;math&gt;P_5+P_7+P_9+...&lt;/math&gt; should be equal to the ratio of &lt;math&gt;\frac{P_{odd}}{P_{even}}&lt;/math&gt;, because in this case the 1st roll no longer matters, so we can disregard the very existence of it in counting how many times of rolls, and thus, 4 rolls, 6 rolls, 8 rolls... would become an odd number of rolls (while 5 rolls, 7 rolls, 9 rolls... would become even number of rolls).<br /> <br /> Notice &lt;math&gt;P_4+P_6+P_8+...=P_{even}&lt;/math&gt;, and also &lt;math&gt;P_5+P_7+P_9+...=P_{odd}-P_3=P_{odd}-\frac{1}{216}&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{P_{even}}{P_{odd}-\frac{1}{216}}=\frac{P_{odd}}{P_{even}}&lt;/math&gt;.<br /> <br /> Finally, we get &lt;math&gt;P_{odd}=\frac{m}{n}=\frac{216}{431}&lt;/math&gt;. <br /> Therefore, &lt;math&gt;m+n = \boxed{647}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Call the probability you win on a certain toss &lt;math&gt;f_n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the toss number.<br /> Obviously, since the sequence has length 3, &lt;math&gt;f_1=0&lt;/math&gt; and &lt;math&gt;f_2=0&lt;/math&gt;.<br /> Additionally, &lt;math&gt;f_3=\left(\frac{1}{6}\right)^3&lt;/math&gt;. We can call this value &lt;math&gt;x&lt;/math&gt;, to keep our further equations looking clean.<br /> We can now write our general form for &lt;math&gt;f&lt;/math&gt; as &lt;math&gt;f_n=x(1-\sum_{i=1}^{n-3}f_i)&lt;/math&gt;. This factors the probability of the last 3 rolls being 1-2-3, and the important probability that the sequence has not been rolled in the past (because then the game would already be over).<br /> Note that &lt;math&gt;\sum_{i=1}^{\infty}f_i=1&lt;/math&gt; since you'll win at some point.<br /> An intermediate step here is figuring out &lt;math&gt;f_n-f_{n+1}&lt;/math&gt;. This is equal to &lt;math&gt;x(1-\sum_{i=1}^{n-3}f_i)-x(1-\sum_{i=1}^{n-2}f_i)=x(\sum_{i=1}^{n-2}f_i-\sum_{i=1}^{n-3}f_i)=xf_{n-2}&lt;/math&gt;.<br /> Adding up all the differences, i.e. &lt;math&gt;\sum_{i=2}^{\infty}(f_{2n-1}-f_{2n})&lt;/math&gt; will give us the amount by which the odds probability exceeds the even probability. Since they sum to 1, that means the odds probability will be half of the difference above one-half. Subbing in our earlier result from the intermediate step, the odd probability is equal to &lt;math&gt;\frac{1}{2}+\frac{1}{2}x\sum_{i=2}^{\infty}f_{2n-3}&lt;/math&gt;.<br /> Another way to find the odd probability is simply summing it up, which turns out to be &lt;math&gt;\sum_{i=1}^{\infty}f_{2n-1}&lt;/math&gt;. Note the infinite sums in both expressions are equal; let's call it &lt;math&gt;P&lt;/math&gt;. Equating them gives &lt;math&gt;\frac{1}{2}+\frac{1}{2}xP=P&lt;/math&gt;, or &lt;math&gt;P=\frac{1}{2-x}&lt;/math&gt;.<br /> Finally, substituting &lt;math&gt;x=\frac{1}{216}&lt;/math&gt;, we find that &lt;math&gt;P=\frac{216}{431}&lt;/math&gt;, giving us a final answer of &lt;math&gt;216 + 431 = \boxed{647}&lt;/math&gt;.<br /> --DanDan0101<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; be the number of strings of length &lt;math&gt;n&lt;/math&gt; containing the digits &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;6&lt;/math&gt; that do not contain the string &lt;math&gt;123&lt;/math&gt;. Then we have &lt;math&gt;S(n) = 6 \cdot S(n-1) - S(n-3)&lt;/math&gt; because we can add any digit to end of a string with length &lt;math&gt;n-1&lt;/math&gt; but we have to subtract all the strings that end in &lt;math&gt;123&lt;/math&gt;. We rewrite this as<br /> &lt;cmath&gt;\begin{align*}<br /> S(n) &amp;= 6 \cdot S(n-1) - S(n-3) \\ &amp;= 6 \cdot (6 \cdot S(n-2) - S(n-4)) - (6 \cdot (S(n-4) - S(n-6)) \\ &amp;= 36 \cdot S(n-2) - 12 \cdot S(n-4) + S(n-6)<br /> \end{align*}&lt;/cmath&gt; We wish to compute &lt;math&gt;P=\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}&lt;/math&gt; since the last three rolls are &lt;math&gt;123&lt;/math&gt; for the game to end. Summing over the recursion, we obtain<br /> &lt;cmath&gt;\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} =36 \cdot \sum_{n=0}^\infty \frac{S(2n-2)}{6^{2n+3}} - 12 \cdot \sum_{n=0}^\infty \frac{S(2n-4)}{6^{2n+3}}+ \sum_{n=0}^\infty \frac{S(2n-6)}{6^{2n+3}} &lt;/cmath&gt;Now shift the indices so that the inside term is the same:<br /> &lt;cmath&gt;\begin{align*}<br /> \sum_{n=3}^\infty \frac{S(2n)}{6^{2n+3}} &amp;= \frac{36}{6^2} \cdot \sum_{n=2}^\infty \frac{S(2n)}{6^{2n+3}} - \frac{12}{6^4} \cdot \sum_{n=1}^\infty \frac{S(2n)}{6^{2n+3}} + \frac{1}{6^6} \cdot \sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} \\ \left(P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5} -\frac{S(4)}{6^7} \right) &amp;= \frac{36}{6^2} \cdot \left( P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{S(0)}{6^3} \right) + \frac{1}{6^6} \cdot P <br /> \end{align*}&lt;/cmath&gt;Note that &lt;math&gt;S(0) = 1, S(2) = 36&lt;/math&gt; and &lt;math&gt;S(4) = 6^4 - 2 \cdot 6 = 1284&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{align*}<br /> \left(P - \frac{1}{6^3} - \frac{36}{6^5} -\frac{1284}{6^7} \right) = \frac{36}{6^2} \cdot \left( P - \frac{1}{6^3} - \frac{36}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{1}{6^3} \right) + \frac{1}{6^6} \cdot P <br /> \end{align*}&lt;/cmath&gt;Solving for &lt;math&gt;P&lt;/math&gt;, we obtain &lt;math&gt;P = \frac{216}{431} \implies m+n = \boxed{647}&lt;/math&gt;. <br /> <br /> -Vfire<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;A=\frac{1}{6} \begin{bmatrix}<br /> 5 &amp; 1 &amp; 0 &amp; 0 \\<br /> 4 &amp; 1 &amp; 1 &amp; 0 \\<br /> 4 &amp; 1 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> \end{bmatrix}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly &lt;math&gt;k&lt;/math&gt; times is &lt;math&gt;(A^k)_{1,4}&lt;/math&gt;. Thus the probability that we roll the dice an odd number of times is &lt;math&gt;1-\left(\sum_{k=0}^\infty A^{2k}\right)_{1,4} = 1-\left((I - A^2)^{-1}\right)_{1,4} = \frac{216}{431}&lt;/math&gt;. Thus the answer is &lt;math&gt;216+431=\boxed{647}&lt;/math&gt;.<br /> ==Solution 5 quick cheat ==<br /> Consider it as a contest of Odd and Even. Let &lt;math&gt;P_o&lt;/math&gt; and &lt;math&gt;P_e&lt;/math&gt; be probability that Odd and Even wins, respectively. If we consider every 3 rolls as an atomic action, then we can have a simple solution. If the rolls is 1-2-3, Odd wins; otherwise, Odd and Even switch the odds of winning. Therefore, we have<br /> &lt;cmath&gt; P_o = \frac{1}{216} + \frac{215}{216}P_e&lt;/cmath&gt;<br /> Plug in &lt;math&gt;P_e = 1 - P_o&lt;/math&gt; and we can easily solve for &lt;math&gt;Po=\frac{216}{431}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\boxed{647}&lt;/math&gt;.<br /> <br /> Of course this is not a rigorous solution. I think it works because the requirement is a strict sequence of pure random events.<br /> <br /> -Mathdummy<br /> <br /> ==Solution 6 Elementary Probability ==<br /> Consider it a contest for Odd or Even to win. Let &lt;math&gt;P_o&lt;/math&gt;, &lt;math&gt;P_e&lt;/math&gt; be the winning probabilities respectively. We call Odd is &quot;in position&quot; when a new sequence of 1-2-3 starts at odd position, and likewise, call Even is &quot;in position&quot; when a new sequence starts at even position. Now consider the situation when the first roll is 1. The conditional probability for Odd or Even to eventually win out depends on whose is in position. So let's denote by &lt;math&gt;P_o(1), P_e(1)&lt;/math&gt; the probabilities of Odd and Even winning out, respectively, both when Odd is in position. Remember that the probabilities simply switch if Even is in position. Similarly, after 1-2 is rolled, we denote by &lt;math&gt;P_o(2), P_e(2)&lt;/math&gt; the conditional probabilities of Odd and Even winning out, when Odd is in position.<br /> <br /> Consider the first roll. If it's not a 1, the sequence restart, but Even is now in position; if it's a 1, then Odd's winning probability becomes &lt;math&gt;P_o(1)&lt;/math&gt;. So,<br /> &lt;cmath&gt;P_o = \frac{1}{6}P_o(1) + \frac{5}{6}P_e&lt;/cmath&gt;<br /> In the next roll, there are 3 outcomes. If the roll is 2, then Odd's winning probability becomes &lt;math&gt;P_o(2)&lt;/math&gt;; if the roll is 1, then we stay in the sequence, but Even is now in position, so the probability of Odd winning now becomes P_e(1); if the rolls is any other number, then the sequence restarts, and Odd is still in position. So,<br /> &lt;cmath&gt; P_o(1) = \frac{1}{6}P_o(2) + \frac{1}{6}P_e(1) + \frac{4}{6}P_o&lt;/cmath&gt;<br /> In the next roll after a 1-2 sequence, there are 3 outcomes. If the roll is a 3, Odd wins; if it's a 1, we go back to the state when 1 is just rolled, and Odd is in position; if it's any other number, then the sequence restarts, and Even is in position. So,<br /> &lt;cmath&gt; P_o(2) = \frac{1}{6} + \frac{1}{6}P_o(1) + \frac{4}{6}P_e&lt;/cmath&gt;<br /> <br /> Plug in &lt;math&gt;P_e = 1-P_o&lt;/math&gt; and &lt;math&gt;P_e(1) = 1 - P_o(1)&lt;/math&gt;, we have a 3-equation linear system which is not hard to solve. The final answer is &lt;math&gt;Po=\frac{216}{431}&lt;/math&gt;. &lt;math&gt;\boxed{647}&lt;/math&gt;.<br /> <br /> -Mathdummy<br /> <br /> {{AIME box|year=2018|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125550 2016 AMC 12A Problems/Problem 17 2020-06-15T21:47:14Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),&lt;/math&gt; and &lt;math&gt;(3, -\sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives &lt;math&gt;EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12<br /> \sqrt{3}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12\sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125472 2016 AMC 12A Problems/Problem 17 2020-06-15T06:57:05Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),&lt;/math&gt; and ,&lt;math&gt;(3, -\sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives &lt;math&gt;EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12<br /> \sqrt{3}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12\sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125470 2016 AMC 12A Problems/Problem 17 2020-06-15T06:44:21Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),&lt;/math&gt; and ,&lt;math&gt;(3, -\sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives &lt;math&gt;EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12<br /> \sqrt{3}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12\sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \frac{2+\sqrt{3}}{3}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125469 2016 AMC 12A Problems/Problem 17 2020-06-15T06:41:49Z <p>Pleaseletmewin: /* Solution(Coordinates) */</p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),&lt;/math&gt; and ,&lt;math&gt;(3, -\sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives &lt;math&gt;EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12<br /> sqrt{3}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12\sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \frac{2+\sqrt{3}}{3}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125468 2016 AMC 12A Problems/Problem 17 2020-06-15T06:37:27Z <p>Pleaseletmewin: I fixed my LaTex</p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),&lt;/math&gt; and ,&lt;math&gt;(3, -\sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives EF = &lt;math&gt;sqrt((3-(-\sqrt{3}))^2+(6+\sqrt{3}-3)^2)=\sqrt(24+12\sqrt{3})&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12\sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12sqrt3}{36} = \frac{2+sqrt3}{3}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125467 2016 AMC 12A Problems/Problem 17 2020-06-15T06:34:01Z <p>Pleaseletmewin: </p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-sqrt{3}, 3), (3, 6+sqrt{3}), (6+sqrt{3}, 3), and (3, -sqrt{3})&lt;/math&gt;.<br /> <br /> Using the first two points, this gives EF = &lt;math&gt;sqrt((3-(-sqrt{3}))^2+(6+sqrt{3}-3)^2)=sqrt(24+12sqrt{3})&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12sqrt{3}&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;[ABCD] = 36&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\frac{[EFGH]}{[ABCD]} = \frac{24+12sqrt3}{36} = \frac{2+sqrt3}{3}&lt;/math&gt;<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=125466 2016 AMC 12A Problems/Problem 17 2020-06-15T06:28:18Z <p>Pleaseletmewin: I added another solution to this page.</p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution(Coordinates)==<br /> <br /> Note that we can also use coordinates to solve this problem. WLOG, set the side length of square &lt;math&gt;ABCD&lt;/math&gt; equal to &lt;math&gt;6&lt;/math&gt;.<br /> <br /> This makes the coordinates of the square &lt;math&gt;EFGH&lt;/math&gt; equal to &lt;math&gt;(-sqrt3, 3), (3, 6+sqrt3), (6+sqrt3, 3), and (3, -sqrt3)&lt;/math&gt;.<br /> <br /> Using the first two points, this gives EF = &lt;math&gt;sqrt((3-(-sqrt3))^2+(6+sqrt3-3)^2)=sqrt(24+12sqrt3)&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;[EFGH]=24+12sqrt3&lt;/math&gt;.<br /> <br /> Because the side length of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, [ABCD] = 36.<br /> <br /> Therefore, $\frac{[EFGH]}{[ABCD]} = \frac{24+12sqrt3}{36} = \frac{2+sqrt3}{3}<br /> <br /> - Pleaseletmewin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_25&diff=125465 2016 AMC 12A Problems/Problem 25 2020-06-15T06:13:25Z <p>Pleaseletmewin: /* Solution 2(Cheap Realization) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + \dots + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider &lt;math&gt;f(2)&lt;/math&gt;. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least &lt;math&gt;100&lt;/math&gt;. Calculus &lt;math&gt;\left(\frac{\text{d}}{\text{d}x} x^2=2x\right)&lt;/math&gt; and a bit of thinking says this first happens at &lt;math&gt;x\ge 100/2 = 50&lt;/math&gt;*. The perfect squares from here go: &lt;math&gt;2500, 2601, 2704, 2809\dots&lt;/math&gt;. Note that the ones and tens also make the perfect squares, &lt;math&gt;1^2,2^2,3^2\dots&lt;/math&gt;. After the ones and tens make &lt;math&gt;100&lt;/math&gt;, the hundreds place will go up by &lt;math&gt;2&lt;/math&gt;, thus reaching our goal. Since &lt;math&gt;10^2=100&lt;/math&gt;, the last perfect square to be written will be &lt;math&gt;\left(50+10\right)^2=60^2=3600&lt;/math&gt;. The missing number is one less than the number of hundreds &lt;math&gt;(k=2)&lt;/math&gt; of &lt;math&gt;3600&lt;/math&gt;, or &lt;math&gt;35&lt;/math&gt;.<br /> <br /> Now consider f(4). Instead of the difference between two squares needing to be &lt;math&gt;100&lt;/math&gt;, the difference must now be &lt;math&gt;10000&lt;/math&gt;. This first happens at &lt;math&gt;x\ge 5000&lt;/math&gt;. After this point, similarly, &lt;math&gt;\sqrt{10000}=100&lt;/math&gt; more numbers are needed to make the &lt;math&gt;10^4&lt;/math&gt; th's place go up by &lt;math&gt;2&lt;/math&gt;. This will take place at &lt;math&gt;\left(5000+100\right)^2=5100^2= 26010000&lt;/math&gt;. Removing the last four digits (the zeros) and subtracting one yields &lt;math&gt;2600&lt;/math&gt; for the skipped value.<br /> <br /> In general, each new value of &lt;math&gt;f(k+2)&lt;/math&gt; will add two digits to the &quot;&lt;math&gt;5&lt;/math&gt;&quot; and one digit to the &quot;&lt;math&gt;1&lt;/math&gt;&quot;. This means that the last number Bernardo writes for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;\left(500000+1000\right)^2&lt;/math&gt;, the last for &lt;math&gt;k = 8&lt;/math&gt; will be &lt;math&gt;\left(50000000+10000\right)^2&lt;/math&gt;, and so on until &lt;math&gt;k=2016&lt;/math&gt;. Removing the last &lt;math&gt;k&lt;/math&gt; digits as Silvia does will be the same as removing &lt;math&gt;k/2&lt;/math&gt; trailing zeroes on the number to be squared. This means that the last number on the board for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;5001^2&lt;/math&gt;, &lt;math&gt;k=8&lt;/math&gt; is &lt;math&gt;50001^2&lt;/math&gt;, and so on. So the first missing number is &lt;math&gt;5001^2-1,50001^2-1\text{ etc.}&lt;/math&gt; The squaring will make a &quot;&lt;math&gt;25&lt;/math&gt;&quot; with two more digits than the last number, a &quot;&lt;math&gt;10&lt;/math&gt;&quot; with one more digit, and a &quot;&lt;math&gt;1&lt;/math&gt;&quot;. The missing number is one less than that, so the &quot;1&quot; will be subtracted from &lt;math&gt;f(k)&lt;/math&gt;. In other words, &lt;math&gt;f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}&lt;/math&gt;.<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;f(2) =35 =25 +10&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4) =2600 =2500 +100&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6) =251000 =250000 +1000&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8) = 25010000 = 25000000 + 10000&lt;/cmath&gt;<br /> <br /> And so on. The sum &lt;math&gt;f(2) + f(4) + f(6) +\dots + f(2016)&lt;/math&gt; is:<br /> <br /> &lt;math&gt;2.52525252525\dots 2525\cdot 10^{2015}&lt;/math&gt; + &lt;math&gt;1.11111\dots 110\cdot 10^{1008}&lt;/math&gt;, with &lt;math&gt;2016&lt;/math&gt; repetitions each of &quot;&lt;math&gt;25&lt;/math&gt;&quot; and &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> There is no carrying in this addition. Therefore each &lt;math&gt;f(k)&lt;/math&gt; adds &lt;math&gt;2 + 5 + 1 = 8&lt;/math&gt; to the sum of the digits. <br /> Since &lt;math&gt;2n = 2016&lt;/math&gt;, &lt;math&gt;n = 1008&lt;/math&gt;, and &lt;math&gt;8n = 8064&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E)}\text{ 8064}}&lt;/math&gt;.<br /> <br /> Addendum: *You could also use the fact that &lt;cmath&gt;(x+1)^2 = x^2 +2x+1&lt;/cmath&gt; <br /> In other words, the difference between &lt;math&gt;x^2&lt;/math&gt; and &lt;math&gt;(x+1)^2&lt;/math&gt; is equal to &lt;math&gt;2x+1&lt;/math&gt;. We can set the inequality &lt;math&gt;2x+1 \geq 100&lt;/math&gt;. Obviously, the first integer &lt;math&gt;x&lt;/math&gt; that satisfies this is 50.<br /> This way, while being longer, is IMO more motivated and doesn't use calculus.<br /> <br /> ==Solution 2(Cheap Realization)==<br /> <br /> If you are one of those people who are willing take educated guesses, then just realize that &lt;math&gt;\textbf{(E)}\text{ 8064}&lt;/math&gt; is the only answer choice that is a multiple of &lt;math&gt;2016&lt;/math&gt;.<br /> <br /> ==See Also==<br /> Related Question: https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6<br /> {{AMC12 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_25&diff=125464 2016 AMC 12A Problems/Problem 25 2020-06-15T06:13:12Z <p>Pleaseletmewin: /* Solution 2(Cheap Realization) */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;k&lt;/math&gt; be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with &lt;math&gt;k+1&lt;/math&gt; digits. Every time Bernardo writes a number, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it. Bernardo then writes the next perfect square, Silvia erases the last &lt;math&gt;k&lt;/math&gt; digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let &lt;math&gt;f(k)&lt;/math&gt; be the smallest positive integer not written on the board. For example, if &lt;math&gt;k = 1&lt;/math&gt;, then the numbers that Bernardo writes are &lt;math&gt;16, 25, 36, 49, 64&lt;/math&gt;, and the numbers showing on the board after Silvia erases are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;f(1) = 5&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;f(2) + f(4)+ f(6) + \dots + f(2016)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider &lt;math&gt;f(2)&lt;/math&gt;. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least &lt;math&gt;100&lt;/math&gt;. Calculus &lt;math&gt;\left(\frac{\text{d}}{\text{d}x} x^2=2x\right)&lt;/math&gt; and a bit of thinking says this first happens at &lt;math&gt;x\ge 100/2 = 50&lt;/math&gt;*. The perfect squares from here go: &lt;math&gt;2500, 2601, 2704, 2809\dots&lt;/math&gt;. Note that the ones and tens also make the perfect squares, &lt;math&gt;1^2,2^2,3^2\dots&lt;/math&gt;. After the ones and tens make &lt;math&gt;100&lt;/math&gt;, the hundreds place will go up by &lt;math&gt;2&lt;/math&gt;, thus reaching our goal. Since &lt;math&gt;10^2=100&lt;/math&gt;, the last perfect square to be written will be &lt;math&gt;\left(50+10\right)^2=60^2=3600&lt;/math&gt;. The missing number is one less than the number of hundreds &lt;math&gt;(k=2)&lt;/math&gt; of &lt;math&gt;3600&lt;/math&gt;, or &lt;math&gt;35&lt;/math&gt;.<br /> <br /> Now consider f(4). Instead of the difference between two squares needing to be &lt;math&gt;100&lt;/math&gt;, the difference must now be &lt;math&gt;10000&lt;/math&gt;. This first happens at &lt;math&gt;x\ge 5000&lt;/math&gt;. After this point, similarly, &lt;math&gt;\sqrt{10000}=100&lt;/math&gt; more numbers are needed to make the &lt;math&gt;10^4&lt;/math&gt; th's place go up by &lt;math&gt;2&lt;/math&gt;. This will take place at &lt;math&gt;\left(5000+100\right)^2=5100^2= 26010000&lt;/math&gt;. Removing the last four digits (the zeros) and subtracting one yields &lt;math&gt;2600&lt;/math&gt; for the skipped value.<br /> <br /> In general, each new value of &lt;math&gt;f(k+2)&lt;/math&gt; will add two digits to the &quot;&lt;math&gt;5&lt;/math&gt;&quot; and one digit to the &quot;&lt;math&gt;1&lt;/math&gt;&quot;. This means that the last number Bernardo writes for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;\left(500000+1000\right)^2&lt;/math&gt;, the last for &lt;math&gt;k = 8&lt;/math&gt; will be &lt;math&gt;\left(50000000+10000\right)^2&lt;/math&gt;, and so on until &lt;math&gt;k=2016&lt;/math&gt;. Removing the last &lt;math&gt;k&lt;/math&gt; digits as Silvia does will be the same as removing &lt;math&gt;k/2&lt;/math&gt; trailing zeroes on the number to be squared. This means that the last number on the board for &lt;math&gt;k=6&lt;/math&gt; is &lt;math&gt;5001^2&lt;/math&gt;, &lt;math&gt;k=8&lt;/math&gt; is &lt;math&gt;50001^2&lt;/math&gt;, and so on. So the first missing number is &lt;math&gt;5001^2-1,50001^2-1\text{ etc.}&lt;/math&gt; The squaring will make a &quot;&lt;math&gt;25&lt;/math&gt;&quot; with two more digits than the last number, a &quot;&lt;math&gt;10&lt;/math&gt;&quot; with one more digit, and a &quot;&lt;math&gt;1&lt;/math&gt;&quot;. The missing number is one less than that, so the &quot;1&quot; will be subtracted from &lt;math&gt;f(k)&lt;/math&gt;. In other words, &lt;math&gt;f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}&lt;/math&gt;.<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;f(2) =35 =25 +10&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4) =2600 =2500 +100&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6) =251000 =250000 +1000&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8) = 25010000 = 25000000 + 10000&lt;/cmath&gt;<br /> <br /> And so on. The sum &lt;math&gt;f(2) + f(4) + f(6) +\dots + f(2016)&lt;/math&gt; is:<br /> <br /> &lt;math&gt;2.52525252525\dots 2525\cdot 10^{2015}&lt;/math&gt; + &lt;math&gt;1.11111\dots 110\cdot 10^{1008}&lt;/math&gt;, with &lt;math&gt;2016&lt;/math&gt; repetitions each of &quot;&lt;math&gt;25&lt;/math&gt;&quot; and &quot;&lt;math&gt;1&lt;/math&gt;&quot;. <br /> There is no carrying in this addition. Therefore each &lt;math&gt;f(k)&lt;/math&gt; adds &lt;math&gt;2 + 5 + 1 = 8&lt;/math&gt; to the sum of the digits. <br /> Since &lt;math&gt;2n = 2016&lt;/math&gt;, &lt;math&gt;n = 1008&lt;/math&gt;, and &lt;math&gt;8n = 8064&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(E)}\text{ 8064}}&lt;/math&gt;.<br /> <br /> Addendum: *You could also use the fact that &lt;cmath&gt;(x+1)^2 = x^2 +2x+1&lt;/cmath&gt; <br /> In other words, the difference between &lt;math&gt;x^2&lt;/math&gt; and &lt;math&gt;(x+1)^2&lt;/math&gt; is equal to &lt;math&gt;2x+1&lt;/math&gt;. We can set the inequality &lt;math&gt;2x+1 \geq 100&lt;/math&gt;. Obviously, the first integer &lt;math&gt;x&lt;/math&gt; that satisfies this is 50.<br /> This way, while being longer, is IMO more motivated and doesn't use calculus.<br /> <br /> ==Solution 2(Cheap Realization)==<br /> <br /> If you are one of those people who are willing take educated guesses, then just realize that &lt;math&gt;\textbf{(E)}\text{ 8064}&lt;/math&gt; is the only answer choice that is a multiple of &lt;math&gt;2016&lt;/math&gt;<br /> <br /> ==See Also==<br /> Related Question: https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6<br /> {{AMC12 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Pleaseletmewin https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems&diff=125414 1982 AHSME Problems 2020-06-14T23:09:51Z <p>Pleaseletmewin: /* Problem 29 */</p> <hr /> <div>{{AHSME Problems<br /> |year = 1982<br /> }}<br /> == Problem 1 ==<br /> <br /> When the polynomial &lt;math&gt;x^3-2&lt;/math&gt; is divided by the polynomial &lt;math&gt;x^2-2&lt;/math&gt;, the remainder is <br /> <br /> &lt;math&gt;\text{(A)} \ 2 \qquad <br /> \text{(B)} \ -2 \qquad <br /> \text{(C)} \ -2x-2 \qquad <br /> \text{(D)} \ 2x+2 \qquad <br /> \text{(E)} \ 2x-2&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> If a number eight times as large as &lt;math&gt;x&lt;/math&gt; is increased by two, then one fourth of the result equals <br /> <br /> &lt;math&gt;\text{(A)} \ 2x + \frac{1}{2} \qquad <br /> \text{(B)} \ x + \frac{1}{2} \qquad <br /> \text{(C)} \ 2x+2 \qquad <br /> \text{(D)}\ 2x+4 \qquad<br /> \text{(E)}\ 2x+16 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Evaluate &lt;math&gt;(x^x)^{(x^x)}&lt;/math&gt; at &lt;math&gt;x = 2&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{(A)} \ 16 \qquad <br /> \text{(B)} \ 64 \qquad <br /> \text{(C)} \ 256 \qquad <br /> \text{(D)} \ 1024 \qquad <br /> \text{(E)} \ 65,536 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, <br /> measured in square centimeters. The radius of the semicircle, measured in centimeters, is <br /> <br /> &lt;math&gt;\text{(A)} \ \pi \qquad <br /> \text{(B)} \ \frac{2}{\pi} \qquad <br /> \text{(C)} \ 1 \qquad <br /> \text{(D)} \ \frac{1}{2}\qquad<br /> \text{(E)} \ \frac{4}{\pi}+2 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Two positive numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are in the ratio &lt;math&gt;a: b&lt;/math&gt; where &lt;math&gt;0 &lt; a &lt; b&lt;/math&gt;. If &lt;math&gt;x+y = c&lt;/math&gt;, then the smaller of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ \frac{ac}{b} \qquad <br /> \text{(B)} \ \frac{bc-ac}{b} \qquad <br /> \text{(C)} \ \frac{ac}{a+b} \qquad <br /> \text{(D)}\ \frac{bc}{a+b}\qquad<br /> \text{(E)}\ \frac{ac}{b-a} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> The sum of all but one of the interior angles of a convex polygon equals &lt;math&gt;2570^\circ&lt;/math&gt;. The remaining angle is <br /> <br /> &lt;math&gt;\text{(A)} \ 90^\circ \qquad <br /> \text{(B)} \ 105^\circ \qquad <br /> \text{(C)} \ 120^\circ \qquad <br /> \text{(D)}\ 130^\circ\qquad<br /> \text{(E)}\ 144^\circ &lt;/math&gt;<br /> <br /> [[1982 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> If the operation &lt;math&gt;x \star y&lt;/math&gt; is defined by &lt;math&gt;x \star y = (x+1)(y+1) - 1&lt;/math&gt;, then which one of the following is FALSE? <br /> <br /> &lt;math&gt;\text{(A)} \ x \star y = y\star x \text{ for all real } x,y. \\<br /> \text{(B)} \ x \star (y + z) = ( x \star y ) + (x \star z) \text{ for all real } x,y, \text{ and } z.\\<br /> \text{(C)} \ (x-1) \star (x+1) = (x \star x) - 1 \text{ for all real } x. \\<br /> \text{(D)} \ x \star 0 = x \text{ for all real } x. \\<br /> \text{(E)} \ x \star (y \star z) = (x \star y) \star z \text{ for all real } x,y, \text{ and } z. &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> By definition, &lt;math&gt;r! = r(r - 1) \cdots 1&lt;/math&gt; and &lt;math&gt;\binom{j}{k} = \frac {j!}{k!(j - k)!}&lt;/math&gt;, where &lt;math&gt;r,j,k&lt;/math&gt; are positive integers and &lt;math&gt;k &lt; j&lt;/math&gt;. <br /> If &lt;math&gt;\binom{n}{1}, \binom{n}{2}, \binom{n}{3}&lt;/math&gt; form an arithmetic progression with &lt;math&gt;n &gt; 3&lt;/math&gt;, then &lt;math&gt;n&lt;/math&gt; equals <br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\qquad <br /> \textbf{(B)}\ 7\qquad <br /> \textbf{(C)}\ 9\qquad <br /> \textbf{(D)}\ 11\qquad <br /> \textbf{(E)}\ 12&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> A vertical line divides the triangle with vertices &lt;math&gt;(0,0), (1,1)&lt;/math&gt;, and &lt;math&gt;(9,1)&lt;/math&gt; in the &lt;math&gt;xy\text{-plane}&lt;/math&gt; into two regions of equal area. <br /> The equation of the line is &lt;math&gt;x=&lt;/math&gt; <br /> <br /> &lt;math&gt;\text {(A)} 2.5 \qquad <br /> \text {(B)} 3.0 \qquad <br /> \text {(C)} 3.5 \qquad <br /> \text {(D)} 4.0\qquad <br /> \text {(E)} 4.5 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> In the adjoining diagram, &lt;math&gt;BO&lt;/math&gt; bisects &lt;math&gt;\angle CBA&lt;/math&gt;, &lt;math&gt;CO&lt;/math&gt; bisects &lt;math&gt;\angle ACB&lt;/math&gt;, and &lt;math&gt;MN&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. <br /> If &lt;math&gt;AB=12, BC=24&lt;/math&gt;, and &lt;math&gt;AC=18&lt;/math&gt;, then the perimeter of &lt;math&gt;\triangle AMN&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18)), O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0));<br /> draw(B--M--O--B--C--O--N--C^^N--A--M);<br /> label(&quot;$A$&quot;, A, dir(90));<br /> label(&quot;$B$&quot;, B, dir(O--B));<br /> label(&quot;$C$&quot;, C, dir(O--C));<br /> label(&quot;$M$&quot;, M, dir(90)*dir(B--A));<br /> label(&quot;$N$&quot;, N, dir(90)*dir(A--C));<br /> label(&quot;$O$&quot;, O, dir(90));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text {(A)} 30 \qquad <br /> \text {(B)} 33 \qquad <br /> \text {(C)} 36 \qquad <br /> \text {(D)} 39 \qquad <br /> \text {(E)} 42 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> How many integers with four different digits are there between &lt;math&gt;1,000&lt;/math&gt; and &lt;math&gt;9,999&lt;/math&gt; such that the absolute value of <br /> the difference between the first digit and the last digit is &lt;math&gt;2&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text {(A)} 672 \qquad <br /> \text {(B)} 784 \qquad <br /> \text {(C)} 840 \qquad <br /> \text {(D)} 896 \qquad <br /> \text {(E)} 1008&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> Let &lt;math&gt;f(x) = ax^7+bx^3+cx-5&lt;/math&gt;, where &lt;math&gt;a,b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are constants. If &lt;math&gt;f(-7) = 7&lt;/math&gt;, the &lt;math&gt;f(7)&lt;/math&gt; equals <br /> <br /> &lt;math&gt;\text {(A)} -17 \qquad <br /> \text {(B)} -7 \qquad<br /> \text {(C)} 14 \qquad <br /> \text {(D)} 21\qquad <br /> \text {(E)} \text{not uniquely determined}&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> If &lt;math&gt;a&gt;1, b&gt;1&lt;/math&gt;, and &lt;math&gt;p=\frac{\log_b(\log_ba)}{\log_ba}&lt;/math&gt;, then &lt;math&gt;a^p&lt;/math&gt; equals <br /> <br /> &lt;math&gt;\text {(A)} 1 \qquad <br /> \text {(B)} b \qquad <br /> \text {(C)} \log_ab \qquad <br /> \text {(D)} \log_ba \qquad <br /> \text {(E)} a^{\log_ba} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> In the adjoining figure, points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on line segment &lt;math&gt;AD&lt;/math&gt;, and &lt;math&gt;AB, BC&lt;/math&gt;, and &lt;math&gt;CD&lt;/math&gt; are diameters of circle <br /> &lt;math&gt;O, N&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt;, respectively. Circles &lt;math&gt;O, N&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; all have radius &lt;math&gt;15&lt;/math&gt; and the line &lt;math&gt;AG&lt;/math&gt; is tangent to circle &lt;math&gt;P&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. <br /> If &lt;math&gt;AG&lt;/math&gt; intersects circle &lt;math&gt;N&lt;/math&gt; at points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, then chord &lt;math&gt;EF&lt;/math&gt; has length<br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10));<br /> pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1)), F=intersectionpoints(A--G, Circle(N,1));<br /> draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7));<br /> dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P);<br /> label(&quot;$A$&quot;, A, W);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, dir(0));<br /> label(&quot;$P$&quot;, P, S);<br /> label(&quot;$N$&quot;, N, S);<br /> label(&quot;$O$&quot;, O, S);<br /> label(&quot;$E$&quot;, E, dir(120));<br /> label(&quot;$F$&quot;, F, NE);<br /> label(&quot;$G&quot;, G, dir(100));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text {(A)} 20 \qquad <br /> \text {(B)} 15\sqrt{2} \qquad <br /> \text {(C)} 24 \qquad <br /> \text{(D)} 25 \qquad <br /> \text {(E)} \text{none of these}&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Let &lt;math&gt;[z]&lt;/math&gt; denote the greatest integer not exceeding &lt;math&gt;z&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy the simultaneous equations <br /> <br /> &lt;cmath&gt;\begin{align*} y&amp;=2[x]+3 \\ y&amp;=3[x-2]+5. \end{align*}&lt;/cmath&gt;<br /> <br /> If &lt;math&gt;x&lt;/math&gt; is not an integer, then &lt;math&gt;x+y&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text {(A) } \text{ an integer} \qquad <br /> \text {(B) } \text{ between 4 and 5} \qquad <br /> \text{(C) }\text{ between -4 and 4}\qquad\\<br /> \text{(D) }\text{ between 15 and 16}\qquad<br /> \text{(E) } 16.5 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> A wooden cube has edges of length &lt;math&gt;3&lt;/math&gt; meters. Square holes, of side one meter, centered in each face are cut through to the opposite face. <br /> The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside, in square meters, is <br /> <br /> &lt;math&gt;\text {(A)} 54 \qquad <br /> \text {(B)} 72 \qquad <br /> \text {(C)} 76 \qquad <br /> \text {(D)} 84\qquad <br /> \text {(E)} 86 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> How many real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;math&gt;3^{2x+2}-3^{x+3}-3^{x}+3=0&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text {(A)} 0 \qquad <br /> \text {(B)} 1 \qquad <br /> \text {(C)} 2 \qquad <br /> \text {(D)} 3 \qquad <br /> \text {(E)} 4 &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> In the adjoining figure of a rectangular solid, &lt;math&gt;\angle DHG=45^\circ&lt;/math&gt; and &lt;math&gt;\angle FHB=60^\circ&lt;/math&gt;. Find the cosine of &lt;math&gt;\angle BHD&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import three;defaultpen(linewidth(0.7)+fontsize(10));<br /> currentprojection=orthographic(1/3+1/10,1-1/10,1/3);<br /> real r=sqrt(3);<br /> triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0);<br /> draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H);<br /> draw(A--E^^G--E^^F--E, linetype(&quot;4 4&quot;));<br /> label(&quot;A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, dir(0));<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$D$&quot;, D, W);<br /> label(&quot;$E$&quot;, E, NW);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, W);<br /> label(&quot;$H$&quot;, H, S);<br /> triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07);<br /> label(&quot;$45^\circ$&quot;, H45, dir(125), fontsize(8));<br /> label(&quot;$60^\circ$&quot;, H60, dir(25), fontsize(8));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text {(A)} \frac{\sqrt{3}}{6} \qquad <br /> \text {(B)} \frac{\sqrt{2}}{6} \qquad <br /> \text {(C)} \frac{\sqrt{6}}{3} \qquad <br /> \text{(D)}\frac{\sqrt{6}}{4}\qquad<br /> \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4} &lt;/math&gt;<br /> <br /> [[1982 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Let &lt;math&gt;f(x)=|x-2|+|x-4|-|2x-6|&lt;/math&gt; for &lt;math&gt;2 \leq x\leq 8&lt;/math&gt;. The sum of the largest and smallest values of &lt;math&gt;f(x)&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text {(A)} 1 \qquad <br /> \text {(B)} 2 \qquad <br /> \text {(C)} 4 \qquad <br /> \text {(D)} 6 \qquad <br /> \text {(E)}\text{none of these} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> The number of pairs of positive integers &lt;math&gt;(x,y)&lt;/math&gt; which satisfy the equation &lt;math&gt;x^2+y^2=x^3&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text {(A)} 0 \qquad <br /> \text {(B)} 1 \qquad <br /> \text {(C)} 2 \qquad <br /> \text {(D)} \text{not finite} \qquad <br /> \text {(E)} \text{none of these} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> In the adjoining figure, the triangle &lt;math&gt;ABC&lt;/math&gt; is a right triangle with &lt;math&gt;\angle BCA=90^\circ&lt;/math&gt;. Median &lt;math&gt;CM&lt;/math&gt; is perpendicular to median &lt;math&gt;BN&lt;/math&gt;, <br /> and side &lt;math&gt;BC=s&lt;/math&gt;. The length of &lt;math&gt;BN&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.7)+fontsize(10));real r=54.72;<br /> pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M);<br /> draw(M--C--A--B--C^^B--N);<br /> pair point=P;<br /> markscalefactor=0.005;<br /> draw(rightanglemark(C,P,B));<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, dir(point--C));<br /> label(&quot;$M$&quot;, M, S);<br /> label(&quot;$N$&quot;, N, dir(C--A)*dir(90));<br /> label(&quot;$s$&quot;, B--C, NW);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text {(A)} s\sqrt 2 \qquad <br /> \text {(B)} \frac 32s\sqrt2 \qquad <br /> \text {(C)} 2s\sqrt2 \qquad <br /> \text{(D)}\frac{1}{2}s\sqrt5\qquad<br /> \text{(E)}\frac{1}{2}s\sqrt6&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> In a narrow alley of width &lt;math&gt;w&lt;/math&gt; a ladder of length a is placed with its foot at point P between the walls. <br /> Resting against one wall at &lt;math&gt;Q&lt;/math&gt;, the distance k above the ground makes a &lt;math&gt;45^\circ&lt;/math&gt; angle with the ground. <br /> Resting against the other wall at &lt;math&gt;R&lt;/math&gt;, a distance h above the ground, the ladder makes a &lt;math&gt;75^\circ&lt;/math&gt; angle with the ground. <br /> The width &lt;math&gt;w&lt;/math&gt; is equal to <br /> <br /> &lt;math&gt; \text{(A)}a\qquad<br /> \text{(B)}RQ\qquad<br /> \text{(C)}k\qquad<br /> \text{(D)}\frac{h+k}{2}\qquad<br /> \text{(E)}h &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. <br /> The cosine of the smallest angle is <br /> <br /> &lt;math&gt; \text{(A)}\frac{3}{4}\qquad<br /> \text{(B)}\frac{7}{10}\qquad<br /> \text{(C)}\frac{2}{3}\qquad<br /> \text{(D)}\frac{9}{14}\qquad<br /> \text{(E)}\text{none of these} &lt;/math&gt;<br /> <br /> [[1982 AHSME Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If &lt;math&gt;AG=2, GF=13, FC=1&lt;/math&gt;, <br /> and &lt;math&gt;HJ=7&lt;/math&gt;, then &lt;math&gt;DE&lt;/math&gt; equals <br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(10));<br /> real r=sqrt(22);<br /> pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J);<br /> dot(A^^B^^C^^D^^E^^F^^G^^H^^J);<br /> draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7));<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, dir(210));<br /> label(&quot;$C$&quot;, C, dir(330));<br /> label(&quot;$D$&quot;, D, SW);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, dir(170));<br /> label(&quot;$G$&quot;, G, dir(250));<br /> label(&quot;$H$&quot;, H, SE);<br /> label(&quot;$J\$&quot;, J, dir(0));<br /> label(&quot;2&quot;, A--G, dir(30));<br /> label(&quot;13&quot;, F--G, dir(180+30));<br /> label(&quot;1&quot;, F--C, dir(30));<br /> label(&quot;7&quot;, H--J, dir(-30));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text {(A)} 2\sqrt{22} \qquad <br /> \text {(B)} 7\sqrt{3} \qquad <br /> \text {(C)} 9 \qquad <br /> \text {(D)} 10 \qquad <br /> \text {(E)} 13&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. <br /> Each morning, a student walks from intersection &lt;math&gt;A&lt;/math&gt; to intersection &lt;math&gt;B&lt;/math&gt;, always walking along streets shown, <br /> and always going east or south. For variety, at each intersection where he has a choice, he chooses with <br /> probability &lt;math&gt;\frac{1}{2}&lt;/math&gt; whether to go east or south. Find the probability that through any given morning, he goes through &lt;math&gt;C&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(8));<br /> size(250);<br /> path p=origin--(5,0)--(5,3)--(0,3)--cycle;<br /> path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle;<br /> int i,j;<br /> for(i=0; i&lt;5; i=i+1) {<br /> for(j=0; j&lt;6; j=j+1) {<br /> draw(shift(6*i, 4*j)*p);<br /> }}<br /> clip((4,2)--(25,2)--(25,21)--(4,21)--cycle);<br /> fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black);<br /> label(&quot;A&quot;, (6,19), SE);<br /> label(&quot;B&quot;, (23,4), NW);<br /> label(&quot;C&quot;, (23,8), NW);<br /> draw((26,11.5)--(30,11.5), Arrows(5));<br /> draw((28,9.5)--(28,13.5), Arrows(5));<br /> label(&quot;N&quot;, (28,13.5), N);<br /> label(&quot;W&quot;, (26,11.5), W);<br /> label(&quot;E&quot;, (30,11.5), E);<br /> label(&quot;S&quot;, (28,9.5), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \text{(A)}\frac{11}{32}\qquad<br /> \text{(B)}\frac{1}{2}\qquad<br /> \text{(C)}\frac{4}{7}\qquad<br /> \text{(D)}\frac{21}{32}\qquad<br /> \text{(E)}\frac{3}{4} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> <br /> If the base &lt;math&gt;8&lt;/math&gt; representation of a perfect square is &lt;math&gt;ab3c&lt;/math&gt;, where &lt;math&gt;a\ne 0&lt;/math&gt;, then &lt;math&gt;c&lt;/math&gt; equals <br /> <br /> &lt;math&gt;\text{(A)} 0\qquad <br /> \text{(B)}1 \qquad <br /> \text{(C)} 3\qquad <br /> \text{(D)} 4\qquad <br /> \text{(E)} \text{not uniquely determined} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> <br /> Suppose &lt;math&gt;z=a+bi&lt;/math&gt; is a solution of the polynomial equation &lt;math&gt;c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0&lt;/math&gt;, where &lt;math&gt;c_0, c_1, c_2, c_3, a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; <br /> are real constants and &lt;math&gt;i^2=-1&lt;/math&gt;. Which of the following must also be a solution? <br /> <br /> &lt;math&gt;\text{(A)} -a-bi\qquad <br /> \text{(B)} a-bi\qquad <br /> \text{(C)} -a+bi\qquad <br /> \text{(D)}b+ai \qquad <br /> \text{(E)} \text{none of these} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> <br /> A set of consecutive positive integers beginning with &lt;math&gt;1&lt;/math&gt; is written on a blackboard. <br /> One number is erased. The average (arithmetic mean) of the remaining numbers is &lt;math&gt;35\frac{7}{17}&lt;/math&gt;. What number was erased? <br /> <br /> &lt;math&gt;\text{(A)} 6\qquad <br /> \text{(B)}7 \qquad <br /> \text{(C)}8 \qquad <br /> \text{(D)} 9\qquad <br /> \text{(E)}\text{cannot be determined} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> <br /> Let &lt;math&gt;x,y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; be three positive real numbers whose sum is &lt;math&gt;1&lt;/math&gt;. If no one of these numbers is more than twice any other, <br /> then the minimum possible value of the product &lt;math&gt;xyz&lt;/math&gt; is <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{32}\qquad<br /> \textbf{(B)}\ \frac{1}{36}\qquad<br /> \textbf{(C)}\ \frac{4}{125}\qquad<br /> \textbf{(D)}\ \frac{1}{127}\qquad<br /> \textbf{(E)}\ \text{none of these} &lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> <br /> Find the units digit of the decimal expansion of <br /> <br /> &lt;math&gt;(15 + \sqrt{220})^{19} + (15 + \sqrt{220})^{82}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad <br /> \textbf{(B)}\ 2\qquad <br /> \textbf{(C)}\ 5\qquad <br /> \textbf{(D)}\ 9\qquad <br /> \textbf{(E)}\ \text{none of these}&lt;/math&gt; <br /> <br /> [[1982 AHSME Problems/Problem 30|Solution]]<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1982|before=[[1981 AHSME]]|after=[[1983 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Pleaseletmewin