https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pickten&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-03T10:47:00Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Shortlist_Problems/A2&diff=69836 2001 IMO Shortlist Problems/A2 2015-04-06T01:15:41Z <p>Pickten: Added a solution.</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;a_0, a_1, a_2, \ldots&lt;/math&gt; be an arbitrary infinite sequence of positive numbers. Show that the inequality &lt;math&gt;1 + a_n &gt; a_{n - 1} \sqrt [n]{2}&lt;/math&gt; holds for infinitely many positive integers &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We proceed with a proof by contradiction. Suppose the statement were false. Then, there exists a sequence &lt;math&gt;a_0, a_1, \ldots&lt;/math&gt; of positive integers for which there are only finitely many &lt;math&gt;a_n&lt;/math&gt; with &lt;math&gt;1+a_n&gt; a_{n-1}\sqrt[n]{2}&lt;/math&gt;. Let the largest such &lt;math&gt;n&lt;/math&gt; be &lt;math&gt;N-1&lt;/math&gt;, so that &lt;math&gt;1+a_n\le a_{n-1}\sqrt[n]{2}&lt;/math&gt; whenever &lt;math&gt;n\le N&lt;/math&gt;. Then, it is clear that &lt;math&gt;1+a_{n+N}\le a_{n+N-1}\sqrt[n]{2}&lt;/math&gt; for all nonnegative &lt;math&gt;n&lt;/math&gt;. Therefore, define &lt;math&gt;b_n=a_{n+N}&lt;/math&gt;. If there does not exist a sequence &lt;math&gt;b_0, b_1, \ldots&lt;/math&gt; of positive integers for which &lt;math&gt;1+b_n\le b_{n-1}\sqrt[n]{2}&lt;/math&gt;, it is clear that there will not exist any sequence &lt;math&gt;a_0, a_1, \ldots&lt;/math&gt; for which that property is eventually true.<br /> <br /> Thus, I claim there does not exist a sequence of positive integers &lt;math&gt;b_0, b_1, \ldots&lt;/math&gt; for which &lt;math&gt;1+b_n\le b_{n-1}\sqrt[n]{2}&lt;/math&gt;. Again, suppose there does exist such a sequence. Then, define &lt;math&gt;x_0=b_0&lt;/math&gt; and &lt;math&gt;x_n=x_{n-1}\sqrt[n]{2} -1&lt;/math&gt;. It is clear that &lt;math&gt;x_n\ge b_n&lt;/math&gt; for all &lt;math&gt;n&lt;/math&gt;. I claim that this sequence will always become eventually negative. Note that<br /> &lt;math&gt;x_n=s_nx_{n-1}-1=s_n(s_{n-1}x_{n-2}-1)-1=\ldots=x_0s_n\cdot s_{n-1}\cdot\ldots\cdot s_0-\sum_{i=1}^{n-1}(\prod_{j=i}^n s_j)-1&lt;/math&gt;,<br /> which becomes negative if and only if &lt;math&gt;\frac{x_n}{\prod_{i=1}^ns_i}=x_0-1-\frac{1}{s_1}-\frac{1}{s_1}{s_2}-\ldots&lt;/math&gt; does. In other words, &lt;math&gt;x_n&lt;/math&gt;<br /> becomes zero if &lt;math&gt;t_k=\sum_{i=1}^{k}\frac{1}{\prod_{j=1}^i s_j}&lt;/math&gt; is unbounded. However, &lt;math&gt;\sqrt[n]{2}&lt;/math&gt; is eventually less than &lt;math&gt;\frac{n+1}{n}&lt;/math&gt;, so this sum is indeed unbounded and the proof is complete.<br /> <br /> == Resources ==<br /> * [[2001 IMO Shortlist Problems]]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119160#119160 Discussion on AOPS/MathLinks]<br /> <br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems&diff=60941 2014 AIME I Problems 2014-03-14T19:33:16Z <p>Pickten: /* Problem 6 */</p> <hr /> <div>{{AIME Problems|year=2014|n=I}}<br /> <br /> ==Problem 1==<br /> The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectrangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.<br /> <br /> [[2014 AIME I Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2== <br /> <br /> An urn contains &lt;math&gt;4&lt;/math&gt; green balls and &lt;math&gt;6&lt;/math&gt; blue balls. A second urn contains &lt;math&gt;16&lt;/math&gt; green balls and &lt;math&gt;N&lt;/math&gt; blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 0.58. Find &lt;math&gt;N&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Find the number of rational numbers &lt;math&gt;r,&lt;/math&gt; &lt;math&gt;0&lt;r&lt;1,&lt;/math&gt; such that when &lt;math&gt;r&lt;/math&gt; is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.<br /> <br /> [[2014 AIME I Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east as 20 miles per hour, and Steve rides west at 20 miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly 1 minute to go past Jon. The westbound train takes 10 times as long as the eastbound train to go past Steve. The length of each train is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Let the set &lt;math&gt;S = \{P_1, P_2, \dots, P_{12}\}&lt;/math&gt; consist of the twelve vertices of a regular 12-gon. A subset &lt;math&gt;Q&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt; is called communal if there is a circle such that all points of &lt;math&gt;Q&lt;/math&gt; are inside the circle, and all points of &lt;math&gt;S&lt;/math&gt; not in &lt;math&gt;Q&lt;/math&gt; are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)<br /> <br /> [[2014 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> The graphs &lt;math&gt;y = 3(x-h)^2 + j&lt;/math&gt; and &lt;math&gt;y = 2(x-h)^2 + k&lt;/math&gt; have y-intercepts of 2013 and 2014, respectively, and each graph has two positive integer x-intercepts. Find &lt;math&gt;h&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Let &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be complex numbers such that &lt;math&gt;|w| = 1&lt;/math&gt; and &lt;math&gt;|z| = 10&lt;/math&gt;. Let &lt;math&gt;\theta = \arg \left(\tfrac{w-z}{z}\right) &lt;/math&gt;. The maximum possible value of &lt;math&gt;\tan^2 \theta&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;. (Note that &lt;math&gt;\arg(w)&lt;/math&gt;, for w &lt;math&gt;\neq 0&lt;/math&gt;, denotes the measure of the angle that the ray from 0 to &lt;math&gt;w&lt;/math&gt; makes with the positive real axis in the complex plane.<br /> <br /> [[2014 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> The positive integers &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;N^2&lt;/math&gt; both end in the same sequence of four digits &lt;math&gt;abcd&lt;/math&gt; when written in base 10, where digit a is not zero. Find the three-digit number &lt;math&gt;abc&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;x_1&lt; x_2 &lt; x_3&lt;/math&gt; be the three real roots of the equation &lt;math&gt;\sqrt{2014} x^3 - 4029x^2 + 2 = 0&lt;/math&gt;. Find &lt;math&gt;x_2(x_1+x_3)&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A disk with radius 1 is externally tangent to a disk with radius 5. Let &lt;math&gt;A&lt;/math&gt; be the point where the disks are tangent, &lt;math&gt;C&lt;/math&gt; be the center of the smaller disk, and &lt;math&gt;E&lt;/math&gt; be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of &lt;math&gt;360^\circ&lt;/math&gt;. That is, if the center of the smaller disk has moved to the point &lt;math&gt;D&lt;/math&gt;, and the point on the smaller disk that began at &lt;math&gt;A&lt;/math&gt; has now moved to point &lt;math&gt;B&lt;/math&gt;, then &lt;math&gt;\overline{AC}&lt;/math&gt; is parallel to &lt;math&gt;\overline{BD}&lt;/math&gt;. Then &lt;math&gt;\sin^2(\angle BEA)=\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> A token starts at the point &lt;math&gt;(0,0)&lt;/math&gt; of an &lt;math&gt;xy&lt;/math&gt;-coordinate grid and them makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of &lt;math&gt;|y|=|x|&lt;/math&gt; is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A=\{1,2,3,4\}&lt;/math&gt;, and &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; be randomly chosen (not necessarily distinct) functions from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;. The probability that the range of &lt;math&gt;f&lt;/math&gt; and the range of &lt;math&gt;g&lt;/math&gt; are disjoint is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> On square &lt;math&gt;ABCD,&lt;/math&gt; points &lt;math&gt;E,F,G&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; lie on sides &lt;math&gt;\overline{AB},\overline{BC},\overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; respectively, so that &lt;math&gt;\overline{EG} \perp \overline{FH}&lt;/math&gt; and &lt;math&gt;EG=FH = 34&lt;/math&gt;. Segments &lt;math&gt;\overline{EG}&lt;/math&gt; and &lt;math&gt;\overline{FH}&lt;/math&gt; intersect at a point &lt;math&gt;P&lt;/math&gt;, and the areas of the quadrilaterals &lt;math&gt;AEPH, BFPE, CGPF,&lt;/math&gt; and &lt;math&gt;DHPG&lt;/math&gt; are in the ratio &lt;math&gt;269:275:405:411.&lt;/math&gt; Find the area of square &lt;math&gt;ABCD.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A = (0,sqrt(850));<br /> pair B = (0,0);<br /> pair C = (sqrt(850),0);<br /> pair D = (sqrt(850),sqrt(850));<br /> draw(A--B--C--D--cycle);<br /> dotfactor = 3;<br /> dot(&quot;$A$&quot;,A,dir(135));<br /> dot(&quot;$B$&quot;,B,dir(215));<br /> dot(&quot;$C$&quot;,C,dir(305));<br /> dot(&quot;$D$&quot;,D,dir(45));<br /> pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));<br /> pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);<br /> dot(&quot;$H$&quot;,H,dir(90));<br /> dot(&quot;$F$&quot;,F,dir(270));<br /> draw(H--F);<br /> pair E = (0,(sqrt(850)-6)/2);<br /> pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);<br /> dot(&quot;$E$&quot;,E,dir(180));<br /> dot(&quot;$G$&quot;,G,dir(0));<br /> draw(E--G);<br /> pair P = extension(H,F,E,G);<br /> dot(&quot;$P$&quot;,P,dir(60));<br /> label(&quot;$w$&quot;, intersectionpoint( A--P, E--H ));<br /> label(&quot;$x$&quot;, intersectionpoint( B--P, E--F ));<br /> label(&quot;$y$&quot;, intersectionpoint( C--P, G--F ));<br /> label(&quot;$z$&quot;, intersectionpoint( D--P, G--H ));&lt;/asy&gt;<br /> <br /> [[2014 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Let &lt;math&gt;m&lt;/math&gt; be the largest real solution to the equation<br /> <br /> &lt;cmath&gt; \dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4&lt;/cmath&gt;<br /> <br /> There are positive integers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; such that &lt;math&gt;m = a + \sqrt{b + \sqrt{c}}&lt;/math&gt;. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> In &lt;math&gt;\triangle ABC, AB = 3, BC = 4,&lt;/math&gt; and &lt;math&gt;CA = 5&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B, \overline{BC}&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;and &lt;math&gt;D,&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;. Given that &lt;math&gt;EF=DF&lt;/math&gt; and &lt;math&gt;\frac{DG}{EG} = \frac{3}{4},&lt;/math&gt; length &lt;math&gt;DE=\frac{a\sqrt{b}}{c},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> [[2014 AIME I Problems/Problem 15|Solution]]<br /> <br /> {{MAA Notice}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems&diff=48833 2010 USAJMO Problems 2012-10-12T13:18:20Z <p>Pickten: /* Problem 2 */</p> <hr /> <div>==Day 1==<br /> ===Problem 1===<br /> A permutation of the set of positive integers &lt;math&gt;[n] = {1,2,\ldots,n}&lt;/math&gt;<br /> is a sequence &lt;math&gt;(a_1,a_2,\ldots,a_n)&lt;/math&gt; such that each element of &lt;math&gt;[n]&lt;/math&gt;<br /> appears precisely one time as a term of the sequence. For example,<br /> &lt;math&gt;(3, 5, 1, 2, 4)&lt;/math&gt; is a permutation of &lt;math&gt;&lt;/math&gt;. Let &lt;math&gt;P(n)&lt;/math&gt; be the number of<br /> permutations of &lt;math&gt;[n]&lt;/math&gt; for which &lt;math&gt;ka_k&lt;/math&gt; is a perfect square for all<br /> &lt;math&gt;1 \le k \le n&lt;/math&gt;. Find with proof the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;P(n)&lt;/math&gt;<br /> is a multiple of &lt;math&gt;2010&lt;/math&gt;.<br /> <br /> [[2010 USAJMO Problems/Problem 1|Solution]]<br /> <br /> ===Problem 2===<br /> Let &lt;math&gt;n &gt; 1&lt;/math&gt; be an integer. Find, with proof, all sequences<br /> &lt;math&gt;x_1, x_2, \ldots, x_{n-1}&lt;/math&gt; of positive integers with the following<br /> three properties:<br /> &lt;ol style=&quot;list-style-type:lower-alpha&quot;&gt;<br /> &lt;li&gt; &lt;math&gt;x_1 &lt; x_2 &lt; \cdots &lt;x_{n-1}&lt;/math&gt;;<br /> &lt;li&gt; &lt;math&gt;x_i +x_{n-i} = 2n&lt;/math&gt; for all &lt;math&gt;i=1,2,\ldots,n-1&lt;/math&gt;;<br /> &lt;li&gt; given any two indices &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; (not necessarily distinct)<br /> for which &lt;math&gt;x_i + x_j &lt; 2n&lt;/math&gt;, there is an index &lt;math&gt;k&lt;/math&gt; such<br /> that &lt;math&gt;x_i+x_j = x_k&lt;/math&gt;.<br /> &lt;/ol&gt;<br /> <br /> [[2010 USAJMO Problems/Problem 2|Solution]]<br /> <br /> ===Problem 3===<br /> Let &lt;math&gt;AXYZB&lt;/math&gt; be a convex pentagon inscribed in a semicircle of diameter<br /> &lt;math&gt;AB&lt;/math&gt;. Denote by &lt;math&gt;P, Q, R, S&lt;/math&gt; the feet of the perpendiculars from &lt;math&gt;Y&lt;/math&gt; onto<br /> lines &lt;math&gt;AX, BX, AZ, BZ&lt;/math&gt;, respectively. Prove that the acute angle<br /> formed by lines &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; is half the size of &lt;math&gt;\angle XOZ&lt;/math&gt;, where<br /> &lt;math&gt;O&lt;/math&gt; is the midpoint of segment &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> [[2010 USAMO Problems/Problem 1|Solution]]<br /> <br /> ==Day 2==<br /> ===Problem 4===<br /> A triangle is called a parabolic triangle if its vertices lie on a<br /> parabola &lt;math&gt;y = x^2&lt;/math&gt;. Prove that for every nonnegative integer &lt;math&gt;n&lt;/math&gt;, there<br /> is an odd number &lt;math&gt;m&lt;/math&gt; and a parabolic triangle with vertices at three<br /> distinct points with integer coordinates with area &lt;math&gt;(2^nm)^2&lt;/math&gt;.<br /> <br /> [[2010 USAJMO Problems/Problem 4|Solution]]<br /> <br /> ===Problem 5===<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> [[2010 USAJMO Problems/Problem 5|Solution]]<br /> <br /> ===Problem 6===<br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle with &lt;math&gt;\angle A = 90^{\circ}&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt;<br /> and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively, such that &lt;math&gt;\angle<br /> ABD = \angle DBC&lt;/math&gt; and &lt;math&gt;\angle ACE = \angle ECB&lt;/math&gt;. Segments &lt;math&gt;BD&lt;/math&gt; and<br /> &lt;math&gt;CE&lt;/math&gt; meet at &lt;math&gt;I&lt;/math&gt;. Determine whether or not it is possible for<br /> segments &lt;math&gt;AB, AC, BI, ID, CI, IE&lt;/math&gt; to all have integer lengths.<br /> <br /> [[2010 USAMO Problems/Problem 4|Solution]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=User:Pickten&diff=47451 User:Pickten 2012-06-22T00:30:53Z <p>Pickten: Blanked the page</p> <hr /> <div></div> Pickten https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=47390 1983 AIME Problems/Problem 15 2012-06-14T15:34:04Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting [[chord]]s in a [[circle]], with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is [[bisect]]ed by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the [[sine]] of the minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this fraction is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> [[Image:1983_AIME-15.png]]<br /> <br /> == Solution ==<br /> Let &lt;math&gt;A&lt;/math&gt; be any [[fixed point]] on [[circle]] &lt;math&gt;O&lt;/math&gt; and let &lt;math&gt;AD&lt;/math&gt; be a [[chord]] of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of [[midpoint]]s &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to BC at point N. <br /> <br /> Let M be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From [[right triangle]] &lt;math&gt;OMB&lt;/math&gt;, &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. Thus, &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos AOM = r(1 + \cos AOM)&lt;/math&gt; (Where &lt;math&gt;r&lt;/math&gt; is the radius of circle P). Evaluating this, &lt;math&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/math&gt;. From &lt;math&gt;\cos \angle AOM&lt;/math&gt;, we see that &lt;math&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/math&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the tangent subtraction formula to obtain , &lt;math&gt;\tan AOB =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/math&gt;. It follows that &lt;math&gt;\sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, resulting in an answer of &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=47387 1983 AIME Problems/Problem 11 2012-06-13T23:47:22Z <p>Pickten: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The solid shown has a [[square]] base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is [[parallel]] to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] &lt;math&gt;ADE&lt;/math&gt;, and one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the [[Pythagorean Theorem]] to find that the height is equal to &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F); <br /> draw(B--Ba--Ca--C,dashed+d);<br /> draw(A--Aa--Da--D,dashed+d);<br /> draw(E--(E.x,E.y,0),dashed+l);<br /> draw(F--(F.x,F.y,0),dashed+l);<br /> draw(Aa--E--Da,dashed+d);<br /> draw(Ba--F--Ca,dashed+d);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> label(&quot;$12\sqrt{2}$&quot;,(E+F)/2,N);<br /> label(&quot;$6\sqrt{2}$&quot;,(A+B)/2);<br /> label(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find the volume, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Extend &lt;math&gt;EA&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt; to meet at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; and &lt;math&gt;FC&lt;/math&gt; to meet at &lt;math&gt;H&lt;/math&gt;. now, we have a regular tetrahedron &lt;math&gt;EFGH&lt;/math&gt;, which has twice the volume of our original solid. This tetrahedron has side length &lt;math&gt;2s = 12\sqrt{2}&lt;/math&gt;. Using the formula for the volume of a regular tetrahedron, which is &lt;math&gt;V = \frac{\sqrt{2}S^3}{12}&lt;/math&gt;, where S is the side length of the tetrahedron, the volume of our original solid is:<br /> <br /> &lt;math&gt;V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=47386 1983 AIME Problems/Problem 11 2012-06-13T23:47:03Z <p>Pickten: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The solid shown has a [[square]] base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is [[parallel]] to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] &lt;math&gt;ADE&lt;/math&gt;, and one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the [[Pythagorean Theorem]] to find that the height is equal to &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F); <br /> draw(B--Ba--Ca--C,dashed+d);<br /> draw(A--Aa--Da--D,dashed+d);<br /> draw(E--(E.x,E.y,0),dashed+l);<br /> draw(F--(F.x,F.y,0),dashed+l);<br /> draw(Aa--E--Da,dashed+d);<br /> draw(Ba--F--Ca,dashed+d);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> label(&quot;$12\sqrt{2}$&quot;,(E+F)/2,N);<br /> label(&quot;$6\sqrt{2}$&quot;,(A+B)/2);<br /> label(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find the volume, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Extend &lt;math&gt;EA&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt; to meet at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; and &lt;math&gt;FC&lt;/math&gt; to meet at &lt;math&gt;H&lt;/math&gt;. now, we have a regular tetrahedron &lt;math&gt;EFGH&lt;/math&gt;, which has twice the volume of our original solid. This tetrahedron has side length &lt;math&gt;2s = 12\sqrt{2}&lt;/math&gt;. Using the formula for the volume of a regular tetrahedron, which is &lt;math&gt;V = \frac{\sqrt{2}S^3}{12}&lt;/math&gt;, where S is the side length of the tetrahedron, the volume of our original solid is:<br /> <br /> &lt;math&gt;V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=47385 1983 AIME Problems/Problem 11 2012-06-13T23:46:29Z <p>Pickten: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The solid shown has a [[square]] base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is [[parallel]] to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] &lt;math&gt;ADE&lt;/math&gt;, and one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the [[Pythagorean Theorem]] to find that the height is equal to &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F); <br /> draw(B--Ba--Ca--C,dashed+d);<br /> draw(A--Aa--Da--D,dashed+d);<br /> draw(E--(E.x,E.y,0),dashed+l);<br /> draw(F--(F.x,F.y,0),dashed+l);<br /> draw(Aa--E--Da,dashed+d);<br /> draw(Ba--F--Ca,dashed+d);<br /> label(&quot;A&quot;,A);<br /> label(&quot;B&quot;,B);<br /> label(&quot;C&quot;,C);<br /> label(&quot;D&quot;,D);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> label(&quot;$12\sqrt{2}$&quot;,(E+F)/2,N);<br /> label(&quot;$6\sqrt{2}$&quot;,(A+B)/2);<br /> label(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find the area, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Extend &lt;math&gt;EA&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt; to meet at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; and &lt;math&gt;FC&lt;/math&gt; to meet at &lt;math&gt;H&lt;/math&gt;. now, we have a regular tetrahedron &lt;math&gt;EFGH&lt;/math&gt;, which has twice the volume of our original solid. This tetrahedron has side length &lt;math&gt;2s = 12\sqrt{2}&lt;/math&gt;. Using the formula for the volume of a regular tetrahedron, which is &lt;math&gt;V = \frac{\sqrt{2}S^3}{12}&lt;/math&gt;, where S is the side length of the tetrahedron, the volume of our original solid is:<br /> <br /> &lt;math&gt;V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_15&diff=45481 2004 AIME I Problems/Problem 15 2012-03-15T18:05:18Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> For all positive integers &lt;math&gt;x&lt;/math&gt;, let<br /> &lt;cmath&gt;<br /> f(x)=\begin{cases}1 &amp; \text{if x = 1}}\\ \frac x{10} &amp; \text{if x is divisible by 10}\\ x+1 &amp; \text{otherwise}\end{cases}<br /> &lt;/cmath&gt;<br /> and define a [[sequence]] as follows: &lt;math&gt;x_1=x&lt;/math&gt; and &lt;math&gt;x_{n+1}=f(x_n)&lt;/math&gt; for all positive integers &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;d(x)&lt;/math&gt; be the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;x_n=1&lt;/math&gt;. (For example, &lt;math&gt;d(100)=3&lt;/math&gt; and &lt;math&gt;d(87)=7&lt;/math&gt;.) Let &lt;math&gt;m&lt;/math&gt; be the number of positive integers &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;d(x)=20&lt;/math&gt;. Find the sum of the distinct prime factors of &lt;math&gt;m&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We backcount the number of ways. Namely, we start at &lt;math&gt;x_{20} = 1&lt;/math&gt;, which can only be reached if &lt;math&gt;x_{19} = 10&lt;/math&gt;, and then we perform &lt;math&gt;18&lt;/math&gt; operations that either consist of &lt;math&gt;A: (-1)&lt;/math&gt; or &lt;math&gt;B: (\times 10)&lt;/math&gt;. We represent these operations in a string format, starting with the operation that sends &lt;math&gt;f(x_{18}) = x_{19}&lt;/math&gt; and so forth downwards. There are &lt;math&gt;2^9&lt;/math&gt; ways to pick the first &lt;math&gt;9&lt;/math&gt; operations; however, not all &lt;math&gt;9&lt;/math&gt; of them may be &lt;math&gt;A: (-1)&lt;/math&gt; otherwise we return back to &lt;math&gt;x_{10} = 1&lt;/math&gt;, contradicting our assumption that &lt;math&gt;20&lt;/math&gt; was the smallest value of &lt;math&gt;n&lt;/math&gt;. Using [[complementary counting]], we see that there are only &lt;math&gt;2^9 - 1&lt;/math&gt; ways.<br /> <br /> Since we performed the operation &lt;math&gt;B: (\times 10)&lt;/math&gt; at least once in the first &lt;math&gt;9&lt;/math&gt; operations, it follows that &lt;math&gt;x_{10} \ge 20&lt;/math&gt;, so that we no longer have to worry about reaching &lt;math&gt;1&lt;/math&gt; again. Thus the remaining &lt;math&gt;9&lt;/math&gt; operations can be picked in &lt;math&gt;2^9&lt;/math&gt; ways, with a total of &lt;math&gt;2^9(2^9 - 1) = 2^{18} - 2^9&lt;/math&gt; strings. <br /> <br /> However, we must also account for a sequence of &lt;math&gt;10&lt;/math&gt; or more &lt;math&gt;A: (-1)&lt;/math&gt;s in a row, because that implies that at least one of those numbers was divisible by &lt;math&gt;10&lt;/math&gt;, yet the &lt;math&gt;\frac{x}{10}&lt;/math&gt; was never used, contradiction. We must use complementary counting again by determining the number of strings of &lt;math&gt;A,B&lt;/math&gt;s of length &lt;math&gt;18&lt;/math&gt; such that there are &lt;math&gt;10&lt;/math&gt; &lt;math&gt;A&lt;/math&gt;s in a row. The first ten are not included since we already accounted for that scenario above, so our string of &lt;math&gt;10&lt;/math&gt; &lt;math&gt;A&lt;/math&gt;s must be preceded by a &lt;math&gt;B&lt;/math&gt;. There are no other restrictions on the remaining seven characters. Letting &lt;math&gt;\square&lt;/math&gt; to denote either of the functions, and &lt;math&gt;^{[k]}&lt;/math&gt; to indicate that the character appears &lt;math&gt;k&lt;/math&gt; times in a row, then our bad strings can take the forms:<br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> &amp;\underbrace{BA^{}}\square \square \square \square \square \square \square \square \\<br /> &amp;\square \underbrace{BA^{}}\square \square \square \square \square \square \square \\<br /> &amp;\qquad \quad \cdots \quad \cdots \\<br /> &amp;\square \square \square \square \square \square \square \underbrace{BA^{}} \square \\<br /> &amp;\square \square \square \square \square \square \square \square \underbrace{BA^{}} \\<br /> \end{align*}&lt;/math&gt;&lt;/center&gt;<br /> <br /> There are &lt;math&gt;2^7&lt;/math&gt; ways to select the operations for the &lt;math&gt;7&lt;/math&gt; &lt;math&gt;\square&lt;/math&gt;s, and &lt;math&gt;8&lt;/math&gt; places to place our &lt;math&gt;BA^{}&lt;/math&gt; block. Thus, our answer is &lt;math&gt;2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{511}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_15&diff=45480 2004 AIME I Problems/Problem 15 2012-03-15T18:04:18Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> For all positive integers &lt;math&gt;x&lt;/math&gt;, let<br /> &lt;cmath&gt;<br /> f(x)=\begin{cases}1 &amp; \text{if x = 1}}\\ \frac x{10} &amp; \text{if x is divisible by 10}\\ x+1 &amp; \text{otherwise}\end{cases}<br /> &lt;/cmath&gt;<br /> and define a [[sequence]] as follows: &lt;math&gt;x_1=x&lt;/math&gt; and &lt;math&gt;x_{n+1}=f(x_n)&lt;/math&gt; for all positive integers &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;d(x)&lt;/math&gt; be the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;x_n=1&lt;/math&gt;. (For example, &lt;math&gt;d(100)=3&lt;/math&gt; and &lt;math&gt;d(87)=7&lt;/math&gt;.) Let &lt;math&gt;m&lt;/math&gt; be the number of positive integers &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;d(x)=20&lt;/math&gt;. Find the sum of the distinct prime factors of &lt;math&gt;m&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We backcount the number of ways. Namely, we start at &lt;math&gt;x_{20} = 1&lt;/math&gt;, which can only be reached if &lt;math&gt;x_{19} = 10&lt;/math&gt;, and then we perform &lt;math&gt;18&lt;/math&gt; operations that either consist of &lt;math&gt;A: (-1)&lt;/math&gt; or &lt;math&gt;B: (\times 10)&lt;/math&gt;. We represent these operations in a string format, starting with the operation that sends &lt;math&gt;f(x_{18}) = x_{19}&lt;/math&gt; and so forth downwards. There are &lt;math&gt;2^9&lt;/math&gt; ways to pick the first &lt;math&gt;9&lt;/math&gt; operations; however, not all &lt;math&gt;9&lt;/math&gt; of them may be &lt;math&gt;A: (-1)&lt;/math&gt; otherwise we return back to &lt;math&gt;x_{10} = 1&lt;/math&gt;, contradicting our assumption that &lt;math&gt;20&lt;/math&gt; was the smallest value of &lt;math&gt;n&lt;/math&gt;. Using [[complementary counting]], we see that there are only &lt;math&gt;2^9 - 1&lt;/math&gt; ways.<br /> <br /> Since we performed the operation &lt;math&gt;B: (\times 10)&lt;/math&gt; at least once in the first &lt;math&gt;9&lt;/math&gt; operations, it follows that &lt;math&gt;x_{10} \ge 20&lt;/math&gt;, so that we no longer have to worry about reaching &lt;math&gt;1&lt;/math&gt; again. Thus the remaining &lt;math&gt;9&lt;/math&gt; operations can be picked in &lt;math&gt;2^9&lt;/math&gt; ways, with a total of &lt;math&gt;2^9(2^9 - 1) = 2^{18} - 2^9&lt;/math&gt; strings. <br /> <br /> However, we must also account for a sequence of &lt;math&gt;10&lt;/math&gt; or more &lt;math&gt;A: (-1)&lt;/math&gt;s in a row, because that implies that at least one of those numbers was divisible by &lt;math&gt;10&lt;/math&gt;, yet the &lt;math&gt;\frac{x}{10}&lt;/math&gt; was never used, contradiction. We must use complement counting again by determining the number of strings of &lt;math&gt;A,B&lt;/math&gt;s of length &lt;math&gt;18&lt;/math&gt; such that there are &lt;math&gt;10&lt;/math&gt; &lt;math&gt;A&lt;/math&gt;s in a row. The first ten are not included since we already accounted for that scenario above, so our string of &lt;math&gt;10&lt;/math&gt; &lt;math&gt;A&lt;/math&gt;s must be preceded by a &lt;math&gt;B&lt;/math&gt;. There are no other restrictions on the remaining seven characters. Letting &lt;math&gt;\square&lt;/math&gt; to denote either of the functions, and &lt;math&gt;^{[k]}&lt;/math&gt; to indicate that the character appears &lt;math&gt;k&lt;/math&gt; times in a row, then our bad strings can take the forms:<br /> &lt;center&gt;&lt;math&gt;\begin{align*}<br /> &amp;\underbrace{BA^{}}\square \square \square \square \square \square \square \square \\<br /> &amp;\square \underbrace{BA^{}}\square \square \square \square \square \square \square \\<br /> &amp;\qquad \quad \cdots \quad \cdots \\<br /> &amp;\square \square \square \square \square \square \square \underbrace{BA^{}} \square \\<br /> &amp;\square \square \square \square \square \square \square \square \underbrace{BA^{}} \\<br /> \end{align*}&lt;/math&gt;&lt;/center&gt;<br /> <br /> There are &lt;math&gt;2^7&lt;/math&gt; ways to select the operations for the &lt;math&gt;7&lt;/math&gt; &lt;math&gt;\square&lt;/math&gt;s, and &lt;math&gt;8&lt;/math&gt; places to place our &lt;math&gt;BA^{}&lt;/math&gt; block. Thus, our answer is &lt;math&gt;2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{511}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_25&diff=42346 2004 AMC 12B Problems/Problem 25 2011-09-18T19:34:55Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Given that &lt;math&gt;2^{2004}&lt;/math&gt; is a &lt;math&gt;604&lt;/math&gt;-[[digit]] number whose first digit is &lt;math&gt;1&lt;/math&gt;, how many [[element]]s of the [[set]] &lt;math&gt;S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}&lt;/math&gt; have a first digit of &lt;math&gt;4&lt;/math&gt;? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198&lt;/math&gt;<br /> == Solution ==<br /> <br /> Given &lt;math&gt;n&lt;/math&gt; digits, there must be a power of &lt;math&gt;2&lt;/math&gt; with &lt;math&gt;n&lt;/math&gt; digits such that the first digit is &lt;math&gt;1&lt;/math&gt;. Thus &lt;math&gt;S&lt;/math&gt; contains &lt;math&gt;603&lt;/math&gt; elements with a first digit of &lt;math&gt;1&lt;/math&gt;. For each number in the form of &lt;math&gt;2^k&lt;/math&gt; such that its first digit is &lt;math&gt;1&lt;/math&gt;, then &lt;math&gt;2^{k+1}&lt;/math&gt; must either have a first digit of &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;2^{k+2}&lt;/math&gt; must have a first digit of &lt;math&gt;4,5,6,7&lt;/math&gt;. Thus there are also &lt;math&gt;603&lt;/math&gt; numbers with first digit either &lt;math&gt;\{2,3\}&lt;/math&gt; or &lt;math&gt;\{4,5,6,7\}&lt;/math&gt;. By using [[complementary counting]], there are &lt;math&gt;2004 - 3 \times 603 = 195&lt;/math&gt; elements of &lt;math&gt;S&lt;/math&gt; with a first digit of &lt;math&gt;\{8,9\}&lt;/math&gt;. Now, &lt;math&gt;2^k&lt;/math&gt; has a first of &lt;math&gt;\{8,9\}&lt;/math&gt; [[iff|if and only if]] the first digit of &lt;math&gt;2^{k-1}&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, so there are &lt;math&gt;\boxed{195} \Rightarrow \mathrm{(B)}&lt;/math&gt; elements of &lt;math&gt;S&lt;/math&gt; with a first digit of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2004|ab=B|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_18&diff=42105 2005 AMC 12A Problems/Problem 18 2011-09-01T23:01:36Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> Call a number ''prime-looking'' if it is [[composite]] but not divisible by &lt;math&gt;2, 3,&lt;/math&gt; or 5. The three smallest prime-looking numbers are &lt;math&gt;49, 77&lt;/math&gt;, and &lt;math&gt;91&lt;/math&gt;. There are &lt;math&gt;168&lt;/math&gt; prime numbers less than &lt;math&gt;1000&lt;/math&gt;. How many prime-looking numbers are there less than &lt;math&gt;1000&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> (\mathrm {A}) \ 100 \qquad (\mathrm {B}) \ 102 \qquad (\mathrm {C})\ 104 \qquad (\mathrm {D}) \ 106 \qquad (\mathrm {E})\ 108<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> The given states that there are &lt;math&gt;168&lt;/math&gt; prime numbers less than &lt;math&gt;1000&lt;/math&gt;, which is a fact we must somehow utilize. Since there seems to be no easy way to directly calculate the number of &quot;prime-looking&quot; numbers, we can apply [[complementary counting]]. We can split the numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;1000&lt;/math&gt; into several groups: &lt;math&gt;\{1\},&lt;/math&gt; &lt;math&gt;\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},&lt;/math&gt; &lt;math&gt; \{\mathrm{numbers\ divisible\ by\ 3 = S_3}\},&lt;/math&gt; &lt;math&gt; \{\mathrm{numbers\ divisible\ by\ 5 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},&lt;/math&gt; &lt;math&gt; \{\mathrm{prime-looking}\}&lt;/math&gt;. Hence, the number of prime-looking numbers is &lt;math&gt;1000 - 165 - 1 - |S_2 \cup S_3 \cup S_5|&lt;/math&gt; (note that &lt;math&gt;2,3,5&lt;/math&gt; are primes).<br /> <br /> We can calculate &lt;math&gt;S_2 \cup S_3 \cup S_5&lt;/math&gt; using the [[Principle of Inclusion-Exclusion]]: (the values of &lt;math&gt;|S_2| \ldots&lt;/math&gt; and their intersections can be found quite easily)<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;|S_2 \cup S_3 \cup S_5| = |S_2| + |S_3| + |S_5| - |S_2 \cap S_3| - |S_3 \cap S_5| - |S_2 \cap S_5| + |S_2 \cap S_3 \cap S_5|&lt;/math&gt;&lt;br /&gt;&lt;math&gt;= 500 + 333 + 200 - 166 - 66 - 100 + 33 = 734&lt;/math&gt;&lt;/div&gt;<br /> <br /> Substituting, we find that our answer is &lt;math&gt;1000 - 165 - 1 - 734 = 100 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_20&diff=41630 2010 AMC 12B Problems/Problem 20 2011-08-19T00:22:50Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem==<br /> A geometric sequence &lt;math&gt;(a_n)&lt;/math&gt; has &lt;math&gt;a_1=\sin x&lt;/math&gt;, &lt;math&gt;a_2=\cos x&lt;/math&gt;, and &lt;math&gt;a_3= \tan x&lt;/math&gt; for some real number &lt;math&gt;x&lt;/math&gt;. For what value of &lt;math&gt;n&lt;/math&gt; does &lt;math&gt;a_n=1+\cos x&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> By defintion of a geometric sequence, we have &lt;math&gt;\cos^2x=\sin x \tan x&lt;/math&gt;. Since &lt;math&gt;\tan x=\frac{\sin x}{\cos x}&lt;/math&gt;, we can rewrite this as &lt;math&gt;\cos^3x=\sin^2x&lt;/math&gt;. <br /> <br /> The common ratio of the sequence is &lt;math&gt;\frac{\cos x}{\sin x}&lt;/math&gt;, so we can write<br /> <br /> &lt;cmath&gt;a_1= \sin x&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2= \cos x&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3= \frac{\cos^2x}{\sin x}&lt;/cmath&gt;<br /> &lt;cmath&gt;a_4=\frac{\cos^3x}{\sin^2x}=1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_5=\frac{\cos x}{\sin x}&lt;/cmath&gt;<br /> &lt;cmath&gt;a_6=\frac{\cos^2x}{\sin^2x}&lt;/cmath&gt;<br /> &lt;cmath&gt;a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}&lt;/cmath&gt;<br /> &lt;cmath&gt;a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}&lt;/cmath&gt;<br /> &lt;cmath&gt;a_9=\frac{\cos x}{\sin x}&lt;/cmath&gt;<br /> <br /> <br /> We can conclude that the sequence from &lt;math&gt;a_4&lt;/math&gt; to &lt;math&gt;a_8&lt;/math&gt; repeats. <br /> <br /> Since &lt;math&gt;\cos^3x=\sin^2x=1-\cos^2x&lt;/math&gt;, we have &lt;math&gt;\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}&lt;/math&gt;, which is &lt;math&gt;a_8&lt;/math&gt; making our answer &lt;math&gt;8 \Rightarrow \boxed{E}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_20&diff=41629 2010 AMC 12B Problems/Problem 20 2011-08-19T00:21:48Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem==<br /> A geometric sequence &lt;math&gt;(a_n)&lt;/math&gt; has &lt;math&gt;a_1=\sin x&lt;/math&gt;, &lt;math&gt;a_2=\cos x&lt;/math&gt;, and &lt;math&gt;a_3= \tan x&lt;/math&gt; for some real number &lt;math&gt;x&lt;/math&gt;. For what value of &lt;math&gt;n&lt;/math&gt; does &lt;math&gt;a_n=1+\cos x&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> By defintion of a geometric sequence, we have &lt;math&gt;\cos^2x=\sin x \tan x&lt;/math&gt;. Since &lt;math&gt;\tan x=\frac{\sin x}{\cos x}&lt;/math&gt;, we can rewrite this as &lt;math&gt;\cos^3x=\sin^2x&lt;/math&gt;. <br /> <br /> The common ratio of the sequence is &lt;math&gt;\frac{\cos x}{\sin x}&lt;/math&gt;, so we can write<br /> <br /> &lt;math&gt;a_1= \sin x&lt;/math&gt;<br /> &lt;math&gt;a_2= \cos x&lt;/math&gt;<br /> &lt;math&gt;a_3= \frac{\cos^2x}{\sin x}&lt;/math&gt;<br /> &lt;math&gt;a_4=\frac{\cos^3x}{\sin^2x}=1&lt;/math&gt;<br /> &lt;math&gt;a_5=\frac{\cos x}{\sin x}&lt;/math&gt;<br /> &lt;math&gt;a_6=\frac{\cos^2x}{\sin^2x}&lt;/math&gt;<br /> &lt;math&gt;a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}&lt;/math&gt;<br /> &lt;math&gt;a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}&lt;/math&gt;<br /> &lt;math&gt;a_9=\frac{\cos x}{\sin x}&lt;/math&gt;<br /> <br /> <br /> We can conclude that the sequence from &lt;math&gt;a_4&lt;/math&gt; to &lt;math&gt;a_8&lt;/math&gt; repeats. <br /> <br /> Since &lt;math&gt;\cos^3x=\sin^2x=1-\cos^2x&lt;/math&gt;, we have &lt;math&gt;\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}&lt;/math&gt;, which is &lt;math&gt;a_8&lt;/math&gt; making our answer &lt;math&gt;8 \Rightarrow \boxed{E}&lt;/math&gt;.<br /> <br /> --Please fix formatting--<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=19|num-a=21|ab=B}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_12&diff=41398 2011 AMC 12B Problems/Problem 12 2011-08-12T01:36:47Z <p>Pickten: /* Solution */</p> <hr /> <div>==Problem==<br /> A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?<br /> <br /> &lt;asy&gt;<br /> unitsize(10mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> draw(A--D);<br /> draw(B--G);<br /> draw(C--F);<br /> draw(E--H);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad \textbf{(E)}\ 2 - \sqrt{2}&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let's assume that the side length of the octagon is &lt;math&gt;x&lt;/math&gt;. The area of the center square is just &lt;math&gt;x^2&lt;/math&gt;. The triangles are all &lt;math&gt;45-45-90&lt;/math&gt; triangles, with a side length ratio of &lt;math&gt;1:1:\sqrt{2}&lt;/math&gt;. The area of each of the &lt;math&gt;4&lt;/math&gt; identical triangles is &lt;math&gt;\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}&lt;/math&gt;, so the total area of all of the triangles is also &lt;math&gt;x^2&lt;/math&gt;. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is &lt;math&gt;x&lt;/math&gt; and the other side length is &lt;math&gt;\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}&lt;/math&gt;, so the area of all of the rectangles is &lt;math&gt;2x^2\sqrt{2}&lt;/math&gt;. The ratio of the area of the square to the area of the octagon is &lt;math&gt;\dfrac{x^2}{2x^2+2x^2\sqrt{2}}&lt;/math&gt;. Cancelling &lt;math&gt;x^2&lt;/math&gt; from the fraction, the ratio becomes &lt;math&gt;\dfrac{1}{2\sqrt2+2}&lt;/math&gt;. Multiplying the numerator and the denominator each by &lt;math&gt;2\sqrt{2}-2&lt;/math&gt; will cancel out the radical, so the fraction is now &lt;math&gt;\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_15&diff=41384 2011 AMC 12A Problems/Problem 15 2011-08-10T00:46:34Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> The circular base of a hemisphere of radius &lt;math&gt;2&lt;/math&gt; rests on the base of a square pyramid of height &lt;math&gt;6&lt;/math&gt;. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3\sqrt{2} \qquad<br /> \textbf{(B)}\ \frac{13}{3} \qquad<br /> \textbf{(C)}\ 4\sqrt{2} \qquad<br /> \textbf{(D)}\ 6 \qquad<br /> \textbf{(E)}\ \frac{13}{2} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt; ABCDE &lt;/math&gt; be the pyramid with &lt;math&gt; ABCD &lt;/math&gt; as the square base. Let &lt;math&gt; O &lt;/math&gt; and &lt;math&gt; M &lt;/math&gt; be the center of square &lt;math&gt; ABCD &lt;/math&gt; and the midpoint of side &lt;math&gt; AB &lt;/math&gt; respectively. Lastly, let the hemisphere be tangent to the triangular face &lt;math&gt; ABE &lt;/math&gt; at &lt;math&gt; P &lt;/math&gt;.<br /> <br /> Notice that &lt;math&gt; \triangle EOM &lt;/math&gt; has a right angle at &lt;math&gt; O &lt;/math&gt;. Since the hemisphere is tangent to the triangular face &lt;math&gt; ABE &lt;/math&gt; at &lt;math&gt; P &lt;/math&gt;, &lt;math&gt;\angle EPO &lt;/math&gt; is also &lt;math&gt; 90^{\circ} &lt;/math&gt;. Hence &lt;math&gt; \triangle EOM &lt;/math&gt; is similar to &lt;math&gt;\triangle EPO &lt;/math&gt;.<br /> <br /> &lt;math&gt; \frac{OM}{2} = \frac{6}{EP} &lt;/math&gt;<br /> <br /> &lt;math&gt; OM = \frac{6}{EP} \times 2 &lt;/math&gt;<br /> <br /> &lt;math&gt; OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2} &lt;/math&gt;<br /> <br /> The length of the square base is thus &lt;math&gt;2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_15&diff=41383 2011 AMC 12A Problems/Problem 15 2011-08-10T00:45:24Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> The circular base of a hemisphere of radius &lt;math&gt;2&lt;/math&gt; rests on the base of a square pyramid of height &lt;math&gt;6&lt;/math&gt;. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3\sqrt{2} \qquad<br /> \textbf{(B)}\ \frac{13}{3} \qquad<br /> \textbf{(C)}\ 4\sqrt{2} \qquad<br /> \textbf{(D)}\ 6 \qquad<br /> \textbf{(E)}\ \frac{13}{2} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt; ABCDE &lt;/math&gt; be the pyramid with &lt;math&gt; ABCD &lt;/math&gt; as the square base. Let &lt;math&gt; O &lt;/math&gt; and &lt;math&gt; M &lt;/math&gt; be the center of square &lt;math&gt; ABCD &lt;/math&gt; and the midpoint of side &lt;math&gt; AB &lt;/math&gt; respectively. Lastly, let the hemisphere be tangent to the triangular face &lt;math&gt; ABE &lt;/math&gt; at &lt;math&gt; P &lt;/math&gt;.<br /> <br /> Notice that &lt;math&gt; \triangle EOM &lt;/math&gt; has a right angle at &lt;math&gt; O &lt;/math&gt;. Since the hemisphere be tangent to the triangular face &lt;math&gt; ABE &lt;/math&gt; at &lt;math&gt; P &lt;/math&gt;, &lt;math&gt;\angle EPO &lt;/math&gt; is also &lt;math&gt; 90^{\circ} &lt;/math&gt;. Hence &lt;math&gt; \triangle EOM &lt;/math&gt; is similar to &lt;math&gt;\triangle EPO &lt;/math&gt;.<br /> <br /> &lt;math&gt; \frac{OM}{2} = \frac{6}{EP} &lt;/math&gt;<br /> <br /> &lt;math&gt; OM = \frac{6}{EP} \times 2 &lt;/math&gt;<br /> <br /> &lt;math&gt; OM = \frac{6}{\sqrt{6^2 - 2^2}} \times 2 = \frac{3\sqrt{2}}{2} &lt;/math&gt;<br /> <br /> The length of the square base is thus &lt;math&gt;2 \times \frac{3\sqrt{2}}{2} = 3\sqrt{2} \rightarrow \boxed{\textbf{A}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=39518 2010 AMC 10B Problems/Problem 16 2011-06-12T22:29:19Z <p>Pickten: </p> <hr /> <div>The radius of circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;<br /> <br /> Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle<br /> <br /> Therefore the picture will look something like this:<br /> <br /> [[Image:squarecircle.png]]<br /> <br /> Then we proceed to find: 4 * (area of the sector marked off by the two radii drawn - area of the triangle with side on the square and the two radii).<br /> <br /> <br /> To find this, we do the following:<br /> <br /> First we realize that the radius perpendicular to the side of the square between the extra lines marking off the sector splits the chord in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Power of a point states that if two chords (AB and CD) intersect at X, then AX&lt;math&gt;\cdot&lt;/math&gt;BX&lt;math&gt;=&lt;/math&gt;CX&lt;math&gt;\cdot&lt;/math&gt;DX. Applying power of a point to this situation, &lt;math&gt;a^2=(\frac{\sqrt{3}}{3}-\frac{1}{2})(\frac{\sqrt{3}}{3}+\frac{1}{2})&lt;/math&gt;. (We know the center of a square to be halfway in each direction, if you know what I mean by direction).<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt;. The significance? This means the chord is equal to the radius of a circle, so the sector has angle &lt;math&gt;60^{\circ}&lt;/math&gt;. Since this is a sixth of the circle, the sector has area &lt;math&gt;\frac{\pi}{6}\cdot \frac{\sqrt{3}}{3}^2=\frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle.<br /> <br /> Since it is equilateral, we know it has area equal to &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; times the square of its sidelength. Therefore, our triangle has area &lt;math&gt;\frac{\frac{\sqrt{3}}{3}^2\sqrt{3}}{4}=\frac{\sqrt{3}}{12}&lt;/math&gt;.<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4( \frac{\pi}{18}-\frac{\sqrt{3}}{12} )=&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\boxed{\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> [[Area of an equilateral triangle]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=39517 2010 AMC 10B Problems/Problem 16 2011-06-12T22:28:57Z <p>Pickten: </p> <hr /> <div>The radius of circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;<br /> <br /> Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle<br /> <br /> Therefore the picture will look something like this:<br /> <br /> [[Image:squarecircle.png]]<br /> <br /> Then we proceed to find: 4 * (area of the sector marked off by the two radii drawn - area of the triangle with side on the square and the two radii).<br /> <br /> <br /> To find this, we do the following:<br /> <br /> First we realize that the radius perpendicular to the side of the square between the extra lines marking off the sector splits the chord in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Power of a point states that if two chords (AB and CD) intersect at X, then AX&lt;math&gt;\cdot&lt;/math&gt;BX&lt;math&gt;=&lt;/math&gt;CX&lt;math&gt;\cdot&lt;/math&gt;DX. Applying power of a point to this situation, &lt;math&gt;a^2=(\frac{\sqrt{3}}{3}-\frac{1}{2})(\frac{\sqrt{3}}{3}+\frac{1}{2})&lt;/math&gt;. (We know the center of a square to be halfway in each direction, if you know what I mean by direction).<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt;. The significance? This means the chord is equal to the radius of a circle, so the sector has angle &lt;math&gt;60^{\circ}&lt;/math&gt;. Since this is a sixth of the circle, the sector has area &lt;math&gt;\frac{\pi}{6}\cdot \frac{\sqrt{3}}{3}^2=\frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle.<br /> <br /> Since it is equilateral, we know it has area equal to &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; times the square of its sidelength. Therefore, our triangle has area &lt;math&gt;\frac{\frac{\sqrt{3}}{3}^2\sqrt{3}}{4}=\frac{\sqrt{3}}{12}&lt;/math&gt;.<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4( \frac{\pi}{18}-\frac{\sqrt{3}{12} )=&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\boxed{\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> [[Area of an equilateral triangle]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=39516 2010 AMC 10B Problems/Problem 16 2011-06-12T22:28:12Z <p>Pickten: </p> <hr /> <div>The radius of circle is &lt;math&gt;\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}&lt;/math&gt;<br /> <br /> Half the diagonal of the square is &lt;math&gt;\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}&lt;/math&gt;. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle<br /> <br /> Therefore the picture will look something like this:<br /> <br /> [[Image:squarecircle.png]]<br /> <br /> Then we proceed to find: 4 * (area of the sector marked off by the two radii drawn - area of the triangle with side on the square and the two radii).<br /> <br /> <br /> To find this, we do the following:<br /> <br /> First we realize that the radius perpendicular to the side of the square between the extra lines marking off the sector splits the chord in half. Let this half-length be &lt;math&gt;a&lt;/math&gt;. Power of a point states that if two chords (AB and CD) intersect at X, then AX&lt;math&gt;\cdot&lt;/math&gt;BX&lt;math&gt;=&lt;/math&gt;CX&lt;math&gt;\cdot&lt;/math&gt;DX. Applying power of a point to this situation, &lt;math&gt;a^2=(\frac{\sqrt{3}}{3}-\frac{1}{2})(\frac{\sqrt{3}}{3}+\frac{1}{2})&lt;/math&gt;. (We know the center of a square to be halfway in each direction, if you know what I mean by direction).<br /> <br /> Solving, &lt;math&gt;a= \frac{\sqrt{3}}{6}&lt;/math&gt;. The significance? This means the chord is equal to the radius of a circle, so the sector has angle &lt;math&gt;60^{\circ}&lt;/math&gt;. Since this is a sixth of the circle, the sector has area &lt;math&gt;\frac{\pi}{6}\cdot \frac{\sqrt{3}}{3}^2=\frac{\pi}{18}&lt;/math&gt;.<br /> <br /> Now we turn to the triangle.<br /> <br /> Since it is equilateral, we know it has area equal to &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; times the square of its sidelength. Therefore, our triangle has area &lt;math&gt;\frac{\frac{\sqrt{3}}{3}^2\sqrt{3}}{4}=\frac{\sqrt{3}}{12}&lt;/math&gt;.<br /> <br /> Putting it together, we get the answer to be &lt;math&gt;4\left( \frac{\pi}{18}-\frac{\sqrt{3}{12} \right)=&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)&lt;/cmath&gt;<br /> <br /> == See also ==<br /> [[Area of an equilateral triangle]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=Area_of_an_equilateral_triangle&diff=39514 Area of an equilateral triangle 2011-06-12T22:23:18Z <p>Pickten: /* Proof */</p> <hr /> <div>The area of an equilateral triangle is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the sidelength of the triangle.<br /> <br /> <br /> == Proof ==<br /> <br /> ''Method 1'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the &quot;short side&quot; is &lt;math&gt;\frac{s}{2}&lt;/math&gt;.<br /> <br /> Using the [[Pythagorean theorem]], we get &lt;math&gt;s^2=h^2+\frac{s^2}{4}&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the height of the triangle. Solving, &lt;math&gt;h=\frac{s\sqrt{3}}{2}&lt;/math&gt;. (note we could use 30-60-90 right triangles.)<br /> <br /> We use the formula for the area of a triangle, &lt;math&gt;{bh \over 2}&lt;/math&gt; (note &lt;math&gt;s&lt;/math&gt; is the length of a base), so the area is &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ''Method 2'' '''warning: uses trig.''' The area of a triangle is &lt;math&gt;\frac{ab\sin{C}}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a=b=s&lt;/math&gt; and &lt;math&gt;C=\frac{\pi}{3}&lt;/math&gt; (the angle at each vertex, in radians), we get the area to be &lt;math&gt;\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=&lt;/math&gt;<br /> &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ----<br /> <br /> == credits: ==<br /> created by [[User:Pickten|Pickten]] 18:22, 12 June 2011 (EDT)</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=Area_of_an_equilateral_triangle&diff=39513 Area of an equilateral triangle 2011-06-12T22:22:45Z <p>Pickten: /* Proof */</p> <hr /> <div>The area of an equilateral triangle is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the sidelength of the triangle.<br /> <br /> <br /> == Proof ==<br /> <br /> ''Method 1'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the &quot;short side&quot; is &lt;math&gt;\frac{s}{2}&lt;/math&gt;.<br /> <br /> Using the [[Pythagorean theorem]], we get &lt;math&gt;s^2=h^2+\frac{s^2}{4}&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the height of the triangle. Solving, &lt;math&gt;h=\frac{s\sqrt{3}}{2}&lt;/math&gt;. (note we could use 30-60-90 right triangles.)<br /> <br /> We use the formula for the area of a triangle, &lt;math&gt;{bh \over 2}&lt;/math&gt; (note &lt;math&gt;s&lt;/math&gt; is the length of a base), so the area is &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ''Method 2'' '''warning: uses trig.''' The area of a triangle is &lt;math&gt;\frac{ab\sin{C}}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a=b=s&lt;/math&gt; and &lt;math&gt;C=\frac{\pi}{3}&lt;/math&gt; (the angle at each vertex, in radians), we get the area to be &lt;math&gt;\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=&lt;/math&gt;<br /> &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ----<br /> credits:<br /> created by [[User:Pickten|Pickten]] 18:22, 12 June 2011 (EDT)</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=Area_of_an_equilateral_triangle&diff=39511 Area of an equilateral triangle 2011-06-12T22:22:09Z <p>Pickten: Created page with &quot;The area of an equilateral triangle is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the sidelength of the triangle. == Proof == ''Method 1'' Dropping the altitu...&quot;</p> <hr /> <div>The area of an equilateral triangle is &lt;math&gt;\frac{s^2\sqrt{3}}{4}&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the sidelength of the triangle.<br /> <br /> <br /> == Proof ==<br /> <br /> ''Method 1'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the &quot;short side&quot; is &lt;math&gt;\frac{s}{2}&lt;/math&gt;.<br /> <br /> Using the [[Pythagorean theorem]], we get &lt;math&gt;s^2=h^2+\frac{s^2}{4}&lt;/math&gt;, where &lt;math&gt;h&lt;/math&gt; is the height of the triangle. Solving, &lt;math&gt;h=\frac{s\sqrt{3}}{2}&lt;/math&gt;. (note we could use 30-60-90 right triangles.)<br /> <br /> We use the formula for the area of a triangle, &lt;math&gt;{bh \over 2}&lt;/math&gt; (note &lt;math&gt;s&lt;/math&gt; is the length of a base), so the area is &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ''Method 2'' '''warning: uses trig.''' The area of a triangle is &lt;math&gt;\frac{ab\sin{C}}{2}&lt;/math&gt;. Plugging in &lt;math&gt;a=b=s&lt;/math&gt; and &lt;math&gt;C=\frac{\pi}{3}&lt;/math&gt; (the angle at each vertex, in radians), we get the area to be &lt;math&gt;\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=&lt;/math&gt;<br /> &lt;cmath&gt;\boxed{\frac{s^2\sqrt{3}}{4}}&lt;/cmath&gt;<br /> <br /> ----<br /> credits:<br /> created by pickten<br /> --[[User:Pickten|Pickten]] 18:22, 12 June 2011 (EDT)</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_25&diff=39508 2010 AMC 10B Problems/Problem 25 2011-06-12T21:38:44Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;a &gt; 0&lt;/math&gt;, and let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial with integer coefficients such that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;P(1) = P(3) = P(5) = P(7) = a&lt;/math&gt;, and&lt;br/&gt;<br /> &lt;math&gt;P(2) = P(4) = P(6) = P(8) = -a&lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> What is the smallest possible value of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!&lt;/math&gt;<br /> <br /> == Solution ==<br /> There must be some polynomial &lt;math&gt;Q(x)&lt;/math&gt; such that &lt;math&gt;P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)&lt;/math&gt;<br /> <br /> Then, plugging in values of &lt;math&gt;2,4,6,8,&lt;/math&gt; we get <br /> <br /> &lt;cmath&gt;P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a&lt;/cmath&gt;<br /> &lt;cmath&gt;P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a&lt;/cmath&gt;<br /> &lt;cmath&gt;P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a&lt;/cmath&gt;<br /> &lt;cmath&gt;P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a&lt;/cmath&gt;<br /> <br /> &lt;math&gt;-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).&lt;/math&gt;<br /> Thus, the least value of &lt;math&gt;a&lt;/math&gt; must be the &lt;math&gt;lcm(15,9,15,105)&lt;/math&gt;.<br /> Solving, we receive &lt;math&gt;315&lt;/math&gt;, so our answer is &lt;math&gt; \boxed{\textbf{(B)}\ 315} &lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=39507 2010 AMC 10B Problems/Problem 24 2011-06-12T21:25:31Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?<br /> <br /> == Solution ==<br /> Represent the teams' scores as: &lt;math&gt;(a, an, an^2, an^3)&lt;/math&gt; and &lt;math&gt;(a, a+m, a+2m, a+3m)&lt;/math&gt;<br /> <br /> We have &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;<br /> Manipulating this, we can get &lt;math&gt;a(1+n+n^2+n^3)=4a+6m+1&lt;/math&gt;, or &lt;math&gt;a(n^4-1)/(n-1)=4a+6m+1&lt;/math&gt;<br /> <br /> Since both are increasing sequences, &lt;math&gt;n&gt;1&lt;/math&gt;. We can check cases up to &lt;math&gt;n=4&lt;/math&gt; because when &lt;math&gt;n=5&lt;/math&gt;, we get &lt;math&gt;156a&gt;100&lt;/math&gt;. When <br /> * &lt;math&gt;n=2, a=[1,6]&lt;/math&gt;<br /> *&lt;math&gt; n=3, a=[1,2]&lt;/math&gt;<br /> *&lt;math&gt; n=4, a=1&lt;/math&gt;<br /> Checking each of these cases individually back into the equation &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find &lt;math&gt;(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34 ~ (E)&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=39506 2010 AMC 10B Problems/Problem 24 2011-06-12T21:24:40Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?<br /> <br /> == Solution ==<br /> Represent the teams' scores as: &lt;math&gt;(a, an, an^2, an^3)&lt;/math&gt; and &lt;math&gt;(a, a+m, a+2m, a+3m)&lt;/math&gt;<br /> <br /> We have &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;<br /> Manipulating this, we can get &lt;math&gt;a(1+n+n^2+n^3)=4a+6m+1&lt;/math&gt;, or &lt;math&gt;a(n^4-1)/(n-1)=4a+6m+1&lt;/math&gt;<br /> <br /> Since both are increasing sequences, &lt;math&gt;n&gt;1&lt;/math&gt;. We can check cases up to &lt;math&gt;n=4&lt;/math&gt; because when &lt;math&gt;n=5&lt;/math&gt;, we get &lt;math&gt;156a&gt;100&lt;/math&gt;. When <br /> * &lt;math&gt;n=2, a=[1,6]&lt;/math&gt;<br /> *&lt;math&gt; n=3, a=[1,2]&lt;/math&gt;<br /> *&lt;math&gt; n=4, a=1&lt;/math&gt;<br /> Checking each of these cases individually back into the equation &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find &lt;math&gt;(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34 ~ (D)&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&diff=39505 2010 AMC 10B Problems/Problem 24 2011-06-12T21:24:28Z <p>Pickten: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?<br /> <br /> == Solution ==<br /> Represent the teams' scores as: &lt;math&gt;(a, an, an^2, an^3)&lt;/math&gt; and &lt;math&gt;(a, a+m, a+2m, a+3m)&lt;/math&gt;<br /> <br /> We have &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;<br /> Manipulating this, we can get &lt;math&gt;a(1+n+n^2+n^3)=4a+6m+1&lt;/math&gt;, or &lt;math&gt;a(n^4-1)/(n-1)=4a+6m+1&lt;/math&gt;<br /> <br /> Since both are increasing sequences, &lt;math&gt;n&gt;1&lt;/math&gt;. We can check cases up to &lt;math&gt;n=4&lt;/math&gt; because when &lt;math&gt;n=5&lt;/math&gt;, we get &lt;math&gt;156a&gt;100&lt;/math&gt;. When <br /> * &lt;math&gt;n=2, a=[1,6]&lt;/math&gt;<br /> *&lt;math&gt; n=3, a=[1,2]&lt;/math&gt;<br /> *&lt;math&gt; n=4, a=1&lt;/math&gt;<br /> Checking each of these cases individually back into the equation &lt;math&gt;a+an+an^2+an^3=4a+6m+1&lt;/math&gt;, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find &lt;math&gt;(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34 (D)&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}</div> Pickten https://artofproblemsolving.com/wiki/index.php?title=User:Pickten&diff=36014 User:Pickten 2010-11-08T19:20:44Z <p>Pickten: very true</p> <hr /> <div>Go Hitchhiker's guide to the galaxy!!!</div> Pickten