https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Lotrgreengrapes7926&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-18T03:48:44Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2005_Alabama_ARML_TST_Problems/Problem_1&diff=21951 2005 Alabama ARML TST Problems/Problem 1 2008-01-06T00:10:15Z <p>Lotrgreengrapes7926: </p> <hr /> <div>==Problem==<br /> Two six-sided dice are constructed such that each face is equally likely to show up when rolled. The numbers on the faces of one of the dice are &lt;math&gt;1, 3, 4, 5, 6,\text{ and }8&lt;/math&gt;. The numbers on the faces of the other die are &lt;math&gt;1, 2, 2, 3, 3,\text{ and }4&lt;/math&gt;. Find the [[probability]] of rolling a sum of &lt;math&gt;9&lt;/math&gt; with these two dice.<br /> <br /> ==Solution==<br /> We use [[generating function]]s to represent the sum of the two dice rolls: &lt;center&gt;&lt;math&gt;(x+x^3+x^4+x^5+x^6+x^8)(x+2x^2+2x^3+x^4)=&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;x^2(1+x^2+x^3+x^4+x^5+x^7)(1+x+x^2)(1+x)&lt;/math&gt;&lt;/center&gt;<br /> The [[coefficient]] of &lt;math&gt;x^9&lt;/math&gt;, that is, the number of ways of rolling a sum of 9, is thus &lt;math&gt;(1+2+1)=4&lt;/math&gt;, out of a total of &lt;math&gt;6^2&lt;/math&gt; possible two-roll combinations, for a probability of &lt;math&gt;\frac 19&lt;/math&gt;.<br /> <br /> Alternatively, just note the possible pairs which work: &lt;math&gt;(5, 4), (6, 3), (6, 3)&lt;/math&gt; and &lt;math&gt;(8, 1)&lt;/math&gt; are all possible combinations that give us a sum of &lt;math&gt;9&lt;/math&gt; (where we count &lt;math&gt;(6, 3)&lt;/math&gt; twice because there are two different &lt;math&gt;3&lt;/math&gt;s to roll). Thus the probability of one of these outcomes is &lt;math&gt;\frac{4}{36} = \frac19&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{ARML box|year=2005|state=Alabama|before=First question|num-a=2}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_19&diff=21950 2006 AMC 12A Problems/Problem 19 2008-01-06T00:07:26Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> [[Circle]]s with [[center]]s &lt;math&gt;(2,4)&lt;/math&gt; and &lt;math&gt;(14,9)&lt;/math&gt; have [[radius | radii]] &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form &lt;math&gt;y=mx+b&lt;/math&gt; with &lt;math&gt;m&gt;0&lt;/math&gt;. What is &lt;math&gt;b&lt;/math&gt;?<br /> &lt;center&gt;[[Image:AMC12_2006A_19.png]]&lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}&lt;/math&gt;&lt;math&gt;\mathrm{(E) \ } \frac{912}{119}&lt;/math&gt;<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> :''This solution needs a clearer explanation and a diagram.''<br /> Notice that both [[circle]]s are [[tangent]] to the [[x-axis]] and each other. Call the circles (respectively) A and B; the [[distance]] between the two centers is &lt;math&gt;4 + 9 = 13&lt;/math&gt;. If we draw the parallel radii that lead to the common [[external tangent]], a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the [[double tangent identity]],<br /> <br /> :&lt;math&gt;\tan (2 \tan ^{-1} \left(\frac{5}{12}\right) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}&lt;/math&gt;<br /> :&lt;math&gt;= \frac{120}{119}&lt;/math&gt;<br /> <br /> To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal):<br /> <br /> :&lt;math&gt;\frac{119}{\sqrt{119^2 + 120^2}}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{119}{169} = \frac{y - 4}{4}&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}&lt;/math&gt;<br /> <br /> :&lt;math&gt;x = \frac{-142}{169}, y = \frac{1152}{169}&lt;/math&gt;<br /> <br /> We can plug this into the equation of the line for the tangent to get:<br /> <br /> :&lt;math&gt;\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b&lt;/math&gt;<br /> :&lt;math&gt;b = \frac{912}{119}&lt;/math&gt; &lt;math&gt;\Rightarrow \mathrm{E}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> By [[User:skiron|skiron]].<br /> <br /> Let &lt;math&gt;L_1&lt;/math&gt; be the line that goes through &lt;math&gt;(2,4)&lt;/math&gt; and &lt;math&gt;(14,9)&lt;/math&gt;, and let &lt;math&gt;L_2&lt;/math&gt; be the line &lt;math&gt;y=mx+b&lt;/math&gt;. If we let &lt;math&gt;\theta&lt;/math&gt; be the measure of the acute angle formed by &lt;math&gt;L_1&lt;/math&gt; and the x-axis, then &lt;math&gt;\tan\theta=\frac{5}{12}&lt;/math&gt;. &lt;math&gt;L_1&lt;/math&gt; clearly bisects the angle formed by &lt;math&gt;L_2&lt;/math&gt; and the x-axis, so &lt;math&gt;m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}&lt;/math&gt;. We also know that &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; intersect at a point on the x-axis. The equation of &lt;math&gt;L_1&lt;/math&gt; is &lt;math&gt;y=\frac{5}{12}x+\frac{19}{6}&lt;/math&gt;, so the coordinate of this point is &lt;math&gt;\left(-\frac{38}{5},0\right)&lt;/math&gt;. Hence the equation of &lt;math&gt;L_2&lt;/math&gt; is &lt;math&gt;y=\frac{120}{119}x+\frac{912}{119}&lt;/math&gt;, so &lt;math&gt;b=\frac{912}{119}&lt;/math&gt;, and our answer choice is &lt;math&gt;\boxed{\mathrm{E}}&lt;/math&gt;. <br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_9&diff=21945 2006 AMC 12A Problems/Problem 9 2008-01-05T23:24:56Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> <br /> Oscar buys &lt;math&gt;13&lt;/math&gt; pencils and &lt;math&gt;3&lt;/math&gt; erasers for &lt;math&gt;1.00&lt;/math&gt;. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(E) \ } 20&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let the price of a pencil be &lt;math&gt;p&lt;/math&gt; and an eraser &lt;math&gt;e&lt;/math&gt;. Then &lt;math&gt;13p + 3e = 100&lt;/math&gt; with &lt;math&gt;p &gt; e &gt; 0&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; are [[positive integer]]s, we must have &lt;math&gt;e \geq 1&lt;/math&gt; and &lt;math&gt;p \geq 2&lt;/math&gt;.<br /> <br /> Considering the [[equation]] &lt;math&gt;13p + 3e = 100&lt;/math&gt; [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have &lt;math&gt;p + 0e \equiv 1 \pmod 3&lt;/math&gt; so &lt;math&gt;p&lt;/math&gt; leaves a remainder of 1 on division by 3.<br /> <br /> Since &lt;math&gt;p \geq 2&lt;/math&gt;, possible values for &lt;math&gt;p&lt;/math&gt; are 4, 7, 10 ....<br /> <br /> Since 13 pencils cost less than 100 cents, &lt;math&gt;13p &lt; 100&lt;/math&gt;. &lt;math&gt;13 \times 10 = 130&lt;/math&gt; is too high, so &lt;math&gt;p&lt;/math&gt; must be 4 or 7.<br /> <br /> If &lt;math&gt;p = 4&lt;/math&gt; then &lt;math&gt;13p = 52&lt;/math&gt; and so &lt;math&gt;3e = 48&lt;/math&gt; giving &lt;math&gt;e = 16&lt;/math&gt;. This contradicts the pencil being more expensive. The only remaining value for &lt;math&gt;p&lt;/math&gt; is 7; then the 13 pencils cost &lt;math&gt;7 \times 13= 91&lt;/math&gt; cents and so the 3 erasers together cost 9 cents and each eraser costs &lt;math&gt;\frac{9}{3} = 3&lt;/math&gt; cents.<br /> <br /> Thus one pencil plus one eraser cost &lt;math&gt;7 + 3 = 10&lt;/math&gt; cents, which is answer choice &lt;math&gt;\mathrm{(A) \ }&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_9&diff=21940 2006 AMC 12A Problems/Problem 9 2008-01-05T23:23:29Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> <br /> Oscar buys &lt;math&gt;13&lt;/math&gt; pencils and &lt;math&gt;3&lt;/math&gt; erasers for &lt;math&gt;\&lt;/math&gt;1.00&lt;math&gt;. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?<br /> <br /> &lt;/math&gt; \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18&lt;math&gt;<br /> <br /> &lt;/math&gt;\mathrm{(E) \ } 20&lt;math&gt;<br /> <br /> == Solution ==<br /> Let the price of a pencil be &lt;/math&gt;p&lt;math&gt; and an eraser &lt;/math&gt;e&lt;math&gt;. Then &lt;/math&gt;13p + 3e = 100&lt;math&gt; with &lt;/math&gt;p &gt; e &gt; 0&lt;math&gt;. Since &lt;/math&gt;p&lt;math&gt; and &lt;/math&gt;e&lt;math&gt; are [[positive integer]]s, we must have &lt;/math&gt;e \geq 1&lt;math&gt; and &lt;/math&gt;p \geq 2&lt;math&gt;.<br /> <br /> Considering the [[equation]] &lt;/math&gt;13p + 3e = 100&lt;math&gt; [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have &lt;/math&gt;p + 0e \equiv 1 \pmod 3&lt;math&gt; so &lt;/math&gt;p&lt;math&gt; leaves a remainder of 1 on division by 3.<br /> <br /> Since &lt;/math&gt;p \geq 2&lt;math&gt;, possible values for &lt;/math&gt;p&lt;math&gt; are 4, 7, 10 ....<br /> <br /> Since 13 pencils cost less than 100 cents, &lt;/math&gt;13p &lt; 100&lt;math&gt;. &lt;/math&gt;13 \times 10 = 130&lt;math&gt; is too high, so &lt;/math&gt;p&lt;math&gt; must be 4 or 7.<br /> <br /> If &lt;/math&gt;p = 4&lt;math&gt; then &lt;/math&gt;13p = 52&lt;math&gt; and so &lt;/math&gt;3e = 48&lt;math&gt; giving &lt;/math&gt;e = 16&lt;math&gt;. This contradicts the pencil being more expensive. The only remaining value for &lt;/math&gt;p&lt;math&gt; is 7; then the 13 pencils cost &lt;/math&gt;7 \times 13= 91&lt;math&gt; cents and so the 3 erasers together cost 9 cents and each eraser costs &lt;/math&gt;\frac{9}{3} = 3&lt;math&gt; cents.<br /> <br /> Thus one pencil plus one eraser cost &lt;/math&gt;7 + 3 = 10&lt;math&gt; cents, which is answer choice &lt;/math&gt;\mathrm{(A) \ }$.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_9&diff=21938 2006 AMC 12A Problems/Problem 9 2008-01-05T23:23:02Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> <br /> Oscar buys &lt;math&gt;13&lt;/math&gt; pencils and &lt;math&gt;3&lt;/math&gt; erasers for &lt;math&gt;&lt;/math&gt;1.00&lt;math&gt;. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?<br /> <br /> &lt;/math&gt; \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18&lt;math&gt;<br /> <br /> &lt;/math&gt;\mathrm{(E) \ } 20&lt;math&gt;<br /> <br /> == Solution ==<br /> Let the price of a pencil be &lt;/math&gt;p&lt;math&gt; and an eraser &lt;/math&gt;e&lt;math&gt;. Then &lt;/math&gt;13p + 3e = 100&lt;math&gt; with &lt;/math&gt;p &gt; e &gt; 0&lt;math&gt;. Since &lt;/math&gt;p&lt;math&gt; and &lt;/math&gt;e&lt;math&gt; are [[positive integer]]s, we must have &lt;/math&gt;e \geq 1&lt;math&gt; and &lt;/math&gt;p \geq 2&lt;math&gt;.<br /> <br /> Considering the [[equation]] &lt;/math&gt;13p + 3e = 100&lt;math&gt; [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have &lt;/math&gt;p + 0e \equiv 1 \pmod 3&lt;math&gt; so &lt;/math&gt;p&lt;math&gt; leaves a remainder of 1 on division by 3.<br /> <br /> Since &lt;/math&gt;p \geq 2&lt;math&gt;, possible values for &lt;/math&gt;p&lt;math&gt; are 4, 7, 10 ....<br /> <br /> Since 13 pencils cost less than 100 cents, &lt;/math&gt;13p &lt; 100&lt;math&gt;. &lt;/math&gt;13 \times 10 = 130&lt;math&gt; is too high, so &lt;/math&gt;p&lt;math&gt; must be 4 or 7.<br /> <br /> If &lt;/math&gt;p = 4&lt;math&gt; then &lt;/math&gt;13p = 52&lt;math&gt; and so &lt;/math&gt;3e = 48&lt;math&gt; giving &lt;/math&gt;e = 16&lt;math&gt;. This contradicts the pencil being more expensive. The only remaining value for &lt;/math&gt;p&lt;math&gt; is 7; then the 13 pencils cost &lt;/math&gt;7 \times 13= 91&lt;math&gt; cents and so the 3 erasers together cost 9 cents and each eraser costs &lt;/math&gt;\frac{9}{3} = 3&lt;math&gt; cents.<br /> <br /> Thus one pencil plus one eraser cost &lt;/math&gt;7 + 3 = 10&lt;math&gt; cents, which is answer choice &lt;/math&gt;\mathrm{(A) \ }$.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_9&diff=21937 2006 AMC 12A Problems/Problem 9 2008-01-05T23:22:47Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> <br /> Oscar buys &lt;math&gt;13&lt;/math&gt; pencils and &lt;math&gt;3&lt;/math&gt; erasers for &lt;math&gt;\&lt;/math&gt;1.00&lt;math&gt;. A pencil costs more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser?<br /> <br /> &lt;/math&gt; \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18&lt;math&gt;<br /> <br /> &lt;/math&gt;\mathrm{(E) \ } 20&lt;math&gt;<br /> <br /> == Solution ==<br /> Let the price of a pencil be &lt;/math&gt;p&lt;math&gt; and an eraser &lt;/math&gt;e&lt;math&gt;. Then &lt;/math&gt;13p + 3e = 100&lt;math&gt; with &lt;/math&gt;p &gt; e &gt; 0&lt;math&gt;. Since &lt;/math&gt;p&lt;math&gt; and &lt;/math&gt;e&lt;math&gt; are [[positive integer]]s, we must have &lt;/math&gt;e \geq 1&lt;math&gt; and &lt;/math&gt;p \geq 2&lt;math&gt;.<br /> <br /> Considering the [[equation]] &lt;/math&gt;13p + 3e = 100&lt;math&gt; [[modulo]] 3 (that is, comparing the [[remainder]]s when both sides are divided by 3) we have &lt;/math&gt;p + 0e \equiv 1 \pmod 3&lt;math&gt; so &lt;/math&gt;p&lt;math&gt; leaves a remainder of 1 on division by 3.<br /> <br /> Since &lt;/math&gt;p \geq 2&lt;math&gt;, possible values for &lt;/math&gt;p&lt;math&gt; are 4, 7, 10 ....<br /> <br /> Since 13 pencils cost less than 100 cents, &lt;/math&gt;13p &lt; 100&lt;math&gt;. &lt;/math&gt;13 \times 10 = 130&lt;math&gt; is too high, so &lt;/math&gt;p&lt;math&gt; must be 4 or 7.<br /> <br /> If &lt;/math&gt;p = 4&lt;math&gt; then &lt;/math&gt;13p = 52&lt;math&gt; and so &lt;/math&gt;3e = 48&lt;math&gt; giving &lt;/math&gt;e = 16&lt;math&gt;. This contradicts the pencil being more expensive. The only remaining value for &lt;/math&gt;p&lt;math&gt; is 7; then the 13 pencils cost &lt;/math&gt;7 \times 13= 91&lt;math&gt; cents and so the 3 erasers together cost 9 cents and each eraser costs &lt;/math&gt;\frac{9}{3} = 3&lt;math&gt; cents.<br /> <br /> Thus one pencil plus one eraser cost &lt;/math&gt;7 + 3 = 10&lt;math&gt; cents, which is answer choice &lt;/math&gt;\mathrm{(A) \ }$.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_7&diff=21934 2006 AMC 12A Problems/Problem 7 2008-01-05T23:16:29Z <p>Lotrgreengrapes7926: </p> <hr /> <div>== Problem ==<br /> <br /> Mary is &lt;math&gt;20\%&lt;/math&gt; older than Sally, and Sally is &lt;math&gt;40\%&lt;/math&gt; younger than Danielle. The sum of their ages is &lt;math&gt;23.2&lt;/math&gt; years. How old will Mary be on her next birthday?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;m&lt;/math&gt; be Mary's age, let &lt;math&gt;s&lt;/math&gt; be Sally's age, and let &lt;math&gt;d&lt;/math&gt; be Danielle's age. We have &lt;math&gt;s=.6d&lt;/math&gt;, and &lt;math&gt;m=1.2s=1.2(.6d)=.72d&lt;/math&gt;. The sum of their ages is &lt;math&gt;m+s+d=.72d+.6d+d=2.32d&lt;/math&gt;. Therefore, &lt;math&gt;2.32d=23.2&lt;/math&gt;, and &lt;math&gt;d=10&lt;/math&gt;. Then &lt;math&gt;m=.72(10)=7.2&lt;/math&gt;. Mary will be &lt;math&gt;8&lt;/math&gt; on her next birthday. The answer is B.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Joining_an_ARML_team&diff=15137 Joining an ARML team 2007-05-26T21:31:52Z <p>Lotrgreengrapes7926: </p> <hr /> <div>Team selection for the [[American Regions Mathematics League]] varies from state to state.<br /> <br /> == Georgia ==<br /> <br /> [[Georgia ARML]] has sent two teams to ARML each year since 1991. In the past, the A team competed in the A division and the B team in the B division, but recently, the B team placed highly in the B division and so will compete in the A division at least until 2009. Usually, three alternates compete as well. Students interested in participation should do well in [[Georgia mathematics competitions | local tournaments]] and the [[AMC]] and [[AIME]]. In addition to schools invited to the annual [[GCTM State Math Tournament | varsity state tournament]], the Georgia ARML coaches invite other individuals that are under serious consideration for the ARML team.The coaches select the team members during the state tournament based on [[AMC | USAMO index]], performance in local tournaments, and score at the state tournament. Underclassmen are given leeway, as they have years to improve: middle schoolers, however, are rarely selected. For more detailed information on team selection, see the [http://paideiaschool.org/TeacherPages/Steve_Sigur/resources/GAARML/GEORGIAARML/Welcome.html Georgia ARML website].<br /> <br /> The team usually practices on Sundays from the state tournament until the trip to ARML. The specific compositions of the A and B teams are not usually determined until immediately before ARML. Whether a team member will be on the A team, will be on the B team, or will be an alternate depends on the person's performance against other team members in practice individual rounds and the coaches's discretion.<br /> <br /> == Maine ==<br /> <br /> The 2 Maine ARML Teams consist of approximately the top 30 scorers of 5 MAML (Maine Association of Math Leagues) Meets. Training includes the problem set &quot;Pete's Fabulous 42.&quot;<br /> <br /> * [http://www.maml.net MAML Website]<br /> <br /> ==Minnesota==<br /> <br /> Minnesota sends two teams to ARML each year, with the Gold and Maroon teams usually competing in divisions A and B, respectively.<br /> <br /> Roughly 35 students are invited to ARML practices, which take place on three consecutive Saturdays in May. There is no practice Memorial Day weekend. Invitations to the ARML team are extended to the top 10 state scorers on the AMC12, the top 10 regular-season scorers on the Minnesota High School Math League, and the top 10 scorers on the Invitational Event at the statewide math league tournament (held in March).<br /> <br /> Since these lists tend to overlap quite a bit, invitations are usually given to students &quot;further down&quot; these lists until enough invites have been given to fill two 15-person teams. <br /> <br /> In addition, an extra 5 or so younger students (typically in grades 8 through 10) are invited to be ARML &quot;students in training&quot;. They may or may not go to ARML, but often serve as alternates (in the event that other students cannot attend). The expectation is that a student in training will learn from the practices, and the following year will be on one of the two teams. Since the creation of the AMC8 and AMC10 exams, the top scorers from these exams have typically been invited to ARML practices, either as team members or students in training.<br /> <br /> The selection of the Gold and Maroon teams is determined by students' performance at practices, and is not announced until the night before the competition.<br /> <br /> Usually at least one student in training is invited to go to ARML. This is to prevent a last-minute no-show (due to illness or emergency, for example) from crippling one of the teams.<br /> <br /> The 2007 MN ARML team has already been selected. <br /> <br /> * [http://www.macalester.edu/mathleague/index.htm Minnesota State High School Math League site]<br /> <br /> ==New York City==<br /> For information about the New York City Math Team, please visit [http://www.nycmathteam.com the NYC Math Team homepage].<br /> <br /> ==Ohio==<br /> <br /> Invitation is based on OCTM (a state-wide competition) and AMC scores. Also, at the OHMIO (second level of OCTM) the offer to join is extended to anyone who is interested. Team placement is based on a combination of OCTM scores, AMC scores, and how well the person does in practices. Ohio normally sends two teams, but is sending three this year because enough students were interested. Also, starting this year, the Ohio A team is competing in division A. The other two teams are competing in division B.<br /> <br /> The first practice is Sunday, April 22, and the second is Saturday, May 19.<br /> <br /> ==San Francisco Bay Area==<br /> <br /> Any student is welcome to join the SFBA ARML team. Practices are held every Sunday at Stanford University, usually from 2:00 until 4:30 or 5:00. Coaches also set up practices at Berkeley on some weeks for students who live closer to there than Stanford. Practices begin in mid-April and run until the weekend before Memorial Day. Team placement is determined sometime in May by a combination of AMC/AIME scores and practice ARML tests.<br /> <br /> To join the team and receive announcements about practices, join the &quot;SFBA-ARML&quot; Yahoo group, or just show up at practice. You can also send an AoPS PM to Sly Si for more information. If you're a Stanford student and interested in coaching, send a PM to Sly Si.<br /> <br /> The April 22 practice at Stanford has been changed to 3:15-5:30.<br /> <br /> ==South Carolina==<br /> <br /> To join the SC All-State Team, one must take a preliminary exam administered through their school. For more information, please contact ronskimomo@hotmail.com.<br /> <br /> The preliminary exam is composed of 25 questions (non multiple choice), and is usually composed of easy to mid range AMC-12 level questions. From this exam, approximately 50-60 (in 2006 it was 49) of the top scorers from the state are selected into the South Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All State Team, he or she is invited to one or two ARML practices which are usually composed of individual tests, team tests, and a power round test.<br /> <br /> <br /> * [http://scall-statemathteam.com/default.aspx Official SC ARML site]<br /> * [[South Carolina ARML | SC ARML wiki page]]<br /> <br /> == Southern California ==<br /> <br /> The Southern California team is open to residents of the following Southern California counties: Santa Barbara, Ventura, Los Angeles, Orange, Kern, San Bernardino, and Riverside. The organization fields three teams (45 students) and competes at the western ARML site in Las Vegas.<br /> <br /> Practices are held throughout the school year, approximately once a month, on the campus of California State University, Long Beach. Practices are normally held on Saturday afternoons. In addition, there is a Santa Barbara area group that meets and practices in Goleta and becomes part of the Southern California team.<br /> <br /> Team selection is based on all of the following criteria:<br /> <br /> * Attendence at practice sessions.<br /> * Performance on problems at practice sessions.<br /> * Performance at ARML itself in previous years.<br /> * Performance on AMC and AIME, including current and previous years.<br /> * Performance at CSULB Math Day at the Beach, a contest held in March.<br /> <br /> The coach is Dr. Kent Merryfield, a professor at CSULB. His AoPS user name is Kent Merryfield. Please contact him for further information.<br /> <br /> * [http://www.csulb.edu/depts/math/mathday/index.htm Math Day at the Beach website.]<br /> * [http://www.csulb.edu/depts/math/arml/index.htm SoCal ARML website.]<br /> * [[Southern California ARML | SoCal ARML wiki page.]]<br /> <br /> ==See Other==<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=820893#820893 Newer discussion on AoPS message boards.]<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40434 Older discussion on AoPS message boards.]<br /> * [[ARML]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=LaTeX&diff=11676 LaTeX 2006-11-12T18:47:29Z <p>Lotrgreengrapes7926: /* Useful Codes */</p> <hr /> <div>'''LaTeX''' is a typesetting language used primarily to type mathematical expressions in an elegant fashion. For example, without LaTeX, &lt;math&gt;\frac{35}{137}&lt;/math&gt; would have to be written as 35/137. To use LaTeX in the forums, enclose your LaTeX code with dollar signs: &lt;nowiki&gt;$your codes here\$&lt;/nowiki&gt;. To use LaTeX on AoPSWiki, enclose your code with math tags instead of dollar signs, like so: &lt;nowiki&gt;&lt;math&gt;your codes here&lt;/math&gt;&lt;/nowiki&gt;<br /> <br /> CHANGE: Dollar signs can now be used to use LaTeX.<br /> <br /> ==Useful Codes==<br /> <br /> '''\boxed{Answer}''' produces a box around your Answer. Cannot be used in Wiki<br /> <br /> '''\frac{a}{b}''' produces a [[fraction]] with [[numerator]] &lt;math&gt;a&lt;/math&gt; and [[denominator]] &lt;math&gt;b&lt;/math&gt;. &lt;math&gt;\frac{a}{b}&lt;/math&gt;<br /> <br /> '''^\circ''' produces the degrees symbol. &lt;math&gt;a^{\circ}&lt;/math&gt;<br /> <br /> '''\text{Your Text Here}''' produces text within LaTeX. &lt;math&gt;\mbox{Your Text Here}&lt;/math&gt;. Cannot be used in the Wiki.<br /> <br /> '''\mbox{Your Text Here}''' Produces text within LaTeX; can be used in the Wiki. &lt;math&gt;\mbox{Your Text Here}&lt;/math&gt;<br /> <br /> '''\sqrt{x}''' produces the square root of &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\sqrt{x}&lt;/math&gt;<br /> <br /> '''\sqrt[n]{x}''' produces the &lt;math&gt;n&lt;/math&gt;th root of &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\sqrt[n]{x}&lt;/math&gt;<br /> <br /> '''a\equiv b \pmod{c}''' produces &lt;math&gt;a&lt;/math&gt; is equivalent to &lt;math&gt;b&lt;/math&gt; mod &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;a\equiv b \pmod{c}&lt;/math&gt; See [[Mods |Modular Arithmetic]]<br /> <br /> '''\binom{9}{3}''' produces 9 choose 3. Cannot be used in Wiki.<br /> <br /> '''{n}\choose{r}''' produces n choose r. &lt;math&gt;{n}\choose{r}&lt;/math&gt;<br /> <br /> '''x^{y}''' produces x to the power of y. &lt;math&gt;x^y&lt;/math&gt;<br /> <br /> '''x_{y}''' produces x with y in subscript. &lt;math&gt;x_y&lt;/math&gt;<br /> <br /> '''\rightarrow''' produces an arrow to the right. &lt;math&gt;\rightarrow&lt;/math&gt;<br /> <br /> '''\leftarrow''' produces an arrow to the left. &lt;math&gt;\leftarrow&lt;/math&gt;<br /> <br /> '''\uparrow''' produces an arrow pointing upwards. &lt;math&gt;\uparrow&lt;/math&gt;<br /> <br /> '''\downarrow''' produces an arrow pointing downwards. &lt;math&gt;\downarrow&lt;/math&gt;<br /> <br /> '''\updownarrow''' produces an arrow pointing up and down. &lt;math&gt;\updownarrow&lt;/math&gt;<br /> <br /> '''\ge''' produces a greater than or equal to sign. &lt;math&gt;\ge&lt;/math&gt;<br /> <br /> '''\le''' produces a less than or equal to sign. &lt;math&gt;\le&lt;/math&gt;<br /> <br /> '''\not&gt;''' produces a not greater than sign. &lt;math&gt;\not&gt;&lt;/math&gt;<br /> <br /> '''\not&lt;''' produces a not less than sign. &lt;math&gt;\not&lt;&lt;/math&gt;<br /> <br /> '''\not\ge''' produces a not greater than or equal to sign. &lt;math&gt;\not\ge&lt;/math&gt;<br /> <br /> '''\not\le''' produces a not less than or equal to sign. &lt;math&gt;\not\le&lt;/math&gt;<br /> <br /> '''\neq''' produces a not equal to sign. &lt;math&gt;\neq&lt;/math&gt;<br /> <br /> '''\infty''' produces an infinity sign. &lt;math&gt;\infty&lt;/math&gt;<br /> <br /> '''\perp''' produces a perpendicular sign. &lt;math&gt;\perp&lt;/math&gt;<br /> <br /> '''\angle''' produces an angle sign. &lt;math&gt;\angle&lt;/math&gt;<br /> <br /> '''\triangle''' produces a triangle. &lt;math&gt;\triangle&lt;/math&gt;<br /> <br /> '''\ldots''' produces three dots at the bottom of a line (ellipsis). &lt;math&gt;\ldots&lt;/math&gt;<br /> <br /> '''\cdots''' produces three dots in the middle of a line (as in a series sum or product). &lt;math&gt;\cdots&lt;/math&gt;<br /> <br /> Note that on AoPSWiki, many codes that work on the AoPS forums do not work. Also, a helpful tip is that if LaTeX fails to render within AoPSWiki, try adding the code \displaystyle to the beginning of the string of LaTeX. This often fixes minor rendering problems.<br /> <br /> <br /> Also note that you do not have to use braces, &quot;{&quot; and &quot;}&quot;, when you only want one character in the operation.<br /> <br /> ===Examples===<br /> * x^y is the same as x^{y}. &lt;math&gt;x^y&lt;/math&gt;<br /> * x_y is the same as x_{y}. &lt;math&gt;x_y&lt;/math&gt;<br /> * BUT x^10 is ''not'' the same as x^{10}. &lt;math&gt;x^10&lt;/math&gt; instead of &lt;math&gt;x^{10}&lt;/math&gt;.<br /> <br /> ==Fonts==<br /> <br /> === Font families ===<br /> <br /> * Roman (default): \textrm{...}<br /> * Sans-serif: \textsf{...}<br /> * Monospace (typewriter): \texttt{...}<br /> <br /> === Font sizes ===<br /> <br /> To activate a font size, write '{\tiny{This text is tiny}}', for example.<br /> <br /> * \tiny (5 pt.)<br /> * \scriptsize (7 pt.)<br /> * \footnotesize (8 pt.)<br /> * \small (9 pt.)<br /> * \normalsize (10 pt.)<br /> * \large (12 pt.)<br /> * \Large (14 pt.)<br /> * \LARGE (18 pt.)<br /> * \huge (20 pt.)<br /> * \Huge (24 pt.)<br /> <br /> === Font styles ===<br /> <br /> * Bold \textbf{...}<br /> * Italics \textit{...}<br /> * Slanted \textsl{...}<br /> * Small capitals \textsc{...}<br /> * Sans-serif \textsf{...}<br /> * Monospace \texttt{...}<br /> * Emphasis \emph{...}<br /> <br /> ==Tutorials &amp; Tools ==<br /> <br /> * [http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php AoPS LaTeX Guide]<br /> *[http://www.artofproblemsolving.com/LaTeX/AoPS_L_HelpIndex.php AoPS LaTeX Help Index]<br /> *[http://en.wikipedia.org/wiki/LaTeX Wikipedia Article]<br /> *[http://sciencesoft.at/index.jsp?link=latex&amp;lang=en&amp;wiki=1 This] is a useful site that will change LaTeX input into a PNG image.<br /> <br /> <br /> <br /> {{tutorial}}</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Set&diff=11675 Set 2006-11-12T18:36:20Z <p>Lotrgreengrapes7926: /* Power set */</p> <hr /> <div>{{stub}}<br /> <br /> The notion of a '''set''' is one of the fundamental notions in mathematics that has to be left undefined. Of course, we have plenty of synonyms for the word &quot;set,&quot; like ''collection, ensemble, group'', etc., but those names really do not define the meaning of the word ''set''; all they can do is replace it in various sentences. So, instead of defining what sets are, one has to define what can be done with them or, in other words, what axioms the sets satisfy. These axioms are chosen to agree with our intuitive concept of a set, on one hand, and to allow various, sometimes quite sophisticated, mathematical constructions on the other hand. For the full collection of these axioms, see [[Zermelo-Fraenkel Axioms]]. In this article we shall present just a brief discussion of the most common properties of sets and operations related to them.<br /> <br /> <br /> <br /> ==Relation of ''belonging''==<br /> <br /> The most important property of sets is that, for every ''object'' &lt;math&gt;x&lt;/math&gt; and a set &lt;math&gt;S&lt;/math&gt;, we can say whether &lt;math&gt;x&lt;/math&gt; belongs to &lt;math&gt;S&lt;/math&gt; (written as &lt;math&gt;x\in S&lt;/math&gt;), or not (written as &lt;math&gt;x\notin S&lt;/math&gt;). Two sets &lt;math&gt;S'&lt;/math&gt; and &lt;math&gt;S''&lt;/math&gt; are equal if they include the same objects, i.e., if for every object &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;x\in S'&lt;/math&gt; if and only if &lt;math&gt;x\in S''&lt;/math&gt;. <br /> <br /> <br /> <br /> ==Specifying a set by listing its objects==<br /> <br /> This means that in order to indentify a particular set, it suffices to tell which objects belong to this set. If the set contains just several such objects, all you need to do is list them. So, you can specify the set consisting of the numbers &lt;math&gt;1,3,5&lt;/math&gt;, and &lt;math&gt;239&lt;/math&gt;, for example. (The standard notation for this set is &lt;math&gt;\{1,3,5,239\}&lt;/math&gt;. Note that the order in which the terms are listed is completely unimportant: we have to follow some order when writing things in one line, but you should actually imagine those numbers flowing freely inside those curly braces with no preference given to any of them. What matters is that these four numbers are in the set and everything else is out). But how do you specify sets that have very many (maybe [[infinite]]ly many) elements? You cannot list them all even if you spend your entire life writing!<br /> <br /> <br /> <br /> ==Specifying a set by the common property of its elements==<br /> <br /> Another way to specify a set is to use some property to tell when an object belongs to this set. For instance, we may try to think (alas, only try!) of the set of all objects with green hair. In this case, we do not even try to list all such objects. We just decide that something belongs to this set if it has green hair and doesn't belong to it otherwise. This is a wonderful way to describe a set. <br /> <br /> Unfortunately, this method has several potential pitfalls. It turns out, counter-intuitively, that not every collection of objects with a certain property is a set. The most famous example of this problem is [[Russell's Paradox]]: consider the property, &quot;is a set and does not contain itself.&quot; (Remember that, given a set, we should be able to tell about '''every''' object whether it belongs to this set or not; in particular, we can ask this question about the set itself). The set &lt;math&gt;S&lt;/math&gt; specified by this property can neither belong, nor not belong to itself. There are a variety of ways to resolve this paradox, but the problem is clear: this way to describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property &lt;math&gt;P&lt;/math&gt;, choose a known set &lt;math&gt;T&lt;/math&gt;. Then the collection &lt;math&gt;S&lt;/math&gt; of elements of &lt;math&gt;T&lt;/math&gt; which have property &lt;math&gt;P&lt;/math&gt; will always be a set. (In particular, for our previous example to lead to a paradox, we would need to choose &lt;math&gt;T = \{\mathrm{all \; sets}\}&lt;/math&gt;. However, it turns out that it can be proven that the set of all sets does not exist -- the collection of all sets is too big to be a set.)<br /> <br /> <br /> <br /> ==Subsets==<br /> We say that a set &lt;math&gt;A&lt;/math&gt; is a [[subset]] of a set &lt;math&gt;S&lt;/math&gt; if every object &lt;math&gt;x&lt;/math&gt; that belongs to &lt;math&gt;A&lt;/math&gt; also belongs to &lt;math&gt;S&lt;/math&gt;. This is denoted &lt;math&gt;A\subseteq S&lt;/math&gt; or &lt;math&gt;A\subset S&lt;/math&gt;. For example, the sets &lt;math&gt;\{1,2\}&lt;/math&gt; and &lt;math&gt;\{2,3\}&lt;/math&gt; are subsets of the set &lt;math&gt;\{1,2,3\}&lt;/math&gt;, but the set &lt;math&gt;\{1,6\}&lt;/math&gt; is not. Thus we can say that two sets are equal if and only if each is a subset of the other.<br /> <br /> <br /> <br /> ==Power set==<br /> The [[power set]] of a set &lt;math&gt;A&lt;/math&gt;, denoted &lt;math&gt;\mathcal{P}(A)&lt;/math&gt; is defined as the set of all its subsets. For example, the power set of &lt;math&gt;\{1,2,3\}&lt;/math&gt; is &lt;math&gt;\{\{\},\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\},\{1,2,3\}\}&lt;/math&gt;. If a &lt;math&gt;A&lt;/math&gt; is a [[finite]] set of size &lt;math&gt;n&lt;/math&gt; then &lt;math&gt;\mathcal{P}(A)&lt;/math&gt; has size &lt;math&gt;2^n&lt;/math&gt;, because &lt;math&gt;\sum_{i=0}^{n} {n\choose i} =2^n&lt;/math&gt;.<br /> <br /> ==Union and intersection==<br /> The [[union]] of two or more sets is the set of all objects that belong to one or more of the sets. The union of A and B is denoted &lt;math&gt;A\cup B&lt;/math&gt;. For example, the union of &lt;math&gt;\{1,2\}&lt;/math&gt; and &lt;math&gt;\{1,3,5\}&lt;/math&gt; is &lt;math&gt;\{1,2,3,5\}&lt;/math&gt;. Unions can also be represented just as sums and products can be. &lt;math&gt;\displaystyle\bigcup_{statement}S&lt;/math&gt; would be the union of all sets &lt;math&gt;S&lt;/math&gt; that satisfy the statement. So, for example, &lt;math&gt;\displaystyle\bigcup_{S\subset\mathbb{N}}S&lt;/math&gt; would be the set of all natural numbers &lt;math&gt;\mathbb{N}&lt;/math&gt;. <br /> <br /> The [[intersection]] of two or more sets is the set of all objects that belong to all of the sets. The intersection of A and B is denoted &lt;math&gt;A\cap B&lt;/math&gt;. For example, the intersection of &lt;math&gt;\{1,2\}&lt;/math&gt; and &lt;math&gt;\{1,3,5\}&lt;/math&gt; is &lt;math&gt;\{1\}&lt;/math&gt;. Just like unions, intersections can be represented as such: &lt;math&gt;\displaystyle\bigcap_{statement}S&lt;/math&gt;. For example, &lt;math&gt;\displaystyle\bigcap_{S\,is\,P(A)\,for\,some\,set\,A}S=\emptyset&lt;/math&gt;, or the empty set defined next. <br /> <br /> <br /> <br /> ==Empty set==<br /> An [[empty set]] denoted &lt;math&gt;\emptyset&lt;/math&gt; is a set with no elements.<br /> <br /> <br /> <br /> ==Infinite sets==<br /> An [[infinite]] set can be defined as a set that has the same [[cardinality]] as one of its proper subsets. Alternatively, infinite sets are those which cannot be put into [[correspondence]] with any set of the form {1, 2, ..., n}.<br /> <br /> <br /> <br /> == Cardinality==<br /> The [[cardinality]] of a set &lt;math&gt;A&lt;/math&gt;, denoted &lt;math&gt;|A|&lt;/math&gt;, is (informally) the size of the set. For a finite set, the cardinality is simply the number of elements. The empty set has cardinality 0. <br /> <br /> &lt;math&gt;|A|=|B|&lt;/math&gt; [[iff]] there is a bijective function &lt;math&gt;f:A\to B&lt;/math&gt; meaning that there is a function &lt;math&gt;f&lt;/math&gt; that maps all elements of &lt;math&gt;A&lt;/math&gt; to all the elements of &lt;math&gt;B&lt;/math&gt; with one-to-one correspondence. <br /> <br /> &lt;math&gt;|A|\le|B|&lt;/math&gt; iff there exists an injective function &lt;math&gt;f:A\to B&lt;/math&gt; meaning there is a function &lt;math&gt;f&lt;/math&gt; that maps all elements of &lt;math&gt;A&lt;/math&gt; to (not necessarily all) elements of &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;|A|\ge|B|&lt;/math&gt; can be defined similarily by expressing it as &lt;math&gt;|B|\le|A|&lt;/math&gt;.<br /> <br /> &lt;math&gt;|A|&lt;|B|&lt;/math&gt; iff there exists an injective function &lt;math&gt;f:A\to B&lt;/math&gt; and there is no bijective function &lt;math&gt;g:A\to B&lt;/math&gt; meaning &lt;math&gt;|A|\le|B|&lt;/math&gt; but &lt;math&gt;|A|\neq|B|&lt;/math&gt;. &lt;math&gt;|A|&gt;|B|&lt;/math&gt; is defined similarly.<br /> <br /> Although showing that &lt;math&gt;A\le B&lt;/math&gt; and &lt;math&gt;B\le A&lt;/math&gt; implies that &lt;math&gt;A=B&lt;/math&gt; is easy to prove when using finite sets, it is more complicated when using infinite sets. This theorem is called the [[Schroder-Bernstein Theorem]] and was proven by Schroder and Bernstein. <br /> <br /> <br /> ''To be continued''</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_4&diff=11674 2004 AMC 10A Problems/Problem 4 2006-11-12T18:18:10Z <p>Lotrgreengrapes7926: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;x&lt;/math&gt; if &lt;math&gt;|x-1|=|x-2|&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt;|x-1|&lt;/math&gt; is equivalent to the distance between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;; &lt;math&gt;|x-2|&lt;/math&gt; is equivalent to the distance between &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;x&lt;/math&gt; is equidistant from &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> *[[2004 AMC 10A Problems]]<br /> <br /> *[[2004 AMC 10A Problems/Problem 3|Previous Problem]]<br /> <br /> *[[2004 AMC 10A Problems/Problem 5|Next Problem]]<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Harmonic_series&diff=11673 Harmonic series 2006-11-12T18:13:29Z <p>Lotrgreengrapes7926: </p> <hr /> <div>Generally, a '''harmonic series''' is a [[series]] whose terms involve the [[reciprocal]]s of the [[positive integer]]s.<br /> <br /> There are several sub-types of '''harmonic series'''.<br /> <br /> The the most basic harmonic series is the infinite sum<br /> &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots&lt;/math&gt; <br /> This sum slowly approaches infinity.<br /> <br /> The alternating harmonic series,<br /> &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots&lt;/math&gt; , though, approaches &lt;math&gt; \displaystyle \ln 2&lt;/math&gt;.<br /> <br /> The [[zeta-function]] is a harmonic series when the input is one.<br /> <br /> == How to solve ==<br /> <br /> === Harmonic Series ===<br /> <br /> It can be shown that the harmonic series converges by grouping the terms. We know that the first term, 1, added to the second term, &lt;math&gt;\frac{1}{2}&lt;/math&gt; is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. We also know that the third and and fourth terms, &lt;math&gt;\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{1}{4}&lt;/math&gt;, add up to something greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. And we continue grouping the terms between powers of two. So we have <br /> &lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty&lt;/math&gt;<br /> <br /> === Alternating Harmonic Series ===<br /> <br /> === General Harmonic Series ===<br /> &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}&lt;/math&gt; is the general harmonic series, where each term is the reciprocal of a term in an arithmetic series.<br /> <br /> '''Case 1:''' &lt;math&gt;a\ge b&lt;/math&gt;<br /> <br /> &lt;math&gt;ai+a\ge ai+b&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{ai+b}\ge\frac{1}{ai+a}=\frac{1}{a}\left(\frac{1}{i+1}\right)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{a} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty&lt;/math&gt;<br /> <br /> '''Case 2:''' &lt;math&gt;a&lt;b&lt;/math&gt;<br /> <br /> &lt;math&gt;ai+b&lt;bi+b&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{ai+b}&gt;\frac{1}{bi+b}=\frac{1}{b}\left(\frac{1}{i+1}\right)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{ai+b}\ge\frac{1}{b} \left(\sum_{i=1}^{\infty}\frac{1}{i+1}\right)\to\infty&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{ai+b}=\infty&lt;/math&gt;</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetico-geometric_series&diff=11672 Arithmetico-geometric series 2006-11-12T17:08:22Z <p>Lotrgreengrapes7926: </p> <hr /> <div>An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: &lt;math&gt;x_n=a_ng_n&lt;/math&gt;, where &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;g_n&lt;/math&gt; are the &lt;math&gt;n&lt;/math&gt;th terms of arithmetic and geometric sequences, respectively.<br /> == Finite Sum ==<br /> The sum of the first n terms of an arithmetico-geometric sequence is &lt;math&gt;\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common ratio of &lt;math&gt;g_n&lt;/math&gt;. Or, &lt;math&gt;\frac{a_ng_{n+1}-x_1-drS_g}{r-1}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the sum of the first &lt;math&gt;n&lt;/math&gt; terms of &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})&lt;/math&gt;<br /> <br /> Let &lt;math&gt;S_n&lt;/math&gt; represent the sum of the first n terms.<br /> &lt;math&gt;S_n=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots+(a_1+(n-1)d)(g_1r^{n-1})&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=a_1g_1+(a_1+dg_1)r+(a_1g_1+2dg_1)r^2+\ldots+(a_1g_1+(n-1)dg_1)r^{n-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+\ldots+(a_1g_1+(n-1)dg_1)r^{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n-S_n=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots-dg_1r^{n-1}+(a_1g_1+(n-1)dg_1)r^n&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n(r-1)=(a_1+(n-1)d)g_1r^n-a_1g_1-\frac{dg_1r(r^{n-1}-1)}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=\frac{(a_1+(n-1)d)g_1r^n}{r-1}-\frac{a_1g_1}{r-1}-\frac{dg_1r(r^{n-1}-1)}{(r-1)^2}=\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;<br /> <br /> == Infinite Sum ==<br /> The sum of an infinite arithmetico-geometric sequence is &lt;math&gt;\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common difference of &lt;math&gt;g_n&lt;/math&gt; (&lt;math&gt;|r|&lt;1&lt;/math&gt;). Or, &lt;math&gt;\frac{drS_g+x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the infinite sum of the &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> &lt;math&gt;S=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;rS=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+.\,.\,.&lt;/math&gt;<br /> <br /> &lt;math&gt;rS-S=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots=-a_1g_1+\frac{dg_1r}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S=\frac{dg_1r}{(r-1)^2}-\frac{a_1g_1}{r-1}=\frac{dg_2}{(r-1)^2}-\frac{x_1}{r-1}=\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;<br /> <br /> == Example Problems ==<br /> * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]<br /> <br /> == See Also ==<br /> * [[Sequence]]<br /> * [[Arithmetic sequence]]<br /> * [[Geometric sequence]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Harmonic_series&diff=11007 Harmonic series 2006-11-05T14:48:54Z <p>Lotrgreengrapes7926: /* Harmonic Series */</p> <hr /> <div>Generally, a '''harmonic series''' is a [[series]] whose terms involve the [[reciprocal]]s of the [[positive integer]]s.<br /> <br /> There are several sub-types of '''harmonic series'''.<br /> <br /> The the most basic harmonic series is the infinite sum<br /> &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots&lt;/math&gt; <br /> This sum slowly approaches infinity.<br /> <br /> The alternating harmonic series,<br /> &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots&lt;/math&gt; , though, approaches &lt;math&gt; \displaystyle \ln 2&lt;/math&gt;.<br /> <br /> The general harmonic series, &lt;math&gt;\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}&lt;/math&gt;, has its value depending on the value of the constants &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> The [[zeta-function]] is a harmonic series when the input is one.<br /> <br /> == How to solve ==<br /> <br /> ===Harmonic Series===<br /> <br /> It can be shown that the harmonic series converges by grouping the terms. We know that the first term, 1, added to the second term, &lt;math&gt;\frac{1}{2}&lt;/math&gt; is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. We also know that the third and and fourth terms, &lt;math&gt;\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{1}{4}&lt;/math&gt;, add up to something greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. And we continue grouping the terms between powers of two. So we have <br /> &lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots \ge \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \to \infty&lt;/math&gt;<br /> <br /> ===Alternating Harmonic Series===</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetico-geometric_series&diff=10978 Arithmetico-geometric series 2006-11-05T03:37:47Z <p>Lotrgreengrapes7926: /* Infinite Sum */</p> <hr /> <div>An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: &lt;math&gt;x_n=a_ng_n&lt;/math&gt;, where &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;g_n&lt;/math&gt; are the &lt;math&gt;n&lt;/math&gt;th terms of arithmetic and geometric sequences, respectively.<br /> == Finite Sum ==<br /> The sum of the first n terms of an arithmetico-geometric sequence is &lt;math&gt;\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common ratio of &lt;math&gt;g_n&lt;/math&gt;. Or, &lt;math&gt;\frac{a_ng_{n+1}-x_1-drS_g}{r-1}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the sum of the first &lt;math&gt;n&lt;/math&gt; terms of &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})&lt;/math&gt;<br /> <br /> Let &lt;math&gt;S_n&lt;/math&gt; represent the sum of the first n terms.<br /> &lt;math&gt;S_n=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots+(a_1+(n-1)d)(g_1r^{n-1})&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=a_1g_1+(a_1+dg_1)r+(a_1g_1+2dg_1)r^2+\ldots+(a_1g_1+(n-1)dg_1)r^{n-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+\ldots+(a_1g_1+(n-1)dg_1)r^{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n-S_n=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots-dg_1r^{n-1}+(a_1g_1+(n-1)dg_1)r^n&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n(r-1)=(a_1+(n-1)d)g_1r^n-a_1g_1-\frac{dg_1r(r^{n-1}-1)}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=\frac{(a_1+(n-1)d)g_1r^n}{r-1}-\frac{a_1g_1}{r-1}-\frac{dg_1r(r^{n-1}-1)}{(r-1)^2}=\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;<br /> <br /> == Infinite Sum ==<br /> The sum of an infinite arithmetico-geometric sequence is &lt;math&gt;\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common difference of &lt;math&gt;g_n&lt;/math&gt; (&lt;math&gt;|r|&lt;1&lt;/math&gt;). Or, &lt;math&gt;\frac{drS_g+x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the infinite sum of the &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> &lt;math&gt;S=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;rS=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+.\,.\,.&lt;/math&gt;<br /> <br /> &lt;math&gt;rS-S=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots=-a_1g_1+\frac{dg_1r}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S=\frac{dg_1r}{(r-1)^2}-\frac{a_1g_1}{r-1}=\frac{dg_2}{(r-1)^2}-\frac{x_1}{r-1}=\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;<br /> <br /> == Example Problems ==<br /> * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetico-geometric_series&diff=10977 Arithmetico-geometric series 2006-11-05T03:37:16Z <p>Lotrgreengrapes7926: </p> <hr /> <div>An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: &lt;math&gt;x_n=a_ng_n&lt;/math&gt;, where &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;g_n&lt;/math&gt; are the &lt;math&gt;n&lt;/math&gt;th terms of arithmetic and geometric sequences, respectively.<br /> == Finite Sum ==<br /> The sum of the first n terms of an arithmetico-geometric sequence is &lt;math&gt;\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common ratio of &lt;math&gt;g_n&lt;/math&gt;. Or, &lt;math&gt;\frac{a_ng_{n+1}-x_1-drS_g}{r-1}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the sum of the first &lt;math&gt;n&lt;/math&gt; terms of &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})&lt;/math&gt;<br /> <br /> Let &lt;math&gt;S_n&lt;/math&gt; represent the sum of the first n terms.<br /> &lt;math&gt;S_n=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots+(a_1+(n-1)d)(g_1r^{n-1})&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=a_1g_1+(a_1+dg_1)r+(a_1g_1+2dg_1)r^2+\ldots+(a_1g_1+(n-1)dg_1)r^{n-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+\ldots+(a_1g_1+(n-1)dg_1)r^{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n-S_n=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots-dg_1r^{n-1}+(a_1g_1+(n-1)dg_1)r^n&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n(r-1)=(a_1+(n-1)d)g_1r^n-a_1g_1-\frac{dg_1r(r^{n-1}-1)}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=\frac{(a_1+(n-1)d)g_1r^n}{r-1}-\frac{a_1g_1}{r-1}-\frac{dg_1r(r^{n-1}-1)}{(r-1)^2}=\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;<br /> <br /> == Infinite Sum ==<br /> The sum of an infinite arithmetico-geometric sequence is &lt;math&gt;\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common difference of &lt;math&gt;g_n&lt;/math&gt; (|r|&lt;1). Or, &lt;math&gt;\frac{drS_g+x_1}{1-r}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the infinite sum of the &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> &lt;math&gt;S=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r^2)+\ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;rS=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+.\,.\,.&lt;/math&gt;<br /> <br /> &lt;math&gt;rS-S=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots=-a_1g_1+\frac{dg_1r}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S=\frac{dg_1r}{(r-1)^2}-\frac{a_1g_1}{r-1}=\frac{dg_2}{(r-1)^2}-\frac{x_1}{r-1}=\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}&lt;/math&gt;<br /> <br /> == Example Problems ==<br /> * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetico-geometric_series&diff=10968 Arithmetico-geometric series 2006-11-05T03:06:25Z <p>Lotrgreengrapes7926: </p> <hr /> <div>An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: &lt;math&gt;x_n=a_ng_n&lt;/math&gt;, where &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;g_n&lt;/math&gt; are the &lt;math&gt;n&lt;/math&gt;th terms of arithmetic and geometric sequences, respectively.<br /> == Finite Sum ==<br /> The sum of the first n terms of an arithmetico-geometric sequence is &lt;math&gt;\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common ratio of &lt;math&gt;g_n&lt;/math&gt;. Or, &lt;math&gt;\frac{a_ng_{n+1}-x_1-drS_g}{r-1}&lt;/math&gt;, where &lt;math&gt;S_g&lt;/math&gt; is the sum of the first &lt;math&gt;n&lt;/math&gt; terms of &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})&lt;/math&gt;<br /> <br /> Let &lt;math&gt;S_n&lt;/math&gt; represent the sum of the first n terms.<br /> &lt;math&gt;S_n=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r)+\ldots+(a_1+(n-1)d)(g_1r^{n-1})&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=a_1g_1+(a_1+dg_1)r+(a_1g_1+2dg_1)r^2+\ldots+(a_1g_1+(n-1)dg_1)r^{n-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+\ldots+(a_1g_1+(n-1)dg_1)r^{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n-S_n=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots-dg_1r^{n-1}+(a_1g_1+(n-1)dg_1)r^n&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n(r-1)=(a_1+(n-1)d)g_1r^n-a_1g_1-\frac{dg_1r(r^{n-1}-1)}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=\frac{(a_1+(n-1)d)g_1r^n}{r-1}-\frac{a_1g_1}{r-1}-\frac{dg_1r(r^{n-1}-1)}{(r-1)^2}=\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;<br /> <br /> == Example Problems ==<br /> * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetico-geometric_series&diff=10965 Arithmetico-geometric series 2006-11-05T03:00:52Z <p>Lotrgreengrapes7926: </p> <hr /> <div>An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined as: &lt;math&gt;x_n=a_ng_n&lt;/math&gt;, where &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;g_n&lt;/math&gt; are the &lt;math&gt;n&lt;/math&gt;th terms of arithmetic and geometric sequences, respectively.<br /> == Finite Sum ==<br /> The sum of the first n terms of an arithmetico-geometric sequence is &lt;math&gt;\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference of &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; is the common ratio of &lt;math&gt;g_n&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})&lt;/math&gt;<br /> <br /> Let &lt;math&gt;S_n&lt;/math&gt; represent the sum of the first n terms.<br /> &lt;math&gt;S_n=a_1g_1+(a_1+d)(g_1r)+(a_1+2d)(g_1r)+\ldots+(a_1+(n-1)d)(g_1r^{n-1})&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=a_1g_1+(a_1+dg_1)r+(a_1g_1+2dg_1)r^2+\ldots+(a_1g_1+(n-1)dg_1)r^{n-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n=a_1g_1r+(a_1+dg_1)r^2+(a_1g_1+2dg_1)r^3+\ldots+(a_1g_1+(n-1)dg_1)r^{n}&lt;/math&gt;<br /> <br /> &lt;math&gt;rS_n-S_n=-a_1g_1-dg_1r-dg_1r^2-dg_1r^3-\ldots-dg_1r^{n-1}+(a_1g_1+(n-1)dg_1)r^n&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n(r-1)=(a_1+(n-1)d)g_1r^n-a_1g_1-\frac{dg_1r(r^{n-1}-1)}{r-1}&lt;/math&gt;<br /> <br /> &lt;math&gt;S_n=\frac{(a_1+(n-1)d)g_1r^n}{r-1}-\frac{a_1g_1}{r-1}-\frac{dg_1r(r^{n-1}-1)}{(r-1)^2}=\frac{a_ng_{n+1}}{r-1}-\frac{x_1}{r-1}-\frac{d(g_{n+1}-g_2)}{(r-1)^2}&lt;/math&gt;<br /> <br /> == Example Problems ==<br /> * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Geometric_sequence&diff=10962 Geometric sequence 2006-11-05T02:03:19Z <p>Lotrgreengrapes7926: </p> <hr /> <div>A '''geometric sequence''' is a [[sequence]] of numbers in which each term is a fixed [[multiple]] of the previous term. For example: 1, 2, 4, 8, 16, 32, ... is a geometric sequence because each term is twice the previous term. In this case, 2 is called the ''common ratio'' of the sequence. More formally, a geometric sequence may be defined [[recursion|recursively]] by:<br /> <br /> &lt;center&gt;&lt;math&gt;a_n = r\cdot a_{n-1}, n &gt; 1&lt;/math&gt;&lt;/center&gt;<br /> <br /> with a fixed first term &lt;math&gt;a_1&lt;/math&gt; and common ratio &lt;math&gt;r&lt;/math&gt;. Using this definition, the &lt;math&gt;n&lt;/math&gt;th term has the closed-form:<br /> <br /> &lt;center&gt;&lt;math&gt;\displaystyle a_n = a_1\cdot r^{n-1}&lt;/math&gt;&lt;/center&gt;<br /> <br /> ==Summing a Geometric Sequence==<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; terms of a geometric sequence is given by<br /> <br /> &lt;center&gt;&lt;math&gt;S_n = a_1 + a_2 + \ldots + a_n = a_1\cdot\frac{r^n-1}{r-1}&lt;/math&gt;&lt;/center&gt;<br /> <br /> where &lt;math&gt;a_1&lt;/math&gt; is the first term in the sequence, and &lt;math&gt;r&lt;/math&gt; is the common ratio.<br /> <br /> ==Infinite Geometric Sequences==<br /> <br /> An [[infinite]] geometric sequence is a geometric sequence with an infinite number of terms. If the common ratio is small, the terms will approach 0 and the sum of the terms will approach a fixed [[limit]]. In this case, &quot;small&quot; means &lt;math&gt;|r|&lt;1&lt;/math&gt;. We say that the sum of the terms of this sequence is a [[convergent|convergent sum]].<br /> <br /> For instance, the series &lt;math&gt;1 + \frac12 + \frac14 + \frac18 + \cdots&lt;/math&gt;, sums to 2. The general formula for the sum of such a sequence is:<br /> <br /> &lt;center&gt;&lt;math&gt;S = \frac{a_1}{1-r}&lt;/math&gt;.&lt;/center&gt; &lt;br&gt;&lt;br&gt;<br /> <br /> Where &lt;math&gt;a_1&lt;/math&gt; is the first term in the sequence, and &lt;math&gt;r&lt;/math&gt; is the common ratio.<br /> <br /> &quot;Proof&quot;: Let the sequence be <br /> <br /> &lt;center&gt;&lt;math&gt;S=a_1+a_1r+a_1r^2+a_1r^3+\cdots&lt;/math&gt;&lt;/center&gt; <br /> <br /> Multiplying by &lt;math&gt;r&lt;/math&gt; yields, <br /> <br /> &lt;center&gt;&lt;math&gt;S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots&lt;/math&gt;&lt;/center&gt; <br /> <br /> We subtract these two equations to obtain: <br /> <br /> &lt;center&gt;&lt;math&gt; S-Sr=a_1&lt;/math&gt;&lt;/center&gt; <br /> <br /> There is only one term on the RHS because the rest of the terms cancel out after subtraction. Finally, we can factor and divide to get <br /> <br /> &lt;center&gt;&lt;math&gt;\displaystyle S(1-r)=a_1&lt;/math&gt;&lt;/center&gt;<br /> <br /> thus, <br /> <br /> &lt;center&gt;&lt;math&gt;S=\frac{a_1}{1-r}&lt;/math&gt;&lt;/center&gt; <br /> <br /> This method of multiplying the sequence and subtracting equations, called telescoping, is a frequently used method to evaluate infinite sequences. In fact, the same method can be used to calculate the sum of a finite geometric sequence (given above).<br /> <br /> One common instance of summing infinite geometric sequences is the [[decimal expansion]] of most [[rational number]]s. For instance, &lt;math&gt;0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots&lt;/math&gt; has first term &lt;math&gt;a_0 = \frac 3{10}&lt;/math&gt; and common ratio &lt;math&gt;\frac1{10}&lt;/math&gt;, so the infinite sum has value &lt;math&gt;S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13&lt;/math&gt;, just as we would have expected.<br /> <br /> ==See Also==<br /> *[[arithmetic sequence|Arithmetic Sequences]]<br /> *[[sequence|Sequence]]<br /> *[[geometric series|Geometric Series]]<br /> *[[Series]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetic_sequence&diff=10961 Arithmetic sequence 2006-11-05T01:56:25Z <p>Lotrgreengrapes7926: /* Sums of Arithmetic Sequences */</p> <hr /> <div>==Definition==<br /> An '''arithmetic sequence''' is a [[sequence]] of numbers in which each term is given by adding a fixed value to the previous term. For example, -2, 1, 4, 7, 10, ... is an arithmetic sequence because each term is three more than the previous term. In this case, 3 is called the ''common difference'' of the sequence. More formally, an arithmetic sequence &lt;math&gt;a_n&lt;/math&gt; is defined [[recursion|recursively]] by a first term &lt;math&gt;a_0&lt;/math&gt; and &lt;math&gt;a_n = a_{n-1} + d&lt;/math&gt; for &lt;math&gt;n \geq 1&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference. Explicitly, it can be defined as &lt;math&gt;a_n=a_0+dn&lt;/math&gt;.<br /> <br /> ==Sums of Arithmetic Sequences==<br /> <br /> There are many ways of calculating the sum of the terms of a [[finite]] arithmetic sequence. Perhaps the simplest is to take the average, or [[arithmetic mean]], of the first and last term and to multiply this by the number of terms. Formally, &lt;math&gt;s_n=\frac{n}{2}(a_1+a_n)&lt;/math&gt;. For example,<br /> <br /> &lt;math&gt;\displaystyle 5 + 7 + 9 + 11 + 13 + 15 + 17 = \frac{5+17}{2} \cdot 7 = 77&lt;/math&gt;<br /> <br /> or<br /> <br /> &lt;math&gt;\frac{7}{2}(5+17)=77&lt;/math&gt;<br /> <br /> == Example Problems and Solutions ==<br /> === Introductory Problems ===<br /> * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]<br /> * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]]<br /> <br /> <br /> ==See Also==<br /> *[[sequence|Sequence]]<br /> *[[series|Series]]<br /> *[[geometric sequence|Geometric Sequences]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Arithmetic_sequence&diff=10960 Arithmetic sequence 2006-11-05T01:54:02Z <p>Lotrgreengrapes7926: /* Definition */</p> <hr /> <div>==Definition==<br /> An '''arithmetic sequence''' is a [[sequence]] of numbers in which each term is given by adding a fixed value to the previous term. For example, -2, 1, 4, 7, 10, ... is an arithmetic sequence because each term is three more than the previous term. In this case, 3 is called the ''common difference'' of the sequence. More formally, an arithmetic sequence &lt;math&gt;a_n&lt;/math&gt; is defined [[recursion|recursively]] by a first term &lt;math&gt;a_0&lt;/math&gt; and &lt;math&gt;a_n = a_{n-1} + d&lt;/math&gt; for &lt;math&gt;n \geq 1&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is the common difference. Explicitly, it can be defined as &lt;math&gt;a_n=a_0+dn&lt;/math&gt;.<br /> <br /> ==Sums of Arithmetic Sequences==<br /> <br /> There are many ways of calculating the sum of the terms of a [[finite]] arithmetic sequence. Perhaps the simplest is to take the average, or [[arithmetic mean]], of the first and last term and to multiply this by the number of terms. For example,<br /> <br /> &lt;math&gt;\displaystyle 5 + 7 + 9 + 11 + 13 + 15 + 17 = \frac{5+17}{2} \cdot 7 = 77&lt;/math&gt; <br /> <br /> <br /> == Example Problems and Solutions ==<br /> === Introductory Problems ===<br /> * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]<br /> * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]]<br /> <br /> <br /> ==See Also==<br /> *[[sequence|Sequence]]<br /> *[[series|Series]]<br /> *[[geometric sequence|Geometric Sequences]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=10957 Logarithm 2006-11-05T01:26:36Z <p>Lotrgreengrapes7926: /* Negative Logarithm */</p> <hr /> <div>'''Logarithms''' and [[exponents]] are very closely related. In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have &lt;math&gt; a^x = b &lt;/math&gt;, then taking the logarithm with base &lt;math&gt; a&lt;/math&gt; on both sides will give us &lt;math&gt;\displaystyle x=\log_a{b}&lt;/math&gt;.<br /> <br /> We would read this as &quot;the logarithm of b, base a, is x&quot;. For example, we know that &lt;math&gt;3^4=81&lt;/math&gt;. To express the same fact in logarithmic notation we would write &lt;math&gt;\log_3 81=4&lt;/math&gt;.<br /> <br /> Depending on the field, the symbol &lt;math&gt;\log&lt;/math&gt; without a base can have different meanings. Typically, in mathematics through the level of [[calculus]], the symbol is used to refer to a base 10 logarithm. Thus, &lt;math&gt;\log(100)&lt;/math&gt; means &lt;math&gt;\log_{10}(100)=2&lt;/math&gt;. Usually, the symbol &lt;math&gt;\ln&lt;/math&gt; (an abbreviation of the French &quot;logarithme normal,&quot; meaning &quot;natural logarithm&quot;) is introduced to refer to the logarithm base [[e]]. However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol &lt;math&gt;\log&lt;/math&gt; is taken to mean the logarithm base e and the symbol &lt;math&gt;\ln&lt;/math&gt; is not used at all. (This is an example of conflicting [[mathematical convention]]s.)<br /> <br /> <br /> ==Logarithmic Properties==<br /> We can use the properties of exponents to build a set of properties for logarithms.<br /> <br /> We know that &lt;math&gt;a^x\cdot a^y=a^{x+y}&lt;/math&gt;. We let &lt;math&gt; a^x=b&lt;/math&gt; and &lt;math&gt; a^y=c &lt;/math&gt;. This also makes &lt;math&gt;\displaystyle a^{x+y}=bc &lt;/math&gt;. From &lt;math&gt; a^x = b&lt;/math&gt;, we have &lt;math&gt; x = \log_a{b}&lt;/math&gt;, and from &lt;math&gt; a^y=c &lt;/math&gt;, we have &lt;math&gt; y=\log_a{c} &lt;/math&gt;. So, &lt;math&gt; x+y = \log_a{b}+\log_a{c}&lt;/math&gt;. But we also have from &lt;math&gt;\displaystyle a^{x+y} = bc&lt;/math&gt; that &lt;math&gt; x+y = \log_a{bc}&lt;/math&gt;. Thus, we have found two expressions for &lt;math&gt; x+y&lt;/math&gt; establishing the identity:<br /> <br /> &lt;center&gt;&lt;math&gt; \log_a{b} + \log_a{c} = \log_a{bc}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Using the laws of exponents, we can derive and prove the following identities:<br /> <br /> *&lt;math&gt;\log_a b^n=n\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_a b+ \log_a c=\log_a bc&lt;/math&gt; (Known as the product property.)<br /> *&lt;math&gt;\log_a b-\log_a c=\log_a \frac{b}{c}&lt;/math&gt;<br /> *&lt;math&gt;(\log_a b)(\log_c d)= (\log_a d)(\log_c b)&lt;/math&gt;<br /> *&lt;math&gt;\frac{\log_a b}{\log_a c}=\log_c b&lt;/math&gt;<br /> *&lt;math&gt;\displaystyle \log_{a^n} b^n=\log_a b&lt;/math&gt;<br /> <br /> Try proving all of these as exercises.<br /> <br /> Here are some other less useful log properties that follow from these previous ones.<br /> *&lt;math&gt;\log_{a}b=\frac{1}{\log_{b}a}&lt;/math&gt; (Known as the inverse property)<br /> *&lt;math&gt;(\log_{a}b)(\log_{b}c)=\log_{a}c&lt;/math&gt; (Known as the chain rule.)<br /> <br /> == Problems ==<br /> <br /> # Evaluate &lt;math&gt;(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})&lt;/math&gt;.<br /> <br /> # Evaluate &lt;math&gt;(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)&lt;/math&gt;.<br /> <br /> # Simplify &lt;math&gt;\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N} &lt;/math&gt; where &lt;math&gt; N=(100!)^3&lt;/math&gt;.<br /> <br /> <br /> == Natural Logarithm ==<br /> The natural logarithm of &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\log_e a=\ln a&lt;/math&gt;. The function &lt;math&gt;f(x)=\ln x&lt;/math&gt; is the inverse of &lt;math&gt;f(x)=e^x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\ln a&lt;/math&gt; can also be defined as the area under the curve &lt;math&gt;y=\frac{1}{x}&lt;/math&gt; between 1 and a, or &lt;math&gt;\int^a_1 \frac{1}{x}\, dx&lt;/math&gt;.<br /> <br /> All logarithms are undefined in nonnegative reals, as they are complex. From the identity &lt;math&gt;e^{i\pi}=-1&lt;/math&gt;, we have &lt;math&gt;\ln (-1)=i\pi&lt;/math&gt;. Additionally, &lt;math&gt;\ln (-n)=\ln n+i\pi&lt;/math&gt; for positive real &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Negative Logarithm ==<br /> We define &lt;math&gt;\log_b (-x)&lt;/math&gt;, for positive real &lt;math&gt;x&lt;/math&gt;, to be &lt;math&gt;\log_b(x)+i\log_b(e^{\pi})&lt;/math&gt;.<br /> <br /> '''Proof:'''<br /> <br /> &lt;math&gt;\log_b (-x)=\log_b(x)+i\log_b(e^\pi)&lt;/math&gt;<br /> <br /> &lt;math&gt;b^{\log_b x}\cdot b^{i\log_b(e^\pi)}=-x&lt;/math&gt;<br /> <br /> &lt;math&gt;x\cdot \left(b^{\log_b(e^\pi)}\right)^i=-x&lt;/math&gt;<br /> <br /> &lt;math&gt;(e^\pi)^i=-1&lt;/math&gt;<br /> <br /> &lt;math&gt;e^{i\pi}=-1&lt;/math&gt; is also known as Euler's formula.<br /> <br /> == Intermediate ==<br /> === Example Problem ===<br /> * [[2006_AIME_I_Problems/Problem_9 | 2006 AIME I Problem 9]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=10952 Logarithm 2006-11-05T01:17:50Z <p>Lotrgreengrapes7926: </p> <hr /> <div>'''Logarithms''' and [[exponents]] are very closely related. In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have &lt;math&gt; a^x = b &lt;/math&gt;, then taking the logarithm with base &lt;math&gt; a&lt;/math&gt; on both sides will give us &lt;math&gt;\displaystyle x=\log_a{b}&lt;/math&gt;.<br /> <br /> We would read this as &quot;the logarithm of b, base a, is x&quot;. For example, we know that &lt;math&gt;3^4=81&lt;/math&gt;. To express the same fact in logarithmic notation we would write &lt;math&gt;\log_3 81=4&lt;/math&gt;.<br /> <br /> Depending on the field, the symbol &lt;math&gt;\log&lt;/math&gt; without a base can have different meanings. Typically, in mathematics through the level of [[calculus]], the symbol is used to refer to a base 10 logarithm. Thus, &lt;math&gt;\log(100)&lt;/math&gt; means &lt;math&gt;\log_{10}(100)=2&lt;/math&gt;. Usually, the symbol &lt;math&gt;\ln&lt;/math&gt; (an abbreviation of the French &quot;logarithme normal,&quot; meaning &quot;natural logarithm&quot;) is introduced to refer to the logarithm base [[e]]. However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol &lt;math&gt;\log&lt;/math&gt; is taken to mean the logarithm base e and the symbol &lt;math&gt;\ln&lt;/math&gt; is not used at all. (This is an example of conflicting [[mathematical convention]]s.)<br /> <br /> <br /> ==Logarithmic Properties==<br /> We can use the properties of exponents to build a set of properties for logarithms.<br /> <br /> We know that &lt;math&gt;a^x\cdot a^y=a^{x+y}&lt;/math&gt;. We let &lt;math&gt; a^x=b&lt;/math&gt; and &lt;math&gt; a^y=c &lt;/math&gt;. This also makes &lt;math&gt;\displaystyle a^{x+y}=bc &lt;/math&gt;. From &lt;math&gt; a^x = b&lt;/math&gt;, we have &lt;math&gt; x = \log_a{b}&lt;/math&gt;, and from &lt;math&gt; a^y=c &lt;/math&gt;, we have &lt;math&gt; y=\log_a{c} &lt;/math&gt;. So, &lt;math&gt; x+y = \log_a{b}+\log_a{c}&lt;/math&gt;. But we also have from &lt;math&gt;\displaystyle a^{x+y} = bc&lt;/math&gt; that &lt;math&gt; x+y = \log_a{bc}&lt;/math&gt;. Thus, we have found two expressions for &lt;math&gt; x+y&lt;/math&gt; establishing the identity:<br /> <br /> &lt;center&gt;&lt;math&gt; \log_a{b} + \log_a{c} = \log_a{bc}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Using the laws of exponents, we can derive and prove the following identities:<br /> <br /> *&lt;math&gt;\log_a b^n=n\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_a b+ \log_a c=\log_a bc&lt;/math&gt; (Known as the product property.)<br /> *&lt;math&gt;\log_a b-\log_a c=\log_a \frac{b}{c}&lt;/math&gt;<br /> *&lt;math&gt;(\log_a b)(\log_c d)= (\log_a d)(\log_c b)&lt;/math&gt;<br /> *&lt;math&gt;\frac{\log_a b}{\log_a c}=\log_c b&lt;/math&gt;<br /> *&lt;math&gt;\displaystyle \log_{a^n} b^n=\log_a b&lt;/math&gt;<br /> <br /> Try proving all of these as exercises.<br /> <br /> Here are some other less useful log properties that follow from these previous ones.<br /> *&lt;math&gt;\log_{a}b=\frac{1}{\log_{b}a}&lt;/math&gt; (Known as the inverse property)<br /> *&lt;math&gt;(\log_{a}b)(\log_{b}c)=\log_{a}c&lt;/math&gt; (Known as the chain rule.)<br /> <br /> == Problems ==<br /> <br /> # Evaluate &lt;math&gt;(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})&lt;/math&gt;.<br /> <br /> # Evaluate &lt;math&gt;(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)&lt;/math&gt;.<br /> <br /> # Simplify &lt;math&gt;\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N} &lt;/math&gt; where &lt;math&gt; N=(100!)^3&lt;/math&gt;.<br /> <br /> <br /> == Natural Logarithm ==<br /> The natural logarithm of &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\log_e a=\ln a&lt;/math&gt;. The function &lt;math&gt;f(x)=\ln x&lt;/math&gt; is the inverse of &lt;math&gt;f(x)=e^x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\ln a&lt;/math&gt; can also be defined as the area under the curve &lt;math&gt;y=\frac{1}{x}&lt;/math&gt; between 1 and a, or &lt;math&gt;\int^a_1 \frac{1}{x}\, dx&lt;/math&gt;.<br /> <br /> All logarithms are undefined in nonnegative reals, as they are complex. From the identity &lt;math&gt;e^{i\pi}=-1&lt;/math&gt;, we have &lt;math&gt;\ln (-1)=i\pi&lt;/math&gt;. Additionally, &lt;math&gt;\ln (-n)=\ln n+i\pi&lt;/math&gt; for positive real &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Negative Logarithm ==<br /> We define &lt;math&gt;\log_b (-x)&lt;/math&gt; to be &lt;math&gt;\log_b(x)+i\log_b(e^\pi)&lt;/math&gt; for positive real &lt;math&gt;x&lt;/math&gt;.<br /> <br /> == Intermediate ==<br /> === Example Problem ===<br /> * [[2006_AIME_I_Problems/Problem_9 | 2006 AIME I Problem 9]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=10947 Logarithm 2006-11-05T01:13:51Z <p>Lotrgreengrapes7926: /* Natural Logarithm */</p> <hr /> <div>'''Logarithms''' and [[exponents]] are very closely related. In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have &lt;math&gt; a^x = b &lt;/math&gt;, then taking the logarithm with base &lt;math&gt; a&lt;/math&gt; on both sides will give us &lt;math&gt;\displaystyle x=\log_a{b}&lt;/math&gt;.<br /> <br /> We would read this as &quot;the logarithm of b, base a, is x&quot;. For example, we know that &lt;math&gt;3^4=81&lt;/math&gt;. To express the same fact in logarithmic notation we would write &lt;math&gt;\log_3 81=4&lt;/math&gt;.<br /> <br /> Depending on the field, the symbol &lt;math&gt;\log&lt;/math&gt; without a base can have different meanings. Typically, in mathematics through the level of [[calculus]], the symbol is used to refer to a base 10 logarithm. Thus, &lt;math&gt;\log(100)&lt;/math&gt; means &lt;math&gt;\log_{10}(100)=2&lt;/math&gt;. Usually, the symbol &lt;math&gt;\ln&lt;/math&gt; (an abbreviation of the French &quot;logarithme normal,&quot; meaning &quot;natural logarithm&quot;) is introduced to refer to the logarithm base [[e]]. However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol &lt;math&gt;\log&lt;/math&gt; is taken to mean the logarithm base e and the symbol &lt;math&gt;\ln&lt;/math&gt; is not used at all. (This is an example of conflicting [[mathematical convention]]s.)<br /> <br /> <br /> ==Logarithmic Properties==<br /> We can use the properties of exponents to build a set of properties for logarithms.<br /> <br /> We know that &lt;math&gt;a^x\cdot a^y=a^{x+y}&lt;/math&gt;. We let &lt;math&gt; a^x=b&lt;/math&gt; and &lt;math&gt; a^y=c &lt;/math&gt;. This also makes &lt;math&gt;\displaystyle a^{x+y}=bc &lt;/math&gt;. From &lt;math&gt; a^x = b&lt;/math&gt;, we have &lt;math&gt; x = \log_a{b}&lt;/math&gt;, and from &lt;math&gt; a^y=c &lt;/math&gt;, we have &lt;math&gt; y=\log_a{c} &lt;/math&gt;. So, &lt;math&gt; x+y = \log_a{b}+\log_a{c}&lt;/math&gt;. But we also have from &lt;math&gt;\displaystyle a^{x+y} = bc&lt;/math&gt; that &lt;math&gt; x+y = \log_a{bc}&lt;/math&gt;. Thus, we have found two expressions for &lt;math&gt; x+y&lt;/math&gt; establishing the identity:<br /> <br /> &lt;center&gt;&lt;math&gt; \log_a{b} + \log_a{c} = \log_a{bc}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Using the laws of exponents, we can derive and prove the following identities:<br /> <br /> *&lt;math&gt;\log_a b^n=n\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_a b+ \log_a c=\log_a bc&lt;/math&gt; (Known as the product property.)<br /> *&lt;math&gt;\log_a b-\log_a c=\log_a \frac{b}{c}&lt;/math&gt;<br /> *&lt;math&gt;(\log_a b)(\log_c d)= (\log_a d)(\log_c b)&lt;/math&gt;<br /> *&lt;math&gt;\frac{\log_a b}{\log_a c}=\log_c b&lt;/math&gt;<br /> *&lt;math&gt;\displaystyle \log_{a^n} b^n=\log_a b&lt;/math&gt;<br /> <br /> Try proving all of these as exercises.<br /> <br /> Here are some other less useful log properties that follow from these previous ones.<br /> *&lt;math&gt;\log_{a}b=\frac{1}{\log_{b}a}&lt;/math&gt; (Known as the inverse property)<br /> *&lt;math&gt;(\log_{a}b)(\log_{b}c)=\log_{a}c&lt;/math&gt; (Known as the chain rule.)<br /> <br /> == Problems ==<br /> <br /> # Evaluate &lt;math&gt;(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})&lt;/math&gt;.<br /> <br /> # Evaluate &lt;math&gt;(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)&lt;/math&gt;.<br /> <br /> # Simplify &lt;math&gt;\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N} &lt;/math&gt; where &lt;math&gt; N=(100!)^3&lt;/math&gt;.<br /> <br /> <br /> == Natural Logarithm ==<br /> The natural logarithm of &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\log_e a=\ln a&lt;/math&gt;. The function &lt;math&gt;f(x)=\ln x&lt;/math&gt; is the inverse of &lt;math&gt;f(x)=e^x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\ln a&lt;/math&gt; can also be defined as the area under the curve &lt;math&gt;y=\frac{1}{x}&lt;/math&gt; between 1 and a, or &lt;math&gt;\int^a_1 \frac{1}{x}\, dx&lt;/math&gt;.<br /> <br /> All logarithms are undefined in nonnegative reals, as they are complex. From the identity &lt;math&gt;e^{i\pi}=-1&lt;/math&gt;, we have &lt;math&gt;\ln (-1)=i\pi&lt;/math&gt;. Additionally, &lt;math&gt;\ln (-n)=\ln n+i\pi&lt;/math&gt; for positive real &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Intermediate ==<br /> === Example Problem ===<br /> * [[2006_AIME_I_Problems/Problem_9 | 2006 AIME I Problem 9]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Derivative/Formulas&diff=10943 Derivative/Formulas 2006-11-05T01:05:36Z <p>Lotrgreengrapes7926: /* List of formulas */</p> <hr /> <div>== List of formulas ==<br /> <br /> {| style=&quot;margin: 1em auto 1em auto; height:1000px&quot;<br /> | &lt;math&gt;\frac d{dx}(cf(x)) = c\left(\frac d{dx} f(x)\right)&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;(f(x)+g(x))' = f'(x) + g'(x)&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;(f(g(x)))' = f'(g(x))g'(x)&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} x^n = n x^{n-1}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \sin x = \cos x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \cos x = -\sin x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \tan x = \sec^2 x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \sec x = \sec x \tan x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \csc x = -\csc x\cot x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \cot x = -\csc^2 x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} e^x = e^x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} a^x = (\ln a) a^x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \ln x = \frac 1x&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \log_b x =\frac{\log_b e}{x} &lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \arcsin x = \frac 1{\sqrt{1-x^2}}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \arccos x = -\frac 1{\sqrt{1-x^2}}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \arctan x = \frac 1{1+x^2}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \mathrm{arcsec \ } x = \frac 1{\mid x \mid\sqrt{x^2-1}}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \mathrm{arccsc \ } x = - \frac 1{x\sqrt{x^2 - 1}}&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;\frac d{dx} \arccot x = - \frac 1{1+x^2}&lt;/math&gt;<br /> |}<br /> <br /> == See also ==<br /> * [[Calculus]]<br /> * [[Derivative]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=10941 Logarithm 2006-11-05T01:02:58Z <p>Lotrgreengrapes7926: /* Natural Logarithm */</p> <hr /> <div>'''Logarithms''' and [[exponents]] are very closely related. In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have &lt;math&gt; a^x = b &lt;/math&gt;, then taking the logarithm with base &lt;math&gt; a&lt;/math&gt; on both sides will give us &lt;math&gt;\displaystyle x=\log_a{b}&lt;/math&gt;.<br /> <br /> We would read this as &quot;the logarithm of b, base a, is x&quot;. For example, we know that &lt;math&gt;3^4=81&lt;/math&gt;. To express the same fact in logarithmic notation we would write &lt;math&gt;\log_3 81=4&lt;/math&gt;.<br /> <br /> Depending on the field, the symbol &lt;math&gt;\log&lt;/math&gt; without a base can have different meanings. Typically, in mathematics through the level of [[calculus]], the symbol is used to refer to a base 10 logarithm. Thus, &lt;math&gt;\log(100)&lt;/math&gt; means &lt;math&gt;\log_{10}(100)=2&lt;/math&gt;. Usually, the symbol &lt;math&gt;\ln&lt;/math&gt; (an abbreviation of the French &quot;logarithme normal,&quot; meaning &quot;natural logarithm&quot;) is introduced to refer to the logarithm base [[e]]. However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol &lt;math&gt;\log&lt;/math&gt; is taken to mean the logarithm base e and the symbol &lt;math&gt;\ln&lt;/math&gt; is not used at all. (This is an example of conflicting [[mathematical convention]]s.)<br /> <br /> <br /> ==Logarithmic Properties==<br /> We can use the properties of exponents to build a set of properties for logarithms.<br /> <br /> We know that &lt;math&gt;a^x\cdot a^y=a^{x+y}&lt;/math&gt;. We let &lt;math&gt; a^x=b&lt;/math&gt; and &lt;math&gt; a^y=c &lt;/math&gt;. This also makes &lt;math&gt;\displaystyle a^{x+y}=bc &lt;/math&gt;. From &lt;math&gt; a^x = b&lt;/math&gt;, we have &lt;math&gt; x = \log_a{b}&lt;/math&gt;, and from &lt;math&gt; a^y=c &lt;/math&gt;, we have &lt;math&gt; y=\log_a{c} &lt;/math&gt;. So, &lt;math&gt; x+y = \log_a{b}+\log_a{c}&lt;/math&gt;. But we also have from &lt;math&gt;\displaystyle a^{x+y} = bc&lt;/math&gt; that &lt;math&gt; x+y = \log_a{bc}&lt;/math&gt;. Thus, we have found two expressions for &lt;math&gt; x+y&lt;/math&gt; establishing the identity:<br /> <br /> &lt;center&gt;&lt;math&gt; \log_a{b} + \log_a{c} = \log_a{bc}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> Using the laws of exponents, we can derive and prove the following identities:<br /> <br /> *&lt;math&gt;\log_a b^n=n\log_a b&lt;/math&gt;<br /> *&lt;math&gt;\log_a b+ \log_a c=\log_a bc&lt;/math&gt; (Known as the product property.)<br /> *&lt;math&gt;\log_a b-\log_a c=\log_a \frac{b}{c}&lt;/math&gt;<br /> *&lt;math&gt;(\log_a b)(\log_c d)= (\log_a d)(\log_c b)&lt;/math&gt;<br /> *&lt;math&gt;\frac{\log_a b}{\log_a c}=\log_c b&lt;/math&gt;<br /> *&lt;math&gt;\displaystyle \log_{a^n} b^n=\log_a b&lt;/math&gt;<br /> <br /> Try proving all of these as exercises.<br /> <br /> Here are some other less useful log properties that follow from these previous ones.<br /> *&lt;math&gt;\log_{a}b=\frac{1}{\log_{b}a}&lt;/math&gt; (Known as the inverse property)<br /> *&lt;math&gt;(\log_{a}b)(\log_{b}c)=\log_{a}c&lt;/math&gt; (Known as the chain rule.)<br /> <br /> == Problems ==<br /> <br /> # Evaluate &lt;math&gt;(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})&lt;/math&gt;.<br /> <br /> # Evaluate &lt;math&gt;(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)&lt;/math&gt;.<br /> <br /> # Simplify &lt;math&gt;\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N} &lt;/math&gt; where &lt;math&gt; N=(100!)^3&lt;/math&gt;.<br /> <br /> <br /> == Natural Logarithm ==<br /> The natural logarithm of &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\log_e a=\ln a&lt;/math&gt;. The function &lt;math&gt;f(x)=\ln x&lt;/math&gt; is the inverse of &lt;math&gt;f(x)=e^x&lt;/math&gt;.<br /> <br /> &lt;math&gt;\ln a&lt;/math&gt; can also be defined as the area under the curve &lt;math&gt;y=\frac{1}{x}&lt;/math&gt; between 1 and a, or &lt;math&gt;\int^a_1 \frac{1}{x}\, dx&lt;/math&gt;.<br /> <br /> == Intermediate ==<br /> === Example Problem ===<br /> * [[2006_AIME_I_Problems/Problem_9 | 2006 AIME I Problem 9]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Real_number&diff=10922 Real number 2006-11-05T00:44:44Z <p>Lotrgreengrapes7926: </p> <hr /> <div>A '''real number''' is a number that falls on the real number line. It can have any value. Some examples of real numbers are:&lt;math&gt;1, 2, -23.25, 0, \frac{\pi}{\phi}&lt;/math&gt;, and so on. Numbers that are not real are &lt;math&gt;\ 3i&lt;/math&gt;, &lt;math&gt;\ 3+2.5i&lt;/math&gt;, &lt;math&gt;\ 3+2i+2j+k&lt;/math&gt;, i.e. [[complex number]]s, and [[quaternion]]s.<br /> <br /> The set of real numbers, denoted by &lt;math&gt;\mathbb{R}&lt;/math&gt;, is a subset of [[complex number]]s(&lt;math&gt;\mathbb{C}&lt;/math&gt;). Commonly used subsets of the real numbers are the [[rational number]]s (&lt;math&gt;\mathbb{Q}&lt;/math&gt;), [[integer]]s (&lt;math&gt;\displaystyle\mathbb{Z}&lt;/math&gt;), [[natural number]]s (&lt;math&gt;\mathbb{N}&lt;/math&gt;) and [[irrational number]]s (sometimes, but not universally, denoted &lt;math&gt;\mathbb{J}&lt;/math&gt;). The real numbers can also be divided between the [[algebraic number]]s and [[transcendental number]]s, although these two classes are best understood as subsets of the [[complex number]]s.<br /> <br /> <br /> == See Also ==<br /> <br /> *[[Natural number]]<br /> *[[Integer]]<br /> *[[Rational number]]<br /> *[[Irrational number]]<br /> *[[Complex number]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Convex_polygon&diff=10912 Convex polygon 2006-11-05T00:38:04Z <p>Lotrgreengrapes7926: </p> <hr /> <div>{{stub}}<br /> <br /> A '''convex polygon''' is a [[polygon]] whose [[interior]] forms a [[convex set]]. If 2 points on the perimeter of the polygon are connected by a segment, no point in that segment will be in the exterior of the polygon.<br /> <br /> [[Category:Definition]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Sigma_notation&diff=10909 Sigma notation 2006-11-05T00:25:41Z <p>Lotrgreengrapes7926: </p> <hr /> <div>{{stub}}<br /> <br /> '''Sigma notation''', also known as ''summation notation'', provides a method for writing long, complicated, sometimes [[infinite]] sums neatly and compactly. Besides being easier to write than the explicit sum, sigma notation is also useful in that it shows the general form of each addend.<br /> <br /> == Definition ==<br /> &lt;math&gt;\sum_{k=m}^n a_k=a_m+a_{m+1}+a_{m+2}+...+a_{n-1}+a_n&lt;/math&gt; for integral &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;.<br /> <br /> <br /> == Description ==<br /> Let's first use an example to demonstrate how to use the notation. Consider the sum 5+10+15+20+25. We'll denote the [[sequence]] &lt;math&gt;a_k&lt;/math&gt; where k is the number of terms being summed. We want to rewrite the sum so that each term is in terms of it's number k in the sequence. [[Factoring]], we get, &lt;math&gt;5+10+15+20+25=5(1+2+3+4+5)&lt;/math&gt;. Now, we see that each term is five times it's number in the sequence k; that is, as k ranges from 1 to 5, each term can be written as 5k. This sum is written in summation notation as &lt;math&gt;\sum_{k=1}^5 5k=5+10+15+20+25&lt;/math&gt;. In this case, 1 is the lower limit of summation, the number the index of summation k starts the sum with, and ends with the upper limit of summation 5. One way to remember the terms is to note that the lower limit of summation is written ''under'' the sigma and the ''upper'' one is written above the sigma. <br /> <br /> <br /> == Properties ==<br /> For any constant c and [[finite]] (or [[absolutely convergent]]) [[series]] &lt;math&gt;a_k&lt;/math&gt; and &lt;math&gt;b_k&lt;/math&gt;, <br /> *&lt;math&gt;\sum_a^b c = c(b-a+1)&lt;/math&gt;<br /> *&lt;math&gt;\sum ca_k = c\sum a_k&lt;/math&gt;<br /> *&lt;math&gt;\sum (a_k + b_k) = \sum a_k + \sum b_k&lt;/math&gt;<br /> <br /> == Examples ==<br /> <br /> <br /> <br /> == See Also ==<br /> * [[Convergence]]<br /> * [[Greek alphabet]]<br /> * [[Geometric sequence]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Pythagorean_Theorem&diff=10895 Pythagorean Theorem 2006-11-05T00:15:09Z <p>Lotrgreengrapes7926: /* Common Pythagorean Triples */</p> <hr /> <div>{{stub}}<br /> <br /> <br /> The '''Pythagorean Theorem''' states that for all [[right triangle|right triangles]], &lt;math&gt;{a}^{2}+{b}^{2}={c}^{2}&lt;/math&gt;, where c is the [[hypotenuse]], and a and b are the legs of the right triangle. This theorem is a classic to prove, and hundreds of proofs have been published. The Pythagorean Theorem is one of the most frequently used theorem in [[geometry]], and is one of the many tools in a good geometer's arsenal.<br /> <br /> This is generalized by the [[Pythagorean Inequality]] (See [[geometric inequalities]]) and the [[Law of Cosines]].)<br /> <br /> <br /> == Introductory ==<br /> === Example Problems ===<br /> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]<br /> <br /> == Common Pythagorean Triples ==<br /> A [[Pythagorean Triple]] is a [[set]] of 3 [[positive integer]]s such that &lt;math&gt;a^{2}+b^{2}=c^{2}&lt;/math&gt;, i.e. the 3 numbers can be the lengths of the sides of a right triangle. Among these, the [[Primitive Pythagorean Triple]]s, those in which the three numbers have no common [[divisor]], are most interesting. A few of them are:<br /> <br /> 3-4-5<br /> <br /> 5-12-13<br /> <br /> 7-24-25<br /> <br /> 8-15-17<br /> <br /> 9-40-41</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Factorial&diff=10893 Factorial 2006-11-05T00:13:15Z <p>Lotrgreengrapes7926: /* Prime factorization */</p> <hr /> <div>The '''factorial''' is an important function in [[combinatorics]] and [[analysis]], used to determine the number of ways to arrange objects.<br /> <br /> === Definition ===<br /> <br /> The factorial is defined for [[positive integer]]s as &lt;math&gt;n!=n \cdot (n-1) \cdots 2 \cdot 1 = \prod_{i=1}^n i&lt;/math&gt;. Alternatively, a [[recursion|recursive definition]] for the factorial is &lt;math&gt;n!=n \cdot (n-1)!&lt;/math&gt;.<br /> <br /> === Additional Information ===<br /> <br /> By [[mathematical convention|convention]], &lt;math&gt;0!&lt;/math&gt; is given the value &lt;math&gt;1&lt;/math&gt;.<br /> <br /> The [[gamma function]] is a generalization of the factorial to values other than [[nonnegative integer]]s.<br /> <br /> ===[[Prime factorization]]===<br /> <br /> Since &lt;math&gt;n!&lt;/math&gt; is the product of all positive integers not exceeding &lt;math&gt;n&lt;/math&gt;, it is clear that it is divisible by all<br /> primes &lt;math&gt;p\le n&lt;/math&gt;, and not divisible by any prime &lt;math&gt;p&gt;n&lt;/math&gt;. But what is the power of a prime &lt;math&gt;p\le n&lt;/math&gt;<br /> in the prime factorization of &lt;math&gt;n!&lt;/math&gt;? We can find it as the sum of powers of &lt;math&gt;p&lt;/math&gt; in all the factors &lt;math&gt;1,2,\dots, n&lt;/math&gt;;<br /> but rather than counting the power of &lt;math&gt;p&lt;/math&gt; in each factor, we shall count the number of factors divisible by a given power of &lt;math&gt;p&lt;/math&gt;. Among the numbers &lt;math&gt;1,2,\dots,n&lt;/math&gt;, exactly &lt;math&gt;\left\lfloor\frac n{p^k}\right\rfloor&lt;/math&gt; are divisible by &lt;math&gt;p^k&lt;/math&gt; (here &lt;math&gt;\lfloor\cdot\rfloor&lt;/math&gt; is the [[floor function]]). The ones divisible by &lt;math&gt;p&lt;/math&gt; give one power of &lt;math&gt;p&lt;/math&gt;. The ones divisible by &lt;math&gt;p^2&lt;/math&gt; give another power of &lt;math&gt;p&lt;/math&gt;. Those divisible by &lt;math&gt;p^3&lt;/math&gt; give yet another power of &lt;math&gt;p&lt;/math&gt;. Continuing in this manner gives<br /> <br /> &lt;math&gt;\left\lfloor\frac n{p}\right\rfloor+<br /> \left\lfloor\frac n{p^2}\right\rfloor+<br /> \left\lfloor\frac n{p^3}\right\rfloor+\dots&lt;/math&gt;<br /> <br /> for the power of &lt;math&gt;p&lt;/math&gt; in the prime factorization of &lt;math&gt;n!&lt;/math&gt;. The series is formally infinite, but the terms converge to &lt;math&gt;0&lt;/math&gt; rapidly, as it is the reciprocal of an [[exponential function]]. For example, the power of &lt;math&gt;7&lt;/math&gt; in &lt;math&gt;100!&lt;/math&gt; is just<br /> &lt;math&gt;\left\lfloor\frac {100}{7}\right\rfloor+<br /> \left\lfloor\frac {100}{49}\right\rfloor=14+2=16&lt;/math&gt;<br /> (&lt;math&gt;7^3=343&lt;/math&gt; is already greater than &lt;math&gt;100&lt;/math&gt;).<br /> <br /> === Uses ===<br /> <br /> The factorial is used in the definitions of [[combinations]] and [[permutations]], as &lt;math&gt;n!&lt;/math&gt; is the number of ways to order &lt;math&gt;n&lt;/math&gt; distinct objects.<br /> <br /> === Examples ===<br /> <br /> * [[2006 AIME II Problems/Problem 3|2006 AIME II Problem 3]] on finding prime powers in a factorial<br /> <br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=508851#p508851 AIME 2003I/1]<br /> <br /> === See also ===<br /> *[[Combinatorics]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_13&diff=10829 2006 AMC 12A Problems/Problem 13 2006-11-04T23:12:14Z <p>Lotrgreengrapes7926: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> [[Image:2006 AMC 12A Problem 13.gif]]<br /> <br /> The vertices of a &lt;math&gt;3-4-5&lt;/math&gt; right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ } 14\pi&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let the radius of the smallest circle be &lt;math&gt; a &lt;/math&gt;. We find that the radius of the largest circle is &lt;math&gt;4-a&lt;/math&gt; and the radius of the second largest circle is &lt;math&gt;3-a&lt;/math&gt;. Thus, &lt;math&gt;4-a+3-a=5\iff a=1&lt;/math&gt;. The radii of the other circles are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The sum of their areas is &lt;math&gt;\pi+9\pi+4\pi=14\pi\iff\mathrm{(E)}&lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[2006 AMC 12A Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_13&diff=10825 2006 AMC 12A Problems/Problem 13 2006-11-04T23:06:07Z <p>Lotrgreengrapes7926: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> [[Image:2006 AMC 12A Problem 13.gif]]<br /> <br /> The vertices of a &lt;math&gt;3-4-5&lt;/math&gt; right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ } 14\pi&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> {{solution}}<br /> <br /> &lt;math&gt;\rm{(E)}\,14\pi&lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[2006 AMC 12A Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_15&diff=10820 2003 AIME I Problems/Problem 15 2006-11-04T22:44:10Z <p>Lotrgreengrapes7926: /* See also */</p> <hr /> <div>== Problem ==<br /> In &lt;math&gt; \triangle ABC, AB = 360, BC = 507, &lt;/math&gt; and &lt;math&gt; CA = 780. &lt;/math&gt; Let &lt;math&gt; M &lt;/math&gt; be the midpoint of &lt;math&gt; \overline{CA}, &lt;/math&gt; and let &lt;math&gt; D &lt;/math&gt; be the point on &lt;math&gt; \overline{CA} &lt;/math&gt; such that &lt;math&gt; \overline{BD} &lt;/math&gt; bisects angle &lt;math&gt; ABC. &lt;/math&gt; Let &lt;math&gt; F &lt;/math&gt; be the point on &lt;math&gt; \overline{BC} &lt;/math&gt; such that &lt;math&gt; \overline{DF} \perp \overline{BD}. &lt;/math&gt; Suppose that &lt;math&gt; \overline{DF} &lt;/math&gt; meets &lt;math&gt; \overline{BM} &lt;/math&gt; at &lt;math&gt; E. &lt;/math&gt; The ratio &lt;math&gt; DE: EF &lt;/math&gt; can be written in the form &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m + n. &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 14 | Previous problem]]<br /> * [[2003 AIME I Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_14&diff=10819 2003 AIME I Problems/Problem 14 2006-11-04T22:43:48Z <p>Lotrgreengrapes7926: /* See also */</p> <hr /> <div>== Problem ==<br /> The decimal representation of &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers and &lt;math&gt; m &lt; n, &lt;/math&gt; contains the digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of &lt;math&gt; n &lt;/math&gt; for which this is possible.<br /> <br /> == Solution ==<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 13 | Previous problem]]<br /> * [[2003 AIME I Problems/Problem 15 | Next problem]]<br /> * [[2003 AIME I Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_13&diff=10818 2003 AIME I Problems/Problem 13 2006-11-04T22:43:17Z <p>Lotrgreengrapes7926: /* See also */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; N &lt;/math&gt; be the number of positive integers that are less than or equal to 2003 and whose base-2 representation has more 1's than 0's. Find the remainder when &lt;math&gt; N &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 12 | Previous problem]]<br /> * [[2003 AIME I Problems/Problem 14 | Next problem]]<br /> * [[2003 AIME I Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_12&diff=10817 2003 AIME I Problems/Problem 12 2006-11-04T22:42:56Z <p>Lotrgreengrapes7926: /* See also */</p> <hr /> <div>== Problem ==<br /> In convex quadrilateral &lt;math&gt; ABCD, \angle A \cong \angle C, AB = CD = 180, &lt;/math&gt; and &lt;math&gt; AD \neq BC. &lt;/math&gt; The perimeter of &lt;math&gt; ABCD &lt;/math&gt; is 640. Find &lt;math&gt; \lfloor 1000 \cos A \rfloor. &lt;/math&gt; (The notation &lt;math&gt; \lfloor x \rfloor &lt;/math&gt; means the greatest integer that is less than or equal to &lt;math&gt; x. &lt;/math&gt;)<br /> <br /> == Solution ==<br /> <br /> {{solution}}<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 11 | Previous problem]]<br /> * [[2003 AIME I Problems/Problem 13 | Next problem]]<br /> * [[2003 AIME I Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_10&diff=10816 2003 AIME I Problems/Problem 10 2006-11-04T22:41:21Z <p>Lotrgreengrapes7926: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Triangle]] &lt;math&gt; ABC &lt;/math&gt; is [[isosceles triangle | isosceles]] with &lt;math&gt; AC = BC &lt;/math&gt; and &lt;math&gt; \angle ACB = 106^\circ. &lt;/math&gt; Point &lt;math&gt; M &lt;/math&gt; is in the interior of the triangle so that &lt;math&gt; \angle MAC = 7^\circ &lt;/math&gt; and &lt;math&gt; \angle MCA = 23^\circ. &lt;/math&gt; Find the number of degrees in &lt;math&gt; \angle CMB. &lt;/math&gt;<br /> <br /> == Solution ==<br /> From the givens, we have the following [[angle]] [[measure]]s: &lt;math&gt;m\angle AMC = 150^\circ&lt;/math&gt;, &lt;math&gt;m\angle MCB = 83^\circ&lt;/math&gt;. If we define &lt;math&gt;m\angle CMB = \theta&lt;/math&gt; then we also have &lt;math&gt;m\angle CBM = 97^\circ - \theta&lt;/math&gt;. Then Apply the [[Law of Sines]] to triangles &lt;math&gt;\triangle AMC&lt;/math&gt; and &lt;math&gt;\triangle BMC&lt;/math&gt; to get<br /> <br /> &lt;math&gt;\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}&lt;/math&gt;<br /> <br /> Clearing [[denominator]]s, evaluating &lt;math&gt;\sin 150^\circ = \frac 12&lt;/math&gt; and applying one of our [[trigonometric identities]] to the result gives<br /> <br /> &lt;math&gt;\frac{1}{2} \cos 7^\circ - \theta = \sin 7^\circ \sin \theta&lt;/math&gt;<br /> <br /> and multiplying through by 2 and applying the [[double angle formula]] gives<br /> <br /> &lt;math&gt;\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta&lt;/math&gt;<br /> <br /> and so &lt;math&gt;\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta&lt;/math&gt;<br /> <br /> and, since &lt;math&gt;0^\circ &lt; \theta &lt; 180^\circ&lt;/math&gt;, we must have &lt;math&gt;\theta = 83^\circ&lt;/math&gt;, so the answer is &lt;math&gt;083&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 9 | Previous problem]]<br /> * [[2003 AIME I Problems/Problem 11 | Next problem]]<br /> * [[2003 AIME I Problems]]<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_4&diff=10811 2003 AIME I Problems/Problem 4 2006-11-04T22:34:43Z <p>Lotrgreengrapes7926: /* Solution */</p> <hr /> <div>== Problem ==<br /> Given that &lt;math&gt; \log_{10} \sin x + \log_{10} \cos x = -1 &lt;/math&gt; and that &lt;math&gt; \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), &lt;/math&gt; find &lt;math&gt; n. &lt;/math&gt;<br /> == Solution ==<br /> &lt;math&gt; \log_{10} \sin x + \log_{10} \cos x = -1 &lt;/math&gt;<br /> <br /> &lt;math&gt; \log_{10}(\sin x \cos x) = -1 &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin x \cos x = \frac{1}{10} &lt;/math&gt;<br /> <br /> &lt;math&gt; \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1) &lt;/math&gt;<br /> <br /> &lt;math&gt; \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) &lt;/math&gt; <br /> <br /> &lt;math&gt; \log_{10} (\sin x + \cos x) = \frac{1}{2}\left(\log_{10} \frac{n}{10}\right) &lt;/math&gt; <br /> <br /> &lt;math&gt; \log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right) &lt;/math&gt; <br /> <br /> &lt;math&gt; \sin x + \cos x = \sqrt{\frac{n}{10}} &lt;/math&gt;<br /> <br /> &lt;math&gt; (\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2 &lt;/math&gt;<br /> <br /> &lt;math&gt; \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} &lt;/math&gt;<br /> <br /> &lt;math&gt; 1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} &lt;/math&gt;<br /> <br /> &lt;math&gt; \frac{12}{10} = \frac{n}{10} &lt;/math&gt;<br /> <br /> &lt;math&gt; n = 12 &lt;/math&gt;<br /> <br /> == See also ==<br /> * [[2003 AIME I Problems/Problem 3|Previous Problem]]<br /> * [[2003 AIME I Problems/Problem 5|Next Problem]]<br /> * [[2003 AIME I Problems]]<br /> * [[Logarithm]]<br /> * [[Trigonometry]]<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=Talk:Mock_AIME_2_2006-2007_Problems/Problem_7&diff=10808 Talk:Mock AIME 2 2006-2007 Problems/Problem 7 2006-11-04T22:26:50Z <p>Lotrgreengrapes7926: </p> <hr /> <div>Either the image or the problem is wrong -- the slant height is given as 34 centimeters but is labelled 51 in the diagram. Which one is it? --[[User:JBL|JBL]] 21:09, 15 September 2006 (EDT)<br /> <br /> I believe the image (51 cm) is correct, as it implies the given answer.</div> Lotrgreengrapes7926 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_4&diff=10804 2004 AIME I Problems/Problem 4 2006-11-04T22:12:10Z <p>Lotrgreengrapes7926: /* Problem */</p> <hr /> <div>== Problem ==<br /> A square has sides of length 2. Set &lt;math&gt; S &lt;/math&gt; is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set &lt;math&gt; S &lt;/math&gt; enclose a region whose area to the nearest hundredth is &lt;math&gt; k. &lt;/math&gt; Find &lt;math&gt; 100k. &lt;/math&gt;<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> * [[2004 AIME I Problems/Problem 3| Previous problem]]<br /> <br /> * [[2004 AIME I Problems/Problem 5| Next problem]]<br /> <br /> * [[2004 AIME I Problems]]</div> Lotrgreengrapes7926