https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Flamefoxx99&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-25T01:58:19Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User:Flamefoxx99&diff=107950 User:Flamefoxx99 2019-07-26T21:12:09Z <p>Flamefoxx99: </p> <hr /> <div>hello<br /> <br /> you aren't supposed to be here</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=User:Flamefoxx99&diff=107949 User:Flamefoxx99 2019-07-26T21:12:00Z <p>Flamefoxx99: </p> <hr /> <div>Hello<br /> you aren't supposed to be here</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_5&diff=93352 2018 AIME I Problems/Problem 5 2018-03-21T02:47:02Z <p>Flamefoxx99: /* Solution */</p> <hr /> <div>For each ordered pair of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfying &lt;cmath&gt;\log_2(2x+y) = \log_4(x^2+xy+7y^2)&lt;/cmath&gt;there is a real number &lt;math&gt;K&lt;/math&gt; such that &lt;cmath&gt;\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).&lt;/cmath&gt;Find the product of all possible values of &lt;math&gt;K&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> Note that &lt;math&gt;(2x+y)^2 = x^2+xy+7y^2&lt;/math&gt;. <br /> That gives &lt;math&gt;x^2+xy-2y^2=0&lt;/math&gt; upon simplification and division by &lt;math&gt;3&lt;/math&gt;. Then, &lt;math&gt;x=y&lt;/math&gt; or &lt;math&gt;x=-2y&lt;/math&gt;.<br /> From the second equation, &lt;math&gt;9x^2+6xy+y^2=3x^2+4xy+Ky^2&lt;/math&gt;. If we take &lt;math&gt;x=y&lt;/math&gt;, we see that &lt;math&gt;K=9&lt;/math&gt;. If we take &lt;math&gt;x=-2y&lt;/math&gt;, we see that &lt;math&gt;K=21&lt;/math&gt;. The product is &lt;math&gt;\boxed{189}&lt;/math&gt;.<br /> <br /> -expiLnCalc<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_10&diff=93351 2006 AIME II Problems/Problem 10 2018-03-21T02:34:46Z <p>Flamefoxx99: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a &lt;math&gt; 50\% &lt;/math&gt; chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team &lt;math&gt; A &lt;/math&gt; beats team &lt;math&gt; B. &lt;/math&gt; The [[probability]] that team &lt;math&gt; A &lt;/math&gt; finishes with more points than team &lt;math&gt; B &lt;/math&gt; is &lt;math&gt; m/n, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The results of the five remaining games are independent of the first game, so by symmetry, the probability that &lt;math&gt;A&lt;/math&gt; scores higher than &lt;math&gt;B&lt;/math&gt; in these five games is equal to the probability that &lt;math&gt;B&lt;/math&gt; scores higher than &lt;math&gt;A&lt;/math&gt;. We let this probability be &lt;math&gt;p&lt;/math&gt;; then the probability that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; end with the same score in these five games is &lt;math&gt;1-2p&lt;/math&gt;.<br /> <br /> Of these three cases (&lt;math&gt;|A| &gt; |B|, |A| &lt; |B|, |A|=|B|&lt;/math&gt;), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). <br /> <br /> There are &lt;math&gt;{5\choose k}&lt;/math&gt; ways to &lt;math&gt;A&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories, and &lt;math&gt;{5\choose k}&lt;/math&gt; ways for &lt;math&gt;B&lt;/math&gt; to have &lt;math&gt;k&lt;/math&gt; victories. Summing for all values of &lt;math&gt;k&lt;/math&gt;,<br /> &lt;center&gt;&lt;math&gt;1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.&lt;/math&gt;&lt;/center&gt;<br /> Thus &lt;math&gt;p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}&lt;/math&gt;. The desired probability is the sum of the cases when &lt;math&gt;|A| \ge |B|&lt;/math&gt;, so the answer is &lt;math&gt;\frac{126}{512} + \frac{193}{512} = \frac{319}{512}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{831}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> You can break this into cases based on how many rounds &lt;math&gt;A&lt;/math&gt; wins out of the remaining &lt;math&gt;5&lt;/math&gt; games.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 0 games, then &lt;math&gt;B&lt;/math&gt; must win 0 games and the probability of this is &lt;math&gt; \frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 1 games, then &lt;math&gt;B&lt;/math&gt; must win 1 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 2 games, then &lt;math&gt;B&lt;/math&gt; must win 2 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 3 games, then &lt;math&gt;B&lt;/math&gt; must win 3 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 4 games, then &lt;math&gt;B&lt;/math&gt; must win 4 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024} &lt;/math&gt;.<br /> <br /> *If &lt;math&gt;A&lt;/math&gt; wins 5 games, then &lt;math&gt;B&lt;/math&gt; must win 5 or less games and the probability of this is &lt;math&gt; \frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024} &lt;/math&gt;.<br /> <br /> Summing these 6 cases, we get &lt;math&gt; \frac{638}{1024} &lt;/math&gt;, which simplifies to &lt;math&gt; \frac{319}{512} &lt;/math&gt;, so our answer is &lt;math&gt;319 + 512 = \boxed{831}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can apply the concept of generating functions here. <br /> <br /> The generating function for &lt;math&gt;B&lt;/math&gt; is &lt;math&gt; (1 + 0x^{1}) &lt;/math&gt; for the first game where &lt;math&gt;x^{n}&lt;/math&gt; is winning n games. Since &lt;math&gt;B&lt;/math&gt; lost the first game, the coefficient for &lt;math&gt;x^{1}&lt;/math&gt; is 0. The generating function for the next 5 games is &lt;math&gt;(1 + x)^{5}&lt;/math&gt;. Thus, the total generating function for number of games he wins is<br /> <br /> &lt;math&gt;{(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})&lt;/math&gt;.<br /> <br /> The generating function for &lt;math&gt;A&lt;/math&gt; is the same except that it is multiplied by &lt;math&gt;x&lt;/math&gt; instead of &lt;math&gt;(1+0x)&lt;/math&gt;. <br /> Thus, the generating function for &lt;math&gt;A&lt;/math&gt; is <br /> <br /> &lt;math&gt;1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}&lt;/math&gt;. <br /> <br /> The probability that &lt;math&gt;B&lt;/math&gt; wins 0 games is &lt;math&gt;\frac{1}{32}&lt;/math&gt;. Since the coefficients for all &lt;math&gt;x^{n}&lt;/math&gt; where <br /> <br /> &lt;math&gt;n \geq 1&lt;/math&gt; sums to 32, the probability that &lt;math&gt;A&lt;/math&gt; wins more games is &lt;math&gt;\frac{32}{32}&lt;/math&gt;. <br /> <br /> Thus, the probability that &lt;math&gt;A&lt;/math&gt; has more wins than &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;319 + 512 = \boxed{831} &lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_1&diff=87797 2001 AMC 8 Problems/Problem 1 2017-10-14T05:08:16Z <p>Flamefoxx99: /* Solution */</p> <hr /> <div>==Problem==<br /> Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 } &lt;/math&gt;<br /> <br /> ==Solution==<br /> It will take him &lt;math&gt;300\cdot2=600&lt;/math&gt; seconds to paint all the dimples. <br /> <br /> This is equivalent to &lt;math&gt;\frac{600}{60}=10&lt;/math&gt; minutes &lt;math&gt;\Rightarrow \boxed{D}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2001 |before=last&lt;br /&gt;Question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_6&diff=78099 2011 USAJMO Problems/Problem 6 2016-04-17T01:25:30Z <p>Flamefoxx99: Redirected page to 2011 USAMO Problems/Problem 4</p> <hr /> <div>#REDIRECT [[2011 USAMO Problems/Problem 4]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAMO_Problems/Problem_1&diff=78098 2011 USAMO Problems/Problem 1 2016-04-17T01:24:02Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a^2 + b^2 + c^2 + (a + b + c)^2 \le 4&lt;/math&gt;. Prove that<br /> &lt;cmath&gt;\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.&lt;/cmath&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> Since<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> (a+b)^2 + (b+c)^2 + (c+a)^2 &amp;= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\<br /> &amp;= a^2 + b^2 + c^2 + (a + b + c)^2,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> it is natural to consider a change of variables:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \alpha &amp;= b + c \\<br /> \beta &amp;= c + a \\<br /> \gamma &amp;= a + b<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> with the inverse mapping given by:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a &amp;= \frac{\beta + \gamma - \alpha}2 \\<br /> b &amp;= \frac{\alpha + \gamma - \beta}2 \\<br /> c &amp;= \frac{\alpha + \beta - \gamma}2<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> With this change of variables, the constraint becomes<br /> &lt;cmath&gt;<br /> \alpha^2 + \beta^2 + \gamma^2 \le 4,<br /> &lt;/cmath&gt;<br /> while the left side of the inequality we need to prove is now<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> &amp; \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ &amp;<br /> \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ &amp;<br /> \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} +<br /> \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} +<br /> \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ &amp;<br /> \frac32 + \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2}.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore it remains to prove that<br /> &lt;cmath&gt;<br /> \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2} \ge \frac32.<br /> &lt;/cmath&gt;<br /> <br /> We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.<br /> <br /> ===Solution 2===<br /> Rearranging the condition yields that<br /> &lt;cmath&gt;a^2 + b^2 + c^2 +ab+bc+ac \le 2&lt;/cmath&gt;<br /> <br /> Now note that<br /> &lt;cmath&gt;\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}&lt;/cmath&gt;<br /> <br /> Summing this for all pairs of &lt;math&gt;\{ a,b,c \}&lt;/math&gt; gives that<br /> &lt;cmath&gt;\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6&lt;/cmath&gt;<br /> <br /> By AM-GM. Dividing by &lt;math&gt;2&lt;/math&gt; gives the desired inequality.<br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}<br /> {{USAJMO newbox|year=2011|num-b=1|num-a=3}}<br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAMO_Problems/Problem_1&diff=78097 2011 USAMO Problems/Problem 1 2016-04-17T01:23:35Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a^2 + b^2 + c^2 + (a + b + c)^2 \le 4&lt;/math&gt;. Prove that<br /> &lt;cmath&gt;\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.&lt;/cmath&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> Since<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> (a+b)^2 + (b+c)^2 + (c+a)^2 &amp;= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\<br /> &amp;= a^2 + b^2 + c^2 + (a + b + c)^2,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> it is natural to consider a change of variables:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \alpha &amp;= b + c \\<br /> \beta &amp;= c + a \\<br /> \gamma &amp;= a + b<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> with the inverse mapping given by:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a &amp;= \frac{\beta + \gamma - \alpha}2 \\<br /> b &amp;= \frac{\alpha + \gamma - \beta}2 \\<br /> c &amp;= \frac{\alpha + \beta - \gamma}2<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> With this change of variables, the constraint becomes<br /> &lt;cmath&gt;<br /> \alpha^2 + \beta^2 + \gamma^2 \le 4,<br /> &lt;/cmath&gt;<br /> while the left side of the inequality we need to prove is now<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> &amp; \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ &amp;<br /> \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ &amp;<br /> \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} +<br /> \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} +<br /> \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ &amp;<br /> \frac32 + \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2}.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore it remains to prove that<br /> &lt;cmath&gt;<br /> \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2} \ge \frac32.<br /> &lt;/cmath&gt;<br /> <br /> We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.<br /> <br /> ===Solution 2===<br /> Rearranging the condition yields that<br /> &lt;cmath&gt;a^2 + b^2 + c^2 +ab+bc+ac \le 2&lt;/cmath&gt;<br /> <br /> Now note that<br /> &lt;cmath&gt;\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}&lt;/cmath&gt;<br /> <br /> Summing this for all pairs of &lt;math&gt;\{ a,b,c \}&lt;/math&gt; gives that<br /> &lt;cmath&gt;\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6&lt;/cmath&gt;<br /> <br /> By AM-GM. Dividing by &lt;math&gt;2&lt;/math&gt; gives the desired inequality.<br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}<br /> {{USAJMO newbox|year=2011|beforetext=|num-b=1|num-a=3}}<br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAMO_Problems/Problem_1&diff=78096 2011 USAMO Problems/Problem 1 2016-04-17T01:23:07Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a^2 + b^2 + c^2 + (a + b + c)^2 \le 4&lt;/math&gt;. Prove that<br /> &lt;cmath&gt;\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.&lt;/cmath&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> Since<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> (a+b)^2 + (b+c)^2 + (c+a)^2 &amp;= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\<br /> &amp;= a^2 + b^2 + c^2 + (a + b + c)^2,<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> it is natural to consider a change of variables:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \alpha &amp;= b + c \\<br /> \beta &amp;= c + a \\<br /> \gamma &amp;= a + b<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> with the inverse mapping given by:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a &amp;= \frac{\beta + \gamma - \alpha}2 \\<br /> b &amp;= \frac{\alpha + \gamma - \beta}2 \\<br /> c &amp;= \frac{\alpha + \beta - \gamma}2<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> With this change of variables, the constraint becomes<br /> &lt;cmath&gt;<br /> \alpha^2 + \beta^2 + \gamma^2 \le 4,<br /> &lt;/cmath&gt;<br /> while the left side of the inequality we need to prove is now<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> &amp; \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ &amp;<br /> \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} +<br /> \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} +<br /> \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ &amp;<br /> \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} +<br /> \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} +<br /> \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ &amp;<br /> \frac32 + \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2}.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore it remains to prove that<br /> &lt;cmath&gt;<br /> \frac{\alpha\beta}{2\gamma^2} +<br /> \frac{\beta\gamma}{2\alpha^2} +<br /> \frac{\gamma\alpha}{2\beta^2} \ge \frac32.<br /> &lt;/cmath&gt;<br /> <br /> We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.<br /> <br /> ===Solution 2===<br /> Rearranging the condition yields that<br /> &lt;cmath&gt;a^2 + b^2 + c^2 +ab+bc+ac \le 2&lt;/cmath&gt;<br /> <br /> Now note that<br /> &lt;cmath&gt;\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}&lt;/cmath&gt;<br /> <br /> Summing this for all pairs of &lt;math&gt;\{ a,b,c \}&lt;/math&gt; gives that<br /> &lt;cmath&gt;\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6&lt;/cmath&gt;<br /> <br /> By AM-GM. Dividing by &lt;math&gt;2&lt;/math&gt; gives the desired inequality.<br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAMO newbox|year=2011|beforetext=|before=First Problem|num-a=2}}<br /> {{USAJMO newbox|year=2011|beforetext=|before=1|num-a=3}}<br /> [[Category:Olympiad Algebra Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_2&diff=78095 2011 USAJMO Problems/Problem 2 2016-04-17T01:22:34Z <p>Flamefoxx99: Redirected page to 2011 USAMO Problems/Problem 1</p> <hr /> <div>#REDIRECT [[2011 USAMO Problems/Problem 1]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=78094 2011 USAJMO Problems/Problem 3 2016-04-17T01:20:39Z <p>Flamefoxx99: </p> <hr /> <div>== Problem ==<br /> <br /> For a point &lt;math&gt;P = (a, a^2)&lt;/math&gt; in the coordinate plane, let &lt;math&gt;\ell(P)&lt;/math&gt; denote the line passing through &lt;math&gt;P&lt;/math&gt; with slope &lt;math&gt;2a&lt;/math&gt;. Consider the set of triangles with vertices of the form &lt;math&gt;P_1 = (a_1, a_1^2)&lt;/math&gt;, &lt;math&gt;P_2 = (a_2, a_2^2)&lt;/math&gt;, &lt;math&gt;P_3 = (a_3, a_3^2)&lt;/math&gt;, such that the intersections of the lines &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle &lt;math&gt;\triangle&lt;/math&gt;. Find the locus of the center of &lt;math&gt;\triangle&lt;/math&gt; as &lt;math&gt;P_1P_2P_3&lt;/math&gt; ranges over all such triangles.<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> Note that all the points &lt;math&gt;P=(a,a^2)&lt;/math&gt; belong to the parabola &lt;math&gt;y=x^2&lt;/math&gt; which we will denote &lt;math&gt;p&lt;/math&gt;. This parabola has a focus &lt;math&gt;F=\left(0,\frac{1}{4}\right)&lt;/math&gt; and directrix &lt;math&gt;y=-\frac{1}{4}&lt;/math&gt; which we will denote &lt;math&gt;d&lt;/math&gt;. We will prove that the desired locus is &lt;math&gt;d&lt;/math&gt;.<br /> <br /> First note that for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt;, the line &lt;math&gt;\ell(P)&lt;/math&gt; is the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;. This is because &lt;math&gt;\ell(P)&lt;/math&gt; contains &lt;math&gt;P&lt;/math&gt; and because &lt;math&gt;[\frac{d}{dx}] x^2=2x&lt;/math&gt;. If you don't like calculus, you can also verify that &lt;math&gt;\ell(P)&lt;/math&gt; has equation &lt;math&gt;y=2a(x-a)+a^2&lt;/math&gt; and does not intersect &lt;math&gt;y=x^2&lt;/math&gt; at any point besides &lt;math&gt;P&lt;/math&gt;. Now for any point &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;p&lt;/math&gt; let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;d&lt;/math&gt;. Then by the definition of parabolas, &lt;math&gt;PP'=PF&lt;/math&gt;. Let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;. Since &lt;math&gt;PP'=PF&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; passes through &lt;math&gt;P&lt;/math&gt;. Suppose &lt;math&gt;K&lt;/math&gt; is any other point on &lt;math&gt;q&lt;/math&gt; and let &lt;math&gt;K'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;K&lt;/math&gt; to &lt;math&gt;d&lt;/math&gt;. Then in right &lt;math&gt;\Delta KK'P'&lt;/math&gt;, &lt;math&gt;KK'&lt;/math&gt; is a leg and so &lt;math&gt;KK'&lt;KP'=KF&lt;/math&gt;. Therefore &lt;math&gt;K&lt;/math&gt; cannot be on &lt;math&gt;p&lt;/math&gt;. This implies that &lt;math&gt;q&lt;/math&gt; is exactly the tangent line to &lt;math&gt;p&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt;, that is &lt;math&gt;q=\ell(P)&lt;/math&gt;. So we have proved Lemma 1: If &lt;math&gt;P&lt;/math&gt; is a point on &lt;math&gt;p&lt;/math&gt; then &lt;math&gt;\ell(P)&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{P'F}&lt;/math&gt;.<br /> <br /> We need another lemma before we proceed. Lemma 2: If &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt; with orthocenter &lt;math&gt;H&lt;/math&gt;, then the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt;, &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, and &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are collinear with &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Proof of Lemma 2: Say the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; are &lt;math&gt;C'&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, and the reflections of &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; are &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt;. Then we angle chase &lt;math&gt;\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2&lt;/math&gt; where &lt;math&gt;m(JZ)&lt;/math&gt; is the measure of minor arc &lt;math&gt;JZ&lt;/math&gt; on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. This implies that &lt;math&gt;J&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;, and similarly &lt;math&gt;I&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. Therefore &lt;math&gt;\angle C'HJ=\angle FJH=m(XF)/2&lt;/math&gt;, and &lt;math&gt;\angle A'HX=\angle FIX=m(FX)/2&lt;/math&gt;. So &lt;math&gt;\angle C'HJ = \angle A'HX&lt;/math&gt;. Since &lt;math&gt;J&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear it follows that &lt;math&gt;C'&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt; and &lt;math&gt;A'&lt;/math&gt; are collinear. Similarly, the reflection of &lt;math&gt;F&lt;/math&gt; over &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; also lies on this line, and so the claim is proved.<br /> <br /> Now suppose &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three points of &lt;math&gt;p&lt;/math&gt; and let &lt;math&gt;\ell(A)\cap\ell(B)=X&lt;/math&gt;, &lt;math&gt;\ell(A)\cap\ell(C)=Y&lt;/math&gt;, and &lt;math&gt;\ell(B)\cap\ell(C)=Z&lt;/math&gt;. Also let &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; be the midpoints of &lt;math&gt;\overline{A'F}&lt;/math&gt;, &lt;math&gt;\overline{B'F}&lt;/math&gt;, and &lt;math&gt;\overline{C'F}&lt;/math&gt; respectively. Then since &lt;math&gt;\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d&lt;/math&gt;, it follows that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are collinear. By Lemma 1, we know that &lt;math&gt;A''&lt;/math&gt;, &lt;math&gt;B''&lt;/math&gt;, and &lt;math&gt;C''&lt;/math&gt; are the feet of the altitudes from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{XY}&lt;/math&gt;, &lt;math&gt;\overline{XZ}&lt;/math&gt;, and &lt;math&gt;\overline{YZ}&lt;/math&gt;. Therefore by the Simson Line Theorem, &lt;math&gt;F&lt;/math&gt; is on the circumcircle of &lt;math&gt;\Delta XYZ&lt;/math&gt;. If &lt;math&gt;H&lt;/math&gt; is the orthocenter of &lt;math&gt;\Delta XYZ&lt;/math&gt;, then by Lemma 2, it follows that &lt;math&gt;H&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{A'C'}=d&lt;/math&gt;. It follows that the locus described in the problem is a subset of &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Since we claim that the locus described in the problem is &lt;math&gt;d&lt;/math&gt;, we still need to show that for any choice of &lt;math&gt;H&lt;/math&gt; on &lt;math&gt;d&lt;/math&gt; there exists an equilateral triangle with center &lt;math&gt;H&lt;/math&gt; such that the lines containing the sides of the triangle are tangent to &lt;math&gt;p&lt;/math&gt;. So suppose &lt;math&gt;H&lt;/math&gt; is any point on &lt;math&gt;d&lt;/math&gt; and let the circle centered at &lt;math&gt;H&lt;/math&gt; through &lt;math&gt;F&lt;/math&gt; be &lt;math&gt;O&lt;/math&gt;. Then suppose &lt;math&gt;A&lt;/math&gt; is one of the intersections of &lt;math&gt;d&lt;/math&gt; with &lt;math&gt;O&lt;/math&gt;. Let &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, and construct the ray through &lt;math&gt;F&lt;/math&gt; on the same halfplane of &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt; as &lt;math&gt;A&lt;/math&gt; that makes an angle of &lt;math&gt;2\theta&lt;/math&gt; with &lt;math&gt;\overleftrightarrow{HF}&lt;/math&gt;. Say this ray intersects &lt;math&gt;O&lt;/math&gt; in a point &lt;math&gt;B&lt;/math&gt; besides &lt;math&gt;F&lt;/math&gt;, and let &lt;math&gt;q&lt;/math&gt; be the perpendicular bisector of &lt;math&gt;\overline{HB}&lt;/math&gt;. Since &lt;math&gt;\angle HFB=2\theta&lt;/math&gt; and &lt;math&gt;\angle HFA=3\theta&lt;/math&gt;, we have &lt;math&gt;\angle BFA=\theta&lt;/math&gt;. By the inscribed angles theorem, it follows that &lt;math&gt;\angle AHB=2\theta&lt;/math&gt;. Also since &lt;math&gt;HF&lt;/math&gt; and &lt;math&gt;HB&lt;/math&gt; are both radii, &lt;math&gt;\Delta HFB&lt;/math&gt; is isosceles and &lt;math&gt;\angle HBF=\angle HFB=2\theta&lt;/math&gt;. Let &lt;math&gt;P_1'&lt;/math&gt; be the reflection of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;q&lt;/math&gt;. Then &lt;math&gt;2\theta=\angle FBH=\angle C'HB&lt;/math&gt;, and so &lt;math&gt;\angle C'HB=\angle AHB&lt;/math&gt;. It follows that &lt;math&gt;P_1'&lt;/math&gt; is on &lt;math&gt;\overleftrightarrow{AH}=d&lt;/math&gt;, which means &lt;math&gt;q&lt;/math&gt; is the perpendicular bisector of &lt;math&gt;\overline{FP_1'}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;q&lt;/math&gt; intersect &lt;math&gt;O&lt;/math&gt; in points &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt; and let &lt;math&gt;X&lt;/math&gt; be the point diametrically opposite to &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;O&lt;/math&gt;. Also let &lt;math&gt;\overline{HB}&lt;/math&gt; intersect &lt;math&gt;q&lt;/math&gt; at &lt;math&gt;M&lt;/math&gt;. Then &lt;math&gt;HM=HB/2=HZ/2&lt;/math&gt;. Therefore &lt;math&gt;\Delta HMZ&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle and so &lt;math&gt;\angle ZHB=60^{\circ}&lt;/math&gt;. So &lt;math&gt;\angle ZHY=120^{\circ}&lt;/math&gt; and by the inscribed angles theorem, &lt;math&gt;\angle ZXY=60^{\circ}&lt;/math&gt;. Since &lt;math&gt;ZX=ZY&lt;/math&gt; it follows that &lt;math&gt;\Delta ZXY&lt;/math&gt; is and equilateral triangle with center &lt;math&gt;H&lt;/math&gt;.<br /> <br /> By Lemma 2, it follows that the reflections of &lt;math&gt;F&lt;/math&gt; across &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt;, call them &lt;math&gt;P_2'&lt;/math&gt; and &lt;math&gt;P_3'&lt;/math&gt;, lie on &lt;math&gt;d&lt;/math&gt;. Let the intersection of &lt;math&gt;\overleftrightarrow{YZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_1'&lt;/math&gt; be &lt;math&gt;P_1&lt;/math&gt;, the intersection of &lt;math&gt;\overleftrightarrow{XY}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_2'&lt;/math&gt; be &lt;math&gt;P_2&lt;/math&gt;, and the intersection of &lt;math&gt;\overleftrightarrow{XZ}&lt;/math&gt; and the perpendicular to &lt;math&gt;d&lt;/math&gt; through &lt;math&gt;P_3'&lt;/math&gt; be &lt;math&gt;P_3&lt;/math&gt;. Then by the definitions of &lt;math&gt;P_1'&lt;/math&gt;, &lt;math&gt;P_2'&lt;/math&gt;, and &lt;math&gt;P_3'&lt;/math&gt; it follows that &lt;math&gt;FP_i=P_iP_i'&lt;/math&gt; for &lt;math&gt;i=1,2,3&lt;/math&gt; and so &lt;math&gt;P_1&lt;/math&gt;, &lt;math&gt;P_2&lt;/math&gt;, and &lt;math&gt;P_3&lt;/math&gt; are on &lt;math&gt;p&lt;/math&gt;. By lemma 1, &lt;math&gt;\ell(P_1)=\overleftrightarrow{YZ}&lt;/math&gt;, &lt;math&gt;\ell(P_2)=\overleftrightarrow{XY}&lt;/math&gt;, and &lt;math&gt;\ell(P_3)=\overleftrightarrow{XZ}&lt;/math&gt;. Therefore the intersections of &lt;math&gt;\ell(P_1)&lt;/math&gt;, &lt;math&gt;\ell(P_2)&lt;/math&gt;, and &lt;math&gt;\ell(P_3)&lt;/math&gt; form an equilateral triangle with center &lt;math&gt;H&lt;/math&gt;, which finishes the proof.<br /> --Killbilledtoucan<br /> <br /> ===Solution 2===<br /> <br /> Note that the lines &lt;math&gt;l(P_1), l(P_2), l(P_3)&lt;/math&gt; are &lt;cmath&gt;y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,&lt;/cmath&gt; respectively. It is easy to deduce that the three points of intersection are &lt;cmath&gt;\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).&lt;/cmath&gt; The slopes of each side of this equilateral triangle are &lt;cmath&gt;2a_1,2a_2,2a_3,&lt;/cmath&gt; and we want to find the locus of &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).&lt;/cmath&gt; We know that &lt;cmath&gt;2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)&lt;/cmath&gt; for some &lt;math&gt;\theta.&lt;/math&gt; Therefore, we can use the tangent addition formula to deduce &lt;cmath&gt;\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}&lt;/cmath&gt; and &lt;cmath&gt;\begin{align*}<br /> \frac{a_1a_2+a_2a_3+a_3a_1}{3}&amp;=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\<br /> &amp;=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\<br /> &amp;=-\frac{1}{4}.\end{align*}&lt;/cmath&gt; Now we show that &lt;math&gt;\frac{a_1+a_2+a_3}{3}&lt;/math&gt; can be any real number. Let's say &lt;cmath&gt;\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k&lt;/cmath&gt; for some real number &lt;math&gt;k.&lt;/math&gt; Multiplying both sides by &lt;math&gt;2-\tan^2\theta&lt;/math&gt; and rearranging yields a cubic in &lt;math&gt;\tan\theta.&lt;/math&gt; Clearly this cubic has at least one real solution. As &lt;math&gt;\tan \theta&lt;/math&gt; can take on any real number, all values of &lt;math&gt;k&lt;/math&gt; are possible, and our answer is the line &lt;cmath&gt;\boxed{y=-\frac{1}{4}.}&lt;/cmath&gt; Of course, as the denominator could equal 0, we must check &lt;math&gt;\tan \theta=\pm \frac{1}{\sqrt{3}}.&lt;/math&gt; &lt;cmath&gt;3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).&lt;/cmath&gt; The left side is nonzero, while the right side is zero, so these values of &lt;math&gt;\theta&lt;/math&gt; do not contribute to any values of &lt;math&gt;k.&lt;/math&gt; So, our answer remains the same. &lt;math&gt;\blacksquare&lt;/math&gt; ~ Benq<br /> == See also ==<br /> {{USAJMO newbox|year=2011|num-b=2|num-a=3}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=68055 2011 AIME I Problems/Problem 7 2015-03-02T03:11:47Z <p>Flamefoxx99: /* Solution 2 */</p> <hr /> <div>== Problem 7 ==<br /> Find the number of positive integers &lt;math&gt;m&lt;/math&gt; for which there exist nonnegative integers &lt;math&gt;x_0&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; , &lt;math&gt;\dots&lt;/math&gt; , &lt;math&gt;x_{2011}&lt;/math&gt; such that<br /> &lt;cmath&gt;m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.&lt;/cmath&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}&lt;/math&gt;. The problem then becomes finding the number of positive integer roots &lt;math&gt;m&lt;/math&gt; for which &lt;math&gt;P(m) = 0&lt;/math&gt; and &lt;math&gt;x_0, x_1, ..., x_{2011}&lt;/math&gt; are nonnegative integers. We plug in &lt;math&gt;m = 1&lt;/math&gt; and see that &lt;math&gt;P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010&lt;/math&gt;. Now, we can say that &lt;math&gt;P(m) = (m-1)Q(m) - 2010&lt;/math&gt; for some polynomial &lt;math&gt;Q(m)&lt;/math&gt; with integer coefficients. Then if &lt;math&gt;P(m) = 0&lt;/math&gt;, &lt;math&gt;(m-1)Q(m) = 2010&lt;/math&gt;. Thus, if &lt;math&gt;P(m) = 0&lt;/math&gt;, then &lt;math&gt;m-1 | 2010&lt;/math&gt; .<br /> Now, we need to show that for all &lt;math&gt;m-1 | 2010&lt;/math&gt;, &lt;math&gt; m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. &lt;/math&gt;. We try with the first few &lt;math&gt;m&lt;/math&gt; that satisfy this.<br /> For &lt;math&gt;m = 2&lt;/math&gt;, we see we can satisfy this if &lt;math&gt;x_0 = 2010&lt;/math&gt;, &lt;math&gt;x_1 = 2009&lt;/math&gt;, &lt;math&gt;x_2 = 2008&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt; , &lt;math&gt;x_{2008} = 2&lt;/math&gt;, &lt;math&gt;x_{2009} = 1&lt;/math&gt;, &lt;math&gt; x_{2010} = 0&lt;/math&gt;, &lt;math&gt;x_{2011} = 0&lt;/math&gt;, because &lt;math&gt;2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots&lt;/math&gt; (based on the idea &lt;math&gt;2^n + 2^n = 2^{n+1}&lt;/math&gt;, leading to a chain of substitutions of this kind) &lt;math&gt;= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}&lt;/math&gt;. Thus &lt;math&gt;2&lt;/math&gt; is a possible value of &lt;math&gt;m&lt;/math&gt;. For other values, for example &lt;math&gt;m = 3&lt;/math&gt;, we can use the same strategy, with &lt;math&gt;x_{2011} = x_{2010} = x_{2009} = 0&lt;/math&gt;, &lt;math&gt;x_{2008} = x_{2007} = 1&lt;/math&gt;, &lt;math&gt;x_{2006} = x_{2005} = 2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;x_2 = x_1 = 1004&lt;/math&gt; and &lt;math&gt;x_0 = 1005&lt;/math&gt;, because <br /> &lt;math&gt;3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004} = \cdots&lt;/math&gt;<br /> &lt;math&gt;=3^{1004} +3^{1004}+3^{1004} = 3^{1005}&lt;/math&gt;.<br /> It's clearly seen we can use the same strategy for all &lt;math&gt;m-1 |2010&lt;/math&gt;. We count all positive &lt;math&gt;m&lt;/math&gt; satisfying &lt;math&gt;m-1 |2010&lt;/math&gt;, and see there are &lt;math&gt;\boxed{016}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> One notices that &lt;math&gt;m-1 \mid 2010&lt;/math&gt; if and only if there exist non-negative integers &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt; such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;.<br /> <br /> To prove the forward case, we proceed by directly finding &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt;. Suppose &lt;math&gt;m&lt;/math&gt; is an integer such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;. We will count how many &lt;math&gt;x_k = 0&lt;/math&gt;, how many &lt;math&gt;x_k = 1&lt;/math&gt;, etc. Suppose the number of &lt;math&gt;x_k = 0&lt;/math&gt; is non-zero. Then, there must be at least &lt;math&gt;m&lt;/math&gt; such &lt;math&gt;x_k&lt;/math&gt; since &lt;math&gt;m&lt;/math&gt; divides all the remaining terms, so &lt;math&gt;m&lt;/math&gt; must also divide the sum of all the &lt;math&gt;m^0&lt;/math&gt; terms. Thus, if we let &lt;math&gt;x_k = 0&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,m&lt;/math&gt;, we have,<br /> &lt;cmath&gt;m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> Well clearly, &lt;math&gt;m^{x_0}&lt;/math&gt; is greater than &lt;math&gt;m&lt;/math&gt;, so &lt;math&gt;m^2 \mid m^{x_0}&lt;/math&gt;. &lt;math&gt;m^2&lt;/math&gt; will also divide every term, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k \geq 2&lt;/math&gt;. So, all the terms, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k &lt; 2&lt;/math&gt; must sum to a multiple of &lt;math&gt;m^2&lt;/math&gt;. If there are exactly &lt;math&gt;m&lt;/math&gt; terms where &lt;math&gt;x_k = 0&lt;/math&gt;, then we must have at least &lt;math&gt;m-1&lt;/math&gt; terms where &lt;math&gt;x_k = 1&lt;/math&gt;. Suppose there are exactly &lt;math&gt;m-1&lt;/math&gt; such terms and &lt;math&gt;x_k = 1&lt;/math&gt; for &lt;math&gt;k = m+1,m+2,2m-1&lt;/math&gt;. Now, we have,<br /> &lt;cmath&gt;m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> One can repeat this process for successive powers of &lt;math&gt;m&lt;/math&gt; until the number of terms reaches 2011. Since there are &lt;math&gt;m + j(m-1)&lt;/math&gt; terms after the &lt;math&gt;j&lt;/math&gt;th power, we will only hit exactly 2011 terms if &lt;math&gt;m-1&lt;/math&gt; is a factor of 2010. To see this, <br /> &lt;cmath&gt;m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.&lt;/cmath&gt;<br /> Thus, when &lt;math&gt;j = 2010/(m-1) - 1&lt;/math&gt; (which is an integer since &lt;math&gt;m-1 \mid 2010&lt;/math&gt; by assumption, there are exactly 2011 terms. To see that these terms sum to a power of &lt;math&gt;m&lt;/math&gt;, we realize that the sum is a geometric series:<br /> &lt;cmath&gt;1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.&lt;/cmath&gt;<br /> Thus, we have found a solution for the case &lt;math&gt;m-1 \mid 2010&lt;/math&gt;.<br /> <br /> Now, for the reverse case, we use the formula &lt;cmath&gt;x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).&lt;/cmath&gt; Suppose &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x^k}&lt;/math&gt; has a solution. Subtract 2011 from both sides to get &lt;cmath&gt;m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).&lt;/cmath&gt; Now apply the formula to get &lt;cmath&gt;(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],&lt;/cmath&gt; where &lt;math&gt;a_k&lt;/math&gt; are some integers. Rearranging this equation, we find &lt;cmath&gt;(m-1)A = 2010,&lt;/cmath&gt; where &lt;math&gt;A = a_0 - \sum_{k=1}^{2011}a_k&lt;/math&gt;. Thus, if &lt;math&gt;m&lt;/math&gt; is a solution, then &lt;math&gt;m-1 \mid 2010&lt;/math&gt;. <br /> <br /> So, there is one positive integer solution corresponding to each factor of 2010. Since &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;, the number of solutions is &lt;math&gt;2^4 = \boxed{016}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=6|num-a=8}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_3&diff=68052 2011 AIME I Problems/Problem 3 2015-03-02T03:01:46Z <p>Flamefoxx99: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;L&lt;/math&gt; be the line with slope &lt;math&gt;\frac{5}{12}&lt;/math&gt; that contains the point &lt;math&gt;A=(24,-1)&lt;/math&gt;, and let &lt;math&gt;M&lt;/math&gt; be the line perpendicular to line &lt;math&gt;L&lt;/math&gt; that contains the point &lt;math&gt;B=(5,6)&lt;/math&gt;. The original coordinate axes are erased, and line &lt;math&gt;L&lt;/math&gt; is made the &lt;math&gt;x&lt;/math&gt;-axis and line &lt;math&gt;M&lt;/math&gt; the &lt;math&gt;y&lt;/math&gt;-axis. In the new coordinate system, point &lt;math&gt;A&lt;/math&gt; is on the positive &lt;math&gt;x&lt;/math&gt;-axis, and point &lt;math&gt;B&lt;/math&gt; is on the positive &lt;math&gt;y&lt;/math&gt;-axis. The point &lt;math&gt;P&lt;/math&gt; with coordinates &lt;math&gt;(-14,27)&lt;/math&gt; in the original system has coordinates &lt;math&gt;(\alpha,\beta)&lt;/math&gt; in the new coordinate system. Find &lt;math&gt;\alpha+\beta&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> Given that &lt;math&gt;L&lt;/math&gt; has slope &lt;math&gt;\frac{5}{12}&lt;/math&gt; and contains the point &lt;math&gt;A=(24,-1)&lt;/math&gt;, we may write the point-slope equation for &lt;math&gt;L&lt;/math&gt; as &lt;math&gt;y+1=\frac{5}{12}(x-24)&lt;/math&gt;.<br /> Since &lt;math&gt;M&lt;/math&gt; is perpendicular to &lt;math&gt;L&lt;/math&gt; and contains the point &lt;math&gt;B=(5,6)&lt;/math&gt;, we have that the slope of &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;-\frac{12}{5}&lt;/math&gt;, and consequently that the point-slope equation for &lt;math&gt;M&lt;/math&gt; is &lt;math&gt;y-6=-\frac{12}{5}(x-5)&lt;/math&gt;.<br /> <br /> <br /> <br /> Converting both equations to the form &lt;math&gt;0=Ax+By+C&lt;/math&gt;, we have that &lt;math&gt;L&lt;/math&gt; has the equation &lt;math&gt;0=5x-12y-132&lt;/math&gt; and that &lt;math&gt;M&lt;/math&gt; has the equation &lt;math&gt;0=12x+5y-90&lt;/math&gt;.<br /> Applying the point-to-line distance formula, &lt;math&gt;\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}&lt;/math&gt;, to point &lt;math&gt;P&lt;/math&gt; and lines &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;, we find that the distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; are &lt;math&gt;\frac{526}{13}&lt;/math&gt; and &lt;math&gt;\frac{123}{13}&lt;/math&gt;, respectively. <br /> <br /> <br /> <br /> Since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the abscissa of &lt;math&gt;P&lt;/math&gt; is negative, and is therefore &lt;math&gt;-\frac{123}{13}&lt;/math&gt;; similarly, the ordinate of &lt;math&gt;P&lt;/math&gt; is positive and is therefore &lt;math&gt;\frac{526}{13}&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\alpha=-\frac{123}{13}&lt;/math&gt; and that &lt;math&gt;\beta=\frac{526}{13}&lt;/math&gt;. It follows that &lt;math&gt;\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=60854 Mass points 2014-03-12T21:56:04Z <p>Flamefoxx99: /* Problems */</p> <hr /> <div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br /> <br /> Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversly proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br /> <br /> == Examples ==<br /> Consider a triangle &lt;math&gt;ABC&lt;/math&gt; with its three [[Median_(geometry)|median]]s drawn, with the intersection points being &lt;math&gt;D, E, F,&lt;/math&gt; corresponding to &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively. Thus, if we label point &lt;math&gt;A&lt;/math&gt; with a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; must also have a weight of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are equidistant to &lt;math&gt;D&lt;/math&gt;. By the same process, we find &lt;math&gt;C&lt;/math&gt; must also have a weight of 1. Now, since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; both have a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; must have a weight of &lt;math&gt;2&lt;/math&gt; (as is true for &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;). Thus, if we label the centroid &lt;math&gt;P&lt;/math&gt;, we can deduce that &lt;math&gt;DP:PC&lt;/math&gt; is &lt;math&gt;1:2&lt;/math&gt; - the inverse ratio of their weights.<br /> <br /> ==Problems==<br /> [[2004 AMC 10B Problems/Problem 20]]<br /> <br /> [[2009 AIME I Problems/Problem 4]]<br /> <br /> [[2001 AIME I Problems/Problem 7 ]]<br /> <br /> [[ 1992 AIME Problems/Problem 14 ]]<br /> <br /> [[ 1988 AIME Problems/Problem 12]]<br /> <br /> [[1989 AIME Problems/Problem 15]]<br /> <br /> == External links ==<br /> *http://mathcircle.berkeley.edu/archivedocs/2007_2008/lectures/0708lecturespdf/MassPointsBMC07.pdf<br /> *http://mathcircle.berkeley.edu/archivedocs/1999_2000/lectures/9900lecturespdf/mpgeo.pdf<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=60853 Mass points 2014-03-12T21:53:08Z <p>Flamefoxx99: /* External links */</p> <hr /> <div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br /> <br /> Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversly proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br /> <br /> == Examples ==<br /> Consider a triangle &lt;math&gt;ABC&lt;/math&gt; with its three [[Median_(geometry)|median]]s drawn, with the intersection points being &lt;math&gt;D, E, F,&lt;/math&gt; corresponding to &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively. Thus, if we label point &lt;math&gt;A&lt;/math&gt; with a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; must also have a weight of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are equidistant to &lt;math&gt;D&lt;/math&gt;. By the same process, we find &lt;math&gt;C&lt;/math&gt; must also have a weight of 1. Now, since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; both have a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; must have a weight of &lt;math&gt;2&lt;/math&gt; (as is true for &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;). Thus, if we label the centroid &lt;math&gt;P&lt;/math&gt;, we can deduce that &lt;math&gt;DP:PC&lt;/math&gt; is &lt;math&gt;1:2&lt;/math&gt; - the inverse ratio of their weights.<br /> <br /> ==Problems==<br /> [[2004 AMC 10B Problems/Problem 20]]<br /> <br /> [[2009 AIME I Problems/Problem 4]]<br /> <br /> [[2001 AIME I Problems/Problem 7 ]]<br /> <br /> == External links ==<br /> *http://mathcircle.berkeley.edu/archivedocs/2007_2008/lectures/0708lecturespdf/MassPointsBMC07.pdf<br /> *http://mathcircle.berkeley.edu/archivedocs/1999_2000/lectures/9900lecturespdf/mpgeo.pdf<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=60852 Mass points 2014-03-12T21:52:59Z <p>Flamefoxx99: /* Problems */</p> <hr /> <div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br /> <br /> Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversly proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br /> <br /> == Examples ==<br /> Consider a triangle &lt;math&gt;ABC&lt;/math&gt; with its three [[Median_(geometry)|median]]s drawn, with the intersection points being &lt;math&gt;D, E, F,&lt;/math&gt; corresponding to &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; respectively. Thus, if we label point &lt;math&gt;A&lt;/math&gt; with a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; must also have a weight of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are equidistant to &lt;math&gt;D&lt;/math&gt;. By the same process, we find &lt;math&gt;C&lt;/math&gt; must also have a weight of 1. Now, since &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; both have a weight of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt; must have a weight of &lt;math&gt;2&lt;/math&gt; (as is true for &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;). Thus, if we label the centroid &lt;math&gt;P&lt;/math&gt;, we can deduce that &lt;math&gt;DP:PC&lt;/math&gt; is &lt;math&gt;1:2&lt;/math&gt; - the inverse ratio of their weights.<br /> <br /> ==Problems==<br /> [[2004 AMC 10B Problems/Problem 20]]<br /> <br /> [[2009 AIME I Problems/Problem 4]]<br /> <br /> [[2001 AIME I Problems/Problem 7 ]]<br /> <br /> == External links ==<br /> *http://mathcircle.berkeley.edu/archivedocs/2007_2008/lectures/0708lecturesps/MassPointsBMC07.ps<br /> *http://mathcircle.berkeley.edu/archivedocs/1999_2000/lectures/9900lecturespdf/mpgeo.pdf<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Descartes_Rule_of_Signs&diff=60847 Descartes Rule of Signs 2014-03-12T20:20:13Z <p>Flamefoxx99: /* See Also */</p> <hr /> <div>Descarte's rule of signs is a method used to determine the number of positive and negative roots of a polynomial. The rule gives an upper bound on the number of positive or negative roots, but does not specify the exact amount.<br /> <br /> ===Positive roots===<br /> If the terms of a polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is equal to the number of sign differences between consecutive nonzero coefficients or is less than that by an even number. Multiple roots are counted separately.<br /> <br /> ===Negative roots===<br /> The bound for negative roots is a corollary of the positive root bound. The number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than that by a positive even number.<br /> <br /> <br /> ==See Also==<br /> * [[Polynomials]]<br /> * [[Factor Theorem]]<br /> * [[Rational Root Theorem]]<br /> *[[ Algebra/Intermediate]]<br /> {{stub}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Descartes_Rule_of_Signs&diff=60846 Descartes Rule of Signs 2014-03-12T20:08:11Z <p>Flamefoxx99: </p> <hr /> <div>Descarte's rule of signs is a method used to determine the number of positive and negative roots of a polynomial. The rule gives an upper bound on the number of positive or negative roots, but does not specify the exact amount.<br /> <br /> ===Positive roots===<br /> If the terms of a polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is equal to the number of sign differences between consecutive nonzero coefficients or is less than that by an even number. Multiple roots are counted separately.<br /> <br /> ===Negative roots===<br /> The bound for negative roots is a corollary of the positive root bound. The number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than that by a positive even number.<br /> <br /> <br /> ==See Also==<br /> * [[Polynomials]]<br /> * [[Factor Theorem]]<br /> * [[Rational Root Theorem]]<br /> {{stub}}<br /> [[ Algebra/Intermediate]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Descartes_Rule_of_Signs&diff=60845 Descartes Rule of Signs 2014-03-12T20:07:34Z <p>Flamefoxx99: added see also</p> <hr /> <div>Descarte's rule of signs is a method used to determine the number of positive and negative roots of a polynomial. The rule gives an upper bound on the number of positive or negative roots, but does not specify the exact amount.<br /> <br /> ==Positive roots==<br /> If the terms of a polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is equal to the number of sign differences between consecutive nonzero coefficients or is less than that by an even number. Multiple roots are counted separately.<br /> <br /> ==Negative roots==<br /> The bound for negative roots is a corollary of the positive root bound. The number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than that by a positive even number.<br /> <br /> <br /> ==See Also==<br /> * [[Polynomials]]<br /> * [[Factor Theorem]]<br /> * [[Rational Root Theorem]]<br /> {{stub}}<br /> [[ Algebra/Intermediate]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Descartes_Rule_of_Signs&diff=60844 Descartes Rule of Signs 2014-03-12T20:05:05Z <p>Flamefoxx99: Created page with &quot;Descarte's rule of signs is a method used to determine the number of positive and negative roots of a polynomial. The rule gives an upper bound on the number of positive or negat...&quot;</p> <hr /> <div>Descarte's rule of signs is a method used to determine the number of positive and negative roots of a polynomial. The rule gives an upper bound on the number of positive or negative roots, but does not specify the exact amount.<br /> <br /> ==Positive roots==<br /> If the terms of a polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is equal to the number of sign differences between consecutive nonzero coefficients or is less than that by an even number. Multiple roots are counted separately.<br /> <br /> ==Negative roots==<br /> The bound for negative roots is a corollary of the positive root bound. The number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than that by a positive even number.</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_3&diff=60796 2004 AIME II Problems/Problem 3 2014-03-10T15:53:39Z <p>Flamefoxx99: /* Problem */</p> <hr /> <div>== Problem ==<br /> A solid rectangular block is formed by gluing together &lt;math&gt; N &lt;/math&gt; [[congruent (geometry) | congruent]] 1-cm [[cube (geometry) | cube]]s [[face]] to face. When the block is viewed so that three of its faces are visible, exactly &lt;math&gt;231&lt;/math&gt; of the 1-cm cubes cannot be seen. Find the smallest possible value of &lt;math&gt; N. &lt;/math&gt;<br /> <br /> == Solution ==<br /> The &lt;math&gt;231&lt;/math&gt; cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions &lt;math&gt;l \times m \times n&lt;/math&gt;, we must have &lt;math&gt;(l - 1)\times(m-1) \times(n - 1) = 231&lt;/math&gt;. The [[prime factorization]] of &lt;math&gt;231 = 3\cdot7\cdot11&lt;/math&gt;, so we have a variety of possibilities; for instance, &lt;math&gt;l - 1 = 1&lt;/math&gt; and &lt;math&gt;m - 1 = 11&lt;/math&gt; and &lt;math&gt;n - 1 = 3 \cdot 7&lt;/math&gt;, among others. However, it should be fairly clear that the way to minimize &lt;math&gt;l\cdot m\cdot n&lt;/math&gt; is to make &lt;math&gt;l&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; as close together as possible, which occurs when the smaller block is &lt;math&gt;3 \times 7 \times 11&lt;/math&gt;. Then the extra layer makes the entire block &lt;math&gt;4\times8\times12&lt;/math&gt;, and &lt;math&gt;N= \boxed{384}&lt;/math&gt;.<br /> <br /> An alternate way to visualize the problem is to count the blocks that can be seen and subtract the blocks that cannot be seen. In the given block with dimensions &lt;math&gt;l\times m \times n&lt;/math&gt;, the three faces have &lt;math&gt;lm&lt;/math&gt;, &lt;math&gt;mn&lt;/math&gt;, and &lt;math&gt;ln&lt;/math&gt; blocks each. However, &lt;math&gt;l&lt;/math&gt; blocks along the first edge, &lt;math&gt;m&lt;/math&gt; blocks along the second edge, and &lt;math&gt;n&lt;/math&gt; blocks along the third edge were counted twice, so they must be subtracted. After subtracting these three edges, 1 block has not been counted - it was added three times on each face, but subtracted three times on each side. Thus, the total number of visible cubes is &lt;math&gt;lm+mn+ln-l-m-n+1&lt;/math&gt;, and the total number of invisible cubes is &lt;math&gt;lmn-lm-mn-ln+l+m+n-1&lt;/math&gt;, which can be factored into &lt;math&gt;(l-1)(m-1)(n-1)&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|num-b=2|num-a=4|n=II}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_13&diff=60765 2009 AIME II Problems/Problem 13 2014-03-09T15:09:17Z <p>Flamefoxx99: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be the endpoints of a semicircular arc of radius &lt;math&gt;2&lt;/math&gt;. The arc is divided into seven congruent arcs by six equally spaced points &lt;math&gt;C_1&lt;/math&gt;, &lt;math&gt;C_2&lt;/math&gt;, &lt;math&gt;\dots&lt;/math&gt;, &lt;math&gt;C_6&lt;/math&gt;. All chords of the form &lt;math&gt;\overline {AC_i}&lt;/math&gt; or &lt;math&gt;\overline {BC_i}&lt;/math&gt; are drawn. Let &lt;math&gt;n&lt;/math&gt; be the product of the lengths of these twelve chords. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. <br /> <br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Let the radius be 1 instead. All lengths will be halved so we will multiply by &lt;math&gt;2^{12}&lt;/math&gt; at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then &lt;math&gt;C_1,\ldots, C_6&lt;/math&gt; are 6 of the 14th roots of unity. Let &lt;math&gt;\omega=\text{cis}\frac{360^{\circ}}{14}&lt;/math&gt;; then &lt;math&gt;C_1,\ldots, C_6&lt;/math&gt; correspond to &lt;math&gt;\omega,\ldots, \omega^6&lt;/math&gt;. Let &lt;math&gt;C_1',\ldots, C_6'&lt;/math&gt; be their reflections across the diameter. These points correspond to &lt;math&gt;\omega^8\ldots, \omega^{13}&lt;/math&gt;. Then the lengths of the segments are &lt;math&gt;|1-\omega|,\ldots, |1-\omega^6|,|1-\omega^8|,\ldots |1-\omega^{13}|&lt;/math&gt;. Noting that &lt;math&gt;B&lt;/math&gt; represents 1 in the complex plane, the desired product is<br /> &lt;math&gt;<br /> \begin{align*}<br /> BC_1\cdots BC_6 \cdot AC_1\cdots AC_6&amp;=<br /> BC_1\cdots BC_6 \cdot BC_1'\cdots BC_6\\<br /> &amp;=<br /> |(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})|<br /> \end{align*}&lt;/math&gt;<br /> <br /> for &lt;math&gt;x=1&lt;/math&gt;.<br /> However, the polynomial &lt;math&gt;(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})&lt;/math&gt; has as its zeros all 14th roots of unity except for &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Hence<br /> &lt;cmath&gt;<br /> (x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})=\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\cdots +x^2+1.<br /> &lt;/cmath&gt;<br /> Thus the product is &lt;math&gt;|x^{12}+\cdots +x^2+1|=7&lt;/math&gt; (&lt;math&gt;x=1&lt;/math&gt;) when the radius is 1, and the product is &lt;math&gt;2^{12}7=28672&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed {672}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the midpoint of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Assume &lt;math&gt;C_1&lt;/math&gt; is closer to &lt;math&gt;A&lt;/math&gt; instead of &lt;math&gt;B&lt;/math&gt;. &lt;math&gt;\angle AOC_1&lt;/math&gt; = &lt;math&gt;\frac {\pi}{7}&lt;/math&gt;. Using the [[Law of Cosines]], <br /> <br /> &lt;math&gt;\overline {AC_1}^2&lt;/math&gt; = &lt;math&gt;8 - 8 \cos \frac {\pi}{7}&lt;/math&gt;, <br /> &lt;math&gt;\overline {AC_2}^2&lt;/math&gt; = &lt;math&gt;8 - 8 \cos \frac {2\pi}{7}&lt;/math&gt;, <br /> .<br /> .<br /> .<br /> &lt;math&gt;\overline {AC_6}^2&lt;/math&gt; = &lt;math&gt;8 - 8 \cos \frac {6\pi}{7}&lt;/math&gt; <br /> <br /> So &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {2\pi}{7})\dots(1 - \cos \frac{6\pi}{7})&lt;/math&gt;. It can be rearranged to form<br /> <br /> &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;(8^6)(1 - \cos \frac {\pi}{7})(1 - \cos \frac {6\pi}{7})\dots(1 - \cos \frac {3\pi}{7})(1 - \cos \frac {4\pi}{7})&lt;/math&gt;. <br /> <br /> &lt;math&gt;\cos a&lt;/math&gt; = - &lt;math&gt;\cos (\pi - a)&lt;/math&gt;, so we have<br /> <br /> &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;(8^6)(1 - \cos \frac {\pi}{7})(1 + \cos \frac {\pi}{7}) \dots (1 - \cos \frac {3\pi}{7})(1 + \cos \frac {3\pi}{7})&lt;/math&gt;<br /> <br /> = &lt;math&gt;(8^6)(1 - \cos^2 \frac {\pi}{7})(1 - \cos^2 \frac {2\pi}{7})(1 - \cos^2 \frac {3\pi}{7})&lt;/math&gt;<br /> <br /> = &lt;math&gt;(8^6)(\sin^2 \frac {\pi}{7})(\sin^2 \frac {2\pi}{7})(\sin^2 \frac {3\pi}{7})&lt;/math&gt;<br /> <br /> It can be shown that &lt;math&gt;\sin \frac {\pi}{7} \sin \frac {2\pi}{7} \sin \frac {3\pi}{7}&lt;/math&gt; = &lt;math&gt;\frac {\sqrt {7}}{8}&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; = &lt;math&gt;8^6(\frac {\sqrt {7}}{8})^2&lt;/math&gt; = &lt;math&gt;7(8^4)&lt;/math&gt; = &lt;math&gt;28672&lt;/math&gt;, so the answer is &lt;math&gt;\boxed {672}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Note that for each &lt;math&gt;k&lt;/math&gt; the triangle &lt;math&gt;ABC_k&lt;/math&gt; is a right triangle. Hence the product &lt;math&gt;AC_k \cdot BC_k&lt;/math&gt; is twice the area of the triangle &lt;math&gt;ABC_k&lt;/math&gt;. Knowing that &lt;math&gt;AB=4&lt;/math&gt;, the area of &lt;math&gt;ABC_k&lt;/math&gt; can also be expressed as &lt;math&gt;2c_k&lt;/math&gt;, where &lt;math&gt;c_k&lt;/math&gt; is the length of the altitude from &lt;math&gt;C_k&lt;/math&gt; onto &lt;math&gt;AB&lt;/math&gt;. Hence we have &lt;math&gt;AC_k \cdot BC_k = 4c_k&lt;/math&gt;.<br /> <br /> By the definition of &lt;math&gt;C_k&lt;/math&gt; we obviously have &lt;math&gt;c_k = 2\sin\frac{k\pi}7&lt;/math&gt;. <br /> <br /> From these two observations we get that the product we should compute is equal to &lt;math&gt; 8^6 \cdot \prod_{k=1}^6 \sin \frac{k\pi}7 &lt;/math&gt;, which is the same identity as in Solution 1.<br /> <br /> === Computing the product of sines ===<br /> <br /> In this section we show one way how to evaluate the product &lt;math&gt;\prod_{k=1}^6 \sin \frac{k\pi}7 &lt;/math&gt;.<br /> <br /> Let &lt;math&gt;\omega_k = \cos \frac{2k\pi}7 + i\sin \frac{2k\pi}7&lt;/math&gt;. The numbers &lt;math&gt;1,\omega_1,\omega_2,\dots,\omega_6&lt;/math&gt; are the &lt;math&gt;7&lt;/math&gt;-th complex roots of unity. In other words, these are the roots of the polynomial &lt;math&gt;x^7-1&lt;/math&gt;. Then the numbers &lt;math&gt;\omega_1,\omega_2,\dots,\omega_6&lt;/math&gt; are the roots of the polynomial &lt;math&gt;\frac{x^7-1}{x-1} = x^6+x^5+\cdots+x+1&lt;/math&gt;.<br /> <br /> We just proved the identity &lt;math&gt;\prod_{k=1}^6 (x - \omega_k) = x^6+x^5+\cdots+x+1&lt;/math&gt;.<br /> Substitute &lt;math&gt;x=1&lt;/math&gt;. The right hand side is obviously equal to &lt;math&gt;7&lt;/math&gt;. Let's now examine the left hand side.<br /> We have:<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> |1-\omega_k| <br /> &amp; = <br /> \left| 1-\cos \frac{2k\pi}7 - i\sin \frac{2k\pi}7 \right| <br /> \\<br /> &amp; = \sqrt{ \left( 1-\cos \frac{2k\pi}7 \right)^2 + \left( \sin \frac{2k\pi}7 \right)^2 } <br /> \\<br /> &amp; = \sqrt{ 2-2\cos \frac{2k\pi}7 } <br /> \\<br /> &amp; = \sqrt{ 2-2 \left( 1 - 2 \left( \sin \frac{k\pi}7 \right)^2 \right) } <br /> \\<br /> &amp; = \sqrt{ 4\left( \sin \frac{k\pi}7 \right)^2 } <br /> \\<br /> &amp; = 2 \sin \frac{k\pi}7 <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore the size of the left hand side in our equation is &lt;math&gt;\prod_{k=1}^6 |1-\omega_k| = \prod_{k=1}^6 2 \sin \frac{k\pi}7 = 2^6 \prod_{k=1}^6 \sin \frac{k\pi}7&lt;/math&gt;. As the right hand side is &lt;math&gt;7&lt;/math&gt;, we get that &lt;math&gt;\prod_{k=1}^6 \sin \frac{k\pi}7 = \frac{7}{2^6}&lt;/math&gt;. However, since sin &lt;math&gt;x&lt;/math&gt; = sin &lt;math&gt;\pi - x&lt;/math&gt;, then <br /> &lt;math&gt;\prod_{k=1}^3 \sin \frac{k\pi}7 &lt;/math&gt; would be the square root of &lt;math&gt;\frac {7}{2^6}&lt;/math&gt;, or &lt;math&gt;\frac {\sqrt {7}}{8}&lt;/math&gt;, which is what we needed to find.<br /> <br /> == See Also ==<br /> <br /> {{AIME box|year=2009|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Mock_Geometry_AIME_2011_Problems/Problem_3&diff=60764 Mock Geometry AIME 2011 Problems/Problem 3 2014-03-09T14:49:49Z <p>Flamefoxx99: /* Solution */</p> <hr /> <div>==Problem== <br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;BC=9.&lt;/math&gt; Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are located on &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BP=PQ=2,&lt;/math&gt; &lt;math&gt;QC=5.&lt;/math&gt; The circumcircle of &lt;math&gt;APQ&lt;/math&gt; cuts &lt;math&gt;AB,AC&lt;/math&gt; at &lt;math&gt;D,E&lt;/math&gt; respectively. If &lt;math&gt;BD=CE,&lt;/math&gt; then the ratio &lt;math&gt;\frac{AB}{AC}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{m}{n},&lt;/math&gt; where &lt;math&gt;m,n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt; <br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> draw((-1,0)--(8,0)--(0,4)--cycle);<br /> draw((0,4)--(1,0));<br /> draw((0,4)--(3,0));<br /> label(&quot;$A$&quot;,(0,4),N);<br /> label(&quot;$B$&quot;,(-1,0),WSW);<br /> label(&quot;$C$&quot;,(8,0),ESE);<br /> label(&quot;$P$&quot;,(1,0),S);<br /> label(&quot;$Q$&quot;,(3,0),S);<br /> draw(circumcircle((0,4),(1,0),(3,0))); <br /> label(&quot;$D$&quot;,(-0.6,2),SW);<br /> label(&quot;$E$&quot;,(4.5,1.7),NE);<br /> &lt;/asy&gt;<br /> <br /> By the Power of a Point Theorem on &lt;math&gt;B&lt;/math&gt;, we have &lt;math&gt;BD \times BA=BP \times BQ=2 \times 4=8&lt;/math&gt;. By the Power of a Point on &lt;math&gt;C&lt;/math&gt;, we have &lt;math&gt;CE \times CA=CQ \times CP=5 \times 7=35&lt;/math&gt;. Dividing these two results yields &lt;math&gt;\frac{BD \times BA}{CE \times CA}=\frac{8}{35}&lt;/math&gt;. We are given &lt;math&gt;BD=CE&lt;/math&gt; and so &lt;math&gt;\frac{BD}{CE}=1&lt;/math&gt;. Then the previous equation simplifies to &lt;math&gt;\frac{AB}{AC}=\frac{8}{35}&lt;/math&gt;. Hence &lt;math&gt;m+n=8+35=\boxed{043}&lt;/math&gt;</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Mock_Geometry_AIME_2011_Problems/Problem_3&diff=60763 Mock Geometry AIME 2011 Problems/Problem 3 2014-03-09T14:49:34Z <p>Flamefoxx99: /* Solution */</p> <hr /> <div>==Problem== <br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;BC=9.&lt;/math&gt; Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are located on &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BP=PQ=2,&lt;/math&gt; &lt;math&gt;QC=5.&lt;/math&gt; The circumcircle of &lt;math&gt;APQ&lt;/math&gt; cuts &lt;math&gt;AB,AC&lt;/math&gt; at &lt;math&gt;D,E&lt;/math&gt; respectively. If &lt;math&gt;BD=CE,&lt;/math&gt; then the ratio &lt;math&gt;\frac{AB}{AC}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{m}{n},&lt;/math&gt; where &lt;math&gt;m,n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt; <br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> draw((-1,0)--(8,0)--(0,4)--cycle);<br /> draw((0,4)--(1,0));<br /> draw((0,4)--(3,0));<br /> label(&quot;$A$&quot;,(0,4),N);<br /> label(&quot;$B$&quot;,(-1,0),WSW);<br /> label(&quot;$C$&quot;,(8,0),ESE);<br /> label(&quot;$P$&quot;,(1,0),S);<br /> label(&quot;$Q$&quot;,(3,0),S);<br /> draw(circumcircle((0,4),(1,0),(3,0))); <br /> label(&quot;$D$&quot;,(-0.6,2),SW);<br /> label(&quot;$E$&quot;,(4.5,1.7),NE);<br /> &lt;/asy&gt;<br /> <br /> By the Power of a Point Theorem on &lt;math&gt;B&lt;/math&gt;, we have &lt;math&gt;BD*BA=BP \times BQ=2 \times 4=8&lt;/math&gt;. By the Power of a Point on &lt;math&gt;C&lt;/math&gt;, we have &lt;math&gt;CE \times CA=CQ \times CP=5 \times 7=35&lt;/math&gt;. Dividing these two results yields &lt;math&gt;\frac{BD \times BA}{CE \times CA}=\frac{8}{35}&lt;/math&gt;. We are given &lt;math&gt;BD=CE&lt;/math&gt; and so &lt;math&gt;\frac{BD}{CE}=1&lt;/math&gt;. Then the previous equation simplifies to &lt;math&gt;\frac{AB}{AC}=\frac{8}{35}&lt;/math&gt;. Hence &lt;math&gt;m+n=8+35=\boxed{043}&lt;/math&gt;</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_2&diff=60725 2009 AIME II Problems/Problem 2 2014-03-08T19:49:47Z <p>Flamefoxx99: </p> <hr /> <div>== Problem ==<br /> Suppose that &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive real numbers such that &lt;math&gt;a^{\log_3 7} = 27&lt;/math&gt;, &lt;math&gt;b^{\log_7 11} = 49&lt;/math&gt;, and &lt;math&gt;c^{\log_{11}25} = \sqrt{11}&lt;/math&gt;. Find<br /> &lt;cmath&gt; a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}. &lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> <br /> First, we have:<br /> &lt;cmath&gt;<br /> x^{(\log_y z)^2}<br /> = x^{\left( (\log_y z)^2 \right) }<br /> = x^{(\log_y z) \cdot (\log_y z) }<br /> = \left( x^{\log_y z} \right)^{\log_y z}<br /> &lt;/cmath&gt;<br /> <br /> Now, let &lt;math&gt;x=y^w&lt;/math&gt;, then we have:<br /> &lt;cmath&gt;<br /> x^{\log_y z} <br /> = \left( y^w \right)^{\log_y z} <br /> = y^{w\log_y z} <br /> = y^{\log_y (z^w)} <br /> = z^w<br /> &lt;/cmath&gt;<br /> <br /> This is all we need to evaluate the given formula. Note that in our case we have &lt;math&gt;27=3^3&lt;/math&gt;, &lt;math&gt;49=7^2&lt;/math&gt;, and &lt;math&gt;\sqrt{11}=11^{1/2}&lt;/math&gt;. We can now compute:<br /> <br /> &lt;cmath&gt;<br /> a^{(\log_3 7)^2}<br /> = \left( a^{\log_3 7} \right)^{\log_3 7}<br /> = 27^{\log_3 7}<br /> = (3^3)^{\log_3 7}<br /> = 7^3<br /> = 343<br /> &lt;/cmath&gt;<br /> <br /> Similarly, we get<br /> &lt;cmath&gt;<br /> b^{(\log_7 11)^2} <br /> = (7^2)^{\log_7 11}<br /> = 11^2 <br /> = 121<br /> &lt;/cmath&gt;<br /> <br /> and<br /> &lt;cmath&gt;<br /> c^{(\log_{11} 25)^2}<br /> = (11^{1/2})^{\log_{11} 25}<br /> = 25^{1/2}<br /> = 5<br /> &lt;/cmath&gt;<br /> <br /> and therefore the answer is &lt;math&gt;343+121+5 = \boxed{469}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> We know from the first three equations that &lt;math&gt;\log_a27&lt;/math&gt; = &lt;math&gt;\log_37&lt;/math&gt;, &lt;math&gt;\log_b49&lt;/math&gt; = &lt;math&gt;\log_711&lt;/math&gt;, and &lt;math&gt;\log_c\sqrt{11}&lt;/math&gt; = &lt;math&gt;\log_{11}25&lt;/math&gt;. Substituting, we get <br /> <br /> &lt;math&gt;a^{(\log_a27)(\log_37)}&lt;/math&gt; + &lt;math&gt;b^{(\log_b49)(\log_711)&lt;/math&gt; + &lt;math&gt;c^{(\log_c\sqrt {11})(\log_{11}25)}&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;x^{\log_xy}&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;, so we get<br /> <br /> &lt;math&gt;27^{\log_37}&lt;/math&gt; + &lt;math&gt;49^{\log_711}&lt;/math&gt; + &lt;math&gt;\sqrt {11}^{\log_{11}25}&lt;/math&gt;<br /> <br /> &lt;math&gt;(3^{\log_37})^3&lt;/math&gt; + &lt;math&gt;(7^{\log_711})^2&lt;/math&gt; + &lt;math&gt;({11^{\log_{11}25})^{1/2}&lt;/math&gt;<br /> <br /> The &lt;math&gt;3&lt;/math&gt; and the &lt;math&gt;\log_37&lt;/math&gt; cancel out to make &lt;math&gt;7&lt;/math&gt;, and we can do this for the other two terms. We obtain <br /> <br /> &lt;math&gt;7^3&lt;/math&gt; + &lt;math&gt;11^2&lt;/math&gt; + &lt;math&gt;25^{1/2}&lt;/math&gt;<br /> <br /> = &lt;math&gt;343&lt;/math&gt; + &lt;math&gt;121&lt;/math&gt; + &lt;math&gt;5&lt;/math&gt;<br /> = &lt;math&gt;\boxed {469}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AIME box|year=2009|n=II|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=User:Flamefoxx99&diff=57699 User:Flamefoxx99 2013-11-16T01:57:16Z <p>Flamefoxx99: Created page with &quot;Hello plz call me Flame&quot;</p> <hr /> <div>Hello<br /> plz call me Flame</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Trigonometry&diff=57336 Trigonometry 2013-10-19T17:17:08Z <p>Flamefoxx99: /* Trigonometery Definitions for non-acute angles */</p> <hr /> <div>Trigonometry seeks to find the lengths of a [[triangle]]'s sides, given 2 [[angle]]s and a side. Trigonometry is closely related to [[analytic geometry]].<br /> <br /> ==Basic definitions==<br /> Usually we call an angle &lt;math&gt;\theta&lt;/math&gt;, read &quot;theta&quot;, but &lt;math&gt;\theta&lt;/math&gt; is just a variable. We could just as well call it &lt;math&gt; a&lt;/math&gt;.<br /> <br /> For the following definitions, the &quot;opposite side&quot; is the side opposite of angle &lt;math&gt;\theta&lt;/math&gt;, and the &quot;adjacent side&quot; is the side that is part of angle &lt;math&gt;\theta&lt;/math&gt;, but is not the hypotenuse. <br /> <br /> i.e. If ABC is a right triangle with right angle C, and angle A = &lt;math&gt;\theta&lt;/math&gt;, then BC is the &quot;opposite side&quot;, AC is the &quot;adjacent side&quot;, and AB is the hypotenuse. <br /> <br /> [[Image:306090triangle.gif]]<br /> <br /> ===Sine===<br /> The sine of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\sin \theta&lt;/math&gt;, is the ratio between the opposite side and the [[hypotenuse]] of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\sin 30^{\circ}=\frac 12&lt;/math&gt;.<br /> <br /> ===Cosine===<br /> The cosine of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\cos \theta&lt;/math&gt;, is the ratio between the adjacent side and the [[hypotenuse]] of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\cos 30^{\circ} =\frac{\sqrt{3}}{2}&lt;/math&gt;.<br /> <br /> ===Tangent===<br /> The tangent of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\tan \theta&lt;/math&gt;, is the ratio between the opposite side and the adjacent side of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\tan 30^{\circ}=\frac{\sqrt{3}}{3}&lt;/math&gt;. (Note that &lt;math&gt; \tan \theta=\frac{\sin\theta}{\cos\theta}&lt;/math&gt;.)<br /> <br /> ===Cosecant===<br /> The cosecant of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\csc \theta&lt;/math&gt;, is the ratio between the [[hypotenuse]] and the opposite side of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\csc 30^{\circ}=2&lt;/math&gt;. (Note that &lt;math&gt; \csc \theta=\frac{1}{\sin \theta}&lt;/math&gt;.)<br /> <br /> ===Secant===<br /> The secant of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\sec \theta&lt;/math&gt;, is the ratio between the [[hypotenuse]] and the adjacent side of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\sec 30^{\circ}=\frac{2\sqrt{3}}{3}&lt;/math&gt;. (Note that &lt;math&gt; \sec \theta=\frac{1}{\cos \theta}&lt;/math&gt;.)<br /> <br /> <br /> ===Cotangent===<br /> The cotangent of an angle &lt;math&gt;\theta&lt;/math&gt;, abbreviated &lt;math&gt;\cot \theta&lt;/math&gt;, is the ratio between the adjacent side and the opposite side of a triangle. For instance, in the 30-60-90 triangle above, &lt;math&gt;\cot 30^{\circ}=\sqrt{3}&lt;/math&gt;. (Note that &lt;math&gt; \cot \theta=\frac{\cos\theta}{\sin\theta}&lt;/math&gt; or &lt;math&gt; \cot \theta = \frac{1}{\tan \theta}&lt;/math&gt;.)<br /> <br /> ==Trigonometery Definitions for non-acute angles==<br /> Consider a [[unit circle]] that is centered at the origin. By picking a point on the circle, and dropping a perpendicular line to the x-axis, a right triangle is formed with a [[hypotenuse]] 1 unit long. Letting the angle at the origin be &lt;math&gt;\theta &lt;/math&gt; and the coordinates of the point we picked to be &lt;math&gt;(x,y) &lt;/math&gt;, we have:<br /> <br /> &lt;cmath&gt;\begin{align*} \sin \theta &amp;= y \\<br /> <br /> \cos \theta &amp;= x \\<br /> <br /> \tan \theta &amp;= \frac{y}{x} \\<br /> <br /> \csc \theta &amp;= \frac{1}{y} \\<br /> <br /> \sec \theta &amp;= \frac{1}{x} \\<br /> <br /> \cot \theta &amp;= \frac{x}{y} \end{align*}&lt;/cmath&gt;<br /> <br /> Note that &lt;math&gt;(x,y) &lt;/math&gt; is the rectangular coordinates for the point &lt;math&gt; (1,\theta) &lt;/math&gt;.<br /> <br /> This is true for all angles, even negative angles and angles greater than 360 degrees. Due to the way trig ratios are defined for non-acute angles, the value of a trig ratio could be positive or negative, or even 0.<br /> <br /> ==Trigonometric Identities==<br /> <br /> There are many identities that are based on trigonometric functions.<br /> <br /> ===Pythagorean Identities===<br /> <br /> *&lt;math&gt;\sin^2\theta+\cos^2\theta=1&lt;/math&gt; <br /> *&lt;math&gt;1+\tan^2\theta=\sec^2\theta&lt;/math&gt; <br /> *&lt;math&gt;1+\cot^2\theta=\csc^2\theta&lt;/math&gt;<br /> <br /> ===Double-Angle Identities===<br /> <br /> *&lt;math&gt;\sin 2\theta=2\sin\theta\cos\theta&lt;/math&gt;<br /> *&lt;math&gt;\cos 2\theta=\cos^2\theta-\sin^2\theta&lt;/math&gt;<br /> *&lt;math&gt;\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}&lt;/math&gt;<br /> <br /> ===Half-Angle Identites===<br /> <br /> *&lt;math&gt;\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}&lt;/math&gt;<br /> *&lt;math&gt;\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}&lt;/math&gt;<br /> *&lt;math&gt;\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}&lt;/math&gt;<br /> <br /> ==See also==<br /> * [[Trigonometric identities]]<br /> * [[Trigonometric substitution]]<br /> * [[Geometry]]<br /> [[Category:Trigonometry]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_4&diff=57314 2013 USAMO Problems/Problem 4 2013-10-16T15:14:27Z <p>Flamefoxx99: /* Solution 1 (Cauchy or AM-GM) */</p> <hr /> <div>Find all real numbers &lt;math&gt;x,y,z\geq 1&lt;/math&gt; satisfying &lt;cmath&gt;\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.&lt;/cmath&gt;<br /> <br /> <br /> == Solution (Cauchy or AM-GM) ==<br /> The key Lemma is:<br /> &lt;cmath&gt;\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}&lt;/cmath&gt; for all &lt;math&gt;a,b \ge 1&lt;/math&gt;. Equality holds when &lt;math&gt;(a-1)(b-1)=1&lt;/math&gt;.<br /> <br /> This is proven easily.<br /> &lt;cmath&gt;\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}&lt;/cmath&gt; by Cauchy.<br /> Equality then holds when &lt;math&gt;a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1&lt;/math&gt;.<br /> <br /> Now assume that &lt;math&gt;x = \min(x,y,z)&lt;/math&gt;. Now note that, by the Lemma,<br /> <br /> &lt;cmath&gt;\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}&lt;/cmath&gt;. So equality must hold.<br /> So &lt;math&gt;(y-1)(z-1) = 1&lt;/math&gt; and &lt;math&gt;(x-1)(yz) = 1&lt;/math&gt;. If we let &lt;math&gt;z = c&lt;/math&gt;, then we can easily compute that &lt;math&gt;y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}&lt;/math&gt;.<br /> Now it remains to check that &lt;math&gt;x \le y, z&lt;/math&gt;.<br /> <br /> But by easy computations, &lt;math&gt;x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0&lt;/math&gt;, which is obvious.<br /> Also &lt;math&gt;x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1&lt;/math&gt;, which is obvious, since &lt;math&gt;c \ge 1&lt;/math&gt;.<br /> <br /> So all solutions are of the form &lt;math&gt;\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}&lt;/math&gt;, and symmetric (or cyclic) permutations for &lt;math&gt;c &gt; 1&lt;/math&gt;.<br /> <br /> '''Remark:''' An alternative proof of the key Lemma is the following:<br /> By AM-GM, &lt;cmath&gt;(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}&lt;/cmath&gt;. Now taking the square root of both sides gives the desired. Equality holds when &lt;math&gt;(a-1)(b-1) = 1&lt;/math&gt;.<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_24&diff=57313 2007 AMC 8 Problems/Problem 24 2013-10-16T15:12:27Z <p>Flamefoxx99: </p> <hr /> <div>==Problem==<br /> A bag contains four pieces of paper, each labeled with one of the digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of &lt;math&gt;3&lt;/math&gt;?<br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} &lt;/math&gt;<br /> <br /> ==Solution==<br /> The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4). The number of ways to choose three digits out of four is 4. Therefore, the probability is &lt;math&gt;\textbf{(C)}\ \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2007|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8&diff=56980 2001 AMC 8 2013-08-22T14:26:13Z <p>Flamefoxx99: </p> <hr /> <div>'''2001 AMC 8''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> <br /> * [[2001 AMC 8 Problems]]<br /> * [[2001 AMC 8 Answer Key]]<br /> ** [[2001 AMC 8 Problems/Problem 1]]<br /> ** [[2001 AMC 8 Problems/Problem 2]]<br /> ** [[2001 AMC 8 Problems/Problem 3]]<br /> ** [[2001 AMC 8 Problems/Problem 4]]<br /> ** [[2001 AMC 8 Problems/Problem 5]]<br /> ** [[2001 AMC 8 Problems/Problem 6]]<br /> ** [[2001 AMC 8 Problems/Problem 7]]<br /> ** [[2001 AMC 8 Problems/Problem 8]]<br /> ** [[2001 AMC 8 Problems/Problem 9]]<br /> ** [[2001 AMC 8 Problems/Problem 10]]<br /> ** [[2001 AMC 8 Problems/Problem 11]]<br /> ** [[2001 AMC 8 Problems/Problem 12]]<br /> ** [[2001 AMC 8 Problems/Problem 13]]<br /> ** [[2001 AMC 8 Problems/Problem 14]]<br /> ** [[2001 AMC 8 Problems/Problem 15]]<br /> ** [[2001 AMC 8 Problems/Problem 16]]<br /> ** [[2001 AMC 8 Problems/Problem 17]]<br /> ** [[2001 AMC 8 Problems/Problem 18]]<br /> ** [[2001 AMC 8 Problems/Problem 19]]<br /> ** [[2001 AMC 8 Problems/Problem 20]]<br /> ** [[2001 AMC 8 Problems/Problem 21]]<br /> ** [[2001 AMC 8 Problems/Problem 22]]<br /> ** [[2001 AMC 8 Problems/Problem 23]]<br /> ** [[2001 AMC 8 Problems/Problem 24]]<br /> ** [[2001 AMC 8 Problems/Problem 25]]<br /> <br /> == See also ==<br /> {{Succession box|<br /> |header=2001 AMC 8 ([[2001 AMC 8 Problems|Problems]],[http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=42&amp;year=2001 Resources])<br /> |before=[[2000 AMC 8]]<br /> |title=[[AMC 8]]<br /> |after=[[2002 AMC 8]]}}<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Math books]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Answer_Key&diff=56979 2001 AMC 8 Answer Key 2013-08-22T14:25:18Z <p>Flamefoxx99: Created page with &quot;1) D 2) D 3). E 4) E 5) C 6) B 7) A 8) E 9) D 10) A 11) C 12) A 13) D 14) C 15) A 16) E 17) B 18) D 19) D 20) A 21) D 22) E 23) D 24) B 25) D&quot;</p> <hr /> <div>1) D<br /> <br /> 2) D<br /> <br /> 3). E <br /> <br /> 4) E<br /> <br /> 5) C<br /> <br /> 6) B<br /> <br /> 7) A<br /> <br /> 8) E<br /> <br /> 9) D<br /> <br /> 10) A<br /> <br /> 11) C<br /> <br /> 12) A<br /> <br /> 13) D<br /> <br /> 14) C<br /> <br /> 15) A<br /> <br /> 16) E<br /> <br /> 17) B<br /> <br /> 18) D<br /> <br /> 19) D<br /> <br /> 20) A<br /> <br /> 21) D<br /> <br /> 22) E<br /> <br /> 23) D<br /> <br /> 24) B<br /> <br /> 25) D</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_5&diff=56899 2013 AIME I Problems/Problem 5 2013-08-12T17:10:01Z <p>Flamefoxx99: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> The real root of the equation &lt;math&gt;8x^3 - 3x^2 - 3x - 1 = 0&lt;/math&gt; can be written in the form &lt;math&gt;\frac{\sqrta + \sqrtb + 1}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive integers. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> We have that &lt;math&gt;9x^3 = (x+1)^3&lt;/math&gt;, so it follows that &lt;math&gt;\sqrt{9}x = x+1&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;\frac{1}{\sqrt{9}-1} = \frac{\sqrt{81}+\sqrt{9}+1}{8}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;r&lt;/math&gt; be the real root of the given [[polynomial]]. Now define the cubic polynomial &lt;math&gt;Q(x)=-x^3-3x^2-3x+8&lt;/math&gt;. Note that &lt;math&gt;1/r&lt;/math&gt; must be a root of &lt;math&gt;Q&lt;/math&gt;. However we can simplify &lt;math&gt;Q&lt;/math&gt; as &lt;math&gt;Q(x)=9-(x+1)^3&lt;/math&gt;, so we must have that &lt;math&gt;(\frac{1}{r}+1)^3=9&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{r}=\sqrt{9}-1&lt;/math&gt;, and &lt;math&gt;r=\frac{1}{\sqrt{9}-1}&lt;/math&gt;. We can then multiply the numerator and denominator of &lt;math&gt;r&lt;/math&gt; by &lt;math&gt;\sqrt{81}+\sqrt{9}+1&lt;/math&gt; to rationalize the denominator, and we therefore have &lt;math&gt;r=\frac{\sqrt{81}+\sqrt{9}+1}{8}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> It is clear that for the algebraic degree of &lt;math&gt;x&lt;/math&gt; to be &lt;math&gt;3&lt;/math&gt; that there exists some cubefree integer &lt;math&gt;p&lt;/math&gt; and positive integers &lt;math&gt;m,n&lt;/math&gt; such that &lt;math&gt;a = m^3p&lt;/math&gt; and &lt;math&gt;b = n^3p^2&lt;/math&gt; (it is possible that &lt;math&gt;b = n^3p&lt;/math&gt;, but then the problem wouldn't ask for both an &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;). Let &lt;math&gt;f_1&lt;/math&gt; be the [[automorphism]] over &lt;math&gt;\mathbb{Q}[\sqrt{a}][\omega]&lt;/math&gt; which sends &lt;math&gt;\sqrt{a} \to \omega \sqrt{a}&lt;/math&gt; and &lt;math&gt;f_2&lt;/math&gt; which sends &lt;math&gt;\sqrt{a} \to \omega^2 \sqrt{a}&lt;/math&gt; (note : &lt;math&gt;\omega&lt;/math&gt; is a cubic [[Roots of unity|root of unity]]).<br /> <br /> Letting &lt;math&gt;r&lt;/math&gt; be the root, we clearly we have &lt;math&gt;r + f_1(r) + f_2(r) = \frac{3}{8}&lt;/math&gt; by Vieta's. Thus it follows &lt;math&gt;c=8&lt;/math&gt;.<br /> Now, note that &lt;math&gt;\sqrt{a} + \sqrt{b} + 1&lt;/math&gt; is a root of &lt;math&gt;x^3 - 3x^2 - 24x - 64 = 0&lt;/math&gt;. Thus &lt;math&gt;(x-1)^3 = 27x + 63&lt;/math&gt; so &lt;math&gt;(\sqrt{a} + \sqrt{b})^3 = 27(\sqrt{a} + \sqrt{b}) + 90&lt;/math&gt;. Checking the non-cubicroot dimension part, we get &lt;math&gt;a + b = 90&lt;/math&gt; so it follows that &lt;math&gt;a + b + c = \boxed{098}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> We proceed by using the [[cubic formula]].<br /> <br /> Let &lt;math&gt;a=8&lt;/math&gt;, &lt;math&gt;b=-3&lt;/math&gt;, &lt;math&gt;c=-3&lt;/math&gt;, and &lt;math&gt;d=-1&lt;/math&gt;. Then let &lt;math&gt;m=\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)&lt;/math&gt; and &lt;math&gt;n=\left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)&lt;/math&gt;. Then the real root of &lt;math&gt;ax^3+bx^2+cx+d&lt;/math&gt; is<br /> &lt;cmath&gt;\sqrt{m+\sqrt{m^2+n^3}}+\sqrt{m-\sqrt{m^2+n^3}}-\dfrac{b}{3a}&lt;/cmath&gt;<br /> Now note that<br /> &lt;cmath&gt;m=\dfrac{27}{27\cdot 512}+\dfrac{3}{128}+\dfrac{1}{16}=\dfrac{1}{512}+\dfrac{12}{512}+\dfrac{32}{512}=\dfrac{45}{512}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;n=\dfrac{-3}{24}-\dfrac{9}{576}=\dfrac{-9}{64}&lt;/cmath&gt;<br /> Thus<br /> &lt;cmath&gt;r=\sqrt{\dfrac{45}{512}+\sqrt{\dfrac{45^2}{512^2}-\dfrac{9^3}{64^3}}}+\sqrt{\dfrac{45}{512}-\sqrt{\dfrac{45^2}{512^2}-\dfrac{9^3}{64^3}}}+\dfrac{3}{24}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sqrt{\dfrac{45}{512}+\sqrt{\dfrac{45^2 - 729}{2^{18}}}}+\sqrt{\dfrac{45}{512}-\sqrt{\dfrac{45^2 - 729}{2^{18}}}}+\dfrac{1}{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\sqrt{\dfrac{45}{512}+\dfrac{36}{512}}+\sqrt{\dfrac{45}{512}-\dfrac{36}{512}}+\dfrac{1}{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;=\dfrac{\sqrt{81}}{8}+\dfrac{\sqrt{9}}{8}+\dfrac{1}{8}=\dfrac{\sqrt{81}+\sqrt{9}+1}{8}&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;81+9+8=\boxed{098}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2013|n=I|num-b=4|num-a=6}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Thales%27_theorem&diff=56644 Thales' theorem 2013-07-14T22:26:29Z <p>Flamefoxx99: Created page with &quot;Thale's Theorem states that if there are three points on a circle, &lt;math&gt;A,B,C&lt;/math&gt; with &lt;math&gt;AC&lt;/math&gt; being a diameter, &lt;math&gt;\angle ABC=90^{\circ}&lt;/math&gt; &lt;asy&gt; dot((5,0))...&quot;</p> <hr /> <div>Thale's Theorem states that if there are three points on a circle, &lt;math&gt;A,B,C&lt;/math&gt; with &lt;math&gt;AC&lt;/math&gt; being a diameter, &lt;math&gt;\angle ABC=90^{\circ}&lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> dot((5,0));<br /> dot((-5,0));<br /> draw(circle((0,0),5));<br /> dot((3,4));<br /> draw((5,0)--(3,4));<br /> draw((-5,0)--(3,4));<br /> draw((-5,0)--(5,0));<br /> label(&quot;A&quot;,(-5,0),W);<br /> label(&quot;B&quot;,(3,4),NE);<br /> label(&quot;C&quot;,(5,0),E);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> {{stub}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Talk:2006_AMC_12A_Problems/Problem_24&diff=53096 Talk:2006 AMC 12A Problems/Problem 24 2013-06-26T22:10:11Z <p>Flamefoxx99: Created page with &quot;random stuff on bottom of Page-- Define such that . Then the expression in the problem becomes: . Expanding this using binomial theorem gives (we may omit the coefficients, a...&quot;</p> <hr /> <div>random stuff on bottom of Page--<br /> <br /> <br /> Define such that . Then the expression in the problem becomes: .<br /> Expanding this using binomial theorem gives (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).<br /> Simplifying gives: . We can also take out all the 2 and all the x terms, as they will not affect the answer.<br /> Thus, we must find the number of terms in this expression:<br /> .<br /> Because , will have n+1 terms, by the binomial theorem. Thus, the expression will have terms.<br /> We can easily find this sum by noting that this is equal to the sum of the first 1004 consecutive odd integers, or , which gives us , or D.<br /> <br /> <br /> <br /> If you are the writer, fill in the LaTeX. Then put it in as another solution</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_24&diff=53095 2006 AMC 12A Problems/Problem 24 2013-06-26T22:09:13Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>== Problem ==<br /> <br /> The expression <br /> <br /> &lt;math&gt;(x+y+z)^{2006}+(x-y-z)^{2006}&lt;/math&gt;<br /> <br /> is simplified by expanding it and combining like terms. How many terms are in the simplified expression?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028&lt;/math&gt;<br /> <br /> == Solution 1==<br /> By the [[Multinomial Theorem]], the summands can be written as<br /> <br /> &lt;cmath&gt;\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},&lt;/cmath&gt;<br /> <br /> respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:<br /> <br /> &lt;cmath&gt;{2006+2\choose 2} = 2015028&lt;/cmath&gt;<br /> <br /> terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; must be opposite. Now we find a pattern:<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 1, the exponent of &lt;math&gt;z&lt;/math&gt; can be all even integers up to 2004, so there are 1003 terms.<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 3, the exponent of &lt;math&gt;z&lt;/math&gt; can go up to 2002, so there are 1002 terms.<br /> <br /> &lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> if the exponent of &lt;math&gt;y&lt;/math&gt; is 2005, then &lt;math&gt;z&lt;/math&gt; can only be 0, so there is 1 term.<br /> <br /> If we add them up, we get &lt;math&gt;\frac{1003\cdot1004}{2}&lt;/math&gt; terms. However, we can switch the exponents of &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; and these terms will still have a negative sign. So there are a total of &lt;math&gt;1003\cdot1004&lt;/math&gt; negative terms.<br /> <br /> By subtracting this number from 2015028, we obtain &lt;math&gt;\boxed{D}&lt;/math&gt; or &lt;math&gt;1008016&lt;/math&gt; as our answer.<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we can use a generating function to solve this problem. <br /> The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of &lt;math&gt;k&lt;/math&gt;). In other words, we want to find &lt;math&gt;f(x)&lt;/math&gt; where the coefficient of &lt;math&gt;x^k&lt;/math&gt; equals the number of unique terms in &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt;.<br /> <br /> <br /> First, we note that all unique terms in the expression have the form, &lt;math&gt;Cx^ay^bz^c&lt;/math&gt;, where &lt;math&gt;a+b+c=k&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt; is<br /> &lt;cmath&gt;(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}&lt;/cmath&gt;<br /> <br /> <br /> Secondly, we note that a certain number of terms of the form, &lt;math&gt;Cx^ay^bz^c&lt;/math&gt;, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when &lt;math&gt;1 \equiv b+c\text{ (mod }2\text{)}&lt;/math&gt; because every unique term is of the form:<br /> &lt;cmath&gt;\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c&lt;/cmath&gt;<br /> for all possible &lt;math&gt;a,b,c&lt;/math&gt;.<br /> <br /> <br /> Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of &lt;math&gt;k&lt;/math&gt;. With some thought, we see that this desired generating function is the following:<br /> &lt;cmath&gt;2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}&lt;/cmath&gt;<br /> <br /> <br /> Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in &lt;math&gt;(x+y+z)^k + (x-y-z)^k&lt;/math&gt;, our initial goal:<br /> &lt;cmath&gt;\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2&lt;/cmath&gt;<br /> <br /> <br /> The coefficient of &lt;math&gt;x^{2006}&lt;/math&gt; of the above expression equals<br /> &lt;cmath&gt;\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a&lt;/cmath&gt;<br /> <br /> <br /> Evaluating the expression, we get &lt;math&gt;1008016&lt;/math&gt;, as expected.<br /> <br /> == Solution 3 ==<br /> <br /> Define &lt;math&gt;P&lt;/math&gt; such that &lt;math&gt;P=y+z&lt;/math&gt;. Then the expression in the problem becomes:<br /> &lt;math&gt;(x+P)^{2006}+(x-P)^{2006}&lt;/math&gt;. <br /> <br /> Expanding this using binomial theorem gives &lt;math&gt;x^n+P*x^{n-1}+...+P^{n-1}*x+P^n+x^n-P*x^{n-1}+...-P^{n-1}*x+P^n&lt;/math&gt; (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves). <br /> <br /> Simplifying gives: &lt;math&gt;2(x^n+x^{n-2}*P^2+...+x^2*P^{n-2}+x^n)&lt;/math&gt;. We can also take out all the 2 and all the x terms, as they will not affect the answer.<br /> <br /> Thus, we must find the number of terms in this expression:<br /> <br /> &lt;math&gt;1+P^2+P^4+...+P^{2004}+P^{2006}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;P=y+z&lt;/math&gt;, &lt;math&gt;P^n&lt;/math&gt; will have n+1 terms, by the binomial theorem. Thus, the expression will have &lt;math&gt;1+3+5+...+2005+2007&lt;/math&gt; terms. <br /> <br /> We can easily find this sum by noting that this is equal to the sum of the first 1004 consecutive odd integers, or &lt;math&gt;1004^2&lt;/math&gt;, which gives us &lt;math&gt;1,008,016&lt;/math&gt;, or D.<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> [[Category:Introductory Combinatorics Problems]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Factoring&diff=53093 Factoring 2013-06-26T16:00:20Z <p>Flamefoxx99: /* Differences and Sums of Powers */</p> <hr /> <div>'''Factoring''' is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an [[integer]] or [[polynomial]]) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.<br /> <br /> ==Differences and Sums of Powers==<br /> <br /> Using the formula for the sum of a [[geometric sequence]], it's easy to derive the general formula for difference of powers:<br /> <br /> &lt;cmath&gt;a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})&lt;/cmath&gt;<br /> <br /> If &lt;math&gt;n=2&lt;/math&gt;, this creates the difference of squares factorization, &lt;cmath&gt;a^2-b^2=(a+b)(a-b)&lt;/cmath&gt;<br /> <br /> This leads to the difference of cubes factorization, &lt;cmath&gt;a^3-b^3=(a-b)(a^2+ab+b^2)&lt;/cmath&gt;<br /> <br /> In addition, if &lt;math&gt;n&lt;/math&gt; is [[odd integer | odd]]:<br /> <br /> &lt;cmath&gt;a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})&lt;/cmath&gt;<br /> <br /> This also leads to the formula for the sum of cubes,<br /> <br /> &lt;cmath&gt;a^3+b^3=(a+b)(a^2-ab+b^2)&lt;/cmath&gt;<br /> <br /> Another way to discover these factorizations is the following: the expression &lt;math&gt;a^n - b^n&lt;/math&gt; is equal to zero if &lt;math&gt;a = b&lt;/math&gt;. If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so &lt;math&gt;a^n - b^n&lt;/math&gt; must have a factor of &lt;math&gt;a - b&lt;/math&gt;. Similarly, we note that the expression &lt;math&gt;a^n + b^n&lt;/math&gt; when &lt;math&gt;n&lt;/math&gt; is odd is equal to zero if &lt;math&gt;a = -b&lt;/math&gt;, so it must have a factor of &lt;math&gt;a - (-b) = a + b&lt;/math&gt;. Note that when &lt;math&gt;n&lt;/math&gt; is [[even integer | even]], &lt;math&gt;(-b)^n + b^n = 2b^n&lt;/math&gt;, rather than 0, so this gives us no useful information.<br /> <br /> == Vieta's/Newton Factorizations ==<br /> These factorizations are useful for problems that could otherwise be solved by [[Newton sums]] or problems that give a polynomial and ask a question about the roots. Combined with [[Vieta's formulas]], these are excellent factorizations that show up everywhere.<br /> <br /> *&lt;math&gt;(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)&lt;/math&gt;<br /> <br /> *&lt;math&gt;(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)&lt;/math&gt;<br /> <br /> *&lt;math&gt;(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) &lt;/math&gt;<br /> == Other Useful Factorizations ==<br /> *&lt;math&gt;a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)&lt;/math&gt;<br /> * [[Binomial theorem]]<br /> * [[Simon's Favorite Factoring Trick]]<br /> * [[Sophie Germain Identity]]<br /> <br /> == Practice Problems ==<br /> * Prove that &lt;math&gt;n^2 + 3n + 5&lt;/math&gt; is never divisible by 121 for any positive integer &lt;math&gt;{n}&lt;/math&gt;.<br /> * Prove that &lt;math&gt;2222^{5555} + 5555^{2222}&lt;/math&gt; is divisible by 7. - USSR Problem Book<br /> * Factor &lt;math&gt;(x-y)^3 + (y-z)^3 + (z-x)^3&lt;/math&gt;.<br /> * Factor &lt;math&gt;x^4 + 1&lt;/math&gt; into two polynomials with real coefficients.<br /> * Given that &lt;math&gt;a+b+c=0&lt;/math&gt;, prove that &lt;math&gt;abc=\dfrac{a^3+b^3+c^3}{3}&lt;/math&gt;.<br /> <br /> == Other Resources ==<br /> * [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Common Factorizations].<br /> <br /> [[Category:Definition]]<br /> [[Category:Elementary algebra]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Factoring&diff=53092 Factoring 2013-06-26T15:59:52Z <p>Flamefoxx99: /* Differences and Sums of Powers */</p> <hr /> <div>'''Factoring''' is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an [[integer]] or [[polynomial]]) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.<br /> <br /> ==Differences and Sums of Powers==<br /> <br /> Using the formula for the sum of a [[geometric sequence]], it's easy to derive the more general formula:<br /> <br /> &lt;cmath&gt;a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})&lt;/cmath&gt;<br /> <br /> If &lt;math&gt;n=2&lt;/math&gt;, this creates the difference of squares factorization, &lt;cmath&gt;a^2-b^2=(a+b)(a-b)&lt;/cmath&gt;<br /> <br /> This leads to the difference of cubes factorization, &lt;cmath&gt;a^3-b^3=(a-b)(a^2+ab+b^2)&lt;/cmath&gt;<br /> <br /> In addition, if &lt;math&gt;n&lt;/math&gt; is [[odd integer | odd]]:<br /> <br /> &lt;cmath&gt;a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})&lt;/cmath&gt;<br /> <br /> This also leads to the formula for the sum of cubes,<br /> <br /> &lt;cmath&gt;a^3+b^3=(a+b)(a^2-ab+b^2)&lt;/cmath&gt;<br /> <br /> Another way to discover these factorizations is the following: the expression &lt;math&gt;a^n - b^n&lt;/math&gt; is equal to zero if &lt;math&gt;a = b&lt;/math&gt;. If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so &lt;math&gt;a^n - b^n&lt;/math&gt; must have a factor of &lt;math&gt;a - b&lt;/math&gt;. Similarly, we note that the expression &lt;math&gt;a^n + b^n&lt;/math&gt; when &lt;math&gt;n&lt;/math&gt; is odd is equal to zero if &lt;math&gt;a = -b&lt;/math&gt;, so it must have a factor of &lt;math&gt;a - (-b) = a + b&lt;/math&gt;. Note that when &lt;math&gt;n&lt;/math&gt; is [[even integer | even]], &lt;math&gt;(-b)^n + b^n = 2b^n&lt;/math&gt;, rather than 0, so this gives us no useful information.<br /> <br /> == Vieta's/Newton Factorizations ==<br /> These factorizations are useful for problems that could otherwise be solved by [[Newton sums]] or problems that give a polynomial and ask a question about the roots. Combined with [[Vieta's formulas]], these are excellent factorizations that show up everywhere.<br /> <br /> *&lt;math&gt;(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)&lt;/math&gt;<br /> <br /> *&lt;math&gt;(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)&lt;/math&gt;<br /> <br /> *&lt;math&gt;(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) &lt;/math&gt;<br /> == Other Useful Factorizations ==<br /> *&lt;math&gt;a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)&lt;/math&gt;<br /> * [[Binomial theorem]]<br /> * [[Simon's Favorite Factoring Trick]]<br /> * [[Sophie Germain Identity]]<br /> <br /> == Practice Problems ==<br /> * Prove that &lt;math&gt;n^2 + 3n + 5&lt;/math&gt; is never divisible by 121 for any positive integer &lt;math&gt;{n}&lt;/math&gt;.<br /> * Prove that &lt;math&gt;2222^{5555} + 5555^{2222}&lt;/math&gt; is divisible by 7. - USSR Problem Book<br /> * Factor &lt;math&gt;(x-y)^3 + (y-z)^3 + (z-x)^3&lt;/math&gt;.<br /> * Factor &lt;math&gt;x^4 + 1&lt;/math&gt; into two polynomials with real coefficients.<br /> * Given that &lt;math&gt;a+b+c=0&lt;/math&gt;, prove that &lt;math&gt;abc=\dfrac{a^3+b^3+c^3}{3}&lt;/math&gt;.<br /> <br /> == Other Resources ==<br /> * [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Common Factorizations].<br /> <br /> [[Category:Definition]]<br /> [[Category:Elementary algebra]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Multiset&diff=53091 Multiset 2013-06-26T15:55:44Z <p>Flamefoxx99: </p> <hr /> <div>A '''multiset''' is a slight generalization of the notion of a [[set]]. A set is defined by whether or not each object is an [[element]]. A multiset is defined not just by its elements, but also by how many times each element is contained. In other words, a multiset is a set where duplication of elements is allowed.<br /> <br /> For example, &lt;math&gt;\{1, 1, 2, 3\} \neq \{1, 2, 3\}&lt;/math&gt; as multisets.<br /> <br /> Note that the order of the elements is unimportant, so &lt;math&gt;\{1, 1, 2, 3\}= \{3, 1, 2, 1\}&lt;/math&gt;.<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Excenter&diff=53074 Excenter 2013-06-21T02:05:52Z <p>Flamefoxx99: Created page with &quot;An excenter, denoted &lt;math&gt;J_i&lt;/math&gt;, is the center of an excircle of a triangle. An excircle is a circle tangent to the extensions of two sides and the third side. It is also ...&quot;</p> <hr /> <div>An excenter, denoted &lt;math&gt;J_i&lt;/math&gt;, is the center of an excircle of a triangle. An excircle is a circle tangent to the extensions of two sides and the third side. It is also known as an escribed circle.<br /> <br /> ==Properties of the Excenter==<br /> *It lies on the angle bisector of the angle opposite to it in the triangle.<br /> *In any given triangle, &lt;math&gt;\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}&lt;/math&gt;. &lt;math&gt;r_i&lt;/math&gt; are the radii of the excircles, and &lt;math&gt;r&lt;/math&gt; is the [[inradius]].<br /> <br /> {{stub}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Zero_(constant)&diff=53073 Zero (constant) 2013-06-21T01:53:41Z <p>Flamefoxx99: /* Operations with 0 */</p> <hr /> <div>'''Zero''', or 0, is the name traditionally given to the additive [[identity]] in number systems such as [[abelian group]]s, [[ring]]s and [[field]]s (especially in the particular examples of the [[integer]]s, [[rational number]]s, [[real number]]s and [[complex number]]s).<br /> <br /> The development of a concept and notation for 0, probably in ancient Indian civilization, and its subsequent transmission to Europe via the Persians and Arabs, was fundamental to the success of western mathematics in fields beyond [[geometry]]. It has suprisingly much relevance due to its significance in [[positional number system]]s. For instance, normal commercial interactions might be seriously slowed if cashiers had to make change on a purchase of LXIV dollars with bills marked L, X, V and I when handed XC dollars.<br /> <br /> == Operations with 0 ==<br /> *If you add a number to 0, the sum is that number. For example, &lt;math&gt;3+0=3&lt;/math&gt;.<br /> *If you subtract 0 from a number, the difference is that number. For example, &lt;math&gt;7-0=7&lt;/math&gt;.<br /> *If you subtract a number from 0, the difference is that number's opposite. For example, &lt;math&gt;0-3=-3&lt;/math&gt;.<br /> *If you multiply any amount of numbers by any amount of 0's, the product is 0. For example, &lt;math&gt;3\times 0 \times 6\times 0=0&lt;/math&gt;.<br /> *You cannot divide a number by 0.<br /> *Dividing any number which is not equal to 0 will result in a quotient of 0. For example, &lt;math&gt;0\div 9=0&lt;/math&gt;.<br /> *There is a special case when you try to compute &lt;math&gt;0!&lt;/math&gt;. The result is 1. Find more by clicking the 1 in brackets: [http://www.artofproblemsolving.com/Wiki/index.php/Factorial#Additional_Information].<br /> *&lt;math&gt;0^a=0&lt;/math&gt; for any positive &lt;math&gt;a\in\mathbb R&lt;/math&gt;.<br /> *&lt;math&gt;0^0&lt;/math&gt; is not defined.<br /> *&lt;math&gt;0&lt;/math&gt; divided by &lt;math&gt;0&lt;/math&gt; is indeterminate.<br /> <br /> == See also ==<br /> * [[Algebraic geometry]]<br /> * [[Constant]]<br /> * [[Number theory]]<br /> <br /> [[Category:Constants]]<br /> <br /> {{stub}}</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Fallacious_proof/2equals1&diff=52958 Fallacious proof/2equals1 2013-06-05T12:29:14Z <p>Flamefoxx99: /* Explanation */</p> <hr /> <div>The following proofs are examples of [[fallacious proof]]s, namely that &lt;math&gt;2 = 1&lt;/math&gt;.<br /> <br /> == Proof 1 ==<br /> Let &lt;math&gt;a=b&lt;/math&gt;.<br /> <br /> Then we have<br /> <br /> &lt;math&gt; a^2 = ab &lt;/math&gt; (since &lt;math&gt;a=b&lt;/math&gt;)<br /> <br /> &lt;math&gt; 2a^2 - 2ab = a^2 - ab &lt;/math&gt; (adding &lt;math&gt;a^2-2ab&lt;/math&gt; to both sides)<br /> <br /> &lt;math&gt; 2(a^2 - ab) = a^2 - ab &lt;/math&gt; (factoring out a 2 on the [[LHS]])<br /> <br /> &lt;math&gt; 2 = 1 &lt;/math&gt; (dividing by &lt;math&gt;a^2-ab&lt;/math&gt;)<br /> <br /> === Explanation ===<br /> The trick in this argument is when we divide by &lt;math&gt;a^{2}-ab&lt;/math&gt;. Since &lt;math&gt;a=b&lt;/math&gt;, &lt;math&gt;a^2-ab = 0&lt;/math&gt;, and dividing by [[zero (constant) | zero]] is undefined.<br /> <br /> == Proof 2 ==<br /> &lt;center&gt;<br /> &lt;math&gt;1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots&lt;/math&gt;<br /> <br /> &lt;math&gt;2 = 1&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> === Explanation ===<br /> The given series does not converge. Therefore, manipulations such as grouping terms before adding are invalid.<br /> <br /> <br /> ''[[Fallacious_proof#2_.3D_1 | Back to main article]]''</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Fallacy&diff=52957 Fallacy 2013-06-05T12:27:28Z <p>Flamefoxx99: /* All horses are the same color */</p> <hr /> <div>A '''fallacy''' is a step an attempted [[proof]] that is [[logic]]ally flawed in some way. The fact that a proof is fallacious says nothing about the validity of the original proposition.<br /> <br /> == Common false proofs ==<br /> The fallacious proofs are stated first and then links to the explanations of their fallacies follow.<br /> <br /> === 2 = 1 ===<br /> Let &lt;math&gt;a=b&lt;/math&gt;.<br /> <br /> Then we have<br /> <br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt; a^2 = ab &lt;/math&gt; (since &lt;math&gt;a=b&lt;/math&gt;)&lt;/div&gt;<br /> <br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt; 2a^2 - 2ab = a^2 - ab &lt;/math&gt; (adding &lt;math&gt;a^2-2ab&lt;/math&gt; to both sides)&lt;/div&gt;<br /> <br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt; 2(a^2 - ab) = a^2 - ab &lt;/math&gt; (factoring out a 2 on the [[LHS]])&lt;/div&gt;<br /> <br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt; 2 = 1 &lt;/math&gt; (dividing by &lt;math&gt;a^2-ab&lt;/math&gt;)&lt;/div&gt;<br /> <br /> [[Fallacious proof/2equals1 | Explanation]]<br /> <br /> === Polya's Proof That All horses Are the Same Color ===<br /> We shall prove that all horses are the same color by [[induction]] on the number of horses.<br /> <br /> First we shall show our base case, that all horses in a group of 1 horse have the same color to be true. Of course, there's only 1 horse in the group so certainly our base case holds.<br /> <br /> Now assume that all the horses in any group of &lt;math&gt;k&lt;/math&gt; horses are the same color. This is our inductive assumption.<br /> <br /> Using our inductive assumption, we will now show that all horses in a group of &lt;math&gt;k+1&lt;/math&gt; horses have the same color. Number the horses 1 through &lt;math&gt;k+1&lt;/math&gt;. Horses 1 through &lt;math&gt;k&lt;/math&gt; must be the same color as must horses &lt;math&gt;2&lt;/math&gt; through &lt;math&gt;k+1&lt;/math&gt;. It follows that all of the horses are the same color.<br /> <br /> [[Fallacious proof/all horses are the same color | Explanation]]<br /> <br /> === All numbers are equal ===<br /> Consider arbitrary reals &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and let &lt;math&gt;t&lt;/math&gt; = &lt;math&gt;a + b&lt;/math&gt;. Then<br /> &lt;cmath&gt;a + b = t&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(a + b)(a - b) = t(a - b)&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a^2 - b^2 = ta - tb&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a^2 - ta = b^2 - tb&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a^2 - ta + \dfrac{t^2}{4} = b^2 - tb + \dfrac{t^2}{4}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;\left(a - \dfrac{t}{2}\right)^2 = \left(b - \dfrac{t}{2}\right)^2&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a - \dfrac{t}{2} = b - \dfrac{t}{2}&lt;/cmath&gt;<br /> <br /> <br /> &lt;cmath&gt;a = b&lt;/cmath&gt;<br /> <br /> [[Fallacious proof/All numbers are equal|Explanation]]<br /> === Bread ===<br /> *Nothing is better than fame, happiness and success.<br /> *A few small crumbs of bread are better than nothing.<br /> *Thus, a few small crumbs of bread are better than fame, happiness, and success.<br /> <br /> <br /> [[Fallacious proof/Bread|Explanation]]<br /> <br /> == See also ==<br /> * [[Proof writing]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Fibonacci_sequence&diff=52943 Fibonacci sequence 2013-06-04T15:28:45Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>The '''Fibonacci sequence''' is a [[sequence]] of [[integer]]s in which the first and second terms are both equal to 1 and each subsequent term is the sum of the two preceding it. The first few terms are &lt;math&gt;1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...&lt;/math&gt;.<br /> <br /> ==Recursion==<br /> The Fibonacci sequence can be written [[recursion|recursively]] as &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt; and &lt;math&gt;F_n=F_{n-1}+F_{n-2}&lt;/math&gt; for &lt;math&gt;n \geq 3&lt;/math&gt;. This is the simplest nontrivial example of a [[linear recursion]] with constant coefficients. There is also an explicit formula [[#Binet's formula|below]].<br /> <br /> Readers should be wary: some authors give the Fibonacci sequence with the [[initial condition]]s &lt;math&gt;F_0 = F_1 = 1&lt;/math&gt; (or equivalently &lt;math&gt;F_1 = 1, F_2 = 2&lt;/math&gt;). This change in [[indexing of a sequence | indexing]] does not affect the actual numbers in the sequence, but it does change which member of the sequence is referred to by the symbol &lt;math&gt;F_n&lt;/math&gt; and so also changes the appearance of certain [[identity | identities]] involving the Fibonacci numbers. <br /> <br /> == Running Backwards ==<br /> As with many linear recursions, we can run the Fibonacci sequence backwards by solving its recursion relation for the term of smallest index, in this case &lt;math&gt;F_{n - 2} = F_{n} - F_{n - 1}&lt;/math&gt;. This allows us to compute, for example, that &lt;math&gt;F_0 = F_2 - F_1 = 0&lt;/math&gt;, &lt;math&gt;F_{-1} = 1&lt;/math&gt;, &lt;math&gt;F_{-2} = -2&lt;/math&gt;, and so on. Because these preceding terms are uniquely defined by the recursion, one frequently sees the definition of the Fibonacci sequence given in the form &lt;math&gt;F_0 = 0&lt;/math&gt;, &lt;math&gt;F_1 = 1&lt;/math&gt; and &lt;math&gt;F_n = F_{n - 1} + F_{n - 2}&lt;/math&gt; for &lt;math&gt;n \geq 2&lt;/math&gt;. In general, one can show that &lt;math&gt;F_n = (-1)^{n+1}F_{-n}&lt;/math&gt;.<br /> <br /> == &lt;math&gt;\phi&lt;/math&gt; and Binet's Formula==<br /> {{main|Binet's formula}}<br /> <br /> The ratios &lt;math&gt;\frac{1}{1}&lt;/math&gt;, &lt;math&gt;\frac{2}{1}&lt;/math&gt;, &lt;math&gt;\frac{3}{2}&lt;/math&gt;, &lt;math&gt;\frac{5}{3}&lt;/math&gt;, &lt;math&gt;\frac{8}{5}&lt;/math&gt;, ..., between successive terms of the sequence tend towards the limit &lt;math&gt;\frac{1 + \sqrt{5}}{2}&lt;/math&gt;, a constant often denoted &lt;math&gt;\varphi&lt;/math&gt; (the Greek letter [[phi]], also written &lt;math&gt;\phi&lt;/math&gt;). &lt;math&gt;\varphi&lt;/math&gt; is a solution of the quadratic equation &lt;math&gt;x^2-x-1=0&lt;/math&gt;. The other root is &lt;math&gt;-\varphi^{-1} = \frac{1-\sqrt{5}}{2}&lt;/math&gt;. One possible explanation for this fact is that the Fibonacci numbers are given explicitly by Binet's formula. It is <br /> &lt;math&gt;F_n = \frac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}}&lt;/math&gt;.<br /> (Note that this formula is valid for all integers &lt;math&gt;n&lt;/math&gt;.)<br /> It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre.<br /> <br /> == Identities ==<br /> The most important identity regarding the Fibonacci sequence is its recursive definition, &lt;math&gt;F_{n+1} = F_n + F_{n-1}&lt;/math&gt;. The following identities involving the Fibonacci numbers can be proved by [[induction]].<br /> <br /> *&lt;math&gt;F_0 + F_1 + \cdots + F_{n} = F_{n+2} - 1&lt;/math&gt;<br /> *&lt;math&gt;F_0 - F_1 + F_2 - \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1&lt;/math&gt;<br /> *&lt;math&gt;F_0^2 + F_1^2 + F_2^2 + \cdots + F_n^2 = F_n \cdot F_{n+1}&lt;/math&gt;<br /> *&lt;math&gt;F_{n-1}\cdot F_{n+1} = F_{n}^2 + (-1)^n&lt;/math&gt;<br /> *&lt;math&gt;F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n&lt;/math&gt;<br /> <br /> ==Problems==<br /> === Introductory ===<br /> # The Fibonacci sequence &lt;math&gt;1,1,2,3,5,8,13,21,\ldots &lt;/math&gt; starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten [[digit]]s is the last to appear in the units position of a number in the Fibonacci sequence?&lt;br&gt;&lt;br&gt;&lt;math&gt; \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 4 } \qquad \mathrm{(C) \ 6 } \qquad \mathrm{(D) \ 7 } \qquad \mathrm{(E) \ 9 } &lt;/math&gt;&lt;div style=&quot;text-align:right&quot;&gt;([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])&lt;/div&gt;<br /> # A colony has &lt;math&gt;1&lt;/math&gt; rabbit. A rabbit produces one offspring every month. An offspring rabbit takes one month to grow up. Find a formula for the number of rabbits (including offspring) in the &lt;math&gt;n&lt;/math&gt;th month.<br /> ## How about if the colony starts with &lt;math&gt;a&lt;/math&gt; rabbits and &lt;math&gt;b&lt;/math&gt; offspring? <br /> ## Use this result to prove the identity &lt;math&gt;F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n&lt;/math&gt;.<br /> # Find &lt;math&gt;\gcd(F_n,F_{n+1})&lt;/math&gt;.<br /> # Prove the above [[#Identities|identites]].<br /> <br /> === Intermediate ===<br /> # Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle. &lt;div style=&quot;text-align:right&quot;&gt;(Manhattan Mathematical Olympiad 2004)&lt;/div&gt;<br /> # Except for the first two terms, each term of the sequence &lt;math&gt;1000, x, 1000 - x,\ldots&lt;/math&gt; is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first [[negative]] term encounted. What positive integer &lt;math&gt;x&lt;/math&gt; produces a sequence of maximum length? &lt;div style=&quot;text-align:right&quot;&gt;([[1998 AIME Problems/Problem 8|1998 AIME, Problem 8]])&lt;/div&gt;<br /> # A fair coin is to be tossed &lt;math&gt;10_{}^{}&lt;/math&gt; times. Let &lt;math&gt;i/j^{}_{}&lt;/math&gt;, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find &lt;math&gt;i+j_{}^{}&lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1990 AIME Problems/Problem 9|1990 AIME, Problem 9]])&lt;/div&gt;<br /> #Find &lt;math&gt;a&lt;/math&gt; if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are [[integer]]s such that &lt;math&gt;x^2 - x - 1&lt;/math&gt; is a factor of &lt;math&gt;ax^{17} + bx^{16} + 1&lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1998 AIME Problems/Problem 13|1998 AIME, Problem 13]])&lt;/div&gt;<br /> <br /> === Olympiad ===<br /> # Determine the maximum value of &lt;math&gt;m^2 + n^2 &lt;/math&gt;, where &lt;math&gt;m &lt;/math&gt; and &lt;math&gt;n &lt;/math&gt; are integers satisfying &lt;math&gt; m, n \in \{ 1,2, \ldots , 1981 \} &lt;/math&gt; and &lt;math&gt;( n^2 - mn - m^2 )^2 = 1 &lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1981 IMO Problems/Problem 3|1981 IMO, Problem 3]])&lt;/div&gt;<br /> <br /> ==See also==<br /> * [[Combinatorics]]<br /> * [[Lucas Numbers]]<br /> <br /> ==External Links==<br /> * [http://www.scriptspedia.org/Fibonacci_Numbers Fibonacci Algorithm implementation in C++, Java and Javascript]<br /> [[Category:Combinatorics]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Fibonacci_sequence&diff=52942 Fibonacci sequence 2013-06-04T15:27:38Z <p>Flamefoxx99: /* See also */</p> <hr /> <div>The '''Fibonacci sequence''' is a [[sequence]] of [[integer]]s in which the first and second terms are both equal to 1 and each subsequent term is the sum of the two preceding it. The first few terms are &lt;math&gt;1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...&lt;/math&gt;.<br /> <br /> ==Recursion==<br /> The Fibonacci sequence can be written [[recursion|recursively]] as &lt;math&gt;F_1 = F_2 = 1&lt;/math&gt; and &lt;math&gt;F_n=F_{n-1}+F_{n-2}&lt;/math&gt; for &lt;math&gt;n \geq 3&lt;/math&gt;. This is the simplest nontrivial example of a [[linear recursion]] with constant coefficients. There is also an explicit formula [[#Binet's formula|below]].<br /> <br /> Readers should be wary: some authors give the Fibonacci sequence with the [[initial condition]]s &lt;math&gt;F_0 = F_1 = 1&lt;/math&gt; (or equivalently &lt;math&gt;F_1 = 1, F_2 = 2&lt;/math&gt;). This change in [[indexing of a sequence | indexing]] does not affect the actual numbers in the sequence, but it does change which member of the sequence is referred to by the symbol &lt;math&gt;F_n&lt;/math&gt; and so also changes the appearance of certain [[identity | identities]] involving the Fibonacci numbers. <br /> <br /> == Running Backwards ==<br /> As with many linear recursions, we can run the Fibonacci sequence backwards by solving its recursion relation for the term of smallest index, in this case &lt;math&gt;F_{n - 2} = F_{n} - F_{n - 1}&lt;/math&gt;. This allows us to compute, for example, that &lt;math&gt;F_0 = F_2 - F_1 = 0&lt;/math&gt;, &lt;math&gt;F_{-1} = 1&lt;/math&gt;, &lt;math&gt;F_{-2} = -2&lt;/math&gt;, and so on. Because these preceding terms are uniquely defined by the recursion, one frequently sees the definition of the Fibonacci sequence given in the form &lt;math&gt;F_0 = 0&lt;/math&gt;, &lt;math&gt;F_1 = 1&lt;/math&gt; and &lt;math&gt;F_n = F_{n - 1} + F_{n - 2}&lt;/math&gt; for &lt;math&gt;n \geq 2&lt;/math&gt;. In general, one can show that &lt;math&gt;F_n = (-1)^{n+1}F_{-n}&lt;/math&gt;.<br /> <br /> == &lt;math&gt;\phi&lt;/math&gt; and Binet's Formula==<br /> {{main|Binet's formula}}<br /> <br /> The ratios &lt;math&gt;\frac{1}{1}&lt;/math&gt;, &lt;math&gt;\frac{2}{1}&lt;/math&gt;, &lt;math&gt;\frac{3}{2}&lt;/math&gt;, &lt;math&gt;\frac{5}{3}&lt;/math&gt;, &lt;math&gt;\frac{8}{5}&lt;/math&gt;, ..., between successive terms of the sequence tend towards the limit &lt;math&gt;\frac{1 + \sqrt{5}}{2}&lt;/math&gt;, a constant often denoted &lt;math&gt;\varphi&lt;/math&gt; (the Greek letter [[phi]], also written &lt;math&gt;\phi&lt;/math&gt;). &lt;math&gt;\varphi&lt;/math&gt; is a solution of the quadratic equation &lt;math&gt;x^2-x-1=0&lt;/math&gt;. The other root is &lt;math&gt;-\varphi^{-1} = \frac{1-\sqrt{5}}{2}&lt;/math&gt;. One possible explanation for this fact is that the Fibonacci numbers are given explicitly by Binet's formula. It is <br /> &lt;math&gt;F_n = \frac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}}&lt;/math&gt;.<br /> (Note that this formula is valid for all integers &lt;math&gt;n&lt;/math&gt;.)<br /> It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre.<br /> <br /> == Identities ==<br /> The most important identity regarding the Fibonacci sequence is its recursive definition, &lt;math&gt;F_{n+1} = F_n + F_{n-1}&lt;/math&gt;. The following identities involving the Fibonacci numbers can be proved by [[induction]].<br /> <br /> *&lt;math&gt;F_0 + F_1 + \cdots + F_{n} = F_{n+2} - 1&lt;/math&gt;<br /> *&lt;math&gt;F_0 - F_1 + F_2 - \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1&lt;/math&gt;<br /> *&lt;math&gt;F_0^2 + F_1^2 + F_2^2 + \cdots + F_n^2 = F_n \cdot F_{n+1}&lt;/math&gt;<br /> *&lt;math&gt;F_{n-1}\cdot F_{n+1} = F_{n}^2 + (-1)^n&lt;/math&gt;<br /> *&lt;math&gt;F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n&lt;/math&gt;<br /> <br /> ==Problems==<br /> === Introductory ===<br /> # The Fibonacci sequence &lt;math&gt;1,1,2,3,5,8,13,21,\ldots &lt;/math&gt; starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten [[digit]]s is the last to appear in the units position of a number in the Fibonacci sequence?&lt;br&gt;&lt;br&gt;&lt;math&gt; \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 4 } \qquad \mathrm{(C) \ 6 } \qquad \mathrm{(D) \ 7 } \qquad \mathrm{(E) \ 9 } &lt;/math&gt;&lt;div style=&quot;text-align:right&quot;&gt;([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])&lt;/div&gt;<br /> # A colony has &lt;math&gt;1&lt;/math&gt; rabbit. A rabbit produces one offspring every month. An offspring rabbit takes one month to grow up. Find a formula for the number of rabbits (including offspring) in the &lt;math&gt;n&lt;/math&gt;th month.<br /> ## How about if the colony starts with &lt;math&gt;a&lt;/math&gt; rabbits and &lt;math&gt;b&lt;/math&gt; offspring? <br /> ## Use this result to prove the identity &lt;math&gt;F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n&lt;/math&gt;.<br /> # Find &lt;math&gt;\gcd(F_n,F_{n+1})&lt;/math&gt;.<br /> # Prove the above [[#Identities|identites]].<br /> <br /> === Intermediate ===<br /> # Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle. &lt;div style=&quot;text-align:right&quot;&gt;(Manhattan Mathematical Olympiad 2004)&lt;/div&gt;<br /> # Except for the first two terms, each term of the sequence &lt;math&gt;1000, x, 1000 - x,\ldots&lt;/math&gt; is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first [[negative]] term encounted. What positive integer &lt;math&gt;x&lt;/math&gt; produces a sequence of maximum length? &lt;div style=&quot;text-align:right&quot;&gt;([[1998 AIME Problems/Problem 8|1998 AIME, Problem 8]])&lt;/div&gt;<br /> # A fair coin is to be tossed &lt;math&gt;10_{}^{}&lt;/math&gt; times. Let &lt;math&gt;i/j^{}_{}&lt;/math&gt;, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find &lt;math&gt;i+j_{}^{}&lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1990 AIME Problems/Problem 9|1990 AIME, Problem 9]])&lt;/div&gt;<br /> #Find &lt;math&gt;a&lt;/math&gt; if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are [[integer]]s such that &lt;math&gt;x^2 - x - 1&lt;/math&gt; is a factor of &lt;math&gt;ax^{17} + bx^{16} + 1&lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1998 AIME Problems/Problem 13|1998 AIME, Problem 13]])&lt;/div&gt;<br /> <br /> === Olympiad ===<br /> # Determine the maximum value of &lt;math&gt;m^2 + n^2 &lt;/math&gt;, where &lt;math&gt;m &lt;/math&gt; and &lt;math&gt;n &lt;/math&gt; are integers satisfying &lt;math&gt; m, n \in \{ 1,2, \ldots , 1981 \} &lt;/math&gt; and &lt;math&gt;( n^2 - mn - m^2 )^2 = 1 &lt;/math&gt;. &lt;div style=&quot;text-align:right&quot;&gt;([[1981 IMO Problems/Problem 3|1981 IMO, Problem 3]])&lt;/div&gt;<br /> <br /> ==See also==<br /> * [[Combinatorics]]<br /> * [[Lucas numbers]]<br /> <br /> ==External Links==<br /> * [http://www.scriptspedia.org/Fibonacci_Numbers Fibonacci Algorithm implementation in C++, Java and Javascript]<br /> [[Category:Combinatorics]]</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Talk:Lucas_Numbers&diff=52941 Talk:Lucas Numbers 2013-06-04T15:26:06Z <p>Flamefoxx99: Created page with &quot;still working on it--need to add closed form, uses, problems, examples...&quot;</p> <hr /> <div>still working on it--need to add closed form, uses, problems, examples...</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Lucas_Numbers&diff=52940 Lucas Numbers 2013-06-04T15:25:36Z <p>Flamefoxx99: Sitll working on it--need to add closed form, uses, problems</p> <hr /> <div>Lucas Numbers are a recursive sequence defined as &lt;cmath&gt;L_0=2\\L_1=1\\L_n=L_n-1+L_n-2&lt;/cmath&gt;.</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Euler%27s_inequality&diff=52939 Euler's inequality 2013-06-04T15:23:34Z <p>Flamefoxx99: </p> <hr /> <div><br /> Euler's Inequality states that &lt;cmath&gt;R \ge 2r&lt;/cmath&gt;<br /> <br /> ==Proof==<br /> Let the circumradius be &lt;math&gt;R&lt;/math&gt; and inradius &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the distance between the circumcenter and the incenter. Then &lt;cmath&gt;d=\sqrt{R(R-2r)}&lt;/cmath&gt; From this formula, Euler's Inequality follows as &lt;cmath&gt;d^2=R(R-2r)&lt;/cmath&gt; By the [[Trivial Inequality]], &lt;math&gt;R(R-2r)&lt;/math&gt; is positive. Since &lt;math&gt;R&lt;/math&gt; has to be positive as it is the circumradius, &lt;cmath&gt;R-2r \ge 0\\R \ge 2r&lt;/cmath&gt; as desired</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Euler%27s_inequality&diff=52938 Euler's inequality 2013-06-04T15:19:51Z <p>Flamefoxx99: /* Proof */</p> <hr /> <div>==Euler's Inequality==<br /> <br /> Euler's Inequality states that &lt;cmath&gt;R \ge 2r&lt;/cmath&gt;<br /> <br /> ==Proof==<br /> Let the circumradius be &lt;math&gt;R&lt;/math&gt; and inradius &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the distance between the circumcenter and the incenter. Then &lt;cmath&gt;d=\sqrt{R(R-2r)}&lt;/cmath&gt; From this formula, Euler's Inequality follows as &lt;cmath&gt;d^2=R(R-2r)&lt;/cmath&gt; By the [[Trivial Inequality]], &lt;math&gt;R(R-2r)&lt;/math&gt; is positive. Since &lt;math&gt;R&lt;/math&gt; has to be positive as it is the circumradius, &lt;cmath&gt;R-2r \ge 0\\R \ge 2r&lt;/cmath&gt; as desired</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Euler%27s_inequality&diff=52937 Euler's inequality 2013-06-04T15:13:29Z <p>Flamefoxx99: /* Euler's Inequality */</p> <hr /> <div>==Euler's Inequality==<br /> <br /> Euler's Inequality states that &lt;cmath&gt;R \ge 2r&lt;/cmath&gt;<br /> <br /> ==Proof==<br /> Let the circumradius be &lt;math&gt;R&lt;/math&gt; and inradius &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the distance between the circumcenter and the incenter. Then &lt;cmath&gt;d=\sqrt{R(R-2r)}&lt;/cmath&gt;. From this formula, Euler's Inequality follows</div> Flamefoxx99 https://artofproblemsolving.com/wiki/index.php?title=Euler%27s_inequality&diff=52936 Euler's inequality 2013-06-04T15:13:04Z <p>Flamefoxx99: Created page with &quot;==Euler's Inequality== Euler's Inequality states that &lt;cmath&gt;R \gt 2r&lt;/cmath&gt; ==Proof== Let the circumradius be &lt;math&gt;R&lt;/math&gt; and inradius &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be...&quot;</p> <hr /> <div>==Euler's Inequality==<br /> <br /> Euler's Inequality states that &lt;cmath&gt;R \gt 2r&lt;/cmath&gt;<br /> ==Proof==<br /> Let the circumradius be &lt;math&gt;R&lt;/math&gt; and inradius &lt;math&gt;r&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the distance between the circumcenter and the incenter. Then &lt;cmath&gt;d=\sqrt{R(R-2r)}&lt;/cmath&gt;. From this formula, Euler's Inequality follows</div> Flamefoxx99