https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Temperal&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-31T18:32:10Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_2&diff=31480 2004 USAMO Problems/Problem 2 2009-04-26T23:12:04Z <p>Temperal: patch</p> <hr /> <div>==Problem==<br /> <br /> Suppose &lt;math&gt;a_1, \dots, a_n&lt;/math&gt; are integers whose greatest common divisor is 1. Let &lt;math&gt;S&lt;/math&gt; be a set of integers with the following properties:<br /> <br /> (a) For &lt;math&gt;i = 1, \dots, n&lt;/math&gt;, &lt;math&gt;a_i \in S&lt;/math&gt;.<br /> (b) For &lt;math&gt;i,j = 1, \dots, n&lt;/math&gt; (not necessarily distinct), &lt;math&gt;a_i - a_j \in S&lt;/math&gt;.<br /> (c) For any integers &lt;math&gt;x,y \in S&lt;/math&gt;, if &lt;math&gt;x + y \in S&lt;/math&gt;, then &lt;math&gt;x - y \in S&lt;/math&gt;.<br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; must be equal to the set of all integers.<br /> <br /> ==Solution==<br /> Suppose &lt;math&gt;a_i&lt;/math&gt; has only one element; then for the greatest common divisor to be 1, &lt;math&gt;1&lt;/math&gt; has to be the sole element. Then &lt;math&gt;1&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (a), &lt;math&gt;0&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (b), &lt;math&gt;0 + 1 = 1\in S\Rightarrow 0 - 1 = - 1\in S&lt;/math&gt; by (c), and we can apply (c) analogously to get that &lt;math&gt;n\cdot 1 \in S&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; and hence &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> [b]Lemma:[/b] If &lt;math&gt;x,y\in a_i&lt;/math&gt;, then &lt;math&gt;ax + by\in S&lt;/math&gt; for integers &lt;math&gt;a,b&lt;/math&gt;.<br /> [b]Proof:[/b] Assume &lt;math&gt;a_i&lt;/math&gt; has at least two elements; &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. By (b), &lt;math&gt;x - y&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;, and by the application of (c) above, we get that &lt;math&gt;n(x - y)&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;. Then apply (c) to &lt;math&gt;n(x - y)&lt;/math&gt; and &lt;math&gt;ny&lt;/math&gt; or &lt;math&gt;nx&lt;/math&gt; to get that &lt;math&gt;ax + by\in S&lt;/math&gt; for all &lt;math&gt;a,b\in \mathbb{Z}&lt;/math&gt;.<br /> <br /> Now let the terms be &lt;math&gt;a_1,a_2,\ldots,a_{n}&lt;/math&gt;. By applying our lemma many times, all numbers in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; for a sequence of integers &lt;math&gt;c_i&lt;/math&gt; are attainable if the sequence is of a length which is a power of 2. If not, we &quot;pad&quot; the sequence with many copies of an existing element of the sequence until it does have a length which is a power of 2 - it is apparent that this will not change &lt;math&gt;S&lt;/math&gt;.<br /> <br /> By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer &lt;math&gt;n&lt;/math&gt; is attainable, and hence there are two members of &lt;math&gt;S&lt;/math&gt; in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in &lt;math&gt;S&lt;/math&gt;, and hence so is their difference; &lt;math&gt;1&lt;/math&gt;. Thus, by the argument at the beginning at this proof, &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=1|num-a=3}}<br /> <br /> * &lt;url&gt;viewtopic.php?p=17440 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_2&diff=31479 2004 USAMO Problems/Problem 2 2009-04-26T22:37:49Z <p>Temperal: fix</p> <hr /> <div>==Problem==<br /> <br /> Suppose &lt;math&gt;a_1, \dots, a_n&lt;/math&gt; are integers whose greatest common divisor is 1. Let &lt;math&gt;S&lt;/math&gt; be a set of integers with the following properties:<br /> <br /> (a) For &lt;math&gt;i = 1, \dots, n&lt;/math&gt;, &lt;math&gt;a_i \in S&lt;/math&gt;.<br /> (b) For &lt;math&gt;i,j = 1, \dots, n&lt;/math&gt; (not necessarily distinct), &lt;math&gt;a_i - a_j \in S&lt;/math&gt;.<br /> (c) For any integers &lt;math&gt;x,y \in S&lt;/math&gt;, if &lt;math&gt;x + y \in S&lt;/math&gt;, then &lt;math&gt;x - y \in S&lt;/math&gt;.<br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; must be equal to the set of all integers.<br /> <br /> ==Solution==<br /> Suppose &lt;math&gt;a_i&lt;/math&gt; has only one element; then for the greatest common divisor to be 1, &lt;math&gt;1&lt;/math&gt; has to be the sole element. Then &lt;math&gt;1&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (a), &lt;math&gt;0&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (b), &lt;math&gt;0 + 1 = 1\in S\Rightarrow 0 - 1 = - 1\in S&lt;/math&gt; by (c), and we can apply (c) analogously to get that &lt;math&gt;n\cdot 1 \in S&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; and hence &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> [b]Lemma:[/b] If &lt;math&gt;x,y\in a_i&lt;/math&gt;, then &lt;math&gt;ax + by\in S&lt;/math&gt; for integers &lt;math&gt;a,b&lt;/math&gt;.<br /> [b]Proof:[/b] Assume &lt;math&gt;a_i&lt;/math&gt; has at least two elements; &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. By (b), &lt;math&gt;x - y&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;, and by the application of (c) above, we get that &lt;math&gt;n(x - y)&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;. Then apply (c) to &lt;math&gt;n(x - y)&lt;/math&gt; and &lt;math&gt;ny&lt;/math&gt; or &lt;math&gt;nx&lt;/math&gt; to get that &lt;math&gt;ax + by\in S&lt;/math&gt; for all &lt;math&gt;a,b\in \mathbb{Z}&lt;/math&gt;.<br /> <br /> Now let the terms be &lt;math&gt;a_1,a_2,\ldots,a_{n}&lt;/math&gt;. By applying our lemma many times, all numbers in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; for a sequence of integers &lt;math&gt;c_i&lt;/math&gt; are attainable. <br /> <br /> By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer &lt;math&gt;n&lt;/math&gt; is attainable, and hence there are two members of &lt;math&gt;S&lt;/math&gt; in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in &lt;math&gt;S&lt;/math&gt;, and hence so is their difference; &lt;math&gt;1&lt;/math&gt;. Thus, by the argument at the beginning at this proof, &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=1|num-a=3}}<br /> <br /> * &lt;url&gt;viewtopic.php?p=17440 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_2&diff=31478 2004 USAMO Problems/Problem 2 2009-04-26T22:36:54Z <p>Temperal: own solution</p> <hr /> <div>==Problem==<br /> <br /> Suppose &lt;math&gt;a_1, \dots, a_n&lt;/math&gt; are integers whose greatest common divisor is 1. Let &lt;math&gt;S&lt;/math&gt; be a set of integers with the following properties:<br /> <br /> (a) For &lt;math&gt;i = 1, \dots, n&lt;/math&gt;, &lt;math&gt;a_i \in S&lt;/math&gt;.<br /> (b) For &lt;math&gt;i,j = 1, \dots, n&lt;/math&gt; (not necessarily distinct), &lt;math&gt;a_i - a_j \in S&lt;/math&gt;.<br /> (c) For any integers &lt;math&gt;x,y \in S&lt;/math&gt;, if &lt;math&gt;x + y \in S&lt;/math&gt;, then &lt;math&gt;x - y \in S&lt;/math&gt;.<br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; must be equal to the set of all integers.<br /> <br /> ==Solution==<br /> Suppose &lt;math&gt;a_i&lt;/math&gt; has only one element; then for the greatest common divisor to be 1, &lt;math&gt;1&lt;/math&gt; has to be the sole element. Then &lt;math&gt;1&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (a), &lt;math&gt;0&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; by (b), &lt;math&gt;0 + 1 = 1\in S\Rightarrow 0 - 1 = - 1\in S&lt;/math&gt; by (c), and we can apply (c) analogously to get that &lt;math&gt;n\cdot 1 \in S&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; and hence &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> [b]Lemma:[/b] If &lt;math&gt;x,y\in a_i&lt;/math&gt;, then &lt;math&gt;ax + by\in S&lt;/math&gt; for integers &lt;math&gt;a,b&lt;/math&gt;.<br /> [b]Proof:[/b] Assume &lt;math&gt;a_i&lt;/math&gt; has at least two elements; &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. By (b), &lt;math&gt;x - y&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;, and by the application of (c) above, we get that &lt;math&gt;n(x - y)&lt;/math&gt; for integers &lt;math&gt;n&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;. Then apply (c) to &lt;math&gt;n(x - y)&lt;/math&gt; and &lt;math&gt;ny&lt;/math&gt; or &lt;math&gt;nx&lt;/math&gt; to get that &lt;math&gt;ax + by\in S&lt;/math&gt; for all &lt;math&gt;a,b\in \mathbb{Z}&lt;/math&gt;.<br /> <br /> Now let the terms be &lt;math&gt;a_1,a_2,\ldots,a_{n}&lt;/math&gt;. By applying our lemma many times, all numbers in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; for a sequence of integers &lt;math&gt;c_i&lt;/math&gt; are attainable. <br /> <br /> By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer &lt;math&gt;n&lt;/math&gt; is attainable, and hence there are two members of &lt;math&gt;S&lt;/math&gt; in the form &lt;math&gt;\sum c_1a_1&lt;/math&gt; which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in &lt;math&gt;S&lt;/math&gt;, and hence so is their difference; &lt;math&gt;1&lt;/math&gt;. Thus, by the argument at the beginning at this proof, &lt;math&gt;S&lt;/math&gt; is the set of all integers, as desired.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=3|num-a=5}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=1478389 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_1&diff=31477 2004 USAMO Problems/Problem 1 2009-04-26T22:36:13Z <p>Temperal: fix</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that<br /> <br /> &lt;cmath&gt;\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.&lt;/cmath&gt;<br /> <br /> When does equality hold?<br /> <br /> ==Solution==<br /> By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence &lt;math&gt;a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|&lt;/math&gt; Now we factor the desired expression into &lt;math&gt;\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)&lt;/math&gt;. Temporarily discarding the case where &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, we can divide through by the &lt;math&gt;|a - b| = |d - c|&lt;/math&gt; to get the simplified expression &lt;math&gt;(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)&lt;/math&gt;.<br /> <br /> Now, draw diagonal &lt;math&gt;BD&lt;/math&gt;. By the law of cosines, &lt;math&gt;c^2 + d^2 + 2cd\cos A = BD&lt;/math&gt;. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that &lt;math&gt;A\in [60^{\circ},120^{\circ}]&lt;/math&gt;. Cosine is monotonously decreasing on this interval, so by setting &lt;math&gt;A&lt;/math&gt; at the extreme values, we see that &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd&lt;/math&gt;. Applying the law of cosines analogously to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt;; we hence have &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt; and &lt;math&gt;a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd&lt;/math&gt;. <br /> <br /> We wrap up first by considering the second inequality. Because &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt;, &lt;math&gt;\text{RHS}\ge 3(a^2 + b^2 - ab)&lt;/math&gt;. This latter expression is of course greater than or equal to &lt;math&gt;a^2 + b^2 + ab&lt;/math&gt; because the inequality can be rearranged to &lt;math&gt;2(a - b)^2\ge 0&lt;/math&gt;, which is always true. Multiply the first inequality by &lt;math&gt;3&lt;/math&gt; and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.<br /> <br /> Equality occurs when &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, or when &lt;math&gt;ABCD&lt;/math&gt; is a kite.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|before=First problem|num-a=2}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=1478389 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_1&diff=31448 2004 USAMO Problems/Problem 1 2009-04-25T15:24:46Z <p>Temperal: fix</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that<br /> <br /> &lt;cmath&gt;\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.&lt;/cmath&gt;<br /> <br /> When does equality hold?<br /> <br /> ==Solution==<br /> By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence &lt;math&gt;a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|&lt;/math&gt; Now we factor the desired expression into &lt;math&gt;\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)&lt;/math&gt;. Temporarily discarding the case where &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, we can divide through by the &lt;math&gt;|a - b| = |d - c|&lt;/math&gt; to get the simplified expression &lt;math&gt;(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)&lt;/math&gt;.<br /> <br /> Now, draw diagonal &lt;math&gt;BD&lt;/math&gt;. By the law of cosines, &lt;math&gt;c^2 + d^2 + 2cd\cos A = BD&lt;/math&gt;. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that &lt;math&gt;A\in [60^{\circ},120^{\circ}]&lt;/math&gt;. Cosine is monotonously decreasing on this interval, so by setting &lt;math&gt;A&lt;/math&gt; at the extreme values, we see that &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le c^2 + d^2 + cd&lt;/math&gt;. Applying the law of cosines analogously to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;a^2 + b^2 - ab\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt;; we hence have &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt; and &lt;math&gt;a^2 + b^2 - ab\le BD^2 \le c^2 + d^2 + cd&lt;/math&gt;. <br /> <br /> We wrap up first by considering the second inequality. Because &lt;math&gt;c^2 + d^2 - cd\le BD^2 \le a^2 + b^2 + ab&lt;/math&gt;, &lt;math&gt;\text{RHS}\ge 3(a^2 + b^2 - ab)&lt;/math&gt;. This latter expression is of course greater than or equal to &lt;math&gt;a^2 + b^2 + ab&lt;/math&gt; because the inequality can be rearranged to &lt;math&gt;2(a - b)^2\ge 0&lt;/math&gt;, which is always true. Multiply the first inequality by &lt;math&gt;3&lt;/math&gt; and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.<br /> <br /> Equality occurs when &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, or when &lt;math&gt;ABCD&lt;/math&gt; is a kite.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=3|num-a=5}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=1478389 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=Main_Page&diff=31447 Main Page 2009-04-25T15:22:14Z <p>Temperal: kookamunga</p> <hr /> <div>__NOTOC__<br /> __NOEDITSECTION__<br /> &lt;div style=&quot;float:right; width:25%; position: relative;&quot;&gt;<br /> {{topics}}<br /> {{statistics}}<br /> <br /> &lt;/div&gt;<br /> {| class=&quot;wikitable&quot; style&quot;border:thin solid gray;background:#eeffe;padding:10px;width:65%&quot;<br /> | style=&quot;border:thin solid gray;padding:10px;&quot;| <br /> Welcome to the AoPSWiki! 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You can also read [[AoPSWiki:What AoPSWiki is not|what AoPSWiki is not]] and [[Art of Problem Solving|about the Art of Problem Solving.]] <br /> <br /> Lastly, the [[AoPSWiki:Policy|policies]] of the AoPSWiki should be read before editing, and [[AoPSWiki:What makes AoPSWiki different|why AoPSWiki is different from other online resources]].<br /> <br /> Also, here are some tasks you can do if you want to help out on the AoPSWiki:<br /> * Write/edit articles.<br /> * Upload an image. ([[Help:Images|Learn how]])<br /> * Add information useful to students of [[problem solving]].<br /> * [[AoPSWiki:Site support|Support the AoPSWiki]].<br /> * Let others know about the AoPSWiki.<br /> <br /> |}<br /> [[Category:AoPSWiki]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_1&diff=31425 2004 USAMO Problems/Problem 1 2009-04-23T21:30:33Z <p>Temperal: fix</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that<br /> <br /> &lt;cmath&gt;\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.&lt;/cmath&gt;<br /> <br /> When does equality hold?<br /> <br /> ==Solution==<br /> By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence &lt;math&gt;a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|&lt;/math&gt; Now we factor the desired expression into &lt;math&gt;\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)&lt;/math&gt;. Temporarily discarding the case where &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, we can divide through by the &lt;math&gt;|a - b| = |d - c|&lt;/math&gt; to get the simplified expression &lt;math&gt;(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)&lt;/math&gt;.<br /> <br /> Now, draw diagonal &lt;math&gt;BD&lt;/math&gt;. By the law of cosines, &lt;math&gt;c^2 + d^2 + 2cd\cos A = BD&lt;/math&gt;. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that &lt;math&gt;A\in [60^{\circ},120^{\circ}]&lt;/math&gt;. Cosine is monotonously decreasing on this interval, so by setting &lt;math&gt;A&lt;/math&gt; at the extreme values, we see that &lt;math&gt;c^2 + d^2 - cd\le BD \le c^2 + d^2 + cd&lt;/math&gt;. Applying the law of cosines analogously to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;a^2 + b^2 - ab\le BD \le a^2 + b^2 + ab&lt;/math&gt;; we hence have &lt;math&gt;c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab&lt;/math&gt; and &lt;math&gt;a^2 + b^2 - ab\le BD \le c^2 + d^2 + cd&lt;/math&gt;. <br /> <br /> We wrap up first by considering the second inequality. Because &lt;math&gt;c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab&lt;/math&gt;, &lt;math&gt;\text{RHS}\ge 3(a^2 + b^2 - ab)&lt;/math&gt;. This latter expression is of course greater than or equal to &lt;math&gt;a^2 + b^2 + ab&lt;/math&gt; because the inequality can be rearranged to &lt;math&gt;2(a - b)^2\ge 0&lt;/math&gt;, which is always true. Multiply the first inequality by &lt;math&gt;3&lt;/math&gt; and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.<br /> <br /> Equality occurs when &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, or when &lt;math&gt;ABCD&lt;/math&gt; is a kite.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=3|num-a=5}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=1478389 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_1&diff=31424 2004 USAMO Problems/Problem 1 2009-04-23T21:30:13Z <p>Temperal: create</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that<br /> $<br /> \frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.<br />$<br /> When does equality hold?<br /> <br /> ==Solution==<br /> By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence &lt;math&gt;a + c = b + d \Rightarrow a - b = d - c \Rightarrow |a - b| = |d - c|&lt;/math&gt; Now we factor the desired expression into &lt;math&gt;\frac {|d - c|(c^2 + d^2 + cd)}{3} \le|a - b|(a^2 + b^2 + ab)\le 3|d - c|(c^2 + d^2 + cd)&lt;/math&gt;. Temporarily discarding the case where &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, we can divide through by the &lt;math&gt;|a - b| = |d - c|&lt;/math&gt; to get the simplified expression &lt;math&gt;(c^2 + d^2 + cd)/3\le a^2 + b^2 + ab\le 3(c^2 + d^2 + cd)&lt;/math&gt;.<br /> <br /> Now, draw diagonal &lt;math&gt;BD&lt;/math&gt;. By the law of cosines, &lt;math&gt;c^2 + d^2 + 2cd\cos A = BD&lt;/math&gt;. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that &lt;math&gt;A\in [60^{\circ},120^{\circ}]&lt;/math&gt;. Cosine is monotonously decreasing on this interval, so by setting &lt;math&gt;A&lt;/math&gt; at the extreme values, we see that &lt;math&gt;c^2 + d^2 - cd\le BD \le c^2 + d^2 + cd&lt;/math&gt;. Applying the law of cosines analogously to &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;a^2 + b^2 - ab\le BD \le a^2 + b^2 + ab&lt;/math&gt;; we hence have &lt;math&gt;c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab&lt;/math&gt; and &lt;math&gt;a^2 + b^2 - ab\le BD \le c^2 + d^2 + cd&lt;/math&gt;. <br /> <br /> We wrap up first by considering the second inequality. Because &lt;math&gt;c^2 + d^2 - cd\le BD \le a^2 + b^2 + ab&lt;/math&gt;, &lt;math&gt;\text{RHS}\ge 3(a^2 + b^2 - ab)&lt;/math&gt;. This latter expression is of course greater than or equal to &lt;math&gt;a^2 + b^2 + ab&lt;/math&gt; because the inequality can be rearranged to &lt;math&gt;2(a - b)^2\ge 0&lt;/math&gt;, which is always true. Multiply the first inequality by &lt;math&gt;3&lt;/math&gt; and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.<br /> <br /> Equality occurs when &lt;math&gt;a = b&lt;/math&gt; and &lt;math&gt;c = d&lt;/math&gt;, or when &lt;math&gt;ABCD&lt;/math&gt; is a kite.<br /> <br /> == Resources ==<br /> {{USAMO newbox|year=2004|num-b=3|num-a=5}}<br /> <br /> * &lt;url&gt;viewtopic.php?t=1478389 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:PI-Dimension&diff=30940 User:PI-Dimension 2009-03-23T01:04:21Z <p>Temperal: bleh</p> <hr /> <div>Hi, I exist to exist.</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=Lebesgue_measure&diff=30499 Lebesgue measure 2009-02-26T23:07:01Z <p>Temperal: a stub for now; will define later</p> <hr /> <div>The '''Lebesgue measure''' is a method of categorizing sets by magnitude devised by French analyst [[Henri Lebesgue]] for his doctoral thesis in the early 20th century.<br /> <br /> {{stub}}<br /> <br /> [[Category:Calculus]]<br /> [[Category:Definitions]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=30434 AoPS Wiki:AoPS Community Awards 2009-02-21T02:26:51Z <p>Temperal: /* Perfect AMC 8 scorers */ way too many aopsers get perfect amc 8 scores to maintain this list</p> <hr /> <div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br /> <br /> <br /> == IMO Participants and Medalists ==<br /> This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br /> <br /> === Participants ===<br /> * Alex Zhai (2008) <br /> * Krishanu Roy Sankar (2008)<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> * Marco Avila (2006)<br /> * Zarathustra Brady (2006)<br /> * Robert Cordwell (2005)<br /> * Sherry Gong (2002, 2003, 2004, 2005, 2007)<br /> * Elyot Grant (2005)<br /> * Darij Grinberg (2006)<br /> * Mahbubul Hasan (2005)<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Viktoriya Krakovna (2006)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br /> * Thomas Mildorf (2005) (AoPS assistant instructor)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Richard Peng (2005, 2006)<br /> * Eric Price (2005)<br /> * David Rhee (2004, 2005, 2006)<br /> * Peng Shi (2004, 2005, 2006)<br /> * Arne Smeets (2003, 2004)<br /> * Arnav Tripathy (2006, 2007)<br /> * [[Naoki Sato]] (AoPS instructor)<br /> * Yi Sun (2006)<br /> * [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * Alex Zhai (2005, 2006, 2007, 2008)<br /> * Yufei Zhao (2004, 2005, 2006)<br /> * Tigran Sloyan(2003,2004,2005,2006,2007)<br /> * Marco Avila (2006)<br /> * Vipul Naik (2003,2004)<br /> * Bhargav Narayanan (2007)<br /> * Tigran Hakobyan (2007)<br /> <br /> ===Perfect Scorers===<br /> *Brian Lawrence (2005)<br /> *Alex Zhai (2008)<br /> <br /> === Gold medalists ===<br /> * Zarathustra Brady (2006)<br /> * Robert Cordwell (2005)<br /> * Darij Grinberg (2006)<br /> * Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2005) ([[WOOT]] instructor)<br /> * Thomas Mildorf (2005) (AoPS assistant instructor)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Arnav Tripathy (2006)<br /> * Eric Price (2005)<br /> * Yufei Zhao (2005)<br /> * Alex Zhai (2007, 2008)<br /> *Sherry Gong (2007)<br /> *Krishanu Sankar (2008)<br /> <br /> === Silver medalists ===<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> * Sherry Gong (2004, 2005)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br /> * Viktoriya Krakovna (2006)<br /> * Hyun Soo Kim (2005) (AoPS assistant instructor)<br /> * Richard Peng (2005)<br /> * David Rhee (2006)<br /> * Naoki Sato (AoPS instructor)<br /> * Peng Shi (2006)<br /> * Arne Smeets (2004)<br /> * Yi Sun (2006)<br /> * [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * Alex Zhai (2006)<br /> * Yufei Zhao (2006)<br /> * Tigran Sloyan(2006,2007)<br /> * Vipul Naik (2003,2004)<br /> <br /> === Bronze medalists ===<br /> * Sherry Gong (2003)<br /> * Elyot Grant (2005)<br /> * Richard Peng (2006)<br /> * [[Naoki Sato]] (AoPS instructor)<br /> * [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br /> * Yufei Zhao (2004)<br /> * Tigran Sloyan(2004;2005)<br /> * Tigran Hakobyan (2007)<br /> <br /> == IPhO Participants and Medalists ==<br /> This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br /> === Participants ===<br /> * Sherry Gong (2006)<br /> * Yi Sun (2004)<br /> * Arnav Tripathy (2006)<br /> <br /> === Gold Medalists ===<br /> * Yi Sun (2004)<br /> * Rahul Singh (2007)<br /> <br /> === Silver Medalists ===<br /> * Sherry Gong (2006)<br /> <br /> == USAMO ==<br /> The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of &quot;winner&quot; has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br /> === Perfect Scorers ===<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2006) ([[WOOT]] instructor)<br /> <br /> === Winners ===<br /> * Yakov Berchenko-Kogan (2006)<br /> * Sherry Gong (2006, 2007)<br /> * Yi Han (2006)<br /> * Adam Hesterberg (2007)<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br /> * Tedrick Leung (2006, 2007)<br /> * Haitao Mao (2007)<br /> * Richard Mccutchen (2006)<br /> * Albert Ni (2005)<br /> * [[David Patrick]] (1988) (AoPS instructor)<br /> * [[Richard Rusczyk]] (1989) (AoPS founder)<br /> * Krishanu Sankar (2007,2008)<br /> * Peng Shi (2006)<br /> * Jacob Steinhardt (2007)<br /> * Yi Sun (2006)<br /> * Arnav Tripathy (2006, 2007)<br /> * [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * Alex Zhai (2006, 2007,2008)<br /> * Yufei Zhao (2006)<br /> <br /> == Putnam Fellows ==<br /> The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br /> * David Ash (1981, 1982, 1983)<br /> * Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br /> * Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br /> * Matthew Ince (2005) (AoPS assistant instructor)<br /> * Alexander Schwartz (2000, 2002)<br /> * Jan Siwanowicz (2001) <br /> * Melanie Wood (2002) ([[WOOT]] instructor)<br /> *Arnav Tripathy (2008)<br /> *Brian Lawrence (2008)<br /> <br /> == Siemens Competition Winners ==<br /> The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br /> * Michael Viscardi (1st Individual,2005)<br /> * Lucia Mocz (2nd Team, 2006)<br /> <br /> ==Intel STS Finalists==<br /> The annual [[Intel Science Talent Search]] is a science competition seeking to find and reward the most scientifically accomplished seniors.<br /> <br /> * Philip Mocz (2008)<br /> * Yihe Dong (2008)<br /> * Qiaochu Yuan (2008)<br /> * Greg Brockman (2007)<br /> <br /> == Clay Junior Fellows ==<br /> Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br /> * Thomas Belulovich (2005) (AoPS assistant instructor)<br /> * Atoshi Chowdhury (2003) (AoPS assistant instructor)<br /> * Robert Cordwell (2005)<br /> * Eve Drucker (2003) (AoPS assistant instructor)<br /> * Matthew Ince (2004) (AoPS assistant instructor)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Hyun Soo Kim (2005) (AoPS assistant instructor)<br /> * Raju Krishnamoorthy (2005)<br /> * Alison Miller (2003) (AoPS assistant instructor)<br /> * Brian Rice (2003) (AoPS assistant instructor)<br /> * Dmitry Taubinski (2005) (AoPS assistant instructor)<br /> * Ameya Velingker (2005)<br /> <br /> <br /> == Perfect AIME Scores ==<br /> Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br /> <br /> * [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br /> * [[Richard Rusczyk]] (1989) (AoPS founder)<br /> * [[Sandor Lehoczky]] (1990) (AoPS author)<br /> * [[Mathew Crawford]] (1992) (AoPS instructor)<br /> * David Benjamin (2006)<br /> * Tedrick Leung (2006)<br /> * Tony Liu (2006)<br /> * Sam Elder (2008)<br /> * Haitao Mao (2008)<br /> <br /> == Perfect AMC Scores ==<br /> === Perfect AMC 12 Scores ===<br /> The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br /> * David Benjamin (2006)<br /> * Zachary Abel (2005) (AoPS assistant instructor)<br /> * Sam Elder (2008)<br /> * Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br /> * Joel Lewis (2003) <br /> * Jonathan Lowd (2003) (AoPS assistant instructor)<br /> * Thomas Mildorf (2004) (AoPS assistant instructor)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Albert Ni (2003) (AoPS instructor)<br /> * Ajay Sharma (2004)<br /> * Arnav Tripathy (2006, 2007)<br /> * Alex Zhai (2007)<br /> * Qiaochu Yuan (2008)<br /> <br /> ==Perfect AMC 10 scorers==<br /> The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br /> * Sergei Bernstein (2007)<br /> * Yifan Cao (2005)<br /> * Kevin Chen (2007, 2008)<br /> * In Young Cho (2007)<br /> * Mario Choi (2007)<br /> * Calvin Deng (2008)<br /> * Billy Dorminy (2007)<br /> * Zhou Fan (2005)<br /> * Albert Gu (2007)<br /> * Robin He (2007)<br /> * Keone Hon (2005)<br /> * Susan Hu (2005)<br /> * Lyndon Ji (2008)<br /> * Sam Keller (2007)<br /> * Vincent Le (2006)<br /> * Daniel Li (2007)<br /> * Johnny Li (2007)<br /> * Patricia Li (2005)<br /> * Carl Lian (2007)<br /> * Thomas Mildorf (2002) (AoPS assistant instructor)<br /> * Anupa Murali (2008)<br /> * Jeffrey Shen (2008)<br /> * Howard Tong (2005)<br /> * Sam Trabucco (2008)<br /> * Brent Woodhouse (2006, 2007)<br /> * Jonathan Zhou (2007)<br /> <br /> === Perfect AHSME Scores ===<br /> The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br /> * Christopher Chang (1994, 1995, 1996)<br /> * [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br /> * [[David Patrick]] (1988) (AoPS instructor)<br /> <br /> == MATHCOUNTS ==<br /> [[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br /> === National Champions ===<br /> * Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br /> * Albert Ni (2002) (AoPS instructor)<br /> * Adam Hesterberg (2003)<br /> * Neal Wu (2005)<br /> * Daesun Yim (2006)<br /> * Kevin Chen (2007)<br /> <br /> === National Top 12 ===<br /> * Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br /> * Andrew Ardito (2005, 2006)<br /> * David Benjamin (2004, 2005)<br /> * Nathan Benjamin (2005, 2006)<br /> * Wenyu Cao (2007)<br /> * Christopher Chang (1991, 1992)<br /> * Kevin Chen (2006, 2007)<br /> * Andrew Chien (2003)<br /> * Peter Chien (2004)<br /> * Mario Choi (2007)<br /> * Joseph Chu (2004)<br /> * [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br /> * Brian Hamrick (2006)<br /> * Adam Hesterberg (2002, 2003)<br /> * Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br /> * Sam Keller (2006)<br /> * Shaunak Kishore (2003, 2004)<br /> * Kiran Kota (2005)<br /> * Brian Lawrence (2003) ([[WOOT]] instructor)<br /> * Karlanna Lewis (2005)<br /> * Daniel Li (2006)<br /> * Patricia Li (2005)<br /> * Poh-Ling Loh (2000)<br /> * Albert Ni (2002) (AoPS assistant instructor)<br /> * Bobby Shen (2008)<br /> * Elizabeth Synge (2007)<br /> * Jason Trigg (2002)<br /> * [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br /> * Neal Wu (2005, 2006)<br /> * Rolland Wu (2006)<br /> * Xiaoyu He (2008)<br /> * Daesun Yim (2006)<br /> * Darren Yin (2002)<br /> * Allen Yuan (2007)<br /> * Alex Zhai (2004)<br /> * Mark Zhang (2005)<br /> <br /> === Masters Round Champions ===<br /> * Christopher Chang (1991)<br /> * Brian Lawrence (2003) ([[WOOT]] instructor)<br /> * Sergei Bernstein (2005)<br /> * Daniel Li (2006)<br /> * Kevin Chen (2007)<br /> * Bobby Shen (2008)<br /> <br /> === National Test Champions ===<br /> * [[Mathew Crawford]] (1990) (AoPS instructor)<br /> * Adam Hesterberg (2003)<br /> * Sergei Bernstein (2005)<br /> * Neal Wu (2006)<br /> * Bobby Shen (2008)<br /> <br /> == Harvard-MIT Math Tournament ==<br /> <br /> The [[HMMT]] 2007 winning team, the &quot;WOOTlings&quot;, consisted entirely of [[WOOT]]ers:<br /> <br /> * Wenyu Cao<br /> * Eric Chang<br /> * Jeremy Hahn<br /> * Alex Kandell<br /> * Adeel Khan<br /> * Sathish Nagappan<br /> * Krishanu Roy Sankar<br /> * Patrick Tenorio<br /> <br /> == ARML ==<br /> <br /> === ARML winners ===<br /> <br /> === ARML Top 10 ===<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> * Seva Tchernov (2007)<br /> * Arnav Tripathy (2007)<br /> * Daesun Yim (2008)<br /> <br /> == See also ==<br /> * [[Academic competitions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Academic scholarships]]<br /> <br /> <br /> <br /> [[Category:Art of Problem Solving]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=Lebesgue_integral&diff=30431 Lebesgue integral 2009-02-21T01:10:00Z <p>Temperal: just scratching this down from memory; correct any errors please</p> <hr /> <div>The '''Lebesgue integral''' was a replacement for the [[Riemann integral]] codified by French analyst [[Henri Lebesgue]] at the turn of the 19th century. Rather than summing the integral by thefunction's domain, as the Riemann integral did, it summed over its range using a concept Lebesgue himself had created - the [[Lebesgue measure]]. <br /> <br /> The Lebesgue integral agrees with the Riemann integral in terms of result on all Riemann integrable functions. It can also integrate all Lebesgue measurable functions, a huge improvement over Riemann's integral. These functions include such pathological functions as [[Dirichlet's function]]. In fact, even if &lt;math&gt;D_f&lt;/math&gt; is an uncountable cardinality, &lt;math&gt;f&lt;/math&gt; is still integrable. Essentially, every function which had been considered up to the 20th century is Lebesgue integrable.<br /> <br /> {{stub}}<br /> <br /> [[Category:Calculus]]<br /> [[Category:Definition]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource5&diff=29870 User:Temperal/The Problem Solver's Resource5 2009-02-01T21:14:15Z <p>Temperal: /* Balls and Urns */ mieh</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 5}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Combinatorics&lt;/span&gt;==<br /> This section cover combinatorics, and some binomial/multinomial facts.<br /> ===Permutations===<br /> The factorial of a number &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n(n-1)(n-2)...(1)&lt;/math&gt; or also as &lt;math&gt;\prod_{a=0}^{n-1}(n-a)&lt;/math&gt;,and is denoted by &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;0!=1&lt;/math&gt;.<br /> <br /> The number of ways of arranging &lt;math&gt;n&lt;/math&gt; ordered distinct objects is &lt;math&gt;n!&lt;/math&gt;. This is also known as a permutation, and can be notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt;. We can see that this is true because there are &lt;math&gt;n&lt;/math&gt; objects which you can place in the first spot; when you've picked one there are &lt;math&gt;n-1&lt;/math&gt; objects to pick from for the second, and so on. <br /> <br /> ===Combinations===<br /> The number of ways of choosing &lt;math&gt;r&lt;/math&gt; objects from a set of &lt;math&gt;n&lt;/math&gt; objects without replacement (i.e. you can't pick an object twice) is &lt;math&gt;\frac{n!}{r!(n-r)!}&lt;/math&gt;, which is notated as either &lt;math&gt;\,_{n}C_{r}&lt;/math&gt; or &lt;math&gt;\binom{n}{r}&lt;/math&gt;. If you allow replacement, then it's notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt; and is given by &lt;math&gt;\frac{n!}{(n-r)!}&lt;/math&gt;. The reader should be able to deduce simple combinatorial arguments for these.<br /> <br /> ===Binomials and Multinomials===<br /> ====Binomial Theorem==== <br /> &lt;math&gt;(x+y)^n=\sum_{r=0}^{n}x^{n-r}y^r&lt;/math&gt;<br /> <br /> ====Multinomial Coefficients====<br /> The number of ways of ordering &lt;math&gt;n&lt;/math&gt; objects when &lt;math&gt;r_1&lt;/math&gt; of them are of one type, &lt;math&gt;r_2&lt;/math&gt; of them are of a second type, ... and &lt;math&gt;r_s&lt;/math&gt; of them of another type so that &lt;math&gt;\sum r_i=n&lt;/math&gt; is &lt;math&gt;\frac{n!}{r_1!r_2!...r_s!}&lt;/math&gt;<br /> <br /> ====Multinomial Theorem====<br /> &lt;math&gt;(x_1+x_2+x_3...+x_s)^n=\sum \frac{n!}{r_1!r_2!...r_s!} x_1+x_2+x_3...+x_s&lt;/math&gt;. The summation is taken over all sequences &lt;math&gt;r_i&lt;/math&gt; so that &lt;math&gt;\sum_{i=1}^{s}r_i=n&lt;/math&gt;.<br /> <br /> ===Balls and Urns===<br /> There are &lt;math&gt;{n-1\choose k-1}&lt;/math&gt; ways to divide &lt;math&gt;k&lt;/math&gt; objects in &lt;math&gt;n&lt;/math&gt; groups such that no group is empty and the objects are indistinguishable. If groups can be empty, then it's &lt;math&gt;\binom{n+k-1}{k-1}&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource4|Back to page 4]] | [[User:Temperal/The Problem Solver's Resource6|Continue to page 6]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource5&diff=29441 User:Temperal/The Problem Solver's Resource5 2009-01-11T03:53:00Z <p>Temperal: notoc</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 5}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Combinatorics&lt;/span&gt;==<br /> This section cover combinatorics, and some binomial/multinomial facts.<br /> ===Permutations===<br /> The factorial of a number &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n(n-1)(n-2)...(1)&lt;/math&gt; or also as &lt;math&gt;\prod_{a=0}^{n-1}(n-a)&lt;/math&gt;,and is denoted by &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;0!=1&lt;/math&gt;.<br /> <br /> The number of ways of arranging &lt;math&gt;n&lt;/math&gt; ordered distinct objects is &lt;math&gt;n!&lt;/math&gt;. This is also known as a permutation, and can be notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt;. We can see that this is true because there are &lt;math&gt;n&lt;/math&gt; objects which you can place in the first spot; when you've picked one there are &lt;math&gt;n-1&lt;/math&gt; objects to pick from for the second, and so on. <br /> <br /> ===Combinations===<br /> The number of ways of choosing &lt;math&gt;r&lt;/math&gt; objects from a set of &lt;math&gt;n&lt;/math&gt; objects without replacement (i.e. you can't pick an object twice) is &lt;math&gt;\frac{n!}{r!(n-r)!}&lt;/math&gt;, which is notated as either &lt;math&gt;\,_{n}C_{r}&lt;/math&gt; or &lt;math&gt;\binom{n}{r}&lt;/math&gt;. If you allow replacement, then it's notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt; and is given by &lt;math&gt;\frac{n!}{(n-r)!}&lt;/math&gt;. The reader should be able to deduce simple combinatorial arguments for these.<br /> <br /> ===Binomials and Multinomials===<br /> ====Binomial Theorem==== <br /> &lt;math&gt;(x+y)^n=\sum_{r=0}^{n}x^{n-r}y^r&lt;/math&gt;<br /> <br /> ====Multinomial Coefficients====<br /> The number of ways of ordering &lt;math&gt;n&lt;/math&gt; objects when &lt;math&gt;r_1&lt;/math&gt; of them are of one type, &lt;math&gt;r_2&lt;/math&gt; of them are of a second type, ... and &lt;math&gt;r_s&lt;/math&gt; of them of another type so that &lt;math&gt;\sum r_i=n&lt;/math&gt; is &lt;math&gt;\frac{n!}{r_1!r_2!...r_s!}&lt;/math&gt;<br /> <br /> ====Multinomial Theorem====<br /> &lt;math&gt;(x_1+x_2+x_3...+x_s)^n=\sum \frac{n!}{r_1!r_2!...r_s!} x_1+x_2+x_3...+x_s&lt;/math&gt;. The summation is taken over all sequences &lt;math&gt;r_i&lt;/math&gt; so that &lt;math&gt;\sum_{i=1}^{s}r_i=n&lt;/math&gt;.<br /> <br /> ===Balls and Urns===<br /> There are &lt;math&gt;{n+k-1\choose n-1}&lt;/math&gt; ways to divide &lt;math&gt;k&lt;/math&gt; objects in &lt;math&gt;n&lt;/math&gt; groups such that no group is empty and the objects are indistinguishable.<br /> <br /> [[User:Temperal/The Problem Solver's Resource4|Back to page 4]] | [[User:Temperal/The Problem Solver's Resource6|Continue to page 6]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource5&diff=29440 User:Temperal/The Problem Solver's Resource5 2009-01-11T03:52:42Z <p>Temperal: improve - still need examples, proofs, other important concepts, etc.</p> <hr /> <div>{{User:Temperal/testtemplate|page 5}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Combinatorics&lt;/span&gt;==<br /> This section cover combinatorics, and some binomial/multinomial facts.<br /> &lt;!-- will fill in later! --&gt;<br /> ===Permutations===<br /> The factorial of a number &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n(n-1)(n-2)...(1)&lt;/math&gt; or also as &lt;math&gt;\prod_{a=0}^{n-1}(n-a)&lt;/math&gt;,and is denoted by &lt;math&gt;n!&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;0!=1&lt;/math&gt;.<br /> <br /> The number of ways of arranging &lt;math&gt;n&lt;/math&gt; ordered distinct objects is &lt;math&gt;n!&lt;/math&gt;. This is also known as a permutation, and can be notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt;. We can see that this is true because there are &lt;math&gt;n&lt;/math&gt; objects which you can place in the first spot; when you've picked one there are &lt;math&gt;n-1&lt;/math&gt; objects to pick from for the second, and so on. <br /> <br /> ===Combinations===<br /> The number of ways of choosing &lt;math&gt;r&lt;/math&gt; objects from a set of &lt;math&gt;n&lt;/math&gt; objects without replacement (i.e. you can't pick an object twice) is &lt;math&gt;\frac{n!}{r!(n-r)!}&lt;/math&gt;, which is notated as either &lt;math&gt;\,_{n}C_{r}&lt;/math&gt; or &lt;math&gt;\binom{n}{r}&lt;/math&gt;. If you allow replacement, then it's notated &lt;math&gt;\,_{n}P_{r}&lt;/math&gt; and is given by &lt;math&gt;\frac{n!}{(n-r)!}&lt;/math&gt;. The reader should be able to deduce simple combinatorial arguments for these.<br /> <br /> ===Binomials and Multinomials===<br /> ====Binomial Theorem==== <br /> &lt;math&gt;(x+y)^n=\sum_{r=0}^{n}x^{n-r}y^r&lt;/math&gt;<br /> <br /> ====Multinomial Coefficients====<br /> The number of ways of ordering &lt;math&gt;n&lt;/math&gt; objects when &lt;math&gt;r_1&lt;/math&gt; of them are of one type, &lt;math&gt;r_2&lt;/math&gt; of them are of a second type, ... and &lt;math&gt;r_s&lt;/math&gt; of them of another type so that &lt;math&gt;\sum r_i=n&lt;/math&gt; is &lt;math&gt;\frac{n!}{r_1!r_2!...r_s!}&lt;/math&gt;<br /> <br /> ====Multinomial Theorem====<br /> &lt;math&gt;(x_1+x_2+x_3...+x_s)^n=\sum \frac{n!}{r_1!r_2!...r_s!} x_1+x_2+x_3...+x_s&lt;/math&gt;. The summation is taken over all sequences &lt;math&gt;r_i&lt;/math&gt; so that &lt;math&gt;\sum_{i=1}^{s}r_i=n&lt;/math&gt;.<br /> <br /> ===Balls and Urns===<br /> There are &lt;math&gt;{n+k-1\choose n-1}&lt;/math&gt; ways to divide &lt;math&gt;k&lt;/math&gt; objects in &lt;math&gt;n&lt;/math&gt; groups such that no group is empty and the objects are indistinguishable.<br /> <br /> [[User:Temperal/The Problem Solver's Resource4|Back to page 4]] | [[User:Temperal/The Problem Solver's Resource6|Continue to page 6]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource3&diff=29434 User:Temperal/The Problem Solver's Resource3 2009-01-11T03:18:52Z <p>Temperal: /* Derivation */ hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 3}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Summations and Products&lt;/span&gt;==<br /> ===Definitions===<br /> *Summations: &lt;math&gt;\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}&lt;/math&gt;<br /> *Products: &lt;math&gt;\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}&lt;/math&gt;<br /> <br /> ===Rules of Summation===<br /> &lt;math&gt;\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i= \frac{n(n+1)}{2}&lt;/math&gt;, and in general &lt;math&gt;\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}&lt;/math&gt;<br /> <br /> The above should all be self-evident and provable by the reader within seconds. <br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}&lt;/math&gt;<br /> <br /> Derivation: We write &lt;math&gt;n^2&lt;/math&gt; as &lt;math&gt;a_1\binom{n}{1}+a_2\binom{n}{2}&lt;/math&gt;. Substituting n=1 gives &lt;math&gt;a_1=1&lt;/math&gt; while substituting n=2 gives &lt;math&gt;a_2=2&lt;/math&gt;. Hence, &lt;math&gt;n^2=\binom{n}{1}+2\binom{n}{2}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\sum_{i=1}^{n} i^2=\sum_{i=1}^n (\binom{i}{1}+2\binom{i}{2})=\sum_{i=1}^n \binom{i}{1}+2\sum_{i=1}^n \binom{i}2=\binom{n+1}{2}+2\binom{n+1}{3}&lt;/math&gt;, where we use the [[Hockey-Stick Identity]]. After some algebra, this comes out to &lt;math&gt;\frac{(n)(n+1)(2n+1)}{6}&lt;/math&gt;. <br /> <br /> This method can be generalized nicely; &lt;math&gt;i^n=\sum_{k=1}^n a_k\binom{i}{k}&lt;/math&gt;.<br /> <br /> Particularly notable is the case &lt;math&gt;n=3&lt;/math&gt;; we get &lt;math&gt;\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2&lt;/math&gt;. The reader can figure this out themselves.<br /> <br /> ===Rules of Products===<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x=x^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}&lt;/math&gt;<br /> <br /> These should be self-evident, as above.<br /> <br /> [[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=Hockey_Stick_Identity&diff=29433 Hockey Stick Identity 2009-01-11T03:18:01Z <p>Temperal: redir</p> <hr /> <div>#REDIRECT [[Combinatorial identity]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=Hockey-Stick_Identity&diff=29432 Hockey-Stick Identity 2009-01-11T03:17:41Z <p>Temperal: redir</p> <hr /> <div>#REDIRECT [[Combinatorial identity]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource11&diff=29427 User:Temperal/The Problem Solver's Resource11 2009-01-11T03:01:13Z <p>Temperal: /* Trivial Inequality */ proof</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 11}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Inequalities&lt;/span&gt;==<br /> My favorite topic, saved for last.<br /> ===Trivial Inequality===<br /> For any real &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;x^2\ge 0&lt;/math&gt;, with equality iff &lt;math&gt;x=0&lt;/math&gt;.<br /> <br /> Proof: We proceed by contradiction. Suppose there exists a real &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;x^2&lt;0&lt;/math&gt;. We can have either &lt;math&gt;x=0&lt;/math&gt;, &lt;math&gt;x&gt;0&lt;/math&gt;, or &lt;math&gt;x&lt;0&lt;/math&gt;. If &lt;math&gt;x=0&lt;/math&gt;, then there is a clear contradiction, as &lt;math&gt;x^2 = 0^2 \not &lt; 0&lt;/math&gt;. If &lt;math&gt;x&gt;0&lt;/math&gt;, then &lt;math&gt;x^2 &lt; 0&lt;/math&gt; gives &lt;math&gt;x &lt; \frac{0}{x} = 0&lt;/math&gt; upon division by &lt;math&gt;x&lt;/math&gt; (which is positive), so this case also leads to a contradiction. Finally, if &lt;math&gt;x&lt;0&lt;/math&gt;, then &lt;math&gt;x^2 &lt; 0&lt;/math&gt; gives &lt;math&gt;x &gt; \frac{0}{x} = 0&lt;/math&gt; upon division by &lt;math&gt;x&lt;/math&gt; (which is negative), and yet again we have a contradiction.<br /> <br /> Therefore, &lt;math&gt;x^2 \ge 0&lt;/math&gt; for all real &lt;math&gt;x&lt;/math&gt;, as claimed.<br /> <br /> ===Arithmetic Mean/Geometric Mean Inequality===<br /> For any set of real numbers &lt;math&gt;S&lt;/math&gt;, &lt;math&gt;\frac{S_1+S_2+S_3....+S_{k-1}+S_k}{k}\ge \sqrt[k]{S_1\cdot S_2 \cdot S_3....\cdot S_{k-1}\cdot S_k}&lt;/math&gt; with equality iff &lt;math&gt;S_1=S_2=S_3...=S_{k-1}=S_k&lt;/math&gt;.<br /> <br /> <br /> ===Cauchy-Schwarz Inequality===<br /> <br /> For any real numbers &lt;math&gt;a_1,a_2,...,a_n&lt;/math&gt; and &lt;math&gt;b_1,b_2,...,b_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;\left(\sum a_i^2\right)\left(\sum b_i^2\right) \ge \left(\sum a_ib_i\right)^2&lt;/math&gt;<br /> <br /> ====Cauchy-Schwarz Variation====<br /> <br /> For any real numbers &lt;math&gt;a_1,a_2,...,a_n&lt;/math&gt; and positive real numbers &lt;math&gt;b_1,b_2,...,b_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;\sum\left({{a_i^2}\over{b_i}}\right) \ge {{\sum a_i^2}\over{\sum b_i}}&lt;/math&gt;.<br /> <br /> ===Power Mean Inequality===<br /> <br /> Take a set of functions &lt;math&gt;m_j(a) = \left({\frac{\sum a_i^j}{n}}\right)^{1/j}&lt;/math&gt;.<br /> <br /> Note that &lt;math&gt;m_0&lt;/math&gt; does not exist. The geometric mean is &lt;math&gt;m_0 = \lim_{k \to 0} m_k&lt;/math&gt;.<br /> For non-negative real numbers &lt;math&gt;a_1,a_2,\ldots,a_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;m_x \le m_y&lt;/math&gt; for reals &lt;math&gt;x&lt;y&lt;/math&gt;.<br /> <br /> , if &lt;math&gt;m_2&lt;/math&gt; is the quadratic mean, &lt;math&gt;m_1&lt;/math&gt; is the arithmetic mean, &lt;math&gt;m_0&lt;/math&gt; the geometric mean, and &lt;math&gt;m_{-1}&lt;/math&gt; the harmonic mean.<br /> <br /> ===RSM-AM-GM-HM Inequality===<br /> For any positive real numbers &lt;math&gt;x_1,\ldots,x_n&lt;/math&gt;:<br /> <br /> &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt;<br /> <br /> with equality iff &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;.<br /> <br /> ===Chebyshev's Inequality===<br /> <br /> Given real numbers &lt;math&gt;a_1 \ge a_2 \ge ... \ge a_n \ge 0&lt;/math&gt; and &lt;math&gt;b_1 \ge b_2 \ge ... \ge b_n&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;{\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}&lt;/math&gt;.<br /> <br /> ===Minkowski's Inequality===<br /> <br /> Given real numbers &lt;math&gt;a_1,a_2,...,a_n&lt;/math&gt; and &lt;math&gt;b_1,b_2,\ldots,b_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;\sqrt{\sum a_i^2} + \sqrt{\sum b_i^2} \ge \sqrt{\sum (a_i+b_i)^2}&lt;/math&gt;<br /> <br /> ===Nesbitt's Inequality===<br /> <br /> For all positive real numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;{\frac{a}{b+c}} + {\frac{b}{c+a}} + {\frac{c}{a+b}} \ge {\frac{3}{2}}&lt;/math&gt;.<br /> <br /> ===Schur's inequality===<br /> <br /> Given positive real numbers &lt;math&gt;a,b,c&lt;/math&gt; and real &lt;math&gt;r&lt;/math&gt;, the following holds:<br /> <br /> &lt;math&gt;a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0&lt;/math&gt;.<br /> <br /> ===Jensen's Inequality===<br /> For a convex function &lt;math&gt;f(x)&lt;/math&gt; and real numbers &lt;math&gt;a_1,a_2,a_3,a_4\ldots,a_n&lt;/math&gt; and &lt;math&gt;x_1,x_2,x_3,x_4\ldots,x_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;cmath&gt;\sum_{i=1}^{n}a_i\cdot f(x_i)\ge f(\sum_{i=1}^{n}a_i\cdot x_i)&lt;/cmath&gt;<br /> <br /> ===Holder's Inequality===<br /> For positive real numbers &lt;math&gt;a_{i_{j}}, 1\le i\le m, 1\le j\le n&lt;/math&gt;, the following holds:<br /> <br /> &lt;cmath&gt;\prod_{i=1}^{m}\left(\sum_{j=1}^{n}a_{i_{j}}\right)\ge\left(\sum_{j=1}^{n}\sqrt[m]{\prod_{i=1}^{m}a_{i_{j}}}\right)^{m}&lt;/cmath&gt;<br /> <br /> ===Muirhead's Inequality===<br /> For a sequence &lt;math&gt;A&lt;/math&gt; that majorizes a sequence &lt;math&gt;B&lt;/math&gt;, then given a set of positive integers &lt;math&gt;x_1,x_2,\ldots,x_n&lt;/math&gt;, the following holds:<br /> <br /> &lt;cmath&gt;\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\ldots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}&lt;/cmath&gt;<br /> ===Rearrangement Inequality===<br /> For any multi sets &lt;math&gt;{a_1,a_2,a_3\ldots,a_n}&lt;/math&gt; and &lt;math&gt;{b_1,b_2,b_3\ldots,b_n}&lt;/math&gt;, &lt;math&gt;a_1b_1+a_2b_2+\ldots+a_nb_n&lt;/math&gt; is maximized when &lt;math&gt;a_k&lt;/math&gt; is greater than or equal to exactly &lt;math&gt;i&lt;/math&gt; of the other members of &lt;math&gt;A&lt;/math&gt;, then &lt;math&gt;b_k&lt;/math&gt; is also greater than or equal to exactly &lt;math&gt;i&lt;/math&gt; of the other members of &lt;math&gt;B&lt;/math&gt;.<br /> ===Newton's Inequality===<br /> For non-negative real numbers &lt;math&gt;x_1,x_2,x_3\ldots,x_n&lt;/math&gt; and &lt;math&gt;0 &lt; k &lt; n&lt;/math&gt; the following holds:<br /> <br /> &lt;cmath&gt;d_k^2 \ge d_{k-1}d_{k+1}&lt;/cmath&gt;,<br /> <br /> with equality exactly iff all &lt;math&gt;x_i&lt;/math&gt; are equivalent. <br /> ===MacLaurin's Inequality===<br /> For non-negative real numbers &lt;math&gt;x_1,x_2,x_3 \ldots, x_n&lt;/math&gt;, and &lt;math&gt;d_1,d_2,d_3 \ldots, d_n&lt;/math&gt; such that<br /> &lt;cmath&gt;d_k = \frac{\sum\limits_{ 1\leq i_1 &lt; i_2 &lt; \cdots &lt; i_k \leq n}x_{i_1} x_{i_2} \cdots x_{i_k}}{{n \choose k}}&lt;/cmath&gt;, for &lt;math&gt;k\subset [1,n]&lt;/math&gt; the following holds:<br /> <br /> &lt;cmath&gt;d_1 \ge \sqrt{d_2} \ge \sqrt{d_3}\ldots \ge \sqrt[n]{d_n}&lt;/cmath&gt;<br /> <br /> with equality iff all &lt;math&gt;x_i&lt;/math&gt; are equivalent.<br /> <br /> [[User:Temperal/The Problem Solver's Resource10|Back to page 10]] | Last page (But also see the <br /> [[User:Temperal/The Problem Solver's Resource Tips and Tricks|tips and tricks page]], and the <br /> [[User:Temperal/The Problem Solver's Resource Proofs|methods of proof]]!</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource4&diff=29426 User:Temperal/The Problem Solver's Resource4 2009-01-11T02:56:25Z <p>Temperal: /* Abstract Algebra */ rmv</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 4}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Algebra&lt;/span&gt;==<br /> This is a collection of algebra laws and definitions. Obviously, there is '''''WAY''''' too much to cover here, but we'll try to give a good overview.<br /> <br /> ===Elementary Algebra===<br /> ====Definitions====<br /> *A polynomial is a function of the form<br /> &lt;cmath&gt;f(x)=a_nx^n+a_{n-1}x^{n-1}\ldots+a_0&lt;/cmath&gt;, where &lt;math&gt;a_n\ne 0&lt;/math&gt;, and &lt;math&gt;a_i&lt;/math&gt; are real numbers, and are called the [[coefficients]].<br /> *A polynomial has degree &lt;math&gt;c&lt;/math&gt; if the highest exponent of a variable is &lt;math&gt;c&lt;/math&gt;. The degree of polynomial &lt;math&gt;P&lt;/math&gt; is expressed as &lt;math&gt;\deg(P)&lt;/math&gt;.<br /> *A quadratic equation is a polynomial of degree &lt;math&gt;2&lt;/math&gt;. A cubic is of degree &lt;math&gt;3&lt;/math&gt;. A quartic is of degree &lt;math&gt;4&lt;/math&gt;. A quintic is of degree &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ====Factor Theorem====<br /> Iff a polynomial &lt;math&gt;P(x)&lt;/math&gt; has roots &lt;math&gt;a,b,c,d,e,\ldots,z&lt;/math&gt;, then &lt;math&gt;(x-a)(x-b)\ldots (x-z)=0&lt;/math&gt;, and &lt;math&gt;(x-a),(x-b)\ldots (x-z)&lt;/math&gt; are all factors of &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> ====Quadratic Formula====<br /> For a quadratic of form &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;, where &lt;math&gt;a,b,c&lt;/math&gt; are constants, the equation has roots &lt;math&gt;\frac{-b\pm\sqrt{b^2-4ac}}{2a}&lt;/math&gt;<br /> <br /> ====Fundamental Theorems of Algebra====<br /> *Every polynomial not in the form &lt;math&gt;f(x)=c&lt;/math&gt; has at least one root, real or complex.<br /> *A polynomial of degree &lt;math&gt;n&lt;/math&gt; has exactly &lt;math&gt;n&lt;/math&gt; roots, real or complex.<br /> <br /> ====Rational Root Theorem====<br /> Given a polynomial &lt;math&gt;f(x)&lt;/math&gt;, with integer coefficients &lt;math&gt;a_i&lt;/math&gt;, all rational roots are in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;|p|&lt;/math&gt; and &lt;math&gt;|q|&lt;/math&gt; are [[coprime]] natural numbers, &lt;math&gt;p|a_0&lt;/math&gt;, and &lt;math&gt;q|a_n&lt;/math&gt;.<br /> <br /> <br /> <br /> ====Determinants====<br /> The determinant of a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; (said to have order &lt;math&gt;2&lt;/math&gt;) matrix &lt;math&gt;\left |\begin{matrix}a&amp;b \\ c&amp;d\end {matrix}\right|&lt;/math&gt; is &lt;math&gt;ad-bc&lt;/math&gt;. <br /> ====General Formula for the Determinant==== <br /> Let &lt;math&gt;A&lt;/math&gt; be a square matrix of order &lt;math&gt;n&lt;/math&gt;. Write &lt;math&gt;A = a_{ij}&lt;/math&gt;, where &lt;math&gt;a_{ij}&lt;/math&gt; is the entry on the row &lt;math&gt;i&lt;/math&gt; and the column &lt;math&gt;j&lt;/math&gt;, for &lt;math&gt;i=1,\cdots,n&lt;/math&gt; and &lt;math&gt;j=1,\cdots,n&lt;/math&gt;. For any &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt;, set &lt;math&gt;A_{ij}&lt;/math&gt; (called the cofactors) to be the determinant of the square matrix of order &lt;math&gt;n-1&lt;/math&gt; obtained from &lt;math&gt;A&lt;/math&gt; by removing the row number &lt;math&gt;i&lt;/math&gt; and the column number &lt;math&gt;j&lt;/math&gt; multiplied by &lt;math&gt;(-1)^{i+j}&lt;/math&gt;. Thus:<br /> <br /> &lt;math&gt;\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}&lt;/math&gt;<br /> <br /> ====Cramer's Law====<br /> Consider a set of three linear equations (i.e. polynomials of degree one)<br /> *&lt;math&gt;ax+by+cz=d&lt;/math&gt;<br /> *&lt;math&gt;ex+fy+gz=h&lt;/math&gt;<br /> *&lt;math&gt;ix+jy+kz=l&lt;/math&gt;<br /> Let &lt;math&gt;D=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}d&amp;h&amp;1\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_y=\left|\begin{matrix}a&amp;e&amp;i\\d&amp;h&amp;l\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\d&amp;h&amp;l\end{matrix}\right|&lt;/math&gt;<br /> &lt;math&gt;x = \frac{D_x}{D}&lt;/math&gt;, &lt;math&gt;y = \frac{D_y}{D}&lt;/math&gt;, and &lt;math&gt;z = \frac{D_z}{D}&lt;/math&gt;.<br /> This can be generalized to any number of linear equations.<br /> <br /> <br /> ====Newton's Sums====<br /> Consider a polynomial &lt;math&gt;P(x)&lt;/math&gt; of degree &lt;math&gt;n&lt;/math&gt;, Let &lt;math&gt;P(x)=0&lt;/math&gt; have roots &lt;math&gt;x_1,x_2,\ldots,x_n&lt;/math&gt;. Define the following sums:<br /> *&lt;math&gt;S_1 = x_1 + x_2 + \cdots + x_n&lt;/math&gt;<br /> *&lt;math&gt;S_2 = x_1^2 + x_2^2 + \cdots + x_n^2&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> *&lt;math&gt;S_k = x_1^k + x_2^k + \cdots + x_n^k&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> The following holds:<br /> *&lt;math&gt;a_nS_1 + a_{n-1} = 0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> ====Vieta's Sums====<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0&lt;/math&gt;,<br /> where the coefficient of &lt;math&gt;x^{i}&lt;/math&gt; is &lt;math&gt;{a}_i&lt;/math&gt; and &lt;math&gt;a_n \neq 0&lt;/math&gt;. <br /> <br /> We have:<br /> &lt;cmath&gt;a_n = a_n&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt;\vdots&lt;/cmath&gt;<br /> &lt;cmath&gt;a_0 = (-1)^n a_n r_1r_2\cdots r_n&lt;/cmath&gt;<br /> <br /> <br /> [[User:Temperal/The Problem Solver's Resource3|Back to page 3]] | [[User:Temperal/The Problem Solver's Resource5|Continue to page 5]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource4&diff=29425 User:Temperal/The Problem Solver's Resource4 2009-01-11T02:56:01Z <p>Temperal: /* Diophantine Equations */ rmv</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 4}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Algebra&lt;/span&gt;==<br /> This is a collection of algebra laws and definitions. Obviously, there is '''''WAY''''' too much to cover here, but we'll try to give a good overview.<br /> <br /> ===Elementary Algebra===<br /> ====Definitions====<br /> *A polynomial is a function of the form<br /> &lt;cmath&gt;f(x)=a_nx^n+a_{n-1}x^{n-1}\ldots+a_0&lt;/cmath&gt;, where &lt;math&gt;a_n\ne 0&lt;/math&gt;, and &lt;math&gt;a_i&lt;/math&gt; are real numbers, and are called the [[coefficients]].<br /> *A polynomial has degree &lt;math&gt;c&lt;/math&gt; if the highest exponent of a variable is &lt;math&gt;c&lt;/math&gt;. The degree of polynomial &lt;math&gt;P&lt;/math&gt; is expressed as &lt;math&gt;\deg(P)&lt;/math&gt;.<br /> *A quadratic equation is a polynomial of degree &lt;math&gt;2&lt;/math&gt;. A cubic is of degree &lt;math&gt;3&lt;/math&gt;. A quartic is of degree &lt;math&gt;4&lt;/math&gt;. A quintic is of degree &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ====Factor Theorem====<br /> Iff a polynomial &lt;math&gt;P(x)&lt;/math&gt; has roots &lt;math&gt;a,b,c,d,e,\ldots,z&lt;/math&gt;, then &lt;math&gt;(x-a)(x-b)\ldots (x-z)=0&lt;/math&gt;, and &lt;math&gt;(x-a),(x-b)\ldots (x-z)&lt;/math&gt; are all factors of &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> ====Quadratic Formula====<br /> For a quadratic of form &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;, where &lt;math&gt;a,b,c&lt;/math&gt; are constants, the equation has roots &lt;math&gt;\frac{-b\pm\sqrt{b^2-4ac}}{2a}&lt;/math&gt;<br /> <br /> ====Fundamental Theorems of Algebra====<br /> *Every polynomial not in the form &lt;math&gt;f(x)=c&lt;/math&gt; has at least one root, real or complex.<br /> *A polynomial of degree &lt;math&gt;n&lt;/math&gt; has exactly &lt;math&gt;n&lt;/math&gt; roots, real or complex.<br /> <br /> ====Rational Root Theorem====<br /> Given a polynomial &lt;math&gt;f(x)&lt;/math&gt;, with integer coefficients &lt;math&gt;a_i&lt;/math&gt;, all rational roots are in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;|p|&lt;/math&gt; and &lt;math&gt;|q|&lt;/math&gt; are [[coprime]] natural numbers, &lt;math&gt;p|a_0&lt;/math&gt;, and &lt;math&gt;q|a_n&lt;/math&gt;.<br /> <br /> <br /> <br /> ====Determinants====<br /> The determinant of a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; (said to have order &lt;math&gt;2&lt;/math&gt;) matrix &lt;math&gt;\left |\begin{matrix}a&amp;b \\ c&amp;d\end {matrix}\right|&lt;/math&gt; is &lt;math&gt;ad-bc&lt;/math&gt;. <br /> ====General Formula for the Determinant==== <br /> Let &lt;math&gt;A&lt;/math&gt; be a square matrix of order &lt;math&gt;n&lt;/math&gt;. Write &lt;math&gt;A = a_{ij}&lt;/math&gt;, where &lt;math&gt;a_{ij}&lt;/math&gt; is the entry on the row &lt;math&gt;i&lt;/math&gt; and the column &lt;math&gt;j&lt;/math&gt;, for &lt;math&gt;i=1,\cdots,n&lt;/math&gt; and &lt;math&gt;j=1,\cdots,n&lt;/math&gt;. For any &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt;, set &lt;math&gt;A_{ij}&lt;/math&gt; (called the cofactors) to be the determinant of the square matrix of order &lt;math&gt;n-1&lt;/math&gt; obtained from &lt;math&gt;A&lt;/math&gt; by removing the row number &lt;math&gt;i&lt;/math&gt; and the column number &lt;math&gt;j&lt;/math&gt; multiplied by &lt;math&gt;(-1)^{i+j}&lt;/math&gt;. Thus:<br /> <br /> &lt;math&gt;\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}&lt;/math&gt;<br /> <br /> ====Cramer's Law====<br /> Consider a set of three linear equations (i.e. polynomials of degree one)<br /> *&lt;math&gt;ax+by+cz=d&lt;/math&gt;<br /> *&lt;math&gt;ex+fy+gz=h&lt;/math&gt;<br /> *&lt;math&gt;ix+jy+kz=l&lt;/math&gt;<br /> Let &lt;math&gt;D=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}d&amp;h&amp;1\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_y=\left|\begin{matrix}a&amp;e&amp;i\\d&amp;h&amp;l\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\d&amp;h&amp;l\end{matrix}\right|&lt;/math&gt;<br /> &lt;math&gt;x = \frac{D_x}{D}&lt;/math&gt;, &lt;math&gt;y = \frac{D_y}{D}&lt;/math&gt;, and &lt;math&gt;z = \frac{D_z}{D}&lt;/math&gt;.<br /> This can be generalized to any number of linear equations.<br /> <br /> <br /> ====Newton's Sums====<br /> Consider a polynomial &lt;math&gt;P(x)&lt;/math&gt; of degree &lt;math&gt;n&lt;/math&gt;, Let &lt;math&gt;P(x)=0&lt;/math&gt; have roots &lt;math&gt;x_1,x_2,\ldots,x_n&lt;/math&gt;. Define the following sums:<br /> *&lt;math&gt;S_1 = x_1 + x_2 + \cdots + x_n&lt;/math&gt;<br /> *&lt;math&gt;S_2 = x_1^2 + x_2^2 + \cdots + x_n^2&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> *&lt;math&gt;S_k = x_1^k + x_2^k + \cdots + x_n^k&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> The following holds:<br /> *&lt;math&gt;a_nS_1 + a_{n-1} = 0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> ====Vieta's Sums====<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0&lt;/math&gt;,<br /> where the coefficient of &lt;math&gt;x^{i}&lt;/math&gt; is &lt;math&gt;{a}_i&lt;/math&gt; and &lt;math&gt;a_n \neq 0&lt;/math&gt;. <br /> <br /> We have:<br /> &lt;cmath&gt;a_n = a_n&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt;\vdots&lt;/cmath&gt;<br /> &lt;cmath&gt;a_0 = (-1)^n a_n r_1r_2\cdots r_n&lt;/cmath&gt;<br /> <br /> <br /> ===Abstract Algebra===<br /> ====Set Theory====<br /> ====Group Theory====<br /> ====Field Theory==== <br /> ====Ring Theory====<br /> ====Graph Theory and Topology====<br /> ====Other Discrete Topics====<br /> <br /> [[User:Temperal/The Problem Solver's Resource3|Back to page 3]] | [[User:Temperal/The Problem Solver's Resource5|Continue to page 5]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource4&diff=29423 User:Temperal/The Problem Solver's Resource4 2009-01-11T02:55:46Z <p>Temperal: useless</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 4}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Algebra&lt;/span&gt;==<br /> This is a collection of algebra laws and definitions. Obviously, there is '''''WAY''''' too much to cover here, but we'll try to give a good overview.<br /> <br /> ===Elementary Algebra===<br /> ====Definitions====<br /> *A polynomial is a function of the form<br /> &lt;cmath&gt;f(x)=a_nx^n+a_{n-1}x^{n-1}\ldots+a_0&lt;/cmath&gt;, where &lt;math&gt;a_n\ne 0&lt;/math&gt;, and &lt;math&gt;a_i&lt;/math&gt; are real numbers, and are called the [[coefficients]].<br /> *A polynomial has degree &lt;math&gt;c&lt;/math&gt; if the highest exponent of a variable is &lt;math&gt;c&lt;/math&gt;. The degree of polynomial &lt;math&gt;P&lt;/math&gt; is expressed as &lt;math&gt;\deg(P)&lt;/math&gt;.<br /> *A quadratic equation is a polynomial of degree &lt;math&gt;2&lt;/math&gt;. A cubic is of degree &lt;math&gt;3&lt;/math&gt;. A quartic is of degree &lt;math&gt;4&lt;/math&gt;. A quintic is of degree &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ====Factor Theorem====<br /> Iff a polynomial &lt;math&gt;P(x)&lt;/math&gt; has roots &lt;math&gt;a,b,c,d,e,\ldots,z&lt;/math&gt;, then &lt;math&gt;(x-a)(x-b)\ldots (x-z)=0&lt;/math&gt;, and &lt;math&gt;(x-a),(x-b)\ldots (x-z)&lt;/math&gt; are all factors of &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> ====Quadratic Formula====<br /> For a quadratic of form &lt;math&gt;ax^2+bx+c=0&lt;/math&gt;, where &lt;math&gt;a,b,c&lt;/math&gt; are constants, the equation has roots &lt;math&gt;\frac{-b\pm\sqrt{b^2-4ac}}{2a}&lt;/math&gt;<br /> <br /> ====Fundamental Theorems of Algebra====<br /> *Every polynomial not in the form &lt;math&gt;f(x)=c&lt;/math&gt; has at least one root, real or complex.<br /> *A polynomial of degree &lt;math&gt;n&lt;/math&gt; has exactly &lt;math&gt;n&lt;/math&gt; roots, real or complex.<br /> <br /> ====Rational Root Theorem====<br /> Given a polynomial &lt;math&gt;f(x)&lt;/math&gt;, with integer coefficients &lt;math&gt;a_i&lt;/math&gt;, all rational roots are in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;|p|&lt;/math&gt; and &lt;math&gt;|q|&lt;/math&gt; are [[coprime]] natural numbers, &lt;math&gt;p|a_0&lt;/math&gt;, and &lt;math&gt;q|a_n&lt;/math&gt;.<br /> <br /> <br /> <br /> ====Determinants====<br /> The determinant of a &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt; (said to have order &lt;math&gt;2&lt;/math&gt;) matrix &lt;math&gt;\left |\begin{matrix}a&amp;b \\ c&amp;d\end {matrix}\right|&lt;/math&gt; is &lt;math&gt;ad-bc&lt;/math&gt;. <br /> ====General Formula for the Determinant==== <br /> Let &lt;math&gt;A&lt;/math&gt; be a square matrix of order &lt;math&gt;n&lt;/math&gt;. Write &lt;math&gt;A = a_{ij}&lt;/math&gt;, where &lt;math&gt;a_{ij}&lt;/math&gt; is the entry on the row &lt;math&gt;i&lt;/math&gt; and the column &lt;math&gt;j&lt;/math&gt;, for &lt;math&gt;i=1,\cdots,n&lt;/math&gt; and &lt;math&gt;j=1,\cdots,n&lt;/math&gt;. For any &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt;, set &lt;math&gt;A_{ij}&lt;/math&gt; (called the cofactors) to be the determinant of the square matrix of order &lt;math&gt;n-1&lt;/math&gt; obtained from &lt;math&gt;A&lt;/math&gt; by removing the row number &lt;math&gt;i&lt;/math&gt; and the column number &lt;math&gt;j&lt;/math&gt; multiplied by &lt;math&gt;(-1)^{i+j}&lt;/math&gt;. Thus:<br /> <br /> &lt;math&gt;\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}&lt;/math&gt;<br /> <br /> ====Cramer's Law====<br /> Consider a set of three linear equations (i.e. polynomials of degree one)<br /> *&lt;math&gt;ax+by+cz=d&lt;/math&gt;<br /> *&lt;math&gt;ex+fy+gz=h&lt;/math&gt;<br /> *&lt;math&gt;ix+jy+kz=l&lt;/math&gt;<br /> Let &lt;math&gt;D=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}d&amp;h&amp;1\\b&amp;f&amp;j\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_y=\left|\begin{matrix}a&amp;e&amp;i\\d&amp;h&amp;l\\c&amp;g&amp;k\end{matrix}\right|&lt;/math&gt;, &lt;math&gt;D_x=\left|\begin{matrix}a&amp;e&amp;i\\b&amp;f&amp;j\\d&amp;h&amp;l\end{matrix}\right|&lt;/math&gt;<br /> &lt;math&gt;x = \frac{D_x}{D}&lt;/math&gt;, &lt;math&gt;y = \frac{D_y}{D}&lt;/math&gt;, and &lt;math&gt;z = \frac{D_z}{D}&lt;/math&gt;.<br /> This can be generalized to any number of linear equations.<br /> <br /> <br /> ====Newton's Sums====<br /> Consider a polynomial &lt;math&gt;P(x)&lt;/math&gt; of degree &lt;math&gt;n&lt;/math&gt;, Let &lt;math&gt;P(x)=0&lt;/math&gt; have roots &lt;math&gt;x_1,x_2,\ldots,x_n&lt;/math&gt;. Define the following sums:<br /> *&lt;math&gt;S_1 = x_1 + x_2 + \cdots + x_n&lt;/math&gt;<br /> *&lt;math&gt;S_2 = x_1^2 + x_2^2 + \cdots + x_n^2&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> *&lt;math&gt;S_k = x_1^k + x_2^k + \cdots + x_n^k&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> <br /> The following holds:<br /> *&lt;math&gt;a_nS_1 + a_{n-1} = 0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_2 + a_{n-1}S_1 + 2a_{n-2}=0&lt;/math&gt;<br /> *&lt;math&gt;a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3}=0&lt;/math&gt;<br /> *&lt;math&gt;\vdots&lt;/math&gt;<br /> ====Vieta's Sums====<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0&lt;/math&gt;,<br /> where the coefficient of &lt;math&gt;x^{i}&lt;/math&gt; is &lt;math&gt;{a}_i&lt;/math&gt; and &lt;math&gt;a_n \neq 0&lt;/math&gt;. <br /> <br /> We have:<br /> &lt;cmath&gt;a_n = a_n&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt; a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)&lt;/cmath&gt;<br /> &lt;cmath&gt;\vdots&lt;/cmath&gt;<br /> &lt;cmath&gt;a_0 = (-1)^n a_n r_1r_2\cdots r_n&lt;/cmath&gt;<br /> ====Diophantine Equations====<br /> ===Abstract Algebra===<br /> ====Set Theory====<br /> ====Group Theory====<br /> ====Field Theory==== <br /> ====Ring Theory====<br /> ====Graph Theory and Topology====<br /> ====Other Discrete Topics====<br /> <br /> [[User:Temperal/The Problem Solver's Resource3|Back to page 3]] | [[User:Temperal/The Problem Solver's Resource5|Continue to page 5]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource3&diff=29419 User:Temperal/The Problem Solver's Resource3 2009-01-11T02:52:55Z <p>Temperal: /* Rules of Products */ explanatiobn</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 3}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Summations and Products&lt;/span&gt;==<br /> ===Definitions===<br /> *Summations: &lt;math&gt;\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}&lt;/math&gt;<br /> *Products: &lt;math&gt;\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}&lt;/math&gt;<br /> <br /> ===Rules of Summation===<br /> &lt;math&gt;\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i= \frac{n(n+1)}{2}&lt;/math&gt;, and in general &lt;math&gt;\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}&lt;/math&gt;<br /> <br /> The above should all be self-evident and provable by the reader within seconds. <br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}&lt;/math&gt;<br /> <br /> &lt;!-- eliminate useless n=4 and n=5; add proof for n=2 which easily generalizes to general n --&gt;<br /> <br /> ===Rules of Products===<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x=x^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}&lt;/math&gt;<br /> <br /> These should be self-evident, as above.<br /> <br /> [[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource3&diff=29418 User:Temperal/The Problem Solver's Resource3 2009-01-11T02:52:32Z <p>Temperal: /* Rules of Summation */ note to self</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 3}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Summations and Products&lt;/span&gt;==<br /> ===Definitions===<br /> *Summations: &lt;math&gt;\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}&lt;/math&gt;<br /> *Products: &lt;math&gt;\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}&lt;/math&gt;<br /> <br /> ===Rules of Summation===<br /> &lt;math&gt;\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i= \frac{n(n+1)}{2}&lt;/math&gt;, and in general &lt;math&gt;\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}&lt;/math&gt;<br /> <br /> The above should all be self-evident and provable by the reader within seconds. <br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}&lt;/math&gt;<br /> <br /> &lt;!-- eliminate useless n=4 and n=5; add proof for n=2 which easily generalizes to general n --&gt;<br /> <br /> ===Rules of Products===<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x=x^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;!-- same as above, there are others.... but no fancy ones like divergence/convergence, please --&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource3&diff=29417 User:Temperal/The Problem Solver's Resource3 2009-01-11T02:50:50Z <p>Temperal: /* Rules of Summation */ improve</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 3}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Summations and Products&lt;/span&gt;==<br /> ===Definitions===<br /> *Summations: &lt;math&gt;\sum_{i=a}^{b}c_i=c_a+c_{a+1}+c_{a+2}...+c_{b-1}+c_{b}&lt;/math&gt;<br /> *Products: &lt;math&gt;\prod_{i=a}^{b}c_i=c_a\cdot c_{a+1}\cdot c_{a+2}...\cdot c_{b-1}\cdot c_{b}&lt;/math&gt;<br /> <br /> ===Rules of Summation===<br /> &lt;math&gt;\sum_{i=a}^{b}f_1(i)+f_2(i)+\ldots f_n(i)=\sum_{i=a}^{b}f(i)+\sum_{i=a}^{b}f_2(i)+\ldots+\sum_{i=a}^{b}f_n(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=a}^{b}c\cdot f(i)=c\cdot \sum_{i=a}^{b}f(i)&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i= \frac{n(n+1)}{2}&lt;/math&gt;, and in general &lt;math&gt;\sum_{i=a}^{b} i= \frac{(b-a+1)(a+b)}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}&lt;/math&gt;<br /> <br /> &lt;math&gt; \sum_{i=1}^n i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}&lt;/math&gt;<br /> <br /> &lt;!--there are others, I forgot what they were. Could someone please fill them in? --&gt;<br /> <br /> ===Rules of Products===<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x=x^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\prod_{i=a}^{b}x\cdot y=x^{(b-a+1)}y^{(b-a+1)}&lt;/math&gt;<br /> <br /> &lt;!-- same as above, there are others.... but no fancy ones like divergence/convergence, please --&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource2|Back to page 2]] | [[User:Temperal/The Problem Solver's Resource4|Continue to page 4]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource2&diff=29416 User:Temperal/The Problem Solver's Resource2 2009-01-11T02:45:34Z <p>Temperal: /* Rules of Exponentiation and Logarithms */ explanations</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 2}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Exponentials and Logarithms&lt;/span&gt;==<br /> This is just a quick review of logarithms and exponents; it's elementary content.<br /> ===Definitions===<br /> *Exponentials: Do you really need this one? If &lt;math&gt;a=\underbrace{b\times b\times b\times \cdots \times b}_{x\text{ }b'\text{s}}&lt;/math&gt;, then &lt;math&gt;a=b^x&lt;/math&gt;<br /> *Logarithms: If &lt;math&gt;b^a=x&lt;/math&gt;, &lt;math&gt;\log_b{x}=a&lt;/math&gt;. Note that a logarithm in base [[e]], i.e. &lt;math&gt;\log_e{x}=a&lt;/math&gt; is denoted as &lt;math&gt;\ln{x}=a&lt;/math&gt;, or the natural logarithm of x. If no base is specified, then a logarithm is assumed to be in base 10.<br /> <br /> ===Rules of Exponentiation===<br /> &lt;math&gt;a^x \cdot a^y=a^{x+y}&lt;/math&gt;<br /> <br /> &lt;math&gt;(a^x)^y=a^{xy}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{a^x}{a^y}=a^{x-y}&lt;/math&gt;<br /> <br /> &lt;math&gt;a^0=1&lt;/math&gt;, where &lt;math&gt;a\ne 0&lt;/math&gt;.<br /> <br /> These should all be trivial and easily proven by the reader.<br /> <br /> ===Rules of Logarithms===<br /> &lt;math&gt;\log_b b=1&lt;/math&gt; <br /> <br /> This can be seen by writing as &lt;math&gt;b^1=b&lt;/math&gt;.<br /> <br /> &lt;math&gt;\log_b xy=\log_b x +\log_b y &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_b x^y=y\cdot \log_b x &lt;/math&gt;<br /> <br /> &lt;math&gt;\log_b \frac{x}{y} =\log_b x-\log_b y&lt;/math&gt;<br /> <br /> &lt;math&gt;\log_b a=\frac{1}{\log_a b}&lt;/math&gt;<br /> <br /> &lt;math&gt;\log_b a=\frac{\log_x a}{\log_x b}&lt;/math&gt;, where x is a constant.<br /> <br /> All of the above should be proven by the reader without too much difficulty - substitution and putting things in exponential form will help.<br /> <br /> &lt;math&gt;\log_1 a&lt;/math&gt; and &lt;math&gt;\log_0 a&lt;/math&gt; are undefined, as there is no &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;1^x=a&lt;/math&gt; except when &lt;math&gt;a=1&lt;/math&gt; (in which case there are infinite &lt;math&gt;x&lt;/math&gt;) and likewise with &lt;math&gt;0&lt;/math&gt;.<br /> <br /> [[User:Temperal/The Problem Solver's Resource1|Back to page 1]] | [[User:Temperal/The Problem Solver's Resource3|Continue to page 3]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29414 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:35:13Z <p>Temperal: /* Law of Sines */ yikes</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> The proof is left as an exercise for the reader. (Hint: Draw a circle with one of the sides as a radius and use the power of a point theorem)<br /> <br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> Proof: In the diagram below, circle &lt;math&gt; O &lt;/math&gt; [[circumscribe]]s triangle &lt;math&gt; ABC &lt;/math&gt;. &lt;math&gt; OD &lt;/math&gt; is [[perpendicular]] to &lt;math&gt; BC &lt;/math&gt;. Since &lt;math&gt; \triangle ODB \cong \triangle ODC &lt;/math&gt;, &lt;math&gt; BD = CD = \frac a2 &lt;/math&gt; and &lt;math&gt; \angle BOD = \angle COD &lt;/math&gt;. But &lt;math&gt; \angle BAC = 2\angle BOC &lt;/math&gt; making &lt;math&gt; \angle BOD = \angle COD = \theta &lt;/math&gt;. Therefore, we can use simple trig in [[right triangle]] &lt;math&gt; BOD &lt;/math&gt; to find that <br /> <br /> &lt;center&gt;&lt;math&gt; \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. &lt;/math&gt; &lt;/center&gt;<br /> <br /> The same holds for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, thus establishing the identity.<br /> <br /> &lt;center&gt;[[Image:Lawofsines.PNG]]&lt;/center&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . &lt;/cmath&gt;<br /> <br /> The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy:<br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the law of sines,<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} &lt;/cmath&gt;<br /> as desired.<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29413 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:35:00Z <p>Temperal: /* Law of Cosines */ proof exercise</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> The proof is left as an exercise for the reader. (Hint: Draw a circle with one of the sides as a radius and use the power of a point theorem)<br /> <br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> Proof: In the diagram below, circle &lt;math&gt; O &lt;/math&gt; [[circumscribe]]s triangle &lt;math&gt; ABC &lt;/math&gt;. &lt;math&gt; OD &lt;/math&gt; is [[perpendicular]] to &lt;math&gt; BC &lt;/math&gt;. Since &lt;math&gt; \triangle ODB \cong \triangle ODC &lt;/math&gt;, &lt;math&gt; BD = CD = \frac a2 &lt;/math&gt; and &lt;math&gt; \angle BOD = \angle COD &lt;/math&gt;. But &lt;math&gt; \angle BAC = 2\angle BOC &lt;/math&gt; making &lt;math&gt; \angle BOD = \angle COD = \theta &lt;/math&gt;. Therefore, we can use simple trig in [[right triangle]] &lt;math&gt; BOD &lt;/math&gt; to find that <br /> <br /> &lt;center&gt;&lt;math&gt; \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. &lt;/math&gt; &lt;/center&gt;<br /> <br /> The same holds for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, thus establishing the identity.<br /> <br /> &lt;center&gt;[[Image:Lawofsines.PNG]]<br /> <br /> ====Law of Tangents====<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . &lt;/cmath&gt;<br /> <br /> The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy:<br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the law of sines,<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} &lt;/cmath&gt;<br /> as desired.<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29412 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:34:11Z <p>Temperal: /* Law of Sines */ proof</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> Proof: In the diagram below, circle &lt;math&gt; O &lt;/math&gt; [[circumscribe]]s triangle &lt;math&gt; ABC &lt;/math&gt;. &lt;math&gt; OD &lt;/math&gt; is [[perpendicular]] to &lt;math&gt; BC &lt;/math&gt;. Since &lt;math&gt; \triangle ODB \cong \triangle ODC &lt;/math&gt;, &lt;math&gt; BD = CD = \frac a2 &lt;/math&gt; and &lt;math&gt; \angle BOD = \angle COD &lt;/math&gt;. But &lt;math&gt; \angle BAC = 2\angle BOC &lt;/math&gt; making &lt;math&gt; \angle BOD = \angle COD = \theta &lt;/math&gt;. Therefore, we can use simple trig in [[right triangle]] &lt;math&gt; BOD &lt;/math&gt; to find that <br /> <br /> &lt;center&gt;&lt;math&gt; \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. &lt;/math&gt; &lt;/center&gt;<br /> <br /> The same holds for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, thus establishing the identity.<br /> <br /> &lt;center&gt;[[Image:Lawofsines.PNG]]<br /> <br /> ====Law of Tangents====<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . &lt;/cmath&gt;<br /> <br /> The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy:<br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the law of sines,<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} &lt;/cmath&gt;<br /> as desired.<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29411 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:25:48Z <p>Temperal: /* Law of Tangents */ hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . &lt;/cmath&gt;<br /> <br /> The proof of this is less trivial than that of the law of sines and cosines, but still fairly easy:<br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the law of sines,<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} &lt;/cmath&gt;<br /> as desired.<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29407 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:23:21Z <p>Temperal: /* Law of Tangents */ from the wiki</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . &lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the law of sines,<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} &lt;/cmath&gt;<br /> as desired.<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource9&diff=29405 User:Temperal/The Problem Solver's Resource9 2009-01-11T02:19:17Z <p>Temperal: hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 9}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Complex Numbers&lt;/span&gt;==<br /> <br /> [[User:Temperal/The Problem Solver's Resource8|Back to page 8]] | [[User:Temperal/The Problem Solver's Resource10|Continue to page 10]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource10&diff=29403 User:Temperal/The Problem Solver's Resource10 2009-01-11T02:18:55Z <p>Temperal: hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 10}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;More Number Theory&lt;/span&gt;==<br /> <br /> <br /> [[User:Temperal/The Problem Solver's Resource9|Back to page 9]] | [[User:Temperal/The Problem Solver's Resource11|Continue to page 11]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource&diff=29402 User:Temperal/The Problem Solver's Resource 2009-01-11T02:18:18Z <p>Temperal: hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Introduction&lt;/span&gt;==<br /> This is a collection of theorems, formulas, and other such things that are useful for competitions ranging from the AMC tests to the IMO, from Mathcounts to the ARML, from the iTest to Putnam. I've always thought that the things you actually need to know for such competitions were much too spread out, so that it was hard to find them when you needed them. Thus, I decided to create this index of useful information. It starts out with basic trigonometric formulas, and ends with theorems from calculus and advanced number theory. If there is anything incorrect or missing here, then feel free to add it in. <br /> <br /> Note that these are just formula sheets with no explanation. This is '''not''' a good place to learn these topics; just a place to refresh your memory on the specific formulas.<br /> <br /> '''Current status: Algebra page needs to be completed, geometry page not started yet.'''<br /> <br /> Pages on analysis and other advanced calculus topics may be added in the future.<br /> <br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Table of Contents&lt;/span&gt;==<br /> *[[User:Temperal/The Problem Solver's Resource|Introduction]]<br /> #[[User:Temperal/The Problem Solver's Resource1|Trigonometry]]<br /> #[[User:Temperal/The Problem Solver's Resource2|Exponentials and Logarithms]]<br /> #[[User:Temperal/The Problem Solver's Resource3|Summations and Products]]<br /> #[[User:Temperal/The Problem Solver's Resource4|Algebra]]<br /> #[[User:Temperal/The Problem Solver's Resource5|Combinatorics]]<br /> #[[User:Temperal/The Problem Solver's Resource6|Number Theory]]<br /> #[[User:Temperal/The Problem Solver's Resource7|Limits]]<br /> #[[User:Temperal/The Problem Solver's Resource8|Geometry]]<br /> #[[User:Temperal/The Problem Solver's Resource9|More Number Theory]]<br /> #[[User:Temperal/The Problem Solver's Resource10|Complex Numbers]]<br /> #[[User:Temperal/The Problem Solver's Resource11|Inequalities]]<br /> *[[User:Temperal/The Problem Solver's Resource Tips and Tricks|Other Tips and Tricks]]<br /> *[[User:Temperal/The Problem Solver's Resource Proofs|Methods of Proof]]<br /> [[User:Temperal/The Problem Solver's Resource1|Continue to page 1]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User_talk:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29398 User talk:Temperal/The Problem Solver's Resource Proofs 2009-01-11T02:09:31Z <p>Temperal: reply</p> <hr /> <div>Yeah, uh, if you noticed the extremely big jump in difficulty, that was Temperal. [[User:Xpmath|Xpmath]] 01:06, 11 January 2009 (UTC)<br /> :Heh, the first one isn't so bad, and the second one even I can solve with some effort, so it isn't too big of a jump. I just basically copied the solutions from their AoPSWiki pages; I might have to clean them up in a bit. [[User:Temperal|Temperal]] 02:09, 11 January 2009 (UTC)</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29395 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:05:10Z <p>Temperal: /* Area of a Triangle */ prooff</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> This can be easily proven by the well-known formula &lt;math&gt;\frac{1}{2}ah_a&lt;/math&gt; - considering one of the triangles which altitude &lt;math&gt;h_a&lt;/math&gt; divides &lt;math&gt;\triangle ABC&lt;/math&gt; into, we see that &lt;math&gt;h_a=b\sin C&lt;/math&gt; and hence &lt;math&gt;[ABC]=\frac 12ab\sin C&lt;/math&gt; as desired.<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29394 User:Temperal/The Problem Solver's Resource1 2009-01-11T02:02:31Z <p>Temperal: /* Sum of Angle Formulas */ &amp;, not @</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (-A) = -\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics &amp; Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2ab\cos C&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> where &lt;math&gt;R&lt;/math&gt; is the radius of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29377 User:Temperal/The Problem Solver's Resource Proofs 2009-01-11T00:26:20Z <p>Temperal: /* Example Problem 3 */ from the wiki</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the methods of proof section.}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Methods of Proof&lt;/span&gt;==<br /> Many competitions involve only short answer or multiple choice questions, with no justification required. However, once you get to the elite, national competitions, mostly national olympiads, you suddenly are expected to justify your solutions. If you've never encountered problems where you must prove your results, you would most likely be overwhelmed. This page attempts to explain some of the most used techniques in writing proofs. Please note that most material on this page is relatively simple; this section is only for introducing beginners to new techniques. If one wishes to improve at more advanced problem solving, solving problems is recommended.<br /> <br /> ===Contradiction===<br /> To many people, contradiction is one of the simplest methods of proof. In this technique, you attempt to prove the desired result in an indirect way. You first assume that the opposite result holds. Then, you attempt to find something that contradicts something that you already know or is given in the problem. Thus, you must've been wrong in assuming that the opposite result is true, so the only possible result is the one you wanted to prove!<br /> <br /> ====Example Problem 1====<br /> Prove that there are infinite primes. (This is an extremely classic and elegant proof)<br /> <br /> =====Solution=====<br /> Many new problem solvers are intimidated by a question that asks to prove anything relating to infinity. Hence, this seems like an excellent time to use contradiction, as we simply need to find a contradiction. Since our desired result is proving that there are an infinite amount of primes, it makes sense to assume that there are only a finite amount. Let the primes be &lt;math&gt;p_1, p_2, p_3,...p_n&lt;/math&gt; where &lt;math&gt;p_1&lt;p_2&lt;p_3...&lt;p_n&lt;/math&gt;. If we can find another prime number that is greater than &lt;math&gt;p_n&lt;/math&gt;, we will have disproved our assumption. Consider the number &lt;math&gt;p_1p_2p_3...p_n+1&lt;/math&gt;. This number is not divisible by any of our known primes, so it itself must be prime. Also, it is evident that this number is greater than &lt;math&gt;p_n&lt;/math&gt; so we have found a new prime number. This violates our assumption, so there must be an infinite number of primes.<br /> <br /> ====Example Problem 2====<br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational. (Also a very classic question)<br /> <br /> =====Solution=====<br /> The assumption should be fairly obvious here. We assume that &lt;math&gt;\sqrt{2}&lt;/math&gt; is rational. From the definition of rational numbers, &lt;math&gt;\sqrt{2}&lt;/math&gt; can then be written as a fraction in lowest terms. Let &lt;math&gt;\sqrt{2}=\frac{p}{q}&lt;/math&gt;, where p and q are relatively prime. Squaring and multiplying by &lt;math&gt;q^2&lt;/math&gt;, we have &lt;math&gt;2q^2=p^2&lt;/math&gt;. As 2 divides the left hand side, it must also divide p (a very common example of divisibility being helpful). Let &lt;math&gt;p=2r&lt;/math&gt;; then &lt;math&gt;2q^2=4r^2\rightarrowq^2=2r^2&lt;/math&gt;. By the same argument as last time, we let &lt;math&gt;q=2s&lt;/math&gt;. However, now we have &lt;math&gt;\sqrt{2}=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}&lt;/math&gt;. Recall though, that we stated that p and q were relatively prime. Now, we see that both are divisible by 2. This contradicts our original assumption, so &lt;math&gt;\sqrt{2}&lt;/math&gt; must indeed be rational.<br /> <br /> ===Induction===<br /> Induction is used when we need to prove a statement true for an infinite sequence of numbers &lt;math&gt;S&lt;/math&gt;; commonly the natural or whole numbers. The procedure is as follows: Show the desired statement true for a '''base case''', the smallest number in the sequence - in the case of the natural numbers, it would be &lt;math&gt;1&lt;/math&gt;. Then we take an '''inductive step'''; proving that if &lt;math&gt;k&lt;/math&gt; satisfies the statement, then &lt;math&gt;f(k)&lt;/math&gt; does as well, where &lt;math&gt;f:S\rightarrow S&lt;/math&gt; is a function that takes one member of the sequence to the next. In the case of the natural numbers, it's &lt;math&gt;k+1&lt;/math&gt;. Now, since the statement is true for &lt;math&gt;1&lt;/math&gt; (in the case of the natural numbers), it's true for &lt;math&gt;1+1&lt;/math&gt;, as well, and &lt;math&gt;1+1+1&lt;/math&gt;, etc. Thus, it's true for every natural number. A common analogy is a row of falling dominoes. <br /> <br /> '''Strong induction''' is a method where we have to use the fact that all &lt;math&gt;k\in S&lt;/math&gt; so that &lt;math&gt;k&lt;f(k)&lt;/math&gt; satisfy the statement to prove that &lt;math&gt;f(k)&lt;/math&gt; satisfies the statement. In terms of the natural numbers, this means we have to use the fact that &lt;math&gt;1,2,3,\ldots,k&lt;/math&gt; satisfy the statement to prove that &lt;math&gt;k+1&lt;/math&gt; does.<br /> <br /> ====Example Problem 3====<br /> The function &lt;math&gt;f(x,y)&lt;/math&gt; satisfies<br /> <br /> (1) &lt;math&gt;f(0,y)=y+1, &lt;/math&gt;<br /> <br /> (2) &lt;math&gt;f(x+1,0)=f(x,1), &lt;/math&gt;<br /> <br /> (3) &lt;math&gt;f(x+1,y+1)=f(x,f(x+1,y)), &lt;/math&gt;<br /> <br /> for all non-negative integers &lt;math&gt;x,y &lt;/math&gt;. Show that &lt;math&gt;f(1,y) = y+2 &lt;/math&gt; for all non-negative integers &lt;math&gt;y&lt;/math&gt;. (Adapted from 1981 IMO, #6)<br /> <br /> =====Solution=====<br /> We observe that &lt;math&gt;f(1,0) = f(0,1) = 2 &lt;/math&gt; and that &lt;math&gt;f(1, y+1) = f(1, f(1,y)) = f(1,y) + 1&lt;/math&gt;, hence by induction, &lt;math&gt;f(1,y) = y+2 &lt;/math&gt;, as desired.<br /> <br /> ====Example Problem 4====<br /> Let &lt;math&gt;1 \le r \le n &lt;/math&gt; and consider all subsets of &lt;math&gt;r &lt;/math&gt; elements of the set &lt;math&gt; \{ 1, 2, \ldots , n \} &lt;/math&gt;. Each of these subsets has a smallest member. Let &lt;math&gt;F(n,r) &lt;/math&gt; denote the arithmetic mean of these smallest numbers; prove that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(n,r) = \frac{n+1}{r+1}.<br /> &lt;/math&gt;<br /> &lt;/center&gt; (1981 IMO, #2)<br /> <br /> =====Solution=====<br /> We proceed by strong induction.<br /> <br /> We define &lt;math&gt;F(k, k-1)&lt;/math&gt; to be zero (the empty sum).<br /> <br /> We consider &lt;math&gt;r&lt;/math&gt; to be fixed. The assertion obviously holds for &lt;math&gt;r = n&lt;/math&gt;. We now assume the problem to hold for values of &lt;math&gt;n&lt;/math&gt; less than or equal to &lt;math&gt;k&lt;/math&gt;. By considering subsets containing &lt;math&gt;k+1&lt;/math&gt; and not containing &lt;math&gt;k+1&lt;/math&gt;, respectively, we conclude that<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;<br /> F(k+1, r) = \frac{{k \choose r-1}F(k,r-1) + {k \choose r}F(k,r)}{{k+1 \choose r}} = 1 + \frac{k-r+1}{r+1} = \frac{k+2}{r+1}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> as desired.<br /> <br /> ===Infinite Descent===<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29376 User:Temperal/The Problem Solver's Resource Proofs 2009-01-11T00:24:00Z <p>Temperal: /* Induction */ start</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the methods of proof section.}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Methods of Proof&lt;/span&gt;==<br /> Many competitions involve only short answer or multiple choice questions, with no justification required. However, once you get to the elite, national competitions, mostly national olympiads, you suddenly are expected to justify your solutions. If you've never encountered problems where you must prove your results, you would most likely be overwhelmed. This page attempts to explain some of the most used techniques in writing proofs. Please note that most material on this page is relatively simple; this section is only for introducing beginners to new techniques. If one wishes to improve at more advanced problem solving, solving problems is recommended.<br /> <br /> ===Contradiction===<br /> To many people, contradiction is one of the simplest methods of proof. In this technique, you attempt to prove the desired result in an indirect way. You first assume that the opposite result holds. Then, you attempt to find something that contradicts something that you already know or is given in the problem. Thus, you must've been wrong in assuming that the opposite result is true, so the only possible result is the one you wanted to prove!<br /> <br /> ====Example Problem 1====<br /> Prove that there are infinite primes. (This is an extremely classic and elegant proof)<br /> <br /> =====Solution=====<br /> Many new problem solvers are intimidated by a question that asks to prove anything relating to infinity. Hence, this seems like an excellent time to use contradiction, as we simply need to find a contradiction. Since our desired result is proving that there are an infinite amount of primes, it makes sense to assume that there are only a finite amount. Let the primes be &lt;math&gt;p_1, p_2, p_3,...p_n&lt;/math&gt; where &lt;math&gt;p_1&lt;p_2&lt;p_3...&lt;p_n&lt;/math&gt;. If we can find another prime number that is greater than &lt;math&gt;p_n&lt;/math&gt;, we will have disproved our assumption. Consider the number &lt;math&gt;p_1p_2p_3...p_n+1&lt;/math&gt;. This number is not divisible by any of our known primes, so it itself must be prime. Also, it is evident that this number is greater than &lt;math&gt;p_n&lt;/math&gt; so we have found a new prime number. This violates our assumption, so there must be an infinite number of primes.<br /> <br /> ====Example Problem 2====<br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational. (Also a very classic question)<br /> <br /> =====Solution=====<br /> The assumption should be fairly obvious here. We assume that &lt;math&gt;\sqrt{2}&lt;/math&gt; is rational. From the definition of rational numbers, &lt;math&gt;\sqrt{2}&lt;/math&gt; can then be written as a fraction in lowest terms. Let &lt;math&gt;\sqrt{2}=\frac{p}{q}&lt;/math&gt;, where p and q are relatively prime. Squaring and multiplying by &lt;math&gt;q^2&lt;/math&gt;, we have &lt;math&gt;2q^2=p^2&lt;/math&gt;. As 2 divides the left hand side, it must also divide p (a very common example of divisibility being helpful). Let &lt;math&gt;p=2r&lt;/math&gt;; then &lt;math&gt;2q^2=4r^2\rightarrowq^2=2r^2&lt;/math&gt;. By the same argument as last time, we let &lt;math&gt;q=2s&lt;/math&gt;. However, now we have &lt;math&gt;\sqrt{2}=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}&lt;/math&gt;. Recall though, that we stated that p and q were relatively prime. Now, we see that both are divisible by 2. This contradicts our original assumption, so &lt;math&gt;\sqrt{2}&lt;/math&gt; must indeed be rational.<br /> <br /> ===Induction===<br /> Induction is used when we need to prove a statement true for an infinite sequence of numbers &lt;math&gt;S&lt;/math&gt;; commonly the natural or whole numbers. The procedure is as follows: Show the desired statement true for a '''base case''', the smallest number in the sequence - in the case of the natural numbers, it would be &lt;math&gt;1&lt;/math&gt;. Then we take an '''inductive step'''; proving that if &lt;math&gt;k&lt;/math&gt; satisfies the statement, then &lt;math&gt;f(k)&lt;/math&gt; does as well, where &lt;math&gt;f:S\rightarrow S&lt;/math&gt; is a function that takes one member of the sequence to the next. In the case of the natural numbers, it's &lt;math&gt;k+1&lt;/math&gt;. Now, since the statement is true for &lt;math&gt;1&lt;/math&gt; (in the case of the natural numbers), it's true for &lt;math&gt;1+1&lt;/math&gt;, as well, and &lt;math&gt;1+1+1&lt;/math&gt;, etc. Thus, it's true for every natural number. A common analogy is a row of falling dominoes. <br /> <br /> '''Strong induction''' is a method where we have to use the fact that all &lt;math&gt;k\in S&lt;/math&gt; so that &lt;math&gt;k&lt;f(k)&lt;/math&gt; satisfy the statement to prove that &lt;math&gt;f(k)&lt;/math&gt; satisfies the statement. In terms of the natural numbers, this means we have to use the fact that &lt;math&gt;1,2,3,\ldots,k&lt;/math&gt; satisfy the statement to prove that &lt;math&gt;k+1&lt;/math&gt; does.<br /> <br /> ====Example Problem 3====<br /> The function &lt;math&gt;f(x,y)&lt;/math&gt; satisfies<br /> <br /> (1) &lt;math&gt;f(0,y)=y+1, &lt;/math&gt;<br /> <br /> (2) &lt;math&gt;f(x+1,0)=f(x,1), &lt;/math&gt;<br /> <br /> (3) &lt;math&gt;f(x+1,y+1)=f(x,f(x+1,y)), &lt;/math&gt;<br /> <br /> for all non-negative integers &lt;math&gt;x,y &lt;/math&gt;. Show that &lt;math&gt;f(1,y) = y+2 &lt;/math&gt; for all non-negative integers &lt;math&gt;y&lt;/math&gt;. (Adapted from 1981 IMO, #6)<br /> <br /> =====Solution=====<br /> We observe that &lt;math&gt;f(1,0) = f(0,1) = 2 &lt;/math&gt; and that &lt;math&gt;f(1, y+1) = f(1, f(1,y)) = f(1,y) + 1&lt;/math&gt;, hence by induction, &lt;math&gt;f(1,y) = y+2 &lt;/math&gt;, as desired.<br /> <br /> ===Infinite Descent===<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29372 User:Temperal/The Problem Solver's Resource Proofs 2009-01-11T00:08:01Z <p>Temperal: /* Solution */ not needed</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the methods of proof section.}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Methods of Proof&lt;/span&gt;==<br /> Many competitions involve only short answer or multiple choice questions, with no justification required. However, once you get to the elite, national competitions, mostly national olympiads, you suddenly are expected to justify your solutions. If you've never encountered problems where you must prove your results, you would most likely be overwhelmed. This page attempts to explain some of the most used techniques in writing proofs. Please note that most material on this page is relatively simple; this section is only for introducing beginners to new techniques. If one wishes to improve at more advanced problem solving, solving problems is recommended.<br /> <br /> ===Contradiction===<br /> To many people, contradiction is one of the simplest methods of proof. In this technique, you attempt to prove the desired result in an indirect way. You first assume that the opposite result holds. Then, you attempt to find something that contradicts something that you already know or is given in the problem. Thus, you must've been wrong in assuming that the opposite result is true, so the only possible result is the one you wanted to prove!<br /> <br /> ====Example Problem 1====<br /> Prove that there are infinite primes. (This is an extremely classic and elegant proof)<br /> <br /> =====Solution=====<br /> Many new problem solvers are intimidated by a question that asks to prove anything relating to infinity. Hence, this seems like an excellent time to use contradiction, as we simply need to find a contradiction. Since our desired result is proving that there are an infinite amount of primes, it makes sense to assume that there are only a finite amount. Let the primes be &lt;math&gt;p_1, p_2, p_3,...p_n&lt;/math&gt; where &lt;math&gt;p_1&lt;p_2&lt;p_3...&lt;p_n&lt;/math&gt;. If we can find another prime number that is greater than &lt;math&gt;p_n&lt;/math&gt;, we will have disproved our assumption. Consider the number &lt;math&gt;p_1p_2p_3...p_n+1&lt;/math&gt;. This number is not divisible by any of our known primes, so it itself must be prime. Also, it is evident that this number is greater than &lt;math&gt;p_n&lt;/math&gt; so we have found a new prime number. This violates our assumption, so there must be an infinite number of primes.<br /> <br /> ====Example Problem 2====<br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational. (Also a very classic question)<br /> <br /> =====Solution=====<br /> The assumption should be fairly obvious here. We assume that &lt;math&gt;\sqrt{2}&lt;/math&gt; is rational. From the definition of rational numbers, &lt;math&gt;\sqrt{2}&lt;/math&gt; can then be written as a fraction in lowest terms. Let &lt;math&gt;\sqrt{2}=\frac{p}{q}&lt;/math&gt;, where p and q are relatively prime. Squaring and multiplying by &lt;math&gt;q^2&lt;/math&gt;, we have &lt;math&gt;2q^2=p^2&lt;/math&gt;. As 2 divides the left hand side, it must also divide p (a very common example of divisibility being helpful). Let &lt;math&gt;p=2r&lt;/math&gt;; then &lt;math&gt;2q^2=4r^2\rightarrowq^2=2r^2&lt;/math&gt;. By the same argument as last time, we let &lt;math&gt;q=2s&lt;/math&gt;. However, now we have &lt;math&gt;\sqrt{2}=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}&lt;/math&gt;. Recall though, that we stated that p and q were relatively prime. Now, we see that both are divisible by 2. This contradicts our original assumption, so &lt;math&gt;\sqrt{2}&lt;/math&gt; must indeed be rational.<br /> <br /> ===Induction===<br /> <br /> ===Infinite Descent===<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29371 User:Temperal/The Problem Solver's Resource Proofs 2009-01-11T00:07:46Z <p>Temperal: notoc, please</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the methods of proof section.}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Methods of Proof&lt;/span&gt;==<br /> Many competitions involve only short answer or multiple choice questions, with no justification required. However, once you get to the elite, national competitions, mostly national olympiads, you suddenly are expected to justify your solutions. If you've never encountered problems where you must prove your results, you would most likely be overwhelmed. This page attempts to explain some of the most used techniques in writing proofs. Please note that most material on this page is relatively simple; this section is only for introducing beginners to new techniques. If one wishes to improve at more advanced problem solving, solving problems is recommended.<br /> <br /> ===Contradiction===<br /> To many people, contradiction is one of the simplest methods of proof. In this technique, you attempt to prove the desired result in an indirect way. You first assume that the opposite result holds. Then, you attempt to find something that contradicts something that you already know or is given in the problem. Thus, you must've been wrong in assuming that the opposite result is true, so the only possible result is the one you wanted to prove!<br /> <br /> ====Example Problem 1====<br /> Prove that there are infinite primes. (This is an extremely classic and elegant proof)<br /> <br /> =====Solution=====<br /> Solution: Many new problem solvers are intimidated by a question that asks to prove anything relating to infinity. Hence, this seems like an excellent time to use contradiction, as we simply need to find a contradiction. Since our desired result is proving that there are an infinite amount of primes, it makes sense to assume that there are only a finite amount. Let the primes be &lt;math&gt;p_1, p_2, p_3,...p_n&lt;/math&gt; where &lt;math&gt;p_1&lt;p_2&lt;p_3...&lt;p_n&lt;/math&gt;. If we can find another prime number that is greater than &lt;math&gt;p_n&lt;/math&gt;, we will have disproved our assumption. Consider the number &lt;math&gt;p_1p_2p_3...p_n+1&lt;/math&gt;. This number is not divisible by any of our known primes, so it itself must be prime. Also, it is evident that this number is greater than &lt;math&gt;p_n&lt;/math&gt; so we have found a new prime number. This violates our assumption, so there must be an infinite number of primes.<br /> <br /> ====Example Problem 2====<br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational. (Also a very classic question)<br /> <br /> =====Solution=====<br /> The assumption should be fairly obvious here. We assume that &lt;math&gt;\sqrt{2}&lt;/math&gt; is rational. From the definition of rational numbers, &lt;math&gt;\sqrt{2}&lt;/math&gt; can then be written as a fraction in lowest terms. Let &lt;math&gt;\sqrt{2}=\frac{p}{q}&lt;/math&gt;, where p and q are relatively prime. Squaring and multiplying by &lt;math&gt;q^2&lt;/math&gt;, we have &lt;math&gt;2q^2=p^2&lt;/math&gt;. As 2 divides the left hand side, it must also divide p (a very common example of divisibility being helpful). Let &lt;math&gt;p=2r&lt;/math&gt;; then &lt;math&gt;2q^2=4r^2\rightarrowq^2=2r^2&lt;/math&gt;. By the same argument as last time, we let &lt;math&gt;q=2s&lt;/math&gt;. However, now we have &lt;math&gt;\sqrt{2}=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}&lt;/math&gt;. Recall though, that we stated that p and q were relatively prime. Now, we see that both are divisible by 2. This contradicts our original assumption, so &lt;math&gt;\sqrt{2}&lt;/math&gt; must indeed be rational.<br /> <br /> ===Induction===<br /> <br /> ===Infinite Descent===<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Proofs&diff=29370 User:Temperal/The Problem Solver's Resource Proofs 2009-01-11T00:07:31Z <p>Temperal: skeleton</p> <hr /> <div>{{User:Temperal/testtemplate|the methods of proof section.}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Methods of Proof&lt;/span&gt;==<br /> Many competitions involve only short answer or multiple choice questions, with no justification required. However, once you get to the elite, national competitions, mostly national olympiads, you suddenly are expected to justify your solutions. If you've never encountered problems where you must prove your results, you would most likely be overwhelmed. This page attempts to explain some of the most used techniques in writing proofs. Please note that most material on this page is relatively simple; this section is only for introducing beginners to new techniques. If one wishes to improve at more advanced problem solving, solving problems is recommended.<br /> <br /> ===Contradiction===<br /> To many people, contradiction is one of the simplest methods of proof. In this technique, you attempt to prove the desired result in an indirect way. You first assume that the opposite result holds. Then, you attempt to find something that contradicts something that you already know or is given in the problem. Thus, you must've been wrong in assuming that the opposite result is true, so the only possible result is the one you wanted to prove!<br /> <br /> ====Example Problem 1====<br /> Prove that there are infinite primes. (This is an extremely classic and elegant proof)<br /> <br /> =====Solution=====<br /> Solution: Many new problem solvers are intimidated by a question that asks to prove anything relating to infinity. Hence, this seems like an excellent time to use contradiction, as we simply need to find a contradiction. Since our desired result is proving that there are an infinite amount of primes, it makes sense to assume that there are only a finite amount. Let the primes be &lt;math&gt;p_1, p_2, p_3,...p_n&lt;/math&gt; where &lt;math&gt;p_1&lt;p_2&lt;p_3...&lt;p_n&lt;/math&gt;. If we can find another prime number that is greater than &lt;math&gt;p_n&lt;/math&gt;, we will have disproved our assumption. Consider the number &lt;math&gt;p_1p_2p_3...p_n+1&lt;/math&gt;. This number is not divisible by any of our known primes, so it itself must be prime. Also, it is evident that this number is greater than &lt;math&gt;p_n&lt;/math&gt; so we have found a new prime number. This violates our assumption, so there must be an infinite number of primes.<br /> <br /> ====Example Problem 2====<br /> Prove that &lt;math&gt;\sqrt{2}&lt;/math&gt; is irrational. (Also a very classic question)<br /> <br /> =====Solution=====<br /> The assumption should be fairly obvious here. We assume that &lt;math&gt;\sqrt{2}&lt;/math&gt; is rational. From the definition of rational numbers, &lt;math&gt;\sqrt{2}&lt;/math&gt; can then be written as a fraction in lowest terms. Let &lt;math&gt;\sqrt{2}=\frac{p}{q}&lt;/math&gt;, where p and q are relatively prime. Squaring and multiplying by &lt;math&gt;q^2&lt;/math&gt;, we have &lt;math&gt;2q^2=p^2&lt;/math&gt;. As 2 divides the left hand side, it must also divide p (a very common example of divisibility being helpful). Let &lt;math&gt;p=2r&lt;/math&gt;; then &lt;math&gt;2q^2=4r^2\rightarrowq^2=2r^2&lt;/math&gt;. By the same argument as last time, we let &lt;math&gt;q=2s&lt;/math&gt;. However, now we have &lt;math&gt;\sqrt{2}=\frac{p}{q}=\frac{2r}{2s}=\frac{r}{s}&lt;/math&gt;. Recall though, that we stated that p and q were relatively prime. Now, we see that both are divisible by 2. This contradicts our original assumption, so &lt;math&gt;\sqrt{2}&lt;/math&gt; must indeed be rational.<br /> <br /> ===Induction===<br /> <br /> ===Infinite Descent===<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Tips_and_Tricks&diff=29369 User:Temperal/The Problem Solver's Resource Tips and Tricks 2009-01-11T00:06:15Z <p>Temperal: numbah?</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the tips and tricks section}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Other Tips and Tricks&lt;/span&gt;==<br /> This is a collection of general techniques for solving problems.<br /> *Don't be afraid to use casework! Sometimes it's the only way. (But be VERY afraid to use brute force.)<br /> *Remember that substitution is a useful technique! Example problem: <br /> ===Example Problem 1===<br /> If &lt;math&gt;\tan x+\tan y=25&lt;/math&gt; and &lt;math&gt;\cot x+\cot y=30&lt;/math&gt;, find &lt;math&gt;\tan(x+y)&lt;/math&gt;.<br /> <br /> ====Solution====<br /> Let &lt;math&gt;X = \tan x&lt;/math&gt;, &lt;math&gt;Y = \tan y&lt;/math&gt;. Thus, &lt;math&gt;X + Y = 25&lt;/math&gt;, &lt;math&gt;\frac{1}{X} + \frac{1}{Y} = 30&lt;/math&gt;, so &lt;math&gt;XY = \frac{5}{6}&lt;/math&gt;, hence &lt;math&gt;\tan(x+y)=\frac{X+Y}{1-XY}&lt;/math&gt;, which turns out to be &lt;math&gt;\boxed{150}&lt;/math&gt;.<br /> <br /> This technique can also be used to solve quadratics of high degrees, i.e. &lt;math&gt;x^{16}+x^4+6=0&lt;/math&gt;; let &lt;math&gt;y=x^4&lt;/math&gt;, and solve from there.<br /> <br /> ----<br /> <br /> *Remember the special properties of odd numbers: For any odd number &lt;math&gt;o&lt;/math&gt;, &lt;math&gt;o=2n\pm 1&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;o=a^2-(a-1)^2&lt;/math&gt; for some positive integer &lt;math&gt;a&lt;/math&gt;.<br /> <br /> ===Example Problem 2===<br /> How many quadruples &lt;math&gt;(a,b,c,d)&lt;/math&gt; are there such that &lt;math&gt;a+b+c+d=98&lt;/math&gt; and &lt;math&gt;a,b,c,d&lt;/math&gt; are all odd?<br /> <br /> ====Solution====<br /> Since they're odd, &lt;math&gt;a, b, c, d&lt;/math&gt; can each be expressed as &lt;math&gt;2n+1&lt;/math&gt; for some positive integer (or zero) &lt;math&gt;n&lt;/math&gt;.<br /> Thus:<br /> &lt;math&gt;2n_1-1+2n_2-1+2n_3+1+2n_4+1=98&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 2(n_1+n_2+n_3+n_4)+4=98&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 2(n_1+n_2+n_3+n_4)=94&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow n_1+n_2+n_3+n_4=47&lt;/math&gt;<br /> Binomial coefficients will yield the answer of &lt;math&gt;\boxed{19600}&lt;/math&gt;. <br /> <br /> ----<br /> <br /> *The AM-GM and Trivial inequalities are more useful than you might imagine!<br /> <br /> *Memorize, memorize, memorize the following things:<br /> #The trigonometric facts. <br /> #Everything on the Combinatorics page.<br /> #Integrals and derivatives, especially integrals.<br /> <br /> Remember, though, don't memorize without understanding!<br /> <br /> *Test your skills on practice [[AIME]]s (&lt;url&gt;resources.php more resources&lt;/url&gt;) often! <br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource_Tips_and_Tricks&diff=29365 User:Temperal/The Problem Solver's Resource Tips and Tricks 2009-01-11T00:04:23Z <p>Temperal: ticks?</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|the tips and tricks section}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Other Tips and Tricks&lt;/span&gt;==<br /> This is a collection of general techniques for solving problems.<br /> *Don't be afraid to use casework! Sometimes it's the only way. (But be VERY afraid to use brute force.)<br /> *Remember that substitution is a useful technique! Example problem: <br /> ===Example Problem Number 1===<br /> If &lt;math&gt;\tan x+\tan y=25&lt;/math&gt; and &lt;math&gt;\cot x+\cot y=30&lt;/math&gt;, find &lt;math&gt;\tan(x+y)&lt;/math&gt;.<br /> <br /> ====Solution====<br /> Let &lt;math&gt;X = \tan x&lt;/math&gt;, &lt;math&gt;Y = \tan y&lt;/math&gt;. Thus, &lt;math&gt;X + Y = 25&lt;/math&gt;, &lt;math&gt;\frac{1}{X} + \frac{1}{Y} = 30&lt;/math&gt;, so &lt;math&gt;XY = \frac{5}{6}&lt;/math&gt;, hence &lt;math&gt;\tan(x+y)=\frac{X+Y}{1-XY}&lt;/math&gt;, which turns out to be &lt;math&gt;\boxed{150}&lt;/math&gt;.<br /> <br /> This technique can also be used to solve quadratics of high degrees, i.e. &lt;math&gt;x^{16}+x^4+6=0&lt;/math&gt;; let &lt;math&gt;y=x^4&lt;/math&gt;, and solve from there.<br /> <br /> ----<br /> <br /> *Remember the special properties of odd numbers: For any odd number &lt;math&gt;o&lt;/math&gt;, &lt;math&gt;o=2n\pm 1&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;o=a^2-(a-1)^2&lt;/math&gt; for some positive integer &lt;math&gt;a&lt;/math&gt;.<br /> <br /> ===Example Problem Number 2===<br /> How many quadruples &lt;math&gt;(a,b,c,d)&lt;/math&gt; are there such that &lt;math&gt;a+b+c+d=98&lt;/math&gt; and &lt;math&gt;a,b,c,d&lt;/math&gt; are all odd?<br /> <br /> ====Solution====<br /> Since they're odd, &lt;math&gt;a, b, c, d&lt;/math&gt; can each be expressed as &lt;math&gt;2n+1&lt;/math&gt; for some positive integer (or zero) &lt;math&gt;n&lt;/math&gt;.<br /> Thus:<br /> &lt;math&gt;2n_1-1+2n_2-1+2n_3+1+2n_4+1=98&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 2(n_1+n_2+n_3+n_4)+4=98&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 2(n_1+n_2+n_3+n_4)=94&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow n_1+n_2+n_3+n_4=47&lt;/math&gt;<br /> Binomial coefficients will yield the answer of &lt;math&gt;\boxed{19600}&lt;/math&gt;. <br /> <br /> ----<br /> <br /> *The AM-GM and Trivial inequalities are more useful than you might imagine!<br /> <br /> *Memorize, memorize, memorize the following things:<br /> #The trigonometric facts. <br /> #Everything on the Combinatorics page.<br /> #Integrals and derivatives, especially integrals.<br /> <br /> Remember, though, don't memorize without understanding!<br /> <br /> *Test your skills on practice [[AIME]]s (&lt;url&gt;resources.php more resources&lt;/url&gt;) often! <br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to Introduction]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29363 User:Temperal/The Problem Solver's Resource1 2009-01-11T00:03:47Z <p>Temperal: /* Sum of Angle Formulas */ proof</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist. The following proof is taken from ''the Art of Problem Solving, Vol. 2'' and is due to Masakazu Nihei of Japan, who originally had it published in ''Mathematics @ Informatics Quarterly'', Vol. 3, No. 2:<br /> <br /> {{asy image|1=&lt;asy&gt;<br /> pair A,B,C;<br /> C=(0,0);<br /> B=(10,0);<br /> A=(6,4);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,W);<br /> draw(A--(6,0));<br /> label(&quot;$\beta$&quot;,A,(-1,-2));<br /> label(&quot;$\alpha$&quot;,A,(1,-2.5));<br /> label(&quot;$H$&quot;,(6,0),S);<br /> draw((6,0)--(5.5,0)--(5.5,0.5)--(6,0.5)--cycle);<br /> &lt;/asy&gt;|2=right|3=Figure 1}}<br /> <br /> We'll find &lt;math&gt;[ABC]&lt;/math&gt; in two different ways: &lt;math&gt;\frac{1}{2}(AB)(AC)(\sin \angle BAC)&lt;/math&gt; and &lt;math&gt;[ABH]+[ACH]&lt;/math&gt;. We let &lt;math&gt;AH=1&lt;/math&gt;. We have:<br /> <br /> &lt;math&gt;[ABC]=[ABH]+[ACH]&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}(AC)(AB)(\sin \angle BAC)=\frac{1}{2}(AH)(BH)+\frac{1}{2}(AH)(CH)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{2}\left(\frac{1}{\cos \beta}\right)\left(\frac{1}{\cos \alpha}\right)(\sin \angle BAC)=\frac{1}{2}(1)(\tan \alpha)(\tan \beta)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin(\alpha+\beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathbb{QED.}&lt;/math&gt;<br /> <br /> ----<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29358 User:Temperal/The Problem Solver's Resource1 2009-01-10T23:39:34Z <p>Temperal: /* Basic Facts */ again</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec (\pi/2-A)=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist.<br /> <br /> &lt;!-- add proof --&gt;<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29357 User:Temperal/The Problem Solver's Resource1 2009-01-10T23:39:10Z <p>Temperal: /* Basic Facts */ parentheses</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec {\pi/2-A}=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist.<br /> <br /> &lt;!-- add proof --&gt;<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29356 User:Temperal/The Problem Solver's Resource1 2009-01-10T23:35:50Z <p>Temperal: /* Pythagorean identities */ hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec{\pi/2-A}=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist.<br /> <br /> &lt;!-- add proof --&gt;<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> These can be easily seen by going back to the unit circle and the definition of these trig functions.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29355 User:Temperal/The Problem Solver's Resource1 2009-01-10T23:35:23Z <p>Temperal: /* Sum of Angle Formulas */ hm</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec{\pi/2-A}=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> If we can prove this one, the other ones can be derived easily using the &quot;Basic Facts&quot; identities above. In fact, we can simply prove the addition case, for plugging &lt;math&gt;A=-B&lt;/math&gt; into the addition case gives the subtraction case.<br /> <br /> As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist.<br /> <br /> &lt;!-- add proof --&gt;<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> The following identities can be easily derived by plugging &lt;math&gt;A=B&lt;/math&gt; into the above:<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal https://artofproblemsolving.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&diff=29352 User:Temperal/The Problem Solver's Resource1 2009-01-10T23:27:53Z <p>Temperal: /* Terminology and Notation */ explanation</p> <hr /> <div>__NOTOC__<br /> {{User:Temperal/testtemplate|page 1}}<br /> ==&lt;span style=&quot;font-size:20px; color: blue;&quot;&gt;Trigonometric Formulas&lt;/span&gt;==<br /> Note that all measurements are in radians.<br /> ===Basic Facts===<br /> &lt;math&gt;\sin (-A)=-\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (-A)=\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (-A)=-\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin (\pi-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi-A) = -\cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (2\pi-A) = \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi+A) = \tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A)=\sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (\pi/2-A)=\cot A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sec{\pi/2-A}=\csc A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (\pi/2-A) = \sin A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cot (\pi/2-A)=\tan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\csc (\pi/2-A)=\sec A&lt;/math&gt;<br /> <br /> The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.<br /> <br /> ===Terminology and Notation===<br /> &lt;math&gt;\cot A=\frac{1}{\tan A}&lt;/math&gt;, but &lt;math&gt;\cot A\ne\tan^{-1} A}&lt;/math&gt;, the former being the reciprocal and the latter the inverse.<br /> <br /> &lt;math&gt;\csc A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\csc A\ne\sin^{-1} A}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sec A=\frac{1}{\sin A}&lt;/math&gt;, but &lt;math&gt;\sec A\ne\cos^{-1} A}&lt;/math&gt;.<br /> <br /> Speaking of inverses:<br /> <br /> &lt;math&gt;\tan^{-1} A=\text{atan } A=\arctan A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos^{-1} A=\text{acos } A=\arccos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin^{-1} A=\text{asin } A=\arcsin A&lt;/math&gt;<br /> <br /> ===Sum of Angle Formulas===<br /> &lt;math&gt;\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sin2A=2\sin A \cos A&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos2A=\cos^2 A - \sin^2 A&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=2\cos^2 A -1&lt;/math&gt;<br /> or<br /> &lt;math&gt;\cos2A=1- 2 \sin^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan2A=\frac{2\tan A}{1-\tan^2 A}&lt;/math&gt;<br /> <br /> ===Pythagorean identities===<br /> <br /> &lt;math&gt;\sin^2 A+\cos^2 A=1&lt;/math&gt; <br /> <br /> &lt;math&gt;1 + \tan^2 A = \sec^2 A&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + \cot^2 A = \csc^2 A&lt;/math&gt;<br /> <br /> for all &lt;math&gt;A&lt;/math&gt;.<br /> <br /> ===Other Formulas===<br /> ====Law of Cosines====<br /> In a triangle with sides &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; opposite angles &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, respectively,<br /> <br /> &lt;math&gt;c^2=a^2+b^2-2bc\cos A&lt;/math&gt;<br /> <br /> and:<br /> ====Law of Sines====<br /> <br /> &lt;math&gt;\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R&lt;/math&gt;<br /> <br /> ====Law of Tangents====<br /> <br /> For any &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;\tan a,\tan b \subset \mathbb{R}&lt;/math&gt;,<br /> &lt;math&gt;\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}&lt;/math&gt;<br /> <br /> ====Area of a Triangle====<br /> The [[area]] of a triangle can be found by <br /> <br /> &lt;math&gt;\frac 12ab\sin C&lt;/math&gt;<br /> <br /> [[User:Temperal/The Problem Solver's Resource|Back to intro]] | [[User:Temperal/The Problem Solver's Resource2|Continue to page 2]]</div> Temperal