https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Fryingpan546&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-20T04:37:02Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Circumradius&diff=98014 Circumradius 2018-10-04T00:45:12Z <p>Fryingpan546: /* Proof */</p> <hr /> <div>{{stub}}<br /> <br /> The '''circumradius''' of a [[cyclic]] [[polygon]] is the radius of the circumscribed circle of that polygon. For a triangle, it is the measure of the [[radius]] of the [[circle]] that [[circumscribe]]s the triangle. Since every triangle is [[cyclic]], every triangle has a circumscribed circle, or a [[circumcircle]].<br /> <br /> ==Formula for a Triangle==<br /> Let &lt;math&gt;a, b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; denote the triangle's three sides and let &lt;math&gt;A&lt;/math&gt; denote the area of the triangle. Then, the measure of the circumradius of the triangle is simply &lt;math&gt;R=\frac{abc}{4A}&lt;/math&gt;. Also, &lt;math&gt;A=\frac{abc}{4R}&lt;/math&gt;<br /> <br /> == Proof ==<br /> &lt;asy&gt;<br /> pair O, A, B, C, D;<br /> O=(0,0);<br /> A=(-5,1);<br /> B=(1,5);<br /> C=(5,1);<br /> dot(O); dot (A); dot (B); dot (C);<br /> draw(circle(O, sqrt(26)));<br /> draw(A--B--C--cycle);<br /> D=-B; dot (D);<br /> draw(B--D--A);<br /> label(&quot;$A$&quot;, A, W);<br /> label(&quot;$B$&quot;, B, N);<br /> label(&quot;$C$&quot;, C, E);<br /> label(&quot;$D$&quot;, D, S);<br /> label(&quot;$O$&quot;, O, W);<br /> pair E;<br /> E=foot(B,A,C);<br /> draw(B--E);<br /> dot(E);<br /> label(&quot;$E$&quot;, E, S);<br /> draw(rightanglemark(B,A,D,20));<br /> draw(rightanglemark(B,E,C,20));<br /> &lt;/asy&gt;<br /> <br /> <br /> We let &lt;math&gt;AB=c&lt;/math&gt;, &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AC=b&lt;/math&gt;, &lt;math&gt;BE=h&lt;/math&gt;, and &lt;math&gt;BO=R&lt;/math&gt;. We know that &lt;math&gt;\angle BAD&lt;/math&gt; is a right angle because &lt;math&gt;BD&lt;/math&gt; is the diameter. Also, &lt;math&gt;\angle ADB = \angle BCA&lt;/math&gt; because they both subtend arc &lt;math&gt;AB&lt;/math&gt;. Therefore, &lt;math&gt;\triangle BAD \sim \triangle BEC&lt;/math&gt; by AA similarity, so we have<br /> &lt;cmath&gt;\frac{BD}{BA} = \frac{BC}{BE},&lt;/cmath&gt; or &lt;cmath&gt; \frac {2R} c = \frac ah.&lt;/cmath&gt;<br /> However, remember that area &lt;math&gt;\triangle ABC = \frac {bh} 2&lt;/math&gt;, so &lt;math&gt;h=\frac{2 \times \text{Area}}b&lt;/math&gt;. Substituting this in gives us<br /> &lt;cmath&gt; \frac {2R} c = \frac a{\frac{2 \times \text{Area}}b},&lt;/cmath&gt; and then bash through the algebra to get<br /> &lt;cmath&gt; R=\frac{abc}{4\times \text{Area}},&lt;/cmath&gt;<br /> and we are done.<br /> <br /> ==Formula for Circumradius==<br /> &lt;math&gt;R = \frac{abc}{4rs}&lt;/math&gt;<br /> Where &lt;math&gt;R&lt;/math&gt; is the Circumradius, &lt;math&gt;r&lt;/math&gt; is the inradius, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are the respective sides of the triangle and &lt;math&gt;s = (a+b+c)/2&lt;/math&gt; is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that &lt;math&gt;A = rs&lt;/math&gt;.<br /> <br /> =====But if you don't know the inradius=====<br /> <br /> But, if you don't know the inradius, you can find the area of the triangle by Heron's Formula:<br /> <br /> &lt;math&gt;A=\sqrt{s(s-a)(s-b)(s-c)}&lt;/math&gt;<br /> <br /> ==Euler's Theorem for a Triangle==<br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; have circumcenter &lt;math&gt;O&lt;/math&gt; and incenter &lt;math&gt;I&lt;/math&gt;.Then &lt;cmath&gt;OI^2=R(R-2r) \implies R \geq 2r&lt;/cmath&gt;<br /> <br /> ==Proof==<br /> <br /> == Right triangles ==<br /> The hypotenuse of the triangle is the diameter of its circumcircle, and the circumcenter is its midpoint, so the circumradius is equal to half of the hypotenuse of the right triangle.<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,I;<br /> A=(0,0);<br /> B=(0,3);<br /> C=(4,0);<br /> draw(A--B--C--cycle);<br /> I=circumcenter(A,B,C);<br /> draw(circumcircle(A,B,C));<br /> label(&quot;$C$&quot;,I,N);<br /> dot(I);<br /> draw(rightanglemark(B,A,C,10));<br /> &lt;/asy&gt;<br /> <br /> == Equilateral triangles ==<br /> <br /> &lt;math&gt;R=\frac{s}{\sqrt3}&lt;/math&gt;<br /> <br /> where &lt;math&gt;s&lt;/math&gt; is the length of a side of the triangle.<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,I;<br /> A=(0,0);<br /> B=(1,0);<br /> C=intersectionpoint(arc(A,1,0,90),arc(B,1,90,180));<br /> draw(A--B--C--cycle);<br /> I=circumcenter(A,B,C);<br /> draw(circumcircle(A,B,C));<br /> label(&quot;$C$&quot;,I,E);<br /> dot(I);<br /> label(&quot;$s$&quot;,A--B,S);<br /> label(&quot;$s$&quot;,A--C,N);<br /> label(&quot;$s$&quot;,B--C,N);<br /> &lt;/asy&gt;<br /> <br /> == If all three sides are known ==<br /> <br /> &lt;math&gt;R=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}&lt;/math&gt;<br /> <br /> And this formula comes from the area of Heron and &lt;math&gt;R=\frac{abc}{4A}&lt;/math&gt;.<br /> <br /> == If you know just one side and its opposite angle ==<br /> <br /> &lt;math&gt;2R=\frac{a}{\sin{A}}&lt;/math&gt;<br /> <br /> (Extended Law of Sines)<br /> <br /> ==See also==<br /> * [[Inradius]]<br /> * [[Semiperimeter]]<br /> <br /> [[Category:Geometry]]</div> Fryingpan546 https://artofproblemsolving.com/wiki/index.php?title=Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality&diff=94525 Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality 2018-05-17T00:36:10Z <p>Fryingpan546: </p> <hr /> <div>The '''Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (RMS-AM-GM-HM), is an [[inequality]] of the [[root-mean square]], [[arithmetic mean]], [[geometric mean]], and [[harmonic mean]] of a set of [[positive]] [[real number]]s &lt;math&gt;x_1,\ldots,x_n&lt;/math&gt; that says:<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 if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;. This inequality can be expanded to the [[power mean inequality]].<br /> <br /> [[Image:RMS-AM-GM-HM.gif|frame|right|The inequality is clearly shown in this diagram for &lt;math&gt;n=2&lt;/math&gt;]]<br /> <br /> As a consequence we can have the following inequality:<br /> If &lt;math&gt;x_1,x_2,\cdots,x_n&lt;/math&gt; are positive reals, then <br /> &lt;math&gt;(x_1+x_2+\cdots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots \frac{1}{x_n}\right) \geq n^2&lt;/math&gt; with equality if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;; which follows directly by cross multiplication from the AM-HM inequality.This is extremely useful in problem solving.<br /> <br /> == Proof ==<br /> The inequality &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/math&gt; is a direct consequence of the [[Cauchy-Schwarz Inequality]]; &lt;math&gt;(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2&lt;/math&gt;, so &lt;math&gt;\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2&lt;/math&gt;, so &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/math&gt;.<br /> <br /> Alternatively, the RMS-AM can be proved using Jensen's inequality:<br /> Suppose we let &lt;math&gt;F(x)=x^2&lt;/math&gt; (We know that &lt;math&gt;F(x)&lt;/math&gt; is convex because &lt;math&gt;F'(x)=2x&lt;/math&gt; and therefore &lt;math&gt;F''(x)=2&gt;0&lt;/math&gt;). <br /> We have:<br /> &lt;math&gt;F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}&lt;/math&gt;;<br /> <br /> Factoring out the &lt;math&gt;\frac{1}{n}&lt;/math&gt; yields:<br /> <br /> &lt;math&gt;F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}&lt;/math&gt;<br /> <br /> <br /> Taking the square root to both sides (remember that both are positive):<br /> <br /> &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.&lt;/math&gt;<br /> <br /> <br /> <br /> The inequality &lt;math&gt;\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}&lt;/math&gt; is called the AM-GM inequality, and proofs can be found [[Proofs of AM-GM|here]].<br /> <br /> <br /> The inequality &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt; is a direct consequence of AM-GM; &lt;math&gt;\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt;.<br /> <br /> Therefore the original inequality is true.<br /> <br /> The Root Mean Square is also known as the [[quadratic mean]], and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality. <br /> <br /> <br /> {{stub}}<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Fryingpan546 https://artofproblemsolving.com/wiki/index.php?title=Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality&diff=94524 Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality 2018-05-17T00:35:17Z <p>Fryingpan546: </p> <hr /> <div>The '''Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (RMS-AM-GM-HM), is an [[inequality]] of the [[root-mean square]], [[arithmetic mean]], [[geometric mean]], and [[harmonic mean]] of a set of [[positive]] [[real number]]s &lt;math&gt;x_1,\ldots,x_n&lt;/math&gt; that says:<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 if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;. This inequality can be expanded to the [[power mean inequality with lots of chess and whatnot]].<br /> <br /> [[Image:RMS-AM-GM-HM.gif|frame|right|The inequality is clearly shown in this diagram for &lt;math&gt;n=2&lt;/math&gt;]]<br /> <br /> As a consequence we can have the following inequality:<br /> If &lt;math&gt;x_1,x_2,\cdots,x_n&lt;/math&gt; are positive reals, then <br /> &lt;math&gt;(x_1+x_2+\cdots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots \frac{1}{x_n}\right) \geq n^2&lt;/math&gt; with equality if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;; which follows directly by cross multiplication from the AM-HM inequality.This is extremely useful in problem solving.<br /> <br /> == Proof ==<br /> The inequality &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/math&gt; is a direct consequence of the [[Cauchy-Schwarz Inequality]]; &lt;math&gt;(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2&lt;/math&gt;, so &lt;math&gt;\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2&lt;/math&gt;, so &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/math&gt;.<br /> <br /> Alternatively, the RMS-AM can be proved using Jensen's inequality:<br /> Suppose we let &lt;math&gt;F(x)=x^2&lt;/math&gt; (We know that &lt;math&gt;F(x)&lt;/math&gt; is convex because &lt;math&gt;F'(x)=2x&lt;/math&gt; and therefore &lt;math&gt;F''(x)=2&gt;0&lt;/math&gt;). <br /> We have:<br /> &lt;math&gt;F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}&lt;/math&gt;;<br /> <br /> Factoring out the &lt;math&gt;\frac{1}{n}&lt;/math&gt; yields:<br /> <br /> &lt;math&gt;F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}&lt;/math&gt;<br /> <br /> <br /> Taking the square root to both sides (remember that both are positive):<br /> <br /> &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.&lt;/math&gt;<br /> <br /> <br /> <br /> The inequality &lt;math&gt;\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}&lt;/math&gt; is called the AM-GM inequality, and proofs can be found [[Proofs of AM-GM|here]].<br /> <br /> <br /> The inequality &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt; is a direct consequence of AM-GM; &lt;math&gt;\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt;.<br /> <br /> Therefore the original inequality is true.<br /> <br /> The Root Mean Square is also known as the [[quadratic mean]], and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality. <br /> <br /> <br /> {{stub}}<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Fryingpan546 https://artofproblemsolving.com/wiki/index.php?title=Asymptote_(Vector_Graphics_Language)&diff=94204 Asymptote (Vector Graphics Language) 2018-04-25T00:49:44Z <p>Fryingpan546: /* See also */</p> <hr /> <div>{{Asymptote}}<br /> <br /> '''Asymptote''' is a powerful vector graphics language designed for creating mathematical diagrams and figures. It can output images in either eps or pdf format, and is compatible with the standard mathematics typesetting language, [[LaTeX]]. It is also a complete programming language, and has cleaner syntax than its predecessor, [http://netlib.bell-labs.com/who/hobby/MetaPost.html MetaPost], which was a language used only for two-dimensional graphics.<br /> <br /> Here is an example of an image that can be produced using Asymptote:<br /> <br /> &lt;center&gt;[[Image:Figure1.jpg]]&lt;/center&gt;<br /> <br /> In a sense, Asymptote is the ruler and compass of typesetting.<br /> <br /> <br /> You can use Asymptote on the AoPSWiki right now, by enclosing the Asymptote code within &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;...&lt;/asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tags. For example, the following code<br /> &lt;pre&gt;&lt;nowiki&gt;&lt;asy&gt;<br /> draw((0,0)--(3,7),red);<br /> dot((0,0));<br /> dot((3,7));<br /> label(&quot;Produced with Asymptote &quot;+version.VERSION,point(S),2S);<br /> &lt;/asy&gt;&lt;/nowiki&gt;&lt;/pre&gt;<br /> created the picture <br /> &lt;center&gt;&lt;asy&gt;<br /> draw((0,0)--(3,7),red);<br /> dot((0,0));<br /> dot((3,7));<br /> label(&quot;Produced with Asymptote &quot;+version.VERSION,point(S),2S);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> And on the AoPS forums you can use &lt;tt&gt;&lt;nowiki&gt;[asy]..[/asy]&lt;/nowiki&gt;&lt;/tt&gt;<br /> <br /> Another example:<br /> <br /> &lt;pre&gt;&lt;nowiki&gt;[asy]<br /> pair A,B,C,X,Y,Z; <br /> A = (0,0);<br /> B = (1,0);<br /> C = (0.3,0.8);<br /> draw(A--B--C--A);<br /> X = (B+C)/2;<br /> Y = (A+C)/2;<br /> Z = (A+B)/2;<br /> draw(A--X, red);<br /> draw(B--Y,red);<br /> draw(C--Z,red);<br /> [/asy]&lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,X,Y,Z;<br /> A = (0,0);<br /> B = (1,0);<br /> C = (0.3,0.8);<br /> draw(A--B--C--A);<br /> X = (B+C)/2;<br /> Y = (A+C)/2;<br /> Z = (A+B)/2;<br /> draw(A--X, red);<br /> draw(B--Y,red);<br /> draw(C--Z,red);&lt;/asy&gt;<br /> === See also ===<br /> *[[LaTeX]]<br /> * [http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php AoPS's Getting Started with LaTeX guide plus a whole new class: Chess!.]<br /> * [http://www.artofproblemsolving.com/community/c68_latex_and_asymptote Asymptote Forum on AoPS]<br /> [[Asymptote: Getting Started | Next: Getting Started]]</div> Fryingpan546