https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Olive0827&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-19T06:23:17Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Circumradius&diff=103332 Circumradius 2019-02-18T06:51:14Z <p>Olive0827: /* Formula for Circumradius */</p> <hr /> <div>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, 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> Olive0827