Difference between revisions of "Routh's Theorem"

(New page: In triangle <math>ABC</math>, <math>D</math>, <math>E</math> and <math>F</math> are points on sides <math>BC</math>, <math>AC</math>, and <math>AB</math>, respectively. Let <math>r=\frac{A...)
 
(please fix image, I gtg)
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<cmath>[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]</cmath>
 
<cmath>[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]</cmath>
 +
 +
<asy>
 +
unitsize(5);
 +
pair A,B,C,D,E,F,G,H,I;
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A=(10,20);
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B=(0,0);
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C=(30,0);
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D=(20,0);
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E=(16.66,13.33);
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F=(5,10);
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G=(14.585,11.6298);
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H=(9.998,8);
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I=(17.5,5);
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draw(A--B);
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draw(B--C);
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draw(C--A);
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draw(A--D);
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draw(B--E);
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draw(C--F);
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label("$A$",A,N);
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label("$B$",B,SW);
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label("$C$",C,SE);
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label("$D$",D,S);
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label("$E$",E,NE);
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label("$F$",F,NW);
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label("$G$",G,N);
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label("$H$",H,N);
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label("$I$",I,SW);</asy>
  
 
==Proof==
 
==Proof==

Revision as of 14:19, 4 August 2008

In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\frac{AF}{AB}$, $s=\frac{BD}{BC}$, and $=\frac{CE}{CA}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Thus

\[[GHI]=\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]\]

[asy] unitsize(5); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$G$",G,N); label("$H$",H,N); label("$I$",I,SW);[/asy]

Proof

Template:Incomplete

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