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Difference between revisions of "The Devil's Triangle"

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=Definition=
 
=Definition=
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==The Devil's Triangle/Routh's Theorem
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==Generalized Wooga Looga Theorem (The Devil's Triangle)==
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For any triangle <math>\triangle ABC</math>, let <math>D, E</math> and <math>F</math> be points on <math>BC, AC</math> and <math>AB</math> respectively. Devil's Triangle Theorem, also known has Routh Theorem, states that if <math>\frac{BD}{CD}=r, \frac{CE}{AE}=s</math> and <math>\frac{AF}{BF}=t</math>, then <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}</math>, or <math>\frac{rst+1}{(r+1)(s+1)(t+1)}</math>(Shoutout to @Gogobao for pointing this out)
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For any triangle <math>\triangle ABC</math>, let <math>D, E</math> and <math>F</math> be points on <math>BC, AC</math> and <math>AB</math> respectively. The Generalized Wooga Looga Theorem (Gwoologth) or the Devil's Triangle Theorem states that if <math>\frac{BD}{CD}=r, \frac{CE}{AE}=s</math> and <math>\frac{AF}{BF}=t</math>, then <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\frac{rst+1}{(r+1)(s+1)(t+1)}</math>.
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(*Simplification found by @Gogobao)
  
 
=Proofs=
 
=Proofs=
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Thus, <math>\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}</math>.
 
Thus, <math>\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}</math>.
  
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Finally, we have <math>\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}</math>, or <math>\frac{rst+1}{(r+1)(s+1)(t+1)}</math>.  
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Finally, we have <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\boxed{\frac{rst+1}{(r+1)(s+1)(t+1)}}</math>.
  
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(Another shoutout to @Gogobao for the simplification, you are very helpful!!)
 
 
~@CoolJupiter
 
~@CoolJupiter
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==Proof 2==
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Proof by math_comb01
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Apply Barycentrics  <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>.  also  <math>D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {s}{s+1},0,\tfrac {1}{s+1}\right),F=\left(\tfrac {1}{t+1},\tfrac {t}{t+1},0\right)</math>
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In the barycentrics, the area formula is <math>[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we  <cmath>\frac{[DEF]}{[ABC]}</cmath>=<math> \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {s}{s+1}&0&\tfrac {1}{s+1}\\  \tfrac {1}{t+1}&\tfrac {t}{t+1}&0 \end{vmatrix}</math>=<math>\frac{1}{[s+1][r+1][t+1]}</math><math>+\frac{rst}{([s+1][r+1][t+1]}</math>=<math>\frac{rst+1}{([s+1][r+1][t+1]}</math>
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~@Math_comb01
  
 
=Other Remarks=
 
=Other Remarks=
 
This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have "rediscovered". The link to the theorem can be found here:
 
This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have "rediscovered". The link to the theorem can be found here:
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https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem
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https://webcache.googleusercontent.com/search?q=cache:Qoyk2gGO6x8J:https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem+&cd=1&hl=en&ct=clnk&gl=us&client=safari
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Essentially, Wooga Looga is a special case of this, specifically when <math>r=s=t</math>.
 
Essentially, Wooga Looga is a special case of this, specifically when <math>r=s=t</math>.
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=Testimonials=
 
=Testimonials=
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The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck
 
  
 
This is Routh's theorem isn't it~ Ilovepizza2020
 
This is Routh's theorem isn't it~ Ilovepizza2020
  
 
Wow this generalization of my theorem is amazing. good job. - Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society
 
Wow this generalization of my theorem is amazing. good job. - Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society
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trivial by <math>\frac{1}{2}ab\sin(C)</math> but ok ~ bissue
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"Very nice theorem" - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:12, 1 February 2021 (EST)
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“I in the o’l days I used this theorem all the when time trying to tame my mammoth my cave buddy told me to ooga booga” - peelybonehead 9,000 B.C.
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who else got redirected here from 2004 AMC 10B Problem 18 smh
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I love the Wooga Looga Theorem! ~ Math-lover1

Latest revision as of 17:05, 23 March 2024

Definition

Generalized Wooga Looga Theorem (The Devil's Triangle)

For any triangle $\triangle ABC$, let $D, E$ and $F$ be points on $BC, AC$ and $AB$ respectively. The Generalized Wooga Looga Theorem (Gwoologth) or the Devil's Triangle Theorem states that if $\frac{BD}{CD}=r, \frac{CE}{AE}=s$ and $\frac{AF}{BF}=t$, then $\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\frac{rst+1}{(r+1)(s+1)(t+1)}$.

(*Simplification found by @Gogobao)

Proofs

Proof 1

Proof by CoolJupiter:

We have the following ratios: $\frac{BD}{BC}=\frac{r}{r+1}, \frac{CD}{BC}=\frac{1}{r+1},\frac{CE}{AC}=\frac{s}{s+1}, \frac{AE}{AC}=\frac{1}{s+1},\frac{AF}{AB}=\frac{t}{t+1}, \frac{BF}{AB}=\frac{1}{t+1}$.

Now notice that $[DEF]=[ABC]-([BDF]+[CDE]+[AEF])$.

We attempt to find the area of each of the smaller triangles.


Notice that $\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}$ using the ratios derived earlier.


Similarly, $\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}$ and $\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}$.


Thus, $\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}$.

Finally, we have $\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\boxed{\frac{rst+1}{(r+1)(s+1)(t+1)}}$.

~@CoolJupiter

Proof 2

Proof by math_comb01 Apply Barycentrics $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. also $D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {s}{s+1},0,\tfrac {1}{s+1}\right),F=\left(\tfrac {1}{t+1},\tfrac {t}{t+1},0\right)$ 

In the barycentrics, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1} &z_{1} \\ x_{2} &y_{2} &z_{2} \\ x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we  \[\frac{[DEF]}{[ABC]}\]=$\begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {s}{s+1}&0&\tfrac {1}{s+1}\\   \tfrac {1}{t+1}&\tfrac {t}{t+1}&0 \end{vmatrix}$=$\frac{1}{[s+1][r+1][t+1]}$$+\frac{rst}{([s+1][r+1][t+1]}$=$\frac{rst+1}{([s+1][r+1][t+1]}$

~@Math_comb01

Other Remarks

This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have "rediscovered". The link to the theorem can be found here: https://webcache.googleusercontent.com/search?q=cache:Qoyk2gGO6x8J:https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem+&cd=1&hl=en&ct=clnk&gl=us&client=safari


Essentially, Wooga Looga is a special case of this, specifically when $r=s=t$.

Testimonials

This is Routh's theorem isn't it~ Ilovepizza2020

Wow this generalization of my theorem is amazing. good job. - Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society

trivial by $\frac{1}{2}ab\sin(C)$ but ok ~ bissue

"Very nice theorem" - RedFireTruck (talk) 12:12, 1 February 2021 (EST)

“I in the o’l days I used this theorem all the when time trying to tame my mammoth my cave buddy told me to ooga booga” - peelybonehead 9,000 B.C.

who else got redirected here from 2004 AMC 10B Problem 18 smh

I love the Wooga Looga Theorem! ~ Math-lover1

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