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Difference between revisions of "The Devil's Triangle"

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==Proof 2==
 
==Proof 2==
 
Proof by math_comb01
 
Proof by math_comb01
−
Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <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>
+
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>
−
  In the barycentric coordinate system, 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 find that <cmath>\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]}.</cmath>+\frac{rst}{([s+1][r+1][t+1]}.\]=\frac{rst+1}{([s+1][r+1][t+1]}.\]
+
  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>
 
~@Math_comb01
 
~@Math_comb01
  

Revision as of 03:29, 18 December 2020

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 Generalizwed Wooga Looga Theorem 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://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem

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

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