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Difference between revisions of "The Devil's Triangle"

(Created page with "=Definition= 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 T...")
 
(→‎Proof 1)
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Now notice that <math>[DEF]=[ABC]-([BDF]+[CDE]+[AEF])</math>.
 
Now notice that <math>[DEF]=[ABC]-([BDF]+[CDE]+[AEF])</math>.
 +
 
We attempt to find the area of each of the smaller triangles.  
 
We attempt to find the area of each of the smaller triangles.  
 +
 +
 
Notice that <math>\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}</math> using the ratios derived earlier.
 
Notice that <math>\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}</math> using the ratios derived earlier.
 +
 +
 
Similarly, <math>\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}</math> and <math>\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}</math>.
 
Similarly, <math>\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}</math> and <math>\frac{[AEF]}{[ABC]}=\frac{t}{(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>.
 
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>.

Revision as of 09:37, 6 November 2020

Definition

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

Proof

Proof 1

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

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 $r=s=t$.

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