Difference between revisions of "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 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>. | 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 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>. | ||
| − | = | + | =Proofs= |
==Proof 1== | ==Proof 1== | ||
| + | Proof by CoolJupiter: | ||
We have the following ratios: | We have the following ratios: | ||
<math>\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}</math>. | <math>\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}</math>. | ||
| Line 23: | Line 24: | ||
~@CoolJupiter | ~@CoolJupiter | ||
| + | ==Proof 2== | ||
| + | Proof by RedFireTruck: | ||
| + | WLOG | ||
=Other Remarks= | =Other Remarks= | ||
Revision as of 09:52, 6 November 2020
Definition
For any triangle 
, let 
and 
be points on 
and 
respectively. Devil's Triangle Theorem states that if 
and 
, then ![$\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}$](http://latex.artofproblemsolving.com/9/9/6/9968c2deab3db080b719289db31b415efae3ad10.png)
.
Proofs
Proof 1
Proof by CoolJupiter:
We have the following ratios:
.
Now notice that ![$[DEF]=[ABC]-([BDF]+[CDE]+[AEF])$](http://latex.artofproblemsolving.com/3/4/8/348cf5258a141132df51ae2e1ec60dd89e37d47f.png)
.
We attempt to find the area of each of the smaller triangles.
Notice that ![$\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}$](http://latex.artofproblemsolving.com/e/1/f/e1f4ef80cad3ed00c9d0bc6fb8b2f56ed0faca5f.png)
using the ratios derived earlier.
Similarly, ![$\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}$](http://latex.artofproblemsolving.com/6/a/d/6ad09b527364466a3f92f10e3170e1600f3fd300.png)
and ![$\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}$](http://latex.artofproblemsolving.com/f/b/d/fbd5231991313efc841a68b5bcbb6794dfec3004.png)
.
Thus, ![$\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}$](http://latex.artofproblemsolving.com/a/0/1/a012303031a664b7d67454d68942d25743528b37.png)
.
Finally, we have ![$\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}$](http://latex.artofproblemsolving.com/5/e/a/5eaa6311987457d807a81db166673e522586049d.png)
.
~@CoolJupiter
Proof 2
Proof by RedFireTruck: WLOG
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 
.
Testimonials
The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck
