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)}=\frac{rst+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)}=\frac{rst+1}{(r+1)(s+1)(t+1)}</math>. | ||
Revision as of 23:02, 6 November 2020
Contents
[hide]Definition
The Devil's Triangle (Generalized Wooga Looga Theorem)
For any triangle , let and be points on and respectively. Devil's Triangle Theorem, states that if and , then .
(*Simplification found by @Gogobao)
Proofs
Proof 1
Proof by CoolJupiter:
We have the following ratios: .
Now notice that .
We attempt to find the area of each of the smaller triangles.
Notice that using the ratios derived earlier.
Similarly, and .
Thus, .
Finally, we have .
~@CoolJupiter
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 .
This is actually the second part of Routh's Theorem as pointed out by many. https://mathworld.wolfram.com/RouthsTheorem.html
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