Difference between revisions of "The Devil's Triangle"
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=Definition= | =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 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>. | + | ==Generalized Wooga Looga Theorem (The Devil's Triangle)== |
+ | 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>. | ||
+ | |||
+ | (*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>. | ||
− | Finally, we have <math>\frac{[DEF]}{[ABC]}= | + | 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>. |
~@CoolJupiter | ~@CoolJupiter | ||
==Proof 2== | ==Proof 2== | ||
− | Proof by | + | Proof by math_comb01 |
− | + | 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 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 | ||
=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: | ||
− | https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem | + | 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 <math>r=s=t</math>. | Essentially, Wooga Looga is a special case of this, specifically when <math>r=s=t</math>. | ||
− | |||
=Testimonials= | =Testimonials= | ||
− | |||
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 | ||
+ | |||
+ | trivial by <math>\frac{1}{2}ab\sin(C)</math> but ok ~ bissue | ||
+ | |||
+ | "Very nice theorem" - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 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 |
Latest revision as of 17:05, 23 March 2024
Contents
[hide]Definition
Generalized Wooga Looga Theorem (The Devil's Triangle)
For any triangle , let and be points on and respectively. The Generalized Wooga Looga Theorem (Gwoologth) or the 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
Proof 2
Proof by math_comb01 Apply Barycentrics . Then . also
In the barycentrics, the area formula is where is a random triangle and is the reference triangle. Using this, we ===
~@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 .
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 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