# Definition

If there is $\triangle ABC$ and points $D,E,F$ on the sides $BC,CA,AB$ respectively such that $\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r$, then the ratio $\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}$.

Created by Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

# Proofs

## Proof 1

Proof by Gogobao:

We have: $\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1}$

We have: $[DEF] = [ABC] - [DCE] - [FAE] - [FBD]$

$[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}$

$[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}$

Therefore $[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})$

So we have $\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}$

## Proof 2

Proof by franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)$. In the barycentric coordinate system, 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 find that $$\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&\tfrac {1}{r+1}&\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&0&\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&\tfrac {r}{r+1}&0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.$$

## Proof 3

Proof by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG we let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$ for $x$, $y\in\mathbb{R}$. We then use Shoelace Forumla to get $[ABC]=\frac12|y|$. We then figure out that $D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)$, $E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)$, and $F=\left(\frac{r}{r+1}, 0\right)$ so we know that by Shoelace Formula $\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|$. We know that $\frac{r^2-r+1}{(r+1)^2}\ge0$ for all $r\in\mathbb{R}$ so $\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}$.

## Proof 4

Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)

# Application 1

## Problem

The Wooga Looga Theorem states that the solution to this problem by franzliszt:

In $\triangle ABC$ points $X,Y,Z$ are on sides $BC,CA,AB$ such that $\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71$. Find the ratio of $[XYZ]$ to $[ABC]$.

## Solution 1

One solution is this one by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

WLOG let $A=(0, 0)$, $B=(1, 0)$, $C=(x, y)$. Then $[ABC]=\frac12|y|$ by Shoelace Theorem and $X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)$, $Y=\left(\frac{x}{8}, \frac{y}{8}\right)$, $Z=\left(\frac78, 0\right)$. Then $[XYZ]=\frac12\left|\frac{43y}{64}\right|$ by Shoelace Theorem. Therefore the answer is $\boxed{\frac{43}{64}}$.

## Solution 2

or this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)$. In the barycentric coordinate system, 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 find that $$\frac{[XYZ]}{[ABC]}=\begin{vmatrix} 0&\tfrac 18&\tfrac 78\\ \tfrac 78&0&\tfrac 18\\ \tfrac18&\tfrac78&0 \end{vmatrix}=\frac{43}{64}.$$ $\blacksquare$

## Solution 3

or this solution by aaja3427:

According the the Wooga Looga Theorem, It is $\frac{49-7+1}{8^2}$. This is $\boxed{\frac{43}{64}}$

## Solution 4

or this solution by AoPS user ilovepizza2020:

We use the $\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}$ to instantly get $\boxed{\frac{43}{64}}$. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)

## Solution 5

or this solution by eduD_looC:

This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being $\boxed{\frac{43}{64}}$. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.

## Solution 6

or this solution by CoolJupiter:

Wow. All of your solutions are slow, compared to my sol:

By math, we have $\boxed{\frac{43}{64}}$.

~CoolJupiter ^ | EVERYONE USE THIS SOLUTION IT'S BRILLIANT ~bsu1

Yes, very BRILLIANT!


~ TheAoPSLebron

## The Best Solution

By the $1+1=\text{BREAD}$ principle, we get $\boxed{\frac{43}{64}}$. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.

## Easiest Solution

The answer is clearly $\boxed{\frac{43}{64}}$. We leave the proof and intermediate steps to the reader as an exercise.

# Application 2

## Problem

The Wooga Looga Theorem states that the solution to this problem by Matholic:

The figure below shows a triangle ABC whose area is $72 \text{cm}^2$. If $\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}$, find $[DEF].$

~LaTeX-ifyed by RP3.1415

## Solution 1

is this solution by franzliszt:

We apply Barycentric Coordinates w.r.t. $\triangle ABC$. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then we find that $D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)$. In the barycentric coordinate system, 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 find that$$\frac{[DEF]}{[72]}=\begin{vmatrix} \tfrac 56&\tfrac 16&0\\ 0&\tfrac 56&\tfrac 16\\ \tfrac16&0&\tfrac56 \end{vmatrix}=\frac{7}{12}$$ so $[DEF]=42$. $\blacksquare$

## Solution 2

or this solution by RedFireTruck (talk) 12:11, 1 February 2021 (EST):

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}$. We are given that $[ABC]=72$ so $[DEF]=\frac{7}{12}\cdot72=\boxed{42}$

# Application 3

## Problem

The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:

Find the ratio $\frac{[GHI]}{[ABC]}$ if $\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12$ and $\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label("A", (0, 0), SW); label("B", (6, 0), SE); label("C", (4, 3), N); label("D", (2, 0), S); label("E", (16/3, 1), NE); label("F", (8/3, 2), NW); label("G", (11/3, 1/2), SE); label("H", (4, 3/2), NE); label("I", (7/3, 1), W); [/asy]$

## Solution 1

is this solution by franzliszt:

By the Wooga Looga Theorem, $\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13$. Notice that $\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\triangle ABC$. So $\frac{[GHI]}{[DEF]}=\frac 14$ and $\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}$ by Chain Rule ideas.

## Solution 2

or this solution by franzliszt:

Apply Barycentrics w.r.t. $\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)$. And $G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)$.

In the barycentric coordinate system, 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 find that$$\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.$$

# Application 4

## Problem

Let $ABC$ be a triangle and $D,E,F$ be points on sides $BC,AC,$ and $AB$ respectively. We have that $\frac{BD}{DC} = 3$ and similar for the other sides. If the area of triangle $ABC$ is $16$, then what is the area of triangle $DEF$? (By ilovepizza2020)

## Solution 1

By Franzliszt

By Wooga Looga, $\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}$ so the answer is $\boxed7$.

## Solution 2

By franzliszt

Apply Barycentrics w.r.t. $\triangle ABC$. Then $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. We can also find that $D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)$. In the barycentric coordinate system, 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 find that$$\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&\tfrac 14&\tfrac 34\\ \tfrac 34&0&\tfrac 14\\ \tfrac 14&\tfrac 34&0 \end{vmatrix}=\frac{7}{16}.$$So the answer is $\boxed{7}$.

## Solution 3

A long story short, the answer must be $\boxed{7}$ by the inverse of the Inverse Wooga Looga Theorem

## Solution 4

By TwoD_Horse

Another solution involves using the proof of contradiction.

We first let a real number $x$ be the solution of the problem, which we set $x\neq 7$. By using induction, we can find out that such $x$ does not exist. Because the process of induction involves the use of calculus, fractals, and chaos theory, it is omitted in order to not make this page too long (to threaten the position of the great Gmass). However, you can find details of the proof here.

Since such $x$ does not exist, it is clearly a contradiction. The solution will be all real numbers that are not in the domain of $x$, which is $\boxed{7}$.

# Testimonials

Pogpr0 = wooga looga - Ladka13 The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020

Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of The Ooga Booga Tribe of The Caveman Society

Franzlist is wooga looga howsopro - volkie boy

this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL

The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm

The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck (talk) 11:00, 1 February 2021 (EST)

The Wooga Looga Theorem is the best. -aaja3427

The Wooga Looga Theorem is needed for everything and it is great-hi..

The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click "about". now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT

This theorem has helped me with school and I am no longer failing my math class. -mchang

I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur

"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter

Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)

Too powerful... ~franzliszt

The Wooga Looga Theorem is so pro ~ ac142931

It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)

This theorem changed my life... ~ samrocksnature

Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3

It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3

This is franzliszt and I endorse this theorem. ~franzliszt

This theorem is too OP. ~bestzack66

This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283

Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415

The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020

It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321

The Wooga Looga Theorem is a special case of Routh's Theorem. So this wiki article is DEFINITELY needed. -peace

I actually thought this was a joke theorem until I read this page - HumanCalculator9

I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825

I ReAlLy don't get it - Senguamar HOW DARE YOU!!!!

The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.

You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan

If only I knew this on some contests that I had done previously... - JacobJB

The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri

"The Wooga Looga Theorem should be used in contests and should be part of geometry books." ~ Aops-g5-gethsemanea2 (talk) 21:56, 21 December 2020 (EST)

The Wooga Looga Theorem is so OP BRUH

thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie

I have no idea what is going on here - awesomeguy856

fuzimiao2013 waz hear

this theorem is bad

poggers theorem - awesomeming327

The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314

person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun

Wooga Looga Theorem is TRASH.

HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USAJMO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.