Wooga Looga Theorem

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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 $\frac{[DEF]}{[ABC]}=\left|\frac{r^2-r+1}{(r+1)^2}\right|$.

Created by the Ooga Booga Tribe of the Caveman Society, https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ

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:

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 $F=\left(\frac{r}{r+1}, 0\right)$, $E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)$, and $D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)$ 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 $\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}$.

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

is this solution by RedFireTruck:

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 ilovepizza2020:

We use the $\mathbf{FUNDEMENTAL~THEOREM~OF~GEOGEBRA}$ 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 tests 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.

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

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 $72cm^2$. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF

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:

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 $7$.

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 $\boxed{7}$.

Testimonials

Franzlist is wooga looga howsopro - volkie boy

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

The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020

The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck

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 3rd 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 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

This theorem changed my life... ~ samrocksnature

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

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