Difference between revisions of "Wooga Looga Theorem"

(Definition)
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According the the Wooga Looga Theorem, It is <math>\frac{49-7+1}{8^2}</math>. This is <math>\boxed{\frac{43}{64}}</math>
 
According the the Wooga Looga Theorem, It is <math>\frac{49-7+1}{8^2}</math>. This is <math>\boxed{\frac{43}{64}}</math>
 +
 +
==Solution 4==
 +
or this solution by ilovepizza2020:
 +
 +
We use the <math>\mathbf{FUNDEMENTAL~THEOREM~OF~GEOGEBRA}</math> to instantly get <math>\boxed{\frac{43}{64}}</math>. (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.)
  
 
=Application 2=
 
=Application 2=

Revision as of 19:08, 29 October 2020

Definition

If there is $\triangle ABC$ and points $X,Y,Z$ on the sides $BC,CA,AB$ such that $\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac mn$, then the ratio $\frac{[XYZ]}{[ABC]}=\frac{m^2-m+n}{(m+n)^2}$

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

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 $X=(\frac{7x+1}{8}, \frac{7y}{8})$, $Y=(\frac{x}{8}, \frac{y}{8})$, $Z=(\frac78, 0)$. Then $[XYZ]=\frac12|\frac{43y}{64}|$ 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 $X=(0,\tfrac 18,\tfrac 78),Y=(\tfrac 78,0,\tfrac 18),Z=(\tfrac18,\tfrac78,0)$. 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 \[\frac{[XYZ]}{[ABC]}=\begin{vmatrix} 0&\tfrac 18&\tfrac 78\\ \tfrac 78&0&\tfrac 18\\ \tfrac18&\tfrac78&0 \end{vmatrix}=\frac{43}{64}.\] $\blacksquare$

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.)

Application 2

Problem

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

The figure below shows a triangle ABC which area is 72cm2. If AD: DB = BE: EC =CF: FA =1: 5, find the area of triangle DEF

Solution

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 $[DEF]=42$. $\blacksquare$

Testimonials

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

"It is the best thing I have ever seen" - Barack Obama, https://youtu.be/TSIAeHO3vxY

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