Difference between revisions of "Wooga Looga Theorem"

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This lemma sucks. Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemme is way better, as it trivializes every problem -Heavytoothpaste
 
This lemma sucks. Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemme is way better, as it trivializes every problem -Heavytoothpaste
 
  @above, this is a Theorem which is different than a Lemma thonk -franzi
 
  @above, this is a Theorem which is different than a Lemma thonk -franzi
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 +
This Theorem is literally false. Fix it.

Revision as of 20:46, 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.)

Solution 5

or this solution by eduD_looC:

This is a perfect application of the Adhiytaha Anweopifanwpuiefhbavpwiuefnapveuihfnpvwheibfpanuwvfaw Lemma, which results in the answer being $\boxed{\frac{43}{64}}$. A very beautiful application, which leaves graders and readers speechless.

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

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\[\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&\tfrac 12&\tfrac 16\\ \tfrac 16&\tfrac 13&\tfrac 12\\ \tfrac 12&\tfrac 16&\tfrac 13 \end{vmatrix}=\frac{1}{12}.\]

Testimonials

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

This lemma sucks. Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemme is way better, as it trivializes every problem -Heavytoothpaste 

@above, this is a Theorem which is different than a Lemma thonk -franzi

This Theorem is literally false. Fix it.

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