Difference between revisions of "Wooga Looga Theorem"

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\tfrac16&0&\tfrac56
 
\tfrac16&0&\tfrac56
 
\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math>
 
\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math>
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The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.
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~ilp2020

Revision as of 11:50, 29 October 2020

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 $[ABC]$.

is this solution by RedFireTruck:

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

and 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

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


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

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