Difference between revisions of "Wooga Looga Theorem"

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=Definition=
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If there is <math>\triangle ABC</math> and points <math>D,E,F</math> on the sides <math>BC,CA,AB</math> respectively such that <math>\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r</math>, then the ratio <math>\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}</math>.
  
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Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
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=Proofs=
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==Proof 1==
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Proof by Gogobao:
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We have: <math>\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} </math>
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We have: <math>[DEF] = [ABC] - [DCE] - [FAE] - [FBD]</math>
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<math>[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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<math>[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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<math>[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}</math>
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Therefore <math>[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})</math>
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So we have <math>\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}</math>
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==Proof 2==
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Proof by franzliszt
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Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>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)</math>. In the barycentric coordinate system, the area formula is <math>[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]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\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}.</cmath>
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==Proof 3==
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Proof by RedFireTruck:
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WLOG we let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math> for <math>x</math>, <math>y\in\mathbb{R}</math>. We then use Shoelace Forumla to get <math>[ABC]=\frac12|y|</math>. We then figure out that <math>D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)</math>, <math>E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)</math>, and <math>F=\left(\frac{r}{r+1}, 0\right)</math> so we know that by Shoelace Formula <math>\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|</math>. We know that <math>\frac{r^2-r+1}{(r+1)^2}\ge0</math> for all <math>r\in\mathbb{R}</math> so <math>\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}</math>.
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==Proof 4==
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Proof by jasperE3:
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The Wooga Looga theorem is a direct application of Routh's Theorem when a=b=c.
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=Application 1=
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==Problem==
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The Wooga Looga Theorem states that the solution to this problem by franzliszt:
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In <math>\triangle ABC</math> points <math>X,Y,Z</math> are on sides <math>BC,CA,AB</math> such that <math>\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71</math>. Find the ratio of <math>[XYZ]</math> to <math>[ABC]</math>.
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==Solution 1==
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is this solution by RedFireTruck:
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WLOG let <math>A=(0, 0)</math>, <math>B=(1, 0)</math>, <math>C=(x, y)</math>. Then <math>[ABC]=\frac12|y|</math> by Shoelace Theorem and <math>X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)</math>, <math>Y=\left(\frac{x}{8}, \frac{y}{8}\right)</math>, <math>Z=\left(\frac78, 0\right)</math>. Then <math>[XYZ]=\frac12\left|\frac{43y}{64}\right|</math> by Shoelace Theorem. Therefore the answer is <math>\boxed{\frac{43}{64}}</math>.
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==Solution 2==
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or this solution by franzliszt:
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We apply Barycentric Coordinates w.r.t. <math>\triangle ABC</math>. Let <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then we find that <math>X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix}
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x_{1} &y_{1}  &z_{1} \\
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x_{2} &y_{2}  &z_{2} \\
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x_{3}& y_{3} & z_{3}
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\end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that <cmath>\frac{[XYZ]}{[ABC]}=\begin{vmatrix}
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0&\tfrac 18&\tfrac 78\\
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\tfrac 78&0&\tfrac 18\\
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\tfrac18&\tfrac78&0
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\end{vmatrix}=\frac{43}{64}.</cmath> <math>\blacksquare</math>
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==Solution 3==
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or this solution by aaja3427:
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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>
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==Solution 5==
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or this solution by eduD_looC:
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This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being <math>\boxed{\frac{43}{64}}</math>. A very beautiful application, which leaves graders and readers speechless.
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==Solution 6==
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or this solution by CoolJupiter:
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Wow. All of your solutions are slow, compared to my sol:
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By math, we have <math>\boxed{\frac{43}{64}}</math>.
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~CoolJupiter
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^
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|
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EVERYONE USE THIS SOLUTION IT'S BRILLIANT
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~bsu1
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Yes, very BRILLIANT!
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~ TheAoPSLebron
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==The Best Solution==
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By the 1+1=2 principle, we get <math>\boxed{\frac{43}{64}}</math>. Definitely the best method. When asked, please say that OlympusHero taught you this method. Cuz he did.
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=Application 2=
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==Problem==
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The Wooga Looga Theorem states that the solution to this problem by Matholic:
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The figure below shows a triangle ABC whose area is <math>72 \text{cm}^2</math>. If <math>\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}</math>, find <math>[DEF].</math>
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~LaTeX-ifyed by RP3.1415
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==Solution 1==
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is this solution by franzliszt:
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We apply Barycentric Coordinates w.r.t. <math>\triangle ABC</math>. Let <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then we find that <math>D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)</math>. In the barycentric coordinate system, the area formula is <math>[XYZ]=\begin{vmatrix}
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x_{1} &y_{1}  &z_{1} \\
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x_{2} &y_{2}  &z_{2} \\
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x_{3}& y_{3} & z_{3}
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\end{vmatrix}\cdot [ABC]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\frac{[DEF]}{[72]}=\begin{vmatrix}
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\tfrac 56&\tfrac 16&0\\
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0&\tfrac 56&\tfrac 16\\
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\tfrac16&0&\tfrac56
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\end{vmatrix}=\frac{7}{12}</cmath> so <math>[DEF]=42</math>. <math>\blacksquare</math>
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==Solution 2==
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or this solution by RedFireTruck:
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By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}</math>. We are given that <math>[ABC]=72</math> so <math>[DEF]=\frac{7}{12}\cdot72=\boxed{42}</math>
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=Application 3=
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==Problem==
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The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:
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Find the ratio <math>\frac{[GHI]}{[ABC]}</math> if <math>\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12</math> and <math>\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1</math> in the diagram below.<asy>
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draw((0, 0)--(6, 0)--(4, 3)--cycle);
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draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);
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draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);
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label("$A$", (0, 0), SW);
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label("$B$", (6, 0), SE);
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label("$C$", (4, 3), N);
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label("$D$", (2, 0), S);
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label("$E$", (16/3, 1), NE);
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label("$F$", (8/3, 2), NW);
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label("$G$", (11/3, 1/2), SE);
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label("$H$", (4, 3/2), NE);
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label("$I$", (7/3, 1), W);
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</asy>
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==Solution 1==
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is this solution by franzliszt:
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By the Wooga Looga Theorem, <math>\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13</math>. Notice that <math>\triangle GHI</math> is the medial triangle of '''Wooga Looga Triangle ''' of <math>\triangle ABC</math>. So <math>\frac{[GHI]}{[DEF]}=\frac 14</math> and <math>\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}</math> by Chain Rule ideas.
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==Solution 2==
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or this solution by franzliszt:
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Apply Barycentrics w.r.t. <math>\triangle ABC</math> so that <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. Then <math>D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)</math>. And <math>G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)</math>.
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In the barycentric coordinate system, the area formula is <math>[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]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\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}.</cmath>
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=Application 4=
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==Problem==
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Let <math>ABC</math> be a triangle and <math>D,E,F</math> be points on sides <math>BC,AC,</math> and <math>AB</math> respectively. We have that <math>\frac{BD}{DC} = 3</math> and similar for the other sides. If the area of triangle <math>ABC</math> is <math>16</math>, then what is the area of triangle <math>DEF</math>? (By ilovepizza2020)
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==Solution 1==
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By Franzliszt
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By Wooga Looga, <math>\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}</math> so the answer is <math>7</math>.
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==Solution 2==
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By franzliszt
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Apply Barycentrics w.r.t. <math>\triangle ABC</math>. Then <math>A=(1,0,0),B=(0,1,0),C=(0,0,1)</math>. We can also find that <math>D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)</math>. In the barycentric coordinate system, the area formula is <math>[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]</math> where <math>\triangle XYZ</math> is a random triangle and <math>\triangle ABC</math> is the reference triangle. Using this, we find that<cmath>\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}.</cmath>So the answer is <math>\boxed{7}</math>.
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=Testimonials=
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Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
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Franzlist is wooga looga howsopro - volkie boy
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The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111  -centslordm
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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
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The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - RedFireTruck
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The Wooga Looga Theorem is the best. -aaja3427
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The Wooga Looga Theorem is needed for everything and it is great-hi..
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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
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This theorem has helped me with school and I am no longer failing my math class. -mchang
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"I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman." ~CoolJupiter
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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)
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Too powerful... ~franzliszt
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The Wooga Looga Theorem is so pro ~ ac142931
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It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)
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This theorem changed my life... ~ samrocksnature
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Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3
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It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3
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This is franzliszt and I endorse this theorem. ~franzliszt
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This theorem is too OP. ~bestzack66
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This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283
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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
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The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020
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It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321
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The Wooga Looga Theorem is a special case of [url=https://en.wikipedia.org/wiki/Routh%27s_theorem]Routh's Theorem.[/url] So this wiki article is DEFINITELY needed. -peace
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I actually thought this was a joke theorem until I read this page - HumanCalculator9
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I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825
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This is <i>almost</i> as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun
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I ReAlLy don't get it - Senguamar
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The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.

Revision as of 13:41, 22 November 2020

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:

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

Proof by jasperE3:

The Wooga Looga theorem is a direct application of Routh's Theorem when a=b=c.

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

Yes, very BRILLIANT!

~ TheAoPSLebron

The Best Solution

By the 1+1=2 principle, we get $\boxed{\frac{43}{64}}$. Definitely the best method. When asked, please say that OlympusHero taught you this method. Cuz he did.

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:

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

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

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 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 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 [url=https://en.wikipedia.org/wiki/Routh%27s_theorem]Routh's Theorem.[/url] 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

This is almost as OP as the Adihaya Jayasharmaramankumarguptareddybavarajugopal Lemma. Needs to be nerfed. -CoolCarsonTheRun

I ReAlLy don't get it - Senguamar

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

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