Revision as of 09:54, 6 November 2020

Definition

For any triangle $\triangle ABC$ , let $D, E$ and $F$ be points on $BC, AC$ and $AB$ respectively. Devil's Triangle Theorem states that if $\frac{BD}{CD}=r, \frac{CE}{AE}=s$ and $\frac{AF}{BF}=t$ , then $\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}$ .

Proofs

Proof 1

Proof by CoolJupiter:

We have the following ratios: $\frac{BD}{BC}=\frac{r}{r+1}, \frac{CD}{BC}=\frac{1}{r+1},\frac{CE}{AC}=\frac{s}{s+1}, \frac{AE}{AC}=\frac{1}{s+1},\frac{AF}{AB}=\frac{t}{t+1}, \frac{BF}{AB}=\frac{1}{t+1}$ .

Now notice that $[DEF]=[ABC]-([BDF]+[CDE]+[AEF])$ .

We attempt to find the area of each of the smaller triangles.

Notice that $\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}$ using the ratios derived earlier.

Similarly, $\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}$ and $\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}$ .

Thus, $\frac{[BDF]+[CDE]+[AEF]}{[ABC]}=\frac{r}{(r+1)(t+1)}+\frac{s}{(r+1)(s+1)}+\frac{t}{(s+1)(t+1)}=\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}$ .

Finally, we have $\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}$ .

~@CoolJupiter

Proof 2

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

Other Remarks

This theorem is a generalization of the Wooga Looga Theorem, which @RedFireTruck claims to have "rediscovered". The link to the theorem can be found here: https://artofproblemsolving.com/wiki/index.php/Wooga_Looga_Theorem

Essentially, Wooga Looga is a special case of this, specifically when $r=s=t$ .

Testimonials

The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck

This is Routh's theorem isn't it~ Ilovepizza2020

@@ Line 28: / Line 28: @@
 Proof by RedFireTruck:
-WLOG 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>
+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>F=\left(\frac{r}{r+1}, 0\right)</math>, <math>E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)</math>, and <math>D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\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>.
 =Other Remarks=

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