Revision as of 09:53, 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 let $A=(0, 0)$ , $B=(1, 0)$ , $C=(x, y)$ for $x$ , $y\in\mathbb{R}$

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

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