Revision as of 22:26, 6 November 2020

Definition

The Devil's Triangle

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 $\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\frac{rst+1}{(r+1)(s+1)(t+1)}$ .

(*Simplification found by @Gogobao)

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

~@CoolJupiter

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 $r=s=t$ .

This is actually the second part of Routh's Theorem as pointed out by many. https://mathworld.wolfram.com/RouthsTheorem.html

Testimonials

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

Wow this generalization of my theorem is amazing. good job. - Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society

@@ Line 25: / Line 25: @@
 Thus, <math>\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)}</math>.
-Finally, we have <math>\frac{[DEF]}{[ABC]}=\boxed{1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}}</math>, or <math>\frac{rst+1}{(r+1)(s+1)(t+1)}</math>.
+Finally, we have <math>\frac{[DEF]}{[ABC]}=1-\frac{r(s+1)+s(t+1)+t(r+1)}{(r+1)(s+1)(t+1)}=\boxed{\frac{rst+1}{(r+1)(s+1)(t+1)}}</math>.
-(Another shoutout to @Gogobao for the simplification, you are very helpful!!)
 ~@CoolJupiter

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