Difference between revisions of "The Devil's Triangle"
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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>. | 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> | + | 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>. |
− | + | (Another shoutout to @Gogobao for the simplification, you are very helpful!!) | |
~@CoolJupiter | ~@CoolJupiter | ||
Revision as of 15:33, 6 November 2020
Contents
[hide]Definition
For any triangle , let and be points on and respectively. Devil's Triangle Theorem, also known has Routh Theorem, states that if and , then , or (Shoutout to @Gogobao for pointing this out)
Proofs
Proof 1
Proof by CoolJupiter:
We have the following ratios: .
Now notice that .
We attempt to find the area of each of the smaller triangles.
Notice that using the ratios derived earlier.
Similarly, and .
Thus, .
Finally, we have , or . (Another shoutout to @Gogobao for the simplification, you are very helpful!!) ~@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 .
Testimonials
The Ooga Booga Tribe would be proud of you. Amazing theorem - RedFireTruck
This is Routh's theorem isn't it~ Ilovepizza2020