Difference between revisions of "The Devil's Triangle"
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Now notice that <math>[DEF]=[ABC]-([BDF]+[CDE]+[AEF])</math>. | Now notice that <math>[DEF]=[ABC]-([BDF]+[CDE]+[AEF])</math>. | ||
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We attempt to find the area of each of the smaller triangles. | We attempt to find the area of each of the smaller triangles. | ||
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Notice that <math>\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}</math> using the ratios derived earlier. | Notice that <math>\frac{[BDF]}{[ABC]}=\frac{BF}{AB}\times \frac{BD}{BC}=\frac{r}{(r+1)(t+1)}</math> using the ratios derived earlier. | ||
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Similarly, <math>\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}</math> and <math>\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}</math>. | Similarly, <math>\frac{[CDE]}{[ABC]}=\frac{s}{(r+1)(s+1)}</math> and <math>\frac{[AEF]}{[ABC]}=\frac{t}{(s+1)(t+1)}</math>. | ||
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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>. |
Revision as of 09:37, 6 November 2020
Contents
[hide]Definition
For any triangle , let and be points on and respectively. Devil's Triangle Theorem states that if and , then .
Proof
Proof 1
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 .
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 .