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2004 AMC 10B Problems/Problem 18
..., 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42. <math> \frac {42}{96} = \boxed{\frac ==Solution 6 (Wooga Looga Theorem)==

6 KB (904 words) - 12:54, 22 October 2023
The Devil's Triangle
==Generalized Wooga Looga Theorem (The Devil's Triangle)== ...y. The Generalized Wooga Looga Theorem (Gwoologth) or the Devil's Triangle Theorem states that if <math>\frac{BD}{CD}=r, \frac{CE}{AE}=s</math> and <math>\fra

3 KB (582 words) - 18:05, 23 March 2024
Bestzack66's AMC10 Study Plan
[[Wooga Looga Theorem]] [[Georgeooga-Harryooga Theorem]]

1,005 bytes (92 words) - 23:11, 17 January 2021
Arrangement Restriction Theorem
...not an alternative to the [[Georgeooga-Harryooga Theorem]] because in this theorem the only situation that is not allowed is that all <math>k</math> objects a By the Arrangement Restriction Theorem, we get <math>\frac{8! - (8 - 3 + 1)!3!}{2} = \boxed{18000}</math> because

2 KB (269 words) - 00:16, 21 December 2020