Difference between revisions of "Mass points"

(Problems)
(Problems)
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==Problems==
 
==Problems==
 
[http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10B_Problems/Problem_20 2004 AMC 10B #20]
 
[http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10B_Problems/Problem_20 2004 AMC 10B #20]
 +
 
[http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_I_Problems/Problem_4 2009 AIME I #4]
 
[http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_I_Problems/Problem_4 2009 AIME I #4]
  

Revision as of 15:43, 1 January 2014

Mass points is a technique in Euclidean geometry that can greatly simplify the proofs of many theorems concerning polygons, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local coordinate system to identify points by the ratios into which they divide line segments. Mass points are generalized by barycentric coordinates.

Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversly proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).

Examples

Consider a triangle $ABC$ with its three medians drawn, with the intersection points being $D, E, F,$ corresponding to $AB, BC,$ and $AC$ respectively. Thus, if we label point $A$ with a weight of $1$, $B$ must also have a weight of $1$ since $A$ and $B$ are equidistant to $D$. By the same process, we find $C$ must also have a weight of 1. Now, since $A$ and $B$ both have a weight of $1$, $D$ must have a weight of $2$ (as is true for $E$ and $F$). Thus, if we label the centroid $P$, we can deduce that $DP:PC$ is $1:2$ - the inverse ratio of their weights.

Problems

2004 AMC 10B #20

2009 AIME I #4

PLEASE ADD MORE!

External links

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