Difference between revisions of "Mass points"
m |
Juliankuang (talk | contribs) (→Problems) |
||
Line 11: | Line 11: | ||
==Problems== | ==Problems== | ||
+ | [[2019 AMC 8 Problems/Problem 24]] | ||
+ | |||
[[2013 AMC 10B Problems/Problem 16]] | [[2013 AMC 10B Problems/Problem 16]] | ||
Revision as of 20:51, 20 November 2019
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 was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems. 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 inversely 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).
The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Any line passing this central point will balance the figure. If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. If two points are balanced, the point on the balancing line used to balance them has a mass of the sum of the masses of the two points.
Examples
Consider a triangle with its three medians drawn, with the intersection points being
corresponding to
and
respectively. Thus, if we label point
with a weight of
,
must also have a weight of
since
and
are equidistant from
. By the same process, we find
must also have a weight of 1. Now, since
and
both have a weight of
,
must have a weight of
(as is true for
and
). Thus, if we label the centroid
, we can deduce that
is
- the inverse ratio of their weights.
has point
on
, point
on
, and point
on
.
,
, and
intersect at point
. The ratio
is
and the ratio
is
. Find the ratio of
Problems
2019 AMC 8 Problems/Problem 24
2013 AMC 10B Problems/Problem 16
2004 AMC 10B Problems/Problem 20
2009 AIME I Problems/Problem 5
2009 AIME I Problems/Problem 4
2001 AIME I Problems/Problem 7