Difference between revisions of "Mass points"

Latest revision as of 13:27, 8 August 2024

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.

How to Use

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. Given a line $AB$ with point $C$ on it, and the mass put on a point P is denoted as $m_P$ , 1. 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. In other words, $\frac{AC}{CB} = \frac{m_B}{m_A}$ . 2. If two points are balanced, the point on the balancing line used to balance them has a mass equal to the sum of the masses of the two points. That is, $m_C = m_A + m_B$ .

Examples:

Example 1

Consider a triangle $ABC$ with its three medians drawn, with the intersection points being $D, E, F,$ corresponding to $AB, BC,$ and $AC$ respectively. Let the centroid of triangle $ABC$ be $P$ . Prove $DP:PC$ is $1:2$ and the corresponding identities for medians from $A$ and $B$ .

Solution to Example 1

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 from $P$ . 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.

Example 2

$\triangle ABC$ has point $D$ on $AB$ , point $E$ on $BC$ , and point $F$ on $AC$ . $AE$ , $CD$ , and $BF$ intersect at point $G$ . The ratio $AD:DB$ is $3:5$ and the ratio $CE:EB$ is $8:3$ . Find the ratio of $FG:GB$

Solution to Example 2

Throughout this solution, let $W_{X}$ denote the weight at point $X$ . Since $\frac{CE}{EB} = \frac{8}{3}$ , let $W_{B} = 8$ , which makes $W_{C} = 3$ . Now, look at $AB$ . Since $W_{B} \cdot BD = W_{A} \cdot AD$ (this is a general property commonly used in many mass points problems, in fact it is the same property we used above to determine $W_{B}$ ), we have $W_{A} = W_{B} \cdot \frac{BD}{AD} = 8 \cdot \frac{5}{3} = \frac{40}{3}$ . Then, $W_{F} = W_{A} + W_{C} = \frac{40}{3} + 3 = \frac{49}{3}$ (another property of mass points). Finally, we have $FG:GB = W_{B}:W_{F} = 8:\frac{49}{3} = \boxed{24:49}$ .

Problems

2019 AMC 8 Problems/Problem 24

2016 AMC 10A Problems/Problem 19

2013 AMC 10B Problems/Problem 16

2004 AMC 10B Problems/Problem 20

2016 AMC 12A Problems/Problem 12

2009 AIME I Problems/Problem 5

2009 AIME I Problems/Problem 4

2001 AIME I Problems/Problem 7

2011 AIME II Problems/Problem 4

1992 AIME Problems/Problem 14

1988 AIME Problems/Problem 12

1989 AIME Problems/Problem 15

1985 AIME Problems/Problem 6

1971 AHSME Problems/Problem 26

More Info

See "Mass point geometry" on Wikipedia

Video Lectures

The Central NC Math Group released a lecture on Mass Points and Barycentric Coordinates, which you can view at https://www.youtube.com/watch?v=KQim7-wrwL0.

Here's another one from Double Donut: https://www.youtube.com/watch?v=VtBgp5WZni8

@@ Line 7: / Line 7: @@
 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.
 Given a line <math>AB</math> with point <math>C</math> on it, and the mass put on a point P is denoted as <math>m_P</math>,
-. 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. In other words, <math>\frac{AB}{CB} = \frac{m_A}{m_B}</math>.
+. 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. In other words, <math>\frac{AC}{CB} = \frac{m_B}{m_A}</math>.
 . If two points are balanced, the point on the balancing line used to balance them has a mass equal to the sum of the masses of the two points. That is, <math>m_C = m_A + m_B</math>.
@@ Line 13: / Line 13: @@
 == Example 1==
-Consider a triangle <math>ABC</math> with its three [[Median_(geometry)|median]]s drawn, with the intersection points being <math>D, E, F,</math> corresponding to <math>AB, BC,</math> and <math>AC</math> respectively.
+Consider a triangle <math>ABC</math> with its three [[Median_(geometry)|median]]s drawn, with the intersection points being <math>D, E, F,</math> corresponding to <math>AB, BC,</math> and <math>AC</math> respectively. Let the centroid of triangle <math>ABC</math> be <math>P</math>. Prove <math>DP:PC</math> is <math>1:2</math> and the corresponding identities for medians from <math>A</math> and <math>B</math>.
 == Solution to Example 1==
-Thus, if we label point <math>A</math> with a weight of <math>1</math>, <math>B</math> must also have a weight of <math>1</math> since <math>A</math> and <math>B</math> are equidistant from <math>D</math>. By the same process, we find <math>C</math> must also have a weight of 1. Now, since <math>A</math> and <math>B</math> both have a weight of <math>1</math>, <math>D</math> must have a weight of <math>2</math> (as is true for <math>E</math> and <math>F</math>). Thus, if we label the centroid <math>P</math>, we can deduce that <math>DP:PC</math> is <math>1:2</math> - the inverse ratio of their weights.
+Thus, if we label point <math>A</math> with a weight of <math>1</math>, <math>B</math> must also have a weight of <math>1</math> since <math>A</math> and <math>B</math> are equidistant from <math>P</math>. By the same process, we find <math>C</math> must also have a weight of 1. Now, since <math>A</math> and <math>B</math> both have a weight of <math>1</math>, <math>D</math> must have a weight of <math>2</math> (as is true for <math>E</math> and <math>F</math>). Thus, if we label the centroid <math>P</math>, we can deduce that <math>DP:PC</math> is <math>1:2</math> - the inverse ratio of their weights.
 == Example 2==

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