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Difference between revisions of "Mass points"

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'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths.  In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s.  Mass points are generalized by [[barycentric coordinates]].
 
'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths.  In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s.  Mass points are generalized by [[barycentric coordinates]].
  
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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.  
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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.  
  
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== How to Use==
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== Uses ==
 
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).
 
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.  
 
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>,
 
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>,
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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, <math>\frac{AC}{CB} = \frac{m_B}{m_A}</math>.
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# 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>.
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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, <math>m_C = m_A + m_B</math>.
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# 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>.
  
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== Examples: ==
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== Examples ==
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adding examples later
 
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,
 
  
 
== Example 1==
 
== 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. 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>.
 
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>.
  
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== Solution to Example 1==
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=== Solution ===
  
 
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.
 
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.
  
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== Example 2==
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== Example 2 ==
 
<math>\triangle ABC</math> has point <math>D</math> on <math>AB</math>, point <math>E</math> on <math>BC</math>, and point <math>F</math> on <math>AC</math>. <math>AE</math>, <math>CD</math>, and <math>BF</math>  intersect at point <math>G</math>. The ratio <math>AD:DB</math> is <math>3:5</math> and the ratio <math>CE:EB</math> is <math>8:3</math>. Find the ratio of <math>FG:GB</math>
 
<math>\triangle ABC</math> has point <math>D</math> on <math>AB</math>, point <math>E</math> on <math>BC</math>, and point <math>F</math> on <math>AC</math>. <math>AE</math>, <math>CD</math>, and <math>BF</math>  intersect at point <math>G</math>. The ratio <math>AD:DB</math> is <math>3:5</math> and the ratio <math>CE:EB</math> is <math>8:3</math>. Find the ratio of <math>FG:GB</math>
  
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== Solution to Example 2==
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=== Solution ===
  
 
Throughout this solution, let <math>W_{X}</math> denote the weight at point <math>X</math>. Since <math>\frac{CE}{EB} = \frac{8}{3}</math>, let <math>W_{B} = 8</math>, which makes <math>W_{C} = 3</math>. Now, look at <math>AB</math>. Since <math>W_{B} \cdot BD = W_{A} \cdot AD</math> (this is a general property commonly used in many mass points problems, in fact it is the same property we used above to determine <math>W_{B}</math>), we have <math>W_{A} = W_{B} \cdot \frac{BD}{AD} = 8 \cdot \frac{5}{3} = \frac{40}{3}</math>. Then, <math>W_{F} = W_{A} + W_{C} = \frac{40}{3} + 3 = \frac{49}{3}</math> (another property of mass points). Finally, we have <math>FG:GB = W_{B}:W_{F} = 8:\frac{49}{3} = \boxed{24:49}</math>.
 
Throughout this solution, let <math>W_{X}</math> denote the weight at point <math>X</math>. Since <math>\frac{CE}{EB} = \frac{8}{3}</math>, let <math>W_{B} = 8</math>, which makes <math>W_{C} = 3</math>. Now, look at <math>AB</math>. Since <math>W_{B} \cdot BD = W_{A} \cdot AD</math> (this is a general property commonly used in many mass points problems, in fact it is the same property we used above to determine <math>W_{B}</math>), we have <math>W_{A} = W_{B} \cdot \frac{BD}{AD} = 8 \cdot \frac{5}{3} = \frac{40}{3}</math>. Then, <math>W_{F} = W_{A} + W_{C} = \frac{40}{3} + 3 = \frac{49}{3}</math> (another property of mass points). Finally, we have <math>FG:GB = W_{B}:W_{F} = 8:\frac{49}{3} = \boxed{24:49}</math>.
  
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==Problems==
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== Problems ==
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[[2019 AMC 8 Problems/Problem 24]]
 
  
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[[2016 AMC 10A Problems/Problem 19]]
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* [[2019 AMC 8 Problems/Problem 24]]
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* [[2016 AMC 10A Problems/Problem 19]]
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[[2013 AMC 10B Problems/Problem 16]]
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* [[2013 AMC 10B Problems/Problem 16]]
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* [[2004 AMC 10B Problems/Problem 20]]
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[[2004 AMC 10B Problems/Problem 20]]
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* [[2016 AMC 12A Problems/Problem 12]]
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* [[2009 AIME I Problems/Problem 5]]
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[[2016 AMC 12A Problems/Problem 12]]
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* [[2009 AIME I Problems/Problem 4]]
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* [[2003 AIME I Problems/Problem 15]]
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[[2009 AIME I Problems/Problem 5]]
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* [[2001 AIME I Problems/Problem 7 ]]
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* [[2011 AIME II Problems/Problem 4]]
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[[2009 AIME I Problems/Problem 4]]
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* [[1992 AIME  Problems/Problem 14 ]]
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* [[1988 AIME Problems/Problem 12]]
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[[2003 AIME I Problems/Problem 15]]
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* [[1989 AIME Problems/Problem 15]]
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* [[1985 AIME Problems/Problem 6]]
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[[2001 AIME I Problems/Problem 7 ]]
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* [[1971 AHSME Problems/Problem 26]]
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[[2011 AIME II Problems/Problem 4]]
 
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[[ 1992 AIME  Problems/Problem 14 ]]
 
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[[ 1988 AIME Problems/Problem 12]]
 
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[[1989 AIME Problems/Problem 15]]
 
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[[1985 AIME Problems/Problem 6]]
 
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<!-- Previous external links led to errors, removed-->
 
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[[1971 AHSME Problems/Problem 26]]
 
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==More Info==
 
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See [https://en.wikipedia.org/wiki/Mass_point_geometry "Mass point geometry" on Wikipedia]
 
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==Video Lectures==
 
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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.
 
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Here's another one from Double Donut: https://www.youtube.com/watch?v=VtBgp5WZni8
 
  
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== Videos ==
  
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* [//www.youtube.com/watch?v=KQim7-wrwL0 The Central NC Math Group's lecture on Mass Points and Barycentric Coordinates]
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* [//www.youtube.com/watch?v=VtBgp5WZni8 Video from Double Donut]
  
 
[[Category:Definition]]
 
[[Category:Definition]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]
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{{stub}}

Latest revision as of 14:43, 1 March 2025

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.

Uses

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

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

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

Videos

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