Mass Point Geometry

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     Anubhav Sehgal, currently pursuing MBA Core at NMIMS 2015- 2017 Batch, also blogs at Lagom - The MBA Blog as a passion for guiding aspirants for quality learning

    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. 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). There are three different situations that exist in MPG problems which needs to be explored first before moving on to some composite problems where in MPG is used as a sub part to the whole solution which will show us its true utility in geometry.

    Case of Concurrent Cevians

    A Cevian is any line segment in a triangle with one endpoint on a vertex of the triangle and the other endpoint on the opposite side. Medians, altitudes, and angle bisectors are special cases of cevians.

    Assign the vertex common to sides for which ratio split is given and set it as the LCM of two ratio points away from it so that all mass points assigned remain whole numbers.

    Case of Splitting Masses

    The lone vertex on one side of transversal (DE here) has split masses. That is to say, it is a common point to two see-saws having different second ends. Hence the mass points add up from both sides or are considered as split masses to balance both sides of the see saw. First assignment is made at any of the two vertices(A,B ) away from the lone vertex(C). The assignment made so that we get whole numbers at each point. So to achieve that we assign LCM(3,5) = 15 to the point A and then following simple assignment of inverse ratio mass points and added sum at the fulcrum point.

    Lone vertex,C gets 10m from the AC see saw and 9m from the BC see saw achieving a total of 19m to balance all see saws.

    Case of Multiple Systems

    In case we have two central points on a single see saw(any cevian) we have on our hands a case of multiple systems. In such a case we remove one line segment(containing one of the central points) at a time and solve the system assuming that line segment is absent. It would leave us a system which one of the above 2 cases and we can easily solve that. Similarly we do it after removing the other line segment with the other central point. Then use the ratios obtained from two separate systems to solve our problem by simple manipulation of ratios calculated.

    In triangle ABC, medians AD and CE intersect at P, PE = 1.5 , PD = 2, and DE = 2.5. What is the area of AEDC?

    a) 13  b) 13.5  c) 14  d) 14.5  e) 15

    Assign B mass m. Thus, because E is the midpoint of AB, A also has a mass of m.

    Similarly, C has a mass of m. D and E each have a mass of 2m because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A, so AP must be twice as long as PD.

    PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE = 1.5, so CP = 3 and CE = 4.5.

    Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE. This means that the quadrilateral AEDC is a kite.

    The area of a kite is half the product of the diagonals, AD and CE. Recall that they are 6 and 4.5 respectively, so the area of AEDC is 6*4.5/2 = 13.5

    Triangle ABC has AB = 21, AC = 22 and BC = 20. Points D and E are located on AB and AC, respectively, such that DE is parallel to BC and contains the centre of inscribed circle of triangle ABC. Then, DE = m/n, where m and n are relatively prime positive integers. Find m + n.

    In the given figure, DE || BC and AD : DB = 5 : 4 , find the ratio of areas of triangles BDE and BCF.

    a) 70 : 81  b) 16 : 25  c) 56 : 81  d) 52 : 81

    Let the area of triangle ABC be A. Then Area of triangle ADE = 5*5*A/[(5 + 4)(5 + 4)] = 25A/81

    Applying MPG as following and getting the shown ratios.

    We can say area of triangle BDE = (4/5)(25A/81) [Areas divided in the ratio of their shared base]

    Then we find the area of CEF similarly and subtract these areas from total to get area of BCF and our answer.


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