MASS POINT GEOMETRY TOM RIKE OAKLAND HIGH SCHOOL

Transcription

1 MSS POINT GOMTRY TOM RIK OKLN HIGH SHOOL. Introduction.. The power of the mass point technique. This session will introduce a technique that simplifies calculations of ratios in geometric figures in an intuitive way by merging algebra, geometry and basic physics. When the method can be applied, it is far faster than the standard techniques of vectors and area addition. The method is as simple as balancing a see-saw. Let us begin with a problem involving an old geometry concept. Given a triangle, a cevian is a line segment from a vertex to an interior point of the opposite side. (The c is pronounced as ch ). igure a illustrates two cevians and in. evians are named in honor of the Italian mathematician Giovanni eva who used them to prove his famous theorem in 68 (cf. Theorem 4). Problem below is not that famous, but it certainly presents a situation that you may stumble upon in everyday life. 4 5 G J H igure. Triangle with Two evians and the entroid of a Tetrahedron Problem. In, side is divided by in a ratio of 5 to and is divided by in a ratio of to 4 as shown in igure a. ind the ratios in which divides the cevians and, i.e. find : and :. It is true that Problem can be successfully attacked by vectors or area addition, thereby reducing it to elementary algebra. Yet, we shall see in this session that a much easier and more intuitive solution would result from assigning the masses to the vertices of in such a way that becomes the balancing point, and the problem is reduced to elementary arithmetic. xamples illustrating the power of our mass-point technique do not need to be constrained to the plane. The next problem for instance is set up in -dimensional space. Recall that a tetrahedron is the mathematical word for what we know as a pyramid: a polyhedron with four triangular faces. Recall that the centroid of a triangle is the point of concurrence of the three medians. (cf. xercise ). Problem. onsider tetrahedron, and mark the centroids of its four faces by,, G and H (cf. ig. b). Prove that the four segments connecting vertices to the centroids of the opposite faces are concurrent, i.e. that,, G and H intersect at a point J, called the centroid of the tetrahedron. In what ratio does this centroid J divide the four segments. or example, what is J : J?.. rchimedes lever. The underlying idea of the mass point technique is the principle of the lever, which rchimedes used to discover many of his results. You may have heard the boast of rchimedes upon discovering the lever, Give me a place to stand on, and I will move the earth. lthough rchimedes knew his results were correct, based on reasoning with the lever, such justification was unacceptable as proof in Greek mathematics, so he was forced to think of very clever proofs using uclidean geometry to convince the mathematical community at the time that his results were correct. These proofs are masterpieces of reasoning and the reader is recommended to read some of them to appreciate the elegance of the mathematics. However, in this session we are going to use rchimedes lever. The basic idea is that of a see-saw with masses at each end. The see-saw will balance if the product of the mass and its distance to the fulcrum is ate: November, 00 at the erkeley Math ircle.

2 TOM RIK OKLN HIGH SHOOL the same for each mass. or example, if a baby elephant of mass 00 kg is 0.5 m from the fulcrum, then an ant of mass gram must be located 50 km on the other side of the fulcrum for the see-saw to balance (cf. ig., not drawn to scale): distance mass = 00 kg 0.5 m = 00, 000 g km = g 50 km. {z } {z } 0.5 m 50 km igure. alancing on a See-saw: lephant and nt (artwork by Zvezda) s we shall see, the uses of the lever are more far-reaching than one might imagine. or instance, here is another problem for you to consider. It extends our mass point technique to transversals: a transversal of lines l and m is a line that joins a point on l and a point on m. Note that a transversal of two sides in a triangle is a generalization of what we defined earlier as a cevian. Problem. In igure, joins points and on the sides of forming a transversal. evian G divides in a ratio of to and intersects the transversal at point. ind the ratios : and : G. 4 G igure. Transversal Problem 5. efinitions and Properties: in the amiliar Setting of uclidean Geometry To begin, let me say that I misunderstood mathematics for a long time and it was not until I realized that the definitions, postulates and theorems were the key to everything, that I finally began making some progress. In the end, it still depends on how clever you can be in using the definitions, postulates and theorems to arrive at conjectures and prove more theorems. ut if you don t understand these fundamentals completely, you will not go very far in mathematics. This session follows the axiomatic approach to mass points found in Hausner s 96 paper [5]... Objects of mass point geometry. When developing a new theory, the objects in it must be clearly defined, so there are no ambiguities later on. or example, in ordinary high school geometry, you defined what triangles are and explained what it means for two of them to be congruent. In the slightly more advanced setting of, say, coordinate geometry, it becomes necessary to define even more basic objects, such as a point as a pair of numbers (x, y) called the coordinates of the points, and a line as the set of points which are solutions to linear equations x + y = ; one would also define what it means for two points P and Q to be the same P = Q if their corresponding coordinates are equal. In this vein, we define below the main objects of our new Mass Point theory. efinition. mass point is a pair (n, P), also written as np, consisting of a positive number n, the mass, and a point P in the plane or in space. efinition. We say that two mass points coincide, np = mq, if and only if n = m and P = Q, i.e. they correspond to the same ordinary point with the same assigned mass... Operations in mass point geometry. What makes our theory so interesting and powerful is that it combines objects and ideas from both geometry and algebra, and hence makes it necessary for us to define from scratch operations on mass points. How can we add two mass points? efinition (ddition). n + m = (n + m) where is on and : = m : n. This is the crucial idea: adding two mass points n and m results in a mass point (n + m) so that (a) is located at the balancing point of the masses on the line segment, and (b) the mass at this location is the sum n + m of the two original masses.

3 MSS POINT GOMTRY efinition 4 (Scalar Multiplication). Given a mass point (n, P) and a scalar m > 0, we define multiplication of a mass point by a positive real number as m(n, P) = (mn, P)... asic properties in mass point geometry. Property (losure). ddition produces a unique sum. Property (ommutativity). np + mq = mq + np. Property (ssociativity). np + (mq + kr) = (np + mq) + kr = np + mq + kr. Property 4 (Idempotent). np + mp = (n + m)p. Property 5 (istributivity). k(np + mq) = knp + kmq..4. More operations on mass points? Property 6 (Subtraction). If n > m then np = mq + xx may be solved for the unknown mass point xx. Namely, xx = (n m)r where P is on RQ and RP : PQ = m : (n m). xample. Given mass points Q and 5P, find the location and mass of their difference 5P Q.. undamental xamples and xercises Let s take a look at the first problem in the introduction. In order to have as the balancing point of we assign a mass of to and a mass of 5 to. Now on side to have as the balancing point we assign /4 = / to. Then at the balancing points on the sides of the triangle, we have + 5 = or + =. (cf. ig. 4a) N G M igure 4. Two evians Solved and Medians oncurrent The center of mass 8.5X of the system {, 5, } is located at the sum + + 5, which can be calculated in two ways according to our associativity property: + 5 = ( + ) + 5 = 8.5X = + ( + 5) = +. Thus, by definition of addition, X is located on the one hand on segment, and on the other hand on segment, i.e. at their intersection point. Hence is the fulcrum of the see-saw balancing and, and of the see-saw balancing 5 and. This means that : = / : = : 4 and : = 5 : / = 0 :. ll of this can be written down immediately on the figure in a matter of seconds. Many of the following exercises and examples in this section are based on those in an article by Sitomer and onrad [6]. Since the article is no longer in print and not available in most libraries, I want to make available to you some of the examples from their presentation which expanded my understanding of this technique. xercise (Warm-up). If G is on Y, find x and G : GY provided that (a) + 4Y = xg; (b) + xy = 9G. xample. In, is the midpoint of and is the trisection point of nearer (i.e. : = : ). Let G =. ind G : G and G : G. xercise (ast ay Mathletes, pril 999). In, is on and is the on. Let =, =, =, = and = 4. ind : in lowest terms. xercise. Show that the medians of a triangle are concurrent and the point of concurrency divides each median in a ratio of :. (Hint: ssign a mass of to each vertex, cf. ig. 4b.) PST. (Problem Solving Technique) Given two triangles with the same altitude, their areas are in the same ratio as their bases. In addition, if the triangles have have equal bases, then they have equal areas. L

4 4 TOM RIK OKLN HIGH SHOOL xample. Show that all six regions obtained by the slicing a triangle via its three medians have the same area. xercise 4 (Varignon s Theorem). If the midpoints of consecutive sides of a quadrilateral are connected, the resulting quadrilateral is a parallelogram. (Hint: ssign mass to each vertex of the original quadrilateral and find the center of mass in two ways: why does this center lie on each of the line segments joining midpoints of opposite sides?) xercise 5. In quadrilateral,,, G, and H are the trisection points of,,, and nearer,,,, respectively. Show that GH is a parallelogram. (Hint: Use the point K = G H.) xercise 6. Generalize xercise 5 to points,, G, and H which divide the quadrilateral sides in corresponding ratios of m : n. 4. ngle isectors, ombining Mass Points and rea, Mass Points in Space The following problems extend the fundamental idea of mass points in several directions. 4.. Using angle bisectors. To start with, you will need the following famous theorem, which you may have heard in a high school geometry class: Theorem (ngle isector Theorem). n angle bisector in a triangle divides the opposite side in the same ratio as the other two sides. More precisely, in, if ray bisects then : = :. xercise. In, let = c, = a and = b. ssign a mass to each vertex equal to the length of the opposite side, resulting in mass points a, b and c. Show that the center of mass of this system is located on each angle bisector at a point corresponding to the mass point (a + b + c)i. xercise 8. Use xercise to prove that the angle bisectors of the angles of a triangle are concurrent. Those who know the definition of sin may recall the following well-known theorem. Theorem (Law of Sines). In where the opposite sides of,, and are a, b, and c, respectively, and R is the circumradius of : a sin = b sin = c sin = R. xercise 9. In with the bisector of intersecting at : (a) Show that : = sin : sin, or equivalently, sin = sin. (b) Let sin = 4/5 and the sin = 4/5. The bisector intersects median M at point. ind : M and :. 4.. ombining mass points and areas. s you attempt to solve the following problem keep in mind that it can be solved via a combination of mass points and area addition. This leads to a generalization known as Routh s Theorem (to be discussed later on). Have fun! J l J K K m L L n igure 5. ombining Mass Points with rea ddition and Routh s Theorem Problem 4. In,,, and are the trisection points of,, and nearer,,, respectively. (cf. ig. 5a) (a) Let = J. Show that J : J = : 4 and J : J = 6 :. (b) Let = K and = L. xtend part (a) of this problem to show that K : KL : L = : : = J : JK : K = L : LJ : J. (c) Use parts (a) and (b) and to show that the area of triangle JKL is one-seventh the area of.

5 MSS POINT GOMTRY 5 (d) Generalize this problem using points which divide the sides in a ratio of : n in place of : to show the ratio of the areas is ( n) : ( n ). Part (d) can be generalized even further using different ratios on each side. It is known as Routh s Theorem.(cf. ig. 5b) Theorem. (Routh) If the sides,, of are divided at,, in the respective ratios of : l, : m, : n, then the cevians,, and form a triangle whose area is (lmn ) (lm + l + )(mn + m + )(nl + n + ) or example, check that when the ratios are all equal, l = m = n, Routh s formula yields the answer in part(d). The proof of this theorem is beyond the scope of the present session. See oxeter [4], Niven [], and Klamkin [8] for various proofs of this theorem. The proof by oxeter uses a generalization of mass points called areal coordinates, or normalized barycentric coordinates. It is only four lines long. 4.. Mass points in space. Going in another direction, an extension of the mass point technique can be used to solve problems in space: in dimensions. This is illustrated in xample 4, and xercises 0 and. Thus, for the rest of this subsection, we shall assume the same definitions and properties of addition of mass points in space as those in the plane. xample 4. Let be a tetrahedron (cf. ig. b). ssign masses of to each of the vertices. Let H be the point in such that + + = H. Let J be the point on H such that +H = 4J. What is the ratio of J to JH? Now let us apply mass points in space to a couple of exercises. xercise 0. ill in the details of the solution above for Problem. In particular, show that the four segments from the vertices to centroids of the opposite faces are concurrent at the point J. In a tetrahedron, opposite edges are those pairs of edges that have no vertex in common. xercise. Show that the three segments joining the midpoints of opposite edges of a tetrahedron bisect each other. (cf. ig. 6b) 5. Splitting Masses, ltitudes, eva and Menelaus G igure 6. Transversal Problem Solution and Tetrahedron in xercise 5.. Splitting masses. Let s now take a look at Problem stated in the Introduction. This is not as intuitive as the two cevian Problem. ut once it is shown to work, we can then solve a whole new class of problems with mass points. So let s do it! Splitting mass points as in mp + np = (m + n)p is the technique to use when dealing with transversals. The actual assignment of masses is as follows. s a first approximation, assign 4 to and to to balance at. Then to balance at G assign 9 to. To balance 9 at point, 8 5 is needed. So we now have ( ). This gives 44 5 as the center of mass for the masses at, and. Indeed, using associativity of addition: 0 8 G + (4 + 5 ) = ( ) = ( + 4) + ( ) = + 9 5, from where the center of mass lies on both and G, i.e. it is located at point. The sought after ratios can now be read directly from the diagram: : = 9/5 : = 9 : 5 and : G = 0/ : 58/5 = 50 : 58 = 5 : 9. Here is another example.

6 6 TOM RIK OKLN HIGH SHOOL xample 5. In, let be on such that : = :, let be on such that : = : 5, and let be on such that : = : 4. inally, let ray intersect at point G. ind G : G and : G. Try to use this technique in the following exercises. xercise. In xample 5, : = :, : = 4 :, : = 5 :. Show that G : G = 4 : and : G = :. 5.. Problems which involve altitudes. Let be an altitude of acute. Note that / = /. So the appropriate masses to assign to and, respectively, are / and / in order to have the balancing point on be at. Those of you who know some trigonometry will recognize / = / tan = cot and / = / tan = cot. Therefore, assigning masses proportional to cot and cot to the points and, respectively, will balance the side at the foot of the altitude. xercise. Let be a right triangle with =, = 5, and = 8. Let be the altitude to the hypotenuse and let the angle bisector at intersect at and at. Show that : = 5 : and : = : 5. Problem 5. The sides of are =, = 5 and = 4. Let be an altitude of the triangle. The angle bisector of intersects the altitude at and at. ind : and :. xercise 4. Prove that the altitudes of an acute triangle are concurrent using mass points. a x b G z y p c u t q r s t igure. The Theorem of eva and the Theorem of Menelaus 5.. eva, Menelaus and Property. Theorem 4 (eva s Theorem). Three cevians of a triangle are concurrent if and only if the products of the lengths of the non-adjacent parts of the three sides are equal. or example, for in igure a, this means that the three cevians are concurrent iff abc = xyz. Theorem 5 (Menelaus Theorem). If a transversal is drawn across three sides of a triangle (extended if necessary), the products of the non-adjacent segments are equal. or example, for with transversal intersecting in, in, and, externally, in, the conclusion is =, or equivalently, =. (cf. ig. b) Just as eva s Theorem is an if and only if statement, the converse of Menelaus Theorem is also true. Use mass point geometry to prove this. Then prove Menelaus Theorem is true using similar triangles. 6. xamples of ontest Problems 6.. Math contests versus research Mathematics. You may still be thinking that the type of problem that yields to a mass point solution is rare, and that it is more like a parlor trick than an important mathematical technique. In this collection of problems that I have assembled, you will see problems that can often be solved by mass point geometry more readily than with the official solution. They come from a wide variety of contests and were often problems the contestants found difficult. Part of the fun of such contests is knowing that a solution which could take you between 5 and 5 minutes exists and trying to find it. 6.. The contests surveyed for this collection. The problems in this section are from are city-wide, regional and national contests(cf. [,, 4, 8]): the New York ity Mathematics League (NYML), the merican Regional Mathematics League (RML), the merican High School Mathematics xamination (HSM), and the merican Invitational Mathematics xamination (IM).

7 MSS POINT GOMTRY 6.. Using the fundamental mass point technique. ontest Problem (HSM 965 #). Point is selected on side of in such a way that : = : and point is selected on side so that : = :. The point of intersection of and is. ind +. ontest Problem (NYML S5 #). In, is on side such that : = :, and is on such that : = : 4. If and intersect at P, and if is the intersection of ray P and then find P : P Using the angle bisector theorem and the transversal method. ontest Problem (RML 989 T4). In, angle bisectors and intersect at P. If the sides of the triangle are a =, b = 5, c =, with P = x, and P = y, compute the ratio x : y, where x and y are relatively prime integers. ontest Problem 4 (HSM 95 #8). In, M is the midpoint of side, = and = 6. Points and are taken on and, respectively, and lines and M intersect at G. If = then find G/G. ontest Problem 5 (RML 99 I8). In, points and are on and, respectively. The angle bisector of intersects at and at T. If =, =, =, and = 4, compute the ratio : T Using ratios of areas via PST. ontest Problem 6 (HSM 980 #). In, =, is the midpoint of side and is a point on side such that = ; and intersect at. ind the ratio of the area of to the area of quadrilateral. ontest Problem (IM 985 #6). In, cevians, and intersect at point P. The areas of s P, P, P and P are 40, 0, 5 and 84, respectively. ind the area of triangle hange your point of view. ontest Problem 8 (NYML 6 #). In, is on such that : = : and is on such that : = :. If ray and ray intersect at, then find :. ontest Problem 9 (NYML S #). In a triangle, segments are drawn from one vertex to the trisection points of the opposite side. median drawn from a second vertex is divided, by these segments, in the continued ratio x : y : z. If x y z then find x : y : z. 6.. Using special triangles and topics from geometry. ontest Problem 0 (NYML S8 #5). In, = 45 and = 0. If altitude H intersects median M at P, then P : PM = : k. ind k. ontest Problem (HSM 964 #5). The sides of a triangle are of lengths, 4, and 5. The altitudes of the triangle meet at point H. If is the altitude to the side of length 4, what is the ratio H : H? 6.8. Some especially challenging problems. ontest Problem (IM 99 #4). In,,, and respectively. Given that,, and are concurrent at the point O, and that O O + O O + O find the value of O O O O O O. Theorem 6. In, if cevians,, and are concurrent at P then P + P + P =. are on sides,,, O = 9, ontest Problem (IM 988 #). Let P be an interior point of and extend lines from the vertices through P to the opposite sides. Let P = a, P = b, P = c and the extensions from P to the opposite sides all have length d. If a + b + c = 4 and d = then find abc. ontest Problem 4 (IM 989 #5). Point P is inside. Line segments P, P, and P are drawn with on, on, and on. Given that P = 6, P = 9, P = 6, P =, and = 0, find the area of triangle.

8 8 TOM RIK OKLN HIGH SHOOL ontest Problem 5 (Larson [9] problem 8..4). In, let and be the trisection points of with between and. Let be the midpoint of, and let G be the midpoint of. Let H be the intersection of G and. ind the ratio H : HG.. History and Sources Mass points were first used by ugustus erdinand Möbius in 8. They didn t catch on right away. auchy was quite critical of his methods and even Gauss in 84 confessed that he found the new ideas of Möbius difficult. I discovered this historical information recently in a little mathematical note by an Pedoe in Mathematics Magazine []. This was long after I learned about the method. I first encountered the idea about 0 years ago in a math workshop session entitled Teeter-totter Geometry given by rother Raphael from Saint Mary s ollege of alifornia in Moraga. pparently he taught one of the courses each year using only original sources, and that year he was reading rchimedes with his students. It was rchimedes principle of the lever that he used on the day of my visit to show how mass points could be used to make deductions about triangles. or a very readable account of the assumptions rchimedes makes about balancing masses and locating the center of gravity, I recommend the new book rchimedes: What id He o esides ry ureka?[] written by Sherman Stein of U..avis. bout twenty-five years ago ill Medigovich, who was then teaching at Redwood High School in Marin ounty, alifornia, sent me a 0-sheet packet [0] that he used for a presentation he gave to high school students. I also found the topic discussed in the appendix of The New York ity ontest Problem ook [5] with a further reference to an article The enter of Mass and ffine Geometry [5] written by Melvin Hausner in 96. Recently, over Publications reissued a book published by Hausner [6] in 965 that was written for a one year course for high school teachers of mathematics at New York University. s I was preparing for this talk in 00, I was going through old issues of ureka which are no longer available and found a key paper on the this subject, Mass Points [6], that was originally written for the NY Senior Mathletes. The authors are Harry Sitomer and Steven R. onrad. Their paper provided me with what I considered to be the most attractive way to present these ideas. This wonderful journal for problem solvers published by the anadian Mathematical Society is presently published as rux Mathematicorum and Mathematical Mayhem. There are a few other articles that I used in preparing for the talk. They are listed in the references as oyd [], Honsberger [], and Pedoe(90) []. References. Ralph rtino, nthony Gaglione, and Neil Shell, The ontest Problem ook IV, 9-98, Mathematical ssociation of merica, 98.. George erzsenyi and Stephen Maurer, The ontest Problem ook V, , Mathematical ssociation of merica, 99.. J. N. oyd and P. N. Raychowdhury, n pplication of onvex oordinates, Two Year ollege Mathematics Journal (98), H. S. M. oxeter, Introduction to Geometry, pp. 6, John Wiley & Sons Inc., Melvin Hausner, The enter of Mass and ffine Geometry, The merican Mathematical Monthly (96), 4. 6., Vector Space pproach to Geometry, over Publications, Ross Honsberger, Mathematical Gems 5: Geometry via Physics, Two Year ollege Mathematics Journal (99), Murray Klamkin and ndy Liu, Three More Proofs of Routh s Theorem, rux Mathematicorum (98), no. 6, Loren Larson, Problem Solving Through Problems, pp , Springer-Verlag New York Inc., ill Medigovich, Mass Point Geometry, or How to Paint a Triangle, unpublished, Ivan Niven, New Proof of Routh s Theorem, The merican Mathematical Monthly (96), 5.. an Pedoe, Thinking Geometrically, The merican Mathematical Monthly (90),.., Notes on the History of Geometrical Ideas I: Homogeneous oordinates, Mathematics Magazine (95), harles Salkind, The M Problem ook II, , Mathematical ssociation of merica, Mark Saul, Gilbert Kessler, Sheila Krilov, and Lawrence Zimmerman, The New York ity ontest Problem ook, , ale Seymour Publications, Harry Sitomer and Steven R. onrad, Mass Points, ureka (96), no. 4, Sherman Stein, rchimedes: What id He o esides ry ureka?, pp. 5, Mathematical ssociation of merica, Lawrence Zimmerman and Gilbert Kessler, RML-NYSML ontests , MathPro Press, 995.