Geometry. The Method of the Center of Mass (mass points): Solving problems using the Law of Lever (mass points). Menelaus theorem. Pappus theorem.

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1 Noveber 13, 2016 Geoetry. The Method of the enter of Mass (ass points): Solving probles using the Law of Lever (ass points). Menelaus theore. Pappus theore. M d Theore (Law of Lever). Masses (weights) balance at distances fro the fulcru, which are inversely proportional to their agnitudes, For coensurate asses,, the Law was proven using the ain trick of the ass points ethod: each of the two asses is split into and saller asses,, respectively, which are then re-positioned in pairs around the original asses so that positions of the center of ass (M) for each of the two original asses do not change, but the M position for the whole syste becoes obvious. In order to prove the Law of Lever for incoensurate asses, we first ake the following observation. Lea. If two coensurate asses and are placed at distances and fro the fulcru, respectively, then goes up if and only if, First, if distances and are incoensurate, we ove ass slightly, to a position which is coensurate with, but such that still rises up. Therefore, we only need to consider case when and are coensurate.

2 Since rises up, we need to increase ass to achieve balance. Let be such that and balance. Using the Law of Lever for coensurate asses we have, (because distances are coensurate, so are the asses). Since, it follows that. onversely, if fro back to we can increase it to, which balances. ecreasing ass will cause it to rise. orollary. The converse stateent iediately follows via excluded iddle, Proof (case of incoensurate asses). Let now two incoensurate asses and, be placed at distances and fro the fulcru, respectively, such that the Law of Lever is satisfied,. ssue that the asses nevertheless do not balance, for exaple, goes down. ecrease ass by a sall aount, turning it into, such that it still goes down, but is now coensurate with. Now and are coensurate, and, which eans that should rise. This contradicts our assuption, so and ust balance. Note that in the above we used a non-trivial fact that a coensurate ass, or distance can be found that differs fro the given incoensurate one by an arbitrarily sall aount. This eans that for any irrational nuber there exists a rational nuber, which differs fro it as little as we want, i. e. that rational nubers are dense.

Solving probles using the Law of Lever. For the objects in the unifor gravitational field, the enter of Gravity and the enter of Mass are equivalent.

Heuristic efinitions of the enter of Mass (enter of Gravity) known to Greeks. 1.

3 Solving probles using the Law of Lever. For the objects in the unifor gravitational field, the enter of Gravity and the enter of Mass are equivalent. rchiedes uses the concept by considering bodies with the unifor density and defining the enter of Gravity based on postulated properties. Heuristic efinitions of the enter of Mass (enter of Gravity) known to Greeks. 1. The point such that if suspended at it, an object will reain otionless in the equilibriu, independent of the position that it is placed. 2. The point coon to all the lines passing through the point at which the object is suspended 3. The point coon to all lines on which the object balances. rchiedes postulates on the properties of the enter of Gravity (M). 1. The M of siilar figures are siilarly situated. 2. The M of a convex figure lies within the figure. 3. If an object is cut in two pieces, then its M lies on the line joining the M s of the pieces, and its position satisfies the Law of Lever. However, the situation is uch sipler if we only consider point asses. Properties of the enter of Mass for a syste of point asses. 1. Every syste of finite nuber of point asses has unique center of ass (M). 2. For two point asses, and, the M belongs to the segent connecting these points; its position is deterined by the rchiedes

4 lever rule: the point s ass ties the distance fro it to the M is the sae for both points,. 3. The position of the syste s center of ass does not change if we ove any subset of point asses in the syste to the center of ass of this subset. In other words, we can replace any nuber of point asses with a single point ass, whose ass equals the su of all these asses and which is positioned at their M. Solving probles using the M. Given a syste of points and lines, one can derive various relations, such as concurrence of particular lines connecting soe of the points, or the ratio of the lengths of different segents by associating certain asses with these points (i.e. placing point asses at their positions) and considering the center of ass of the obtained syste of ass points. Exercise. Prove that the edians of an arbitrary triangle are concurrent (cross at the sae point ). Exercise. Prove that the bisectors of an arbitrary triangle are concurrent (cross at the sae point ). M M solutions of the selected hoework probles. 1. Proble. Prove that edians of a triangle divide one another in the ratio :1 in other words the edians of a triangle trisect one another (oxeter, Gretzer, p.8). Solution. Load vertices, and with equal asses,. Then, the center of ass (M) of the three asses is at the intersection of the three edians, because it has to belong to each segent connecting the ass at the vertex of the triangle with the M of the other two asses, i.e. the iddle of the opposite side. M this belongs to all three edians and is the centroid, of the triangle. It divides each edian in

5 the 2:1 ratio because it is a M of ass at the iddle of the opposite side. at the vertex and a ass 2. Proble. In isosceles triangle point divides the side into segents such that : 1:. If is the altitude of the triangle and point is the intersection of and, find the ratio to. Solution. a. Using the siilarity and Thales theore. First, let us perfor a suppleentary construction by drawing the segent parallel to,, where point belongs to the side, and point to and the altitude. Notice the siilar triangles,, which iplies,. y Thales theore, 1, and, so that., because 1 1. Therefore, the sought ratio is,. b. Using the Method of the enter of Mass. Load vertices, and with asses,, and, respectively. Then, is the M of asses at and, and is the M of asses at and, and is the M of all 3 asses in the vertices of the triangle. Therefore, : : : 1, : 1:. 3. Proble. Point belongs to the continuation of side of the triangle such that. Point belongs to side, and. Segent intercepts side at point. Find the ratio :. 1 2 H H F E F

6 Solution. a. Using the siilarity and Thales theore. First, let us perfor a suppleentary construction by drawing the segent parallel to,, where belongs to the side of the triangle. is the id-line of the triangle, and, by Thales, also of and. Therefore,, and, so. n the other hand, again, by Thales, or, noting siilar triangles,. b. Using the Method of the enter of Mass. Load vertices, and with asses, and, respectively. Then, is the center of ass (M) of and, is the M of and, and is the M of the triangle, : : :. E G F

7 Theore (Extended eva). Segents (evians) connecting vertices, and, with points, and on the sides, or on the lines that suitably extend the sides,, and, of triangle, are concurrent if and only if, 1 Proof. We have already proven this theore for the case when points, and lie on the sides, but not on the lines extending the sides as it is shown in the figure. Let us now consider this latter case. Let us first load points, and with asses, and such that point is the center of ass for and,, and point is the M for and,. Then, the M of all three asses at the vertices of the triangle is at the point, which is the intersection of and. Let cross side at point. dding ass to vertex would ove the M of the three asses along line, because the M of the initial 3 asses is at. Let us add another ass to vertex, so that the total ass at this vertex is.the resulting syste of asses then has the sae M as two asses, and at points and, respectively. This M is coon to and, and therefore is at point, so. Hence, we obtain,

8 Theore (Menelaus). Points, and on the sides, or on the lines that suitably extend the sides,, and, of triangle, are collinear (belong to the sae line) if and only if, 1 h Menelaus's theore provides a criterion for collinearity, just as eva's theore provides a criterion for concurrence. h Proof (siilarity). The stateent could be proven with, or without using the ethod of point asses. h First, assue the points are collinear and consider rectangular triangles obtained by drawing perpendiculars onto the line. Using their siilarity, one has h h h Wherefro the stateent of the theore is obtained by ultiplication (oxeter & Greitzer). Proof (point asses). lternatively, let us load points, and in the upper Figure with the point asses, and, respectively. We select, and such that is the M of and, and is the M of and. The M of all 3 asses belongs to both segents and, which eans that it is at point. Then, Wherefro the Menelaus theore is obtained by ultiplication. The case shown in the lower figure is considered in a siilar way.

9 Theore (Pappus). If,, E are three points on one line,, and F on another, and if three lines,,, EF, eet E, F,, respectively, then the three points of intersection, L, M, N, are collinear. This is one of the ost iportant theores in planietry, and plays iportant role in the foundations of projective geoetry. There are a nuber of ways to prove it. For exaple, one can consider five triads of points, LE, MF, N, E and F, and apply Menelaus theore to each triad. Then, appropriately dividing all 5 thus obtained equations, we can obtain the equation proving that LMN are collinear, too, also by the Menelaus theore. However, one can prove the Pappus theore directly, using the ethod of point asses. Instead of siply proving the theore, consider the following proble. Proble. Using only pencil and straightedge, continue the line to the right of the drop of ink on the paper without touching the drop. Solution by the Method of the enter of Mass. onstruct a triangle, which encloses the drop, and with the vertex on the given line (). Let 1 be the crossing point of () and the side. Let us now load vertices and of the triangle with point asses and, such that their center of ass (M) is at the point 1. Then, each point of the (evian) segent 1 is the center of ass of the triangle for soe point ass loaded on the vertex. The (evian) segents fro vertices and, which pass through the center of ass of the triangle, connect each of these vertices with the center of ass of the other two vertices on the opposite side of the triangle, and, respectively. For the ass 1 loaded on the vertex, the center of ass of the triangle is 1, and the centers of ass of the sides and are 1 and 1, respectively.

10 Siilarly, 2, 2 and 2 are those for the ass 2 on the vertex. The center of ass of the side is always at the point 1, independent of ass. If we can show that segents 12 and 21 cross the given line () at the sae point,, then our proble is solved, as we can draw evians 2 and 2, whose crossing points are on the segent 1 on the other side of the drop, by sequentially drawing evians 1 and 1 and segents 12, 12, Figure 1(a). Let us load vertices, and with asses 1+ 2, 2 and 2, respectively, Figure 1(b). The center of ass of is now at soe point, in-between 1 and 2 (actually, it is not iportant where it is on the line 1). Let us now ove point asses 1 and to their center of ass 1 on the side, 2 and to their center of ass 2 on the side, and and to their center of ass 1 on the side. Now asses are at the vertices of the triangle 121 with the sae center of ass,, Figure 1(c). onsequently, the crossing point of segents 12 and 1 is the center of ass for asses 1+ and 2+ placed at points 1 and 2, respectively. Point then is the center of ass for at point and + at point 1, Figure 1(e). Repeating siilar arguents for the triangle 211, Figure 1(d,f), we see that point is also the crossing point of segents 12 and 1. Therefore, is the crossing point of all three segents, 12, 21 and 1, which copletes the proof.

11 (a) 2 (b) 2 2 1, (c) (d) (e) 2 (f)

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