Physics helps geometry

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Physics
helps
geometry

B. Yu. Kogan

When studying physics, of course, mathematics is used. But it turns out that the other way around is also true: physics benefits mathematics. Here are some examples of how the concept of center of gravity can be used in geometry.

1. Theorem on medians of a triangle. Let us place material points of equal mass at the vertices of the triangle (Fig. 1). The center of gravity of the system of points A and B is located at point K - the middle of the segment AB. This means that the center of gravity of the system of points A, B and C lies on the median of the SC. Reasoning in a similar way, we will come to the conclusion that this center of gravity lies on the median AL and on the median BM. But the system of points A, B and C has only one center of gravity. Therefore, the medians of a triangle intersect at one point.

Further, since point C has mass m, and point K “represents” a system of points A and B, having mass 2m, then KS is half the size of SC. Therefore, the point of intersection of the medians divides each of them in a ratio of 2:1.

2. Theorem on the midlines of a quadrilateral. Let us place material points of the same mass (Fig. 2) at the vertices of the quadrilateral ABCD. Since the center of gravity of the system of points A

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or about points A and D, we will come to the conclusion that the center of gravity of the system of points A, B, C and D lies at the intersection of the mean lines KM and LN. Further, since the system of points A and B has the same mass as the system of points C and D, then KS = SM. Reasoning in a similar way, we find that LS=SN. Consequently, the midlines of the quadrilateral, intersecting, are divided in half.

Note that our reasoning does not use the fact that points A, B, C and D lie in the same plane. This means that the proven theorem is also true for a spatial quadrilateral. (It follows from this that the midlines of such a quadrilateral necessarily intersect.)

3. Generalization of the previous theorem. Consider an arbitrary quadrilateral ABCD (Fig. 3). Let us place material points at its vertices: at vertex A - masses m, at vertex B - masses αm, at vertex D - masses βm and at vertex C - masses αβm, where α and β are arbitrary positive numbers. Then the center of gravity of the system of points A, B, C and D will lie at the intersection of straight lines KM and LN, where K is the center of gravity of the system of points A and B, M is the center of gravity of the system of points C and D, L is the center of gravity of the system of points B and C, and N is the center of gravity of the system of points A and D. Further, since lengthwise arbitrary correspond to weights, then

$$\frac{AK}{KB} = \frac{\alpha m}{m} = \alpha, \quad \frac{DM}{MC} = \frac{\alpha \beta m}{\beta m} = \alpha,$$

$$\frac{NS}{SL} = \frac{\alpha m + \alpha \beta m}{m + \beta m} = \alpha, \quad \frac{BL}{LC} = \frac{\alpha \beta m}{\alpha m} = \beta,$$

$$\frac{AN}{ND} = \frac{\beta m}{m} = \beta,$$

$$\frac{KS}{SM} = \frac{\alpha \beta m + \beta m}{\alpha m + m} = \beta.$$

Therefore, if the sides of this quadrilateral are divided so that

$$\frac{AK}{KB} = \frac{DM}{MC} = \alpha, \quad \frac{BL}{LC} = \frac{AN}{ND} = \beta,$$

then the segments NL and KM will separate in that

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For example, on figure 4 each “vertical” segment is divided into three identical sections, and each “horizontal” segment is divided into four identical sections *).

4. Finding matches. Using the concept of the center of gravity, problems of the following type can be easily solved. On the sides of the triangle $ABC$ there are points $K$ and $L$ (Fig. 5). Given that $\frac{AK}{KB} = \frac{1}{2}$ and $\frac{BL}{LC} = \frac{3}{5}$, find the ratios of $\frac{AS}{SL}$ and $\frac{CS}{SK}$.

Solution. Let us place material points at the vertices $A, B, C$ so that the correspondence holds
$$\frac{m_B}{m_A} = \frac{AB}{KB} = \frac{1}{2}, \quad \frac{m_C}{m_B} = \frac{BL}{LC} = \frac{3}{5},$$
for example, let’s put
$$m_A = 10m, \quad m_B = 5m, \quad m_C = 3m.$$
Then the center of gravity of the system of points $A$ and $B$ will lie at point $K$, and the center of gravity of the system of points $B$ and $C$ will lie at point $L$. Consequently, the center of gravity of the system of points $A, B$ and $C$ lies on the straight line $CK$ and at the same time on the straight line $AL$, that is, it is located at the intersection point $S$. From here it is easy to find that
$$\frac{AS}{SL} = \frac{m_B + m_C}{m_A} = \frac{5m + 3m}{10m} = \frac{4}{5},$$
$$\frac{CS}{SK} = \frac{m_A + m_B}{m_C} = \frac{10m + 5m}{3m} = \frac{5}{1}.$$
Exercise

1. Each vertex of the tetrahedron is connected to the intersection point of the medians of the opposite face. Prove that the four resulting segments intersect at one point.
2. In triangle $ABC$ the median $AK$ is drawn. The line connecting the vertex $C$ with the midpoint of the median intersects the side $AB$ at the point $L$. Find the ratio $AL:LB$.
3. In the triangle $ABC$ shown at figure 6,
   $$\frac{AK}{AB} = \frac{BL}{BC} = \frac{CM}{CA} = \frac{1}{3}.$$
   What part of the area of ​​triangle $ABC$ is the sum of the areas of the protected triangles?
   *) This theorem is also true for a non-planar quadrilateral.