A problem by Sakharov

Andrei Sakharov (1921-1989) was a Soviet physicist. He is well known (among other things) for his work on the origin of baryon asymmetry in the Universe, his contributions to the Soviet hydrogen bomb program, and his political activism.

Here is one of his math problems.

A pizza his cut by straight lines into small pieces. The number of lines is very large and the cuts are drawn completely randomly. Most of the pizza pieces are then polygons with straight edges, with varying number of vertices.



Question 1: What is the average number of vertices of a pizza piece?

Question 2: Let S be the average area and p be the average perimeter of a pizza piece. What is the value of S/p2?

(In the original version, the pizza is replaced by a slice of cabbage. It is said that Sakharov made up this problem while he was cutting cabbage to help his wife with pirogi making.)

Source: Сахаровский сборник, изд. Хроника, Нью-Йорк, 1981, с. 140.


2 responses to “A problem by Sakharov

  1. Lời giải cho bài toán này, có lẽ là em sẽ chờ đáp án của Giáo sư. Ở đây em giải quyết phiên bản đơn giản của bài toán bằng cách thêm hai giả định:
    -Các đường cong bên ngoài sẽ được coi là thẳng khi các miếng bánh cắt tới một giới hạn nào đó.
    -Các lát cắt thứ 3 cắt qua lát cắt thứ 1, 2. Lát cắt thứ 4 cắt qua lát cắt thứ 1, 2, 3. Lát cắt thứ 5 đi qua lát cắt thứ 1,2,3,4. Và tiếp tục…(Cắt như trong ảnh).

    Gọi n là số đường thằng đã cắt, d là số đỉnh, c là số cạnh và m là số mặt.

    Với :n=0 thì d= 0
    n=1 thì d= 2*1 + 0
    n=2 thì d= 2*2 + 0 + 1
    n=3 thì d= 2*3 + 0 + 1 + 2.
    Với giá trị n thì chuỗi là 2*n +1+2+3…= 2*n + n*(n-1)/2

    Với: n=0 thì c= 1
    n=1 thì c= 2*1 + 1*1
    n=2 thì c= 2*2 + 2*2
    n=3 thì c= 2*3 + 3*3
    Với giá trị n thì chuỗi là 2*n +n*n = 2*n +n*n

    Ta có công thức Euler: d-c+m=2=>m=c-d+2= n*(n+1)/2 +2

    a, Để ý là với đỉnh trên đường tròn thì 1 đỉnh là chung của 2 mặt, đỉnh bên trong đường tròn thì 1 đỉnh là chung của 4 mặt do vậy ta có số đỉnh trung bình là (2*2*n+4*(n*(n-1)/2))/(n*(n+1)/2+2). Với giá trị n rất lớn thì số đỉnh trung bình của mỗi miếng bánh là 4.

    b, Gọi r là bán kính chiếc bánh.

    Với việc các lát cắt là ngẫu nhiên thì độ dài trung bình của một lát cắt khi n rất lớn là đường kính + 0 chia 2 = (pi)r.

    Tổng chu vi của các cạnh là (pi)*r*n/2 + 2(pi)r(thành phần đầu tiên chia 2 là do các cạnh bên trong được dùng chung bởi 2 mặt). Chu vi trung bình của một mặt là ((pi)r*n/2+2(pi)r)/(n*(n+1)/2+2).

    Diện tích trung bình của 1 mặt là ((pi)*r*r)/(n*(n+1)/2+2).

    Vậy, tỉ số S/p^2 khi n rất lớn là (pi)/4.

  2. Here is my solution to Sakharov’s problem.

    For the first part of the problem, the intuition is as follows: the average angle made by two intersecting cuts is π/2. A polygon with all angles equal to π/2 is a quadrangle. Thus, the average number of vertices in a pizza slice is 4.

    To formalize this argument, and to solve the second part of the problem, it is most convenient to imagine that the pizza is not located on a plane, but on a large sphere (for example, on the surface of the Earth). Then each cut corresponds to a large circle on the surface of the sphere. The “random cuts” can be imagine as a collection of many large circles drawn randomly on the surface of the sphere. Of course, most of these large circle will miss the pizza, but for the sake of simplicity, we will consider a completely random collection of large circles drawn on a sphere. These large circles will form a mesh on the surface of the sphere.

    Pizza on Earth's surface

    Now we note that the patch of the sphere covered by the pizza is by no mean special (the “cosmological principle”), so we can forget about the pizza completely and ask about the properties of the small patches of the sphere that are formed by the mesh of large circles. (Equivalently, we imagine a pizza that covers the whole surface of the Earth.) For simplicity we will regard these patches as “polygon”, though, strictly speaking, they are slightly curved.

    Each pair of large circles intersects in 2 points. Therefore, the total number of intersection points (the vertices of the polygons) is N2N, which can be taken to be N2 for large N.

    Let call Nn the number of patches that are n-polygons. Since each vertex is shared by 4 polygons, we conclude that

    \sum\limits_n n N_n = 4 N^2

    The sum of all angles of a n-polygon is (n–2)/π. If we take the sum of all angles of all patches, then the angles around each vertex sum up to 2π. Thus we have

    \sum\limits_n (n-2)\pi N_n = 2\pi N^2.

    From these two equations we then fine

    \sum\limits_n N_n = N^2.

    This is the total number of patches.

    From the two equation that we have obtain, we find that the average number of vertices in a patch is 4.

    Now let’s denote the radius of the sphere as R. The surface of the sphere is 4πR2, divided into N2 pieces. Thus the average area of a patch is

    S = \displaystyle{\frac{4\pi R^2}{N^2}}

    The sum of the perimeters of all patches is twice the total length of all $N$ large circles (each edge is shared between two adjacent patches). The average perimeter is then

    p = \displaystyle{\frac{2\times 2\pi R\times N}{N^2}} = \displaystyle{\frac{4\pi R}{N}}


    \displaystyle{\frac S{p^2}} = \displaystyle{\frac1{4\pi}}

    Interestingly, this is the ratio of the area to the square of the perimeter for a circle. This result does not contradict the fact that 1/(4π) is achieved only for a circle, since we are not computing the average of the ratio S/p2 for all patches, but rather the average of S over the square of the average of the p.

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