Mathematical and Physical Journal
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Solutions for advanced problems "A" in November, 2001

In this page only the sketch of the solutions are published; in some cases only the final results. To achieve the maximum score in the competition more detailed solutions needed.

A. 275.  The numbers 12, 22, 32, ..., (n2)2 are written in increasing order into an nxn array:

12 22 ... n2
(n+1)2 (n+2)2 ... (2n)2
... ...   ...
(n2-n+1)2 (n2-n+2)2 ... (n2)2

A number is selected from each row, such that no two numbers belong to the same column. What are the possible values of the sum of the selected numbers?

From the idea of T. Terpai

Solution. Suppose we have chosen the aith element out of the i-th row, where i=1,2,...,n. The numbers a1,...,an represent a permutation of 1,2,...,n, and the sum of the selected numbers is

\(\displaystyle \sum_{i=1}^n\big((i-1)n+a_i\big)^2=n^2\sum_{i=1}^n(i-1)^2+\sum_{i=1}^na_i^2+2n\sum_{i=1}^n(i-1)a_i.\)

The values of the first two sums do not depend on the choice of the numbers ai:

\(\displaystyle n^2\sum_{i=1}^n(i-1)^2=n^2\cdot{(n-1)n(2n-1)\over6}={n^3(n-1)(2n-1)\over6},\)

\(\displaystyle \sum_{i=1}^na_i^2=\sum_{i=1}^ni^2={n(n+1)(2n+1)\over6}.\)

According to the ordering theorem, the third sum is minimum if the numbers ai form a decreasing sequence, and maximum if they form an increasing sequence. Thus the lowest and highest values are

\(\displaystyle 2n\sum_{i=1}^n(i-1)(n-i+1)=2n\cdot{(n-1)n(n+1)\over6},\)


\(\displaystyle 2n\sum_{i=1}^n(i-1)i=2n\cdot{(n-1)n(n+1)\over3}.\)

It is easy to check with a little calculation that for small values of n, the possible values of \(\displaystyle \sum_{i=1}^n(i-1)a_i\) are as follows:

npossible values of \sum_{i=1}^n(i-1)a_i
21, 2
34, 5, 7, 8
410, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

For all n\(\displaystyle ge\)5, \(\displaystyle \sum_{i=1}^n(i-1)a_i\) assumes every integer value between \(\displaystyle {(n-1)n(n+1)\over6}\) and \(\displaystyle {(n-1)n(n+1)\over3}\). The proof is by induction. As shown above, the statement is true for n=4.

If the statement is true for some n=m, then in the case n=m+1, consider the sequences a1,...,am+1, such that am+1=m+1 or a1=m+1.

If am+1=m+1 then

\(\displaystyle \sum{i=1}^n(i-1)a_i=\sum_{i=1}^ma_i+(m-1)m,\)

and we get the same sums as in the case n=m, but incremented by m(m+1). These sums are, therefore, the integers between \(\displaystyle A={(m-1)m(m+1)\over6}+m(m+1)\) and \(\displaystyle B={(m-1)m(m+1)\over3}+m(m+1)={m(m+1)(m+2)\over3}\).

If a1=m+1, then

\(\displaystyle \sum_{i=1}^n(i-1)a_i=\sum_{i=2}^{m+1}(i-1)a_i=\sum_{i=1}^mia_{i+1}=\)

\(\displaystyle =\sum_{i=1}^m(i-1)a_{i+1}+\sum_{i=1}^na_{i+1}=\sum_{i=1}^m(i-1)a_{i+1}+{m(m+1)\over2},\)

thus here we get the same sums again, but this time incremented by \(\displaystyle {m(m+1)\over2}\). The resulting sums are, therefore, the integers between \(\displaystyle C={(m-1)m(m+1)\over6}+{m(m+1)\over2}={m(m+1)(m+2)\over6}\) and \(\displaystyle D={(m-1)m(m+1)\over3}+{m(m+1)\over2}\).

It is easy to check that if m\(\displaystyle ge\)4 then C<A\(\displaystyle le\)D<B, and thus all integers between C and B are represented. This completes the proof.

Answer: For n\(\displaystyle ne\)3, the possible values of the sum of the elements selected from the table are in arithmetic progression, where the lowest member is

\(\displaystyle {n^3(n-1)(2n-1)\over6}+{n(n+1)(2n+1)\over6}+2n\cdot{(n-1)n(n+1)\over6}={n(2n^4-n^3+3n^2+n+1)\over6},\)

the largest member is

\(\displaystyle {n^3(n-1)(2n-1)\over6}+{n(n+1)(2n+1)\over6}+2n\cdot{(n-1)n(n+1)\over3}={n(2n^4+n^3+3n^2-n+1)\over6},\)

and the common difference is 2n.

Fof n=3, the result is similar, but with the middle term of the sequence, 95 missing; the possible values are 83, 89, 101 and 107.

A. 276.  The product of the positive numbers x1, ..., xn is (n^\alpha-1)^n, where alphage1 is a real number. Prove that

\(\displaystyle \sum_{i=1}^n{1\over(x_i+1)^{1/\alpha}}\ge1.\)

Solution. Let \(\displaystyle f(t)={1\over(t+1)^{1/\alpha}}\). We shall prove the following two lemmas:

Lemma 1. If a\(\displaystyle le\)b\(\displaystyle le\)c\(\displaystyle le\)d are positive numbers, and ad=bc, then

f(a)+f(d)\(\displaystyle ge\)min(f(b)+f(c),1).

Lemma 2. If the geometric mean of the numbers x1,x2,...,xn is m, then

f(x1)+f(x2)+...+f(xn)\(\displaystyle ge\)min(n.f(m),1).

The statement of the problem is a special case of lemma 2, because if the product of the positive numbers x1,...,xn is (n\(\displaystyle alpha\)-1)n, i.e. their geometric mean is m=n\(\displaystyle alpha\)-1, then n.f(m)=1.

Proof for lemma 1. Let \(\displaystyle m=\sqrt{ad}=\sqrt{bc}\), and

\(\displaystyle g(t)=f(mt)+f(m/t)=\bigg(mt+1\bigg)^{-{1\over\alpha}}+\bigg({m\over t}+1\bigg)^{-{1\over\alpha}}\)

for all positive t, and t1=c/m, t2=d/m. Then 1\(\displaystyle le\)t1\(\displaystyle le\)t2, and the statement is that

g(t2)\(\displaystyle ge\)min(g(t1),1).

It follows from the definition that \(\displaystyle \lim\limits_\infty g=1.\)

Let us examine the monotonicity of the function g on the interval [1,\(\displaystyle infty\)). The derivative of the function g is

\(\displaystyle g'(t)={m\over\alpha}\left(-\bigg(mt+1\bigg)^{-{\alpha+1\over\alpha}}+{1\over t^2}\bigg({m\over t}+1\bigg)^{-{\alpha+1\over\alpha}}\right),\)

which is positive if and only if

\(\displaystyle \bigg(mt+1\bigg)^{\alpha+1\over\alpha\)t^2\bigg({m\over t}+1\bigg)^{\alpha+1\over\alpha}.">

By raising to the \(\displaystyle {\alpha\over\alpha+1}\)th power and rearranging, we have

\(\displaystyle t^{2\alpha\over\alpha+1}-mt+mt^{\alpha-1\over\alpha+1}-1<0.\)

Let h(t) denote the left-hand side. For further investigation, we will need the first two derivatives of the function h, and the values of h(1) and h'(1):

\(\displaystyle h(t)=t^{2\alpha\over\alpha+1}-mt+mt^{\alpha-1\over\alpha+1}-1,\qquad h(1)=0;\)

\(\displaystyle h'(t)={2\alpha\over\alpha+1}t^{\alpha-1\over\alpha+1}-m+{\alpha-1\over\alpha+1}mt^{-{2\over\alpha+1}},\qquad h'(1)={2\over\alpha+1}(\alpha-m);\)

\(\displaystyle h''(t)={2\alpha(\alpha-1)\over(\alpha+1)^2}t^{-{\alpha+3\over\alpha+1}}\bigg(t-{m\over\alpha}\bigg).\)

Now let us distinguish the following cases, depending on the values of m and \(\displaystyle alpha\):

Case 1. : \(\displaystyle alpha\)=1 and m\(\displaystyle le\)1.

Case 2. : \(\displaystyle alpha\)=1 and m>1.

Case 3. : \(\displaystyle alpha\)>1 and m\(\displaystyle le\)alpha.

Case 4. : alpha>1 and m>alpha.

In case 1, h(t)=(1-m)(t-1)ge0 for all t>1, thus hge0 in the interval (1,infty).

In case 2, h(t)=(1-m)(t-1)<0 for all t>1, thus h<0 in the interval (1,infty).

In case 3, h''>0 for h'(1)ge0 and t>1, thus h'>0 in the entire interval (1,infty). As h(1)=0, it follows that h>0 in the interval (1,infty).

In case 4, h'(1)<0, and h''<0 in the interval (1,m/alpha). Thus h'<0 in that interval. As h(1)=0, it follows that h<0 in the interval (1,m/alpha]. In the interval (m/alpha,infty), h''>0, that is, the function h is convex down. As h(m/alpha)<0 and \lim\limits_\infty h=\infty, it follows that there exists a real number varrho>1, such that h<0 on the interval (1,varrho), and h>0 on the interval (varrho,infty).

In either case, if h(t2)ge0, then hge0 in the entire interval (t2,infty), that is, g'le0. Hence the function g monotonically decreases in the interval [t2,infty), and g(t_2)\ge\lim\limits_\infty g=1. If h(t2)le0, then hle0 in the interval (1,t2), that is, g'ge0. Thus g imonotonically increases, and és g(t1)leg(t2). Lemma 1 is proved.

Proof for lemma 2. The proof is by induction. If n=1 then only x1=m is possible, and the statement is meaningless. Assume that nge2, and the statement is true if the number of variables is less than that.

The sequence x1,...,xn contains both a number not greater than m, and a number not smaller than m. We can suppose, for example, that x1lemlex2. Let y1 denote the smaller number out of m and x1x2/m, and let y2 denote the larger number. Then x1ley1ley2lex2 and y1y2=x1x2. With lemma 1, it follows that


The geometric mean of the numbers y2,x3,...,xn is also m. The assumption can be applied, and we get


If f(x1)+f(x2)>1, then f(x1)+...+f(xn)gef(x1)+f(x2)>1, and hence the statement is true. If f(x1)+f(x2)le1, then (2) states that f(x1)+f(x2)gef(m)+f(y2). Using this result, and (3), we have



Thus lemma 2 is proved, and so is the statement of the problem

A. 277.  Let H1 be an n-sided polygon. Construct the sequence H1, H2, ..., Hn of polygons as follows. Having constructed the polygon Hk, Hk+1 is obtained by reflecting each vertex of Hk through its k-th neighbour in the counterclockwise direction. Prove that if n is a prime, then the polygons H1 and Hn are similar.

Solution 1. In order to be able to formulate auxiliary statements more accurately, the following solution uses arrays of n points rather than n-sided polygons.

For every integer 0<k<n, let tauk denote the transformation that maps an array of n points onto another by reflecting each point in its kth successor. The definition of the transformation can be extended to any integer k if k is considered modulo n.

The statement is that Hk+1=tauk(Hk), and we need to prove that the n-gon taun-1(taun-2(...tau2(tau1(H1))...)) is similar to H1.

First we shall prove that the transformations commute, that is, for all integers k,m and every array P of n numbers, taum(tauk(P))=tauk(taum(P)). Let the points of P be p1, ..., pn. Then, with all the indices taken modulo n,

- the ith point of the polygon tauk(P) is 2pi+k-pi;

- the ith point of the polygon taum(P) is 2pi+m-pi;

- the ith point of the polygon taum(tauk(P)) is 2(2pi+m+k-pi+m)-(2pi+k-pi)=4pi+m+k-2pi+k-2pi+m+pi;

- the ith point of the polygon tauk(taum(P)) is 2(2pi+k+m-pi+k)-(2pi+m-pi)=4pi+m+k-2pi+k-2pi+m+pi.

Thus the arrays taum(tauk(P)) and tauk(taum(P)) are identical.

It follows from the commutativity of the composition of transformations that the transformations tau1, tau2, ..., taun-1 can be carried out in any order, the result will always be the same.

Let H1=(h1,...,hn) and Hn=(g1,...,gn). As every transformation replaces the points by a linear combination, each of the points g1, ..., gn can be expressed as an appropriate linear combination of the points h1, ..., hn. As the sequence of transformations is symmetrical with respect to the cyclic permutation of the indices, the coefficients of the various linear combinations can be obtained from each other by cyclic permutations, that is, there exist such constants a0, a1, ..., an-1 that for all i,


Now let us prove that a1=a2=...=an-1.

For any integer 0<m<n, let pim denote the permutation (x1,...,xn)to(xm,x2m,...,xnm). (This is really a permutation if n is a prime.) By substituting the definition, it can be seen that for every integer k and every array P of n numbers,


By applying this identity (n-1) times, we get


As n is a prime, each of tau1, ..., taun-1 occurs exactly once among the transformations taum, tau2m, ..., tau(n-1)m. Thus the right-hand side is pim(Hn)=(gm,g2m,...,gnm).

Apply the transformations on the left-hand side to the polygon pim(H1)=(hm,h2m,...,hnm). It follows from identity (4) that the ith member of the resulting array of n points is


Putting the results together, we have that for all i,



The coefficient of h(i+1)m is a1 on the left-hand side, and am on the right-hand side, therefore am=a1. As this is true for all 0<m<n, a1=a2=...=an-1.

The ith edge vector of the polygons H1 and Hn is hi+1-hi, and gi+1-gi=(a0-a1)(hi+1-hi). Thus the two polygons are similar, with a ratio of a0-a1.

Solution 2. This solution uses complex numbers, and the so- called finite Fourier transform . The polygons are considered to be arrays of n complex numbers. In order to make calculations simpler, let us place H1 onto the complex plane so that its centroid should be 0.

If X is an array of n numbers, let Xj denote its jth element. The indices are always meant modulo n, for example, Xn=X0 or X-1=Xn-1.

Arrays of n compex numbers can be added term by term, subtracted, and multiplied by scalars. In addition, define the finite Fourier transform of an array of n numbers as follows. Let \varepsilon=\cos{2\pi\over n}+i\sin{2\pi\over n} be the first complex nth root of unity. The Fourier transform of the array is the array varphi(X), such that for each j,

\varphi(X)_j=\sum_{m=0}^{n-1}\varepsilon^{mj}\cdot X_m.

There are several interesting properties and applications of the Fourier transform. See, for example, Algorithms, by P. Gács and L. Lovász. Now we are going to use the following properties:

- For any number c and array X of n numbers, varphi(c.X)=c.varphi(X).

- The Fourier transform has an inverse. It is easy to check that,

X_j={1\over n}\sum_{m=0}^{n-1}\varepsilon^{-mj}\cdot\varphi(X)_m.

Like in solution 1, let tauk be the transformation that reflects every element of an array in its kth successor, that is,


It follows from the definitions of the Fourier transform and tauk that for any arrayx of n numbers,


=2\varepsilon^{-kj}\sum_{m=0}^{n-1}\varepsilon^{j(m+k)}\cdot x_{m+k}-\sum_{m=0}^{n-1}\varepsilon^{jm}\cdot x_m=(2\varepsilon^{-jk}-1)\varphi(x)_j,

that is,


Applying this to every 0<k<n, we have


If j=0, then \prod_{k=1}^{n-1}(2\varepsilon^{-kj}-1)=1. If 0<j<n, then -- as n is a prime -- \prod_{k=1}^{n-1}(2\varepsilon^{-kj}-1)=\prod_{k=1}^{n-1}(2\varepsilon^k-1). Let C denote this number.

Thus we obtain


Consider the arrays varphi(H1) and varphi(Hn) of n numbers. Because of the choice of the centroid, varphi(H1)0=0, and thus (5) can also be written in the following form:


Hence Hn=C.H1, that is, Hn is obtained from H1 by an enlargement from 0 with a scale factor of C.

Remark. With the polynomial


we get


(Using that n is an odd prime, and n(n-1)/2 is divisible by n.)