11th International Mathematical Competition for University StudentsSkopje, 25–26 July 2004 Solutions for problems on Day 2 1.. This is clear since both functions are convex, their graphs
Trang 111th International Mathematical Competition for University Students
Skopje, 25–26 July 2004 Solutions for problems on Day 2
1 Let A be a real 4 × 2 matrix and B be a real 2 × 4 matrix such that
AB =
Find BA [20 points]
Solution Let A =A1
A2
and B = B1 B2 where A1, A2, B1, B2 are 2 × 2 matrices Then
=A1
A2
B1 B2 =A1B1 A1B2
A2B1 A2B2
therefore, A1B1 = A2B2 = I2 and A1B2 = A2B1 = −I2 Then B1 = A−11 , B2 = −A−11 and A2 = B−12 =
−A1 Finally,
BA = B1 B2A1
A2
= B1A1+ B2A2 = 2I2 =2 0
0 2
2 Let f, g : [a, b] → [0, ∞) be continuous and non-decreasing functions such that for each x ∈ [a, b] we have
Z x a
p
f (t) dt ≤
Z x a
p g(t) dt and Rabpf(t) dt = Rb
apg(t) dt
Prove thatRb
ap1 + f(t) dt ≥ Rb
a p1 + g(t) dt [20 points]
Solution Let F (x) = Rx
a pf(t) dt and G(x) = Rx
a pg(t) dt The functions F, G are convex, F (a) = 0 = G(a) and F (b) = G(b) by the hypothesis We are supposed to show that
Z b a
q
1 + F0(t)2dt ≥
Z b a
q
1 + G0(t)2dt
i.e The length ot the graph of F is ≥ the length of the graph of G This is clear since both functions are convex, their graphs have common ends and the graph of F is below the graph of G — the length of the graph of F is the least upper bound of the lengths of the graphs of piecewise linear functions whose values
at the points of non-differentiability coincide with the values of F , if a convex polygon P1 is contained in
a polygon P2 then the perimeter of P1 is ≤ the perimeter of P2
3 Let D be the closed unit disk in the plane, and let p1, p2, , pn be fixed points in D Show that there exists a point p in D such that the sum of the distances of p to each of p1, p2, , pn is greater than or equal to 1 [20 points]
Solution considering as vectors, thoose p to be the unit vector which points into the opposite direction as n
P
i=1
pi Then, by the triangle inequality,
n X i=1
|p − pi| ≥
np −
n X i=1
pi
= n +
n X i=1
pi
≥ n
Trang 24 For n ≥ 1 let M be an n × n complex matrix with distinct eigenvalues λ1, λ2, , λk, with multiplicities
m1, m2, , mk, respectively Consider the linear operator LM defined by LM(X) = M X + XMT, for any complex n × n matrix X Find its eigenvalues and their multiplicities (MT denotes the transpose of M ; that is, if M = (mk,l), then MT = (ml,k).) [20 points]
Solution We first solve the problem for the special case when the eigenvalues of M are distinct and all sums
λr+ λs are different Let λr and λs be two eigenvalues of M and ~vr, ~vs eigenvectors associated to them, i.e M~vj = λ~vj for j = r, s We have M~vr(~vs)T+~vr(~vs)TMT = (M~vr)(~vs)T+~vr M~vsT
= λr~vr(~vs)T+λs~vr(~vs)T,
so ~vr(~vs) is an eigenmatrix of LM with the eigenvalue λr+ λs
Notice that if λr 6= λs then vectors ~u, ~w are linearly independent and matrices ~u( ~w)T and ~w(~u)T are linearly independent, too This implies that the eigenvalue λr+ λs is double if r 6= s
The map LM maps n2–dimensional linear space into itself, so it has at most n2 eigenvalues We already found n2 eigenvalues, so there exists no more and the problem is solved for the special case
In the general case, matrix M is a limit of matrices M1, M2, such that each of them belongs to the special case above By the continuity of the eigenvalues we obtain that the eigenvalues of LM are
• 2λr with multiplicity m2
r (r = 1, , k);
• λr+ λs with multiplicity 2mrms (1 ≤ r < s ≤ k)
(It can happen that the sums λr+ λs are not pairwise different; for those multiple values the multiplicities should be summed up.)
5 Prove that
Z 1 0
Z 1 0
dx dy
xư1+ | ln y| ư 1 ≤ 1 [20 points]
Solution 1 First we use the inequality
xư1ư 1 ≥ | ln x|, x ∈ (0, 1], which follows from
(xư1ư 1)
x=1 = | ln x||x=1 = 0, (xư1ư 1)0 = ư1
x2 ≤ ư1
x = | ln x|
0 , x ∈ (0, 1]
Therefore
Z 1 0
Z 1 0
dx dy
xư1+ | ln y| ư 1 ≤
Z 1 0
Z 1 0
dx dy
| ln x| + | ln y| =
Z 1 0
Z 1 0
dx dy
| ln(x · y)|. Substituting y = u/x, we obtain
Z 1 0
Z 1 0
dx dy
| ln(x · y)| =
Z 1 0
Z 1 u
dx x
du
| ln u| =
Z 1 0
| ln u| · du
| ln u| = 1.
Solution 2 Substituting s = xư1ư 1 and u = s ư ln y,
Z 1
0
Z 1 0
dx dy
xư1+ | ln y| ư 1 =
Z ∞ 0
Z ∞ s
esưu (s + 1)2ududs =
Z ∞ 0
Z u 0
es (s + 1)2ds eưu
u dsdu.
Since the function (s+1)es 2 is convex,
Z u 0
es (s + 1)2ds ≤ u
2
eu (u + 1)2 + 1
so
Z 1
0
Z 1
0
dx dy
xư1+ | ln y| ư 1 ≤
Z ∞ 0
u 2
eu (u + 1)2 + 1 eưu
u du =
1 2
Z ∞ 0
du (u + 1)2 +
Z ∞ 0
eưudu
= 1
Trang 36 For n ≥ 0 define matrices An and Bn as follows: A0 = B0 = (1) and for every n > 0
An =An−1 An−1
An−1 Bn−1
and Bn =An−1 An−1
An−1 0
Denote the sum of all elements of a matrix M by S(M ) Prove that S(Ak−1
n ) = S(An−1k ) for every n, k ≥ 1 [20 points]
Solution The quantity S(Ak−1
n ) has a special combinatorical meaning Consider an n × k table filled with 0’s and 1’s such that no 2 × 2 contains only 1’s Denote the number of such fillings by Fnk The filling of each row of the table corresponds to some integer ranging from 0 to 2n− 1 written in base 2 Fnk equals
to the number of k-tuples of integers such that every two consecutive integers correspond to the filling of
n × 2 table without 2 × 2 squares filled with 1’s
Consider binary expansions of integers i and j inin−1 i1 and jnjn−1 j1 There are two cases:
1 If injn = 0 then i and j can be consecutive iff in−1 i1 and jn−1 j1 can be consequtive
2 If in = jn = 1 then i and j can be consecutive iff in−1jn−1 = 0 and in−2 i1 and jn−2 j1 can be consecutive
Hence i and j can be consecutive iff (i + 1, j + 1)-th entry of An is 1 Denoting this entry by ai,j, the sum S(Ak−1
n ) =P2 n −1
i 1 =0 · · ·P2 n −1
i k =0 ai1i2ai2i3· · · aik−1ik counts the possible fillings Therefore Fnk = S(Ak−1
n ) The the obvious statement Fnk = Fkn completes the proof
Trang 411th International Mathematical Competition for University Students
Skopje, 25–26 July 2004 Solutions for problems on Day 1
Problem 1 Let S be an infinite set of real numbers such that |s1 + s2 + · · · + sk| < 1 for every finite subset {s1, s2, , sk} ⊂ S Show that S is countable [20 points]
Solution Let Sn= S ∩ (1n, ∞) for any integer n > 0 It follows from the inequality that |Sn| < n Similarly, if we define S−n= S ∩ (−∞, −n1), then |S−n| < n Any nonzero x ∈ S is an element of some Sn or S−n, because there exists an n such that x > n1, or x < −1n Then S ⊂ {0} ∪ S
n∈N
(Sn∪ S−n), S is a countable union of finite sets, and hence countable
Problem 2 Let P (x) = x2− 1 How many distinct real solutions does the following equation have:
P (P ( (P
2004
(x)) )) = 0 ? [20 points]
Solution Put Pn(x) = P (P ( (P
n
(x)) )) As P1(x) ≥ −1, for each x ∈ R, it must be that Pn+1(x) = P1(Pn(x)) ≥
−1, for each n ∈ N and each x ∈ R Therefore the equation Pn(x) = a, where a < −1 has no real solutions Let us prove that the equation Pn(x) = a, where a > 0, has exactly two distinct real solutions To this end we use mathematical induction by n If n = 1 the assertion follows directly Assuming that the assertion holds for a
n ∈ N we prove that it must also hold for n + 1 Since Pn+1(x) = a is equivalent to P1(Pn(x)) = a, we conclude that Pn(x) =√a + 1 or Pn(x) = −√a + 1 The equation Pn(x) =√a + 1, as√a + 1 > 1, has exactly two distinct real solutions by the inductive hypothesis, while the equation Pn(x) = −√a + 1 has no real solutions (because
−√a + 1 < −1) Hence the equation Pn+1(x) = a, has exactly two distinct real solutions
mathematical induction If n = 1 the solutions are x = ±1, and if n = 2 the solutions are x = 0 and x = ±√2,
so in both cases the number of solutions is equal to n + 1 Suppose that the assertion holds for some n ∈ N Note that Pn+2(x) = P2(Pn(x)) = Pn2(x)(Pn2(x) − 2), so the set of all real solutions of the equation Pn+2 = 0 is exactly the union of the sets of all real solutions of the equations Pn(x) = 0, Pn(x) = √2 and Pn(x) = −√2
By the inductive hypothesis the equation Pn(x) = 0 has exactly n + 1 distinct real solutions, while the equations
Pn(x) =√2 and Pn(x) = −√2 have two and no distinct real solutions, respectively Hence, the sets above being pairwise disjoint, the equation Pn+2(x) = 0 has exactly n + 3 distinct real solutions Thus we have proved that, for each n ∈ N , the equation Pn(x) = 0 has exactly n + 1 distinct real solutions, so the answer to the question posed in this problem is 2005
Problem 3 Let Sn be the set of all sums
n
P
k=1
xk, where n ≥ 2, 0 ≤ x1, x2, , xn≤ π
2 and
n
X
k=1
sin xk= 1
a) Show that Sn is an interval [10 points]
b) Let ln be the length of Sn Find lim
n→∞ln [10 points]
Solution (a) Equivalently, we consider the set
Y = {y = (y1, y2, , yn)| 0 ≤ y1, y2, , yn≤ 1, y1+ y2+ + yn= 1} ⊂ Rn and the image f (Y ) of Y under
f (y) = arcsin y1+ arcsin y2+ + arcsin yn Note that f (Y ) = Sn Since Y is a connected subspace of Rn and f is a continuous function, the image f (Y ) is also connected, and we know that the only connected subspaces of R are intervals Thus Sn is an interval
Trang 5(b) We prove that
n arcsin1
n ≤ x1+ x2+ + xn≤
π
2. Since the graph of sin x is concave down for x ∈ [0,π2], the chord joining the points (0, 0) and (π2, 1) lies below the graph Hence
2x
π ≤ sin x for all x ∈ [0,
π
2] and we can deduce the right-hand side of the claim:
2
π(x1+ x2+ + xn) ≤ sin x1+ sin x2+ + sin xn= 1.
The value 1 can be reached choosing x1 = π2 and x2 = · · · = xn= 0
The left-hand side follows immediately from Jensen’s inequality, since sin x is concave down for x ∈ [0,π2] and
0 ≤ x1 +x 2 + +x n
n < π2
1
sin x1+ sin x2+ + sin xn
x1+ x2+ + xn
Equality holds if x1= · · · = xn= arcsinn1
Now we have computed the minimum and maximum of interval Sn; we can conclude that Sn= [n arcsinn1,π2] Thus ln= π2 − n arcsin1n and
lim
n→∞ln= π
2 − limn→∞
arcsin(1/n)
π
2 − 1.
Problem 4 Suppose n ≥ 4 and let M be a finite set of n points in R3, no four of which lie in a plane Assume that the points can be coloured black or white so that any sphere which intersects M in at least four points has the property that exactly half of the points in the intersection of M and the sphere are white Prove that all of the points in M lie on one sphere [20 points]
Solution Define f : M → {−1, 1}, f (X) =
−1, if X is white
1, if X is black The given condition becomes
P
X∈Sf (X) = 0 for any sphere S which passes through at least 4 points of M For any 3 given points A, B, C in M , denote by
S (A, B, C) the set of all spheres which pass through A, B, C and at least one other point of M and by |S (A, B, C)|
We have
S∈S(A,B,C)
X
X∈S
since the values of A, B, C appear |S (A, B, C)| times each and the other values appear only once
If there are 3 points A, B, C such that |S (A, B, C)| = 1, the proof is finished
If |S (A, B, C)| > 1 for any distinct points A, B, C in M , we will prove at first thatP = 0
Assume that P > 0 From (1) it follows that f (A) + f (B) + f (C) < 0 and summing by all n
3 possible choices of (A, B, C) we obtain that n3 P < 0, which means P < 0 (contradicts the starting assumption) The same reasoning is applied when assuming P < 0
Now, from P = 0 and (1), it follows that f (A) + f (B) + f (C) = 0 for any distinct points A, B, C in M Taking another point D ∈ M , the following equalities take place
f (A) + f (B) + f (C) = 0
f (A) + f (B) + f (D) = 0
f (A) + f (C) + f (D) = 0
f (B) + f (C) + f (D) = 0 which easily leads to f (A) = f (B) = f (C) = f (D) = 0, which contradicts the definition of f
Problem 5 Let X be a set of 2k−4k−2 + 1 real numbers, k ≥ 2 Prove that there exists a monotone sequence {xi}k
i=1⊆ X such that
|xi+1− x1| ≥ 2|xi− x1| for all i = 2, , k − 1 [20 points]
Trang 6Solution We prove a more general statement:
Lemma Let k, l ≥ 2, let X be a set of k+l−4k−2 +1 real numbers Then either X contains an increasing sequence {xi}k
i=1⊆ X of length k and
|xi+1− x1| ≥ 2|xi− x1| ∀i = 2, , k − 1,
or X contains a decreasing sequence {xi}l
i=1⊆ X of length l and
|xi+1− x1| ≥ 2|xi− x1| ∀i = 2, , l − 1
Proof of the lemma We use induction on k + l In case k = 2 or l = 2 the lemma is obviously true
Now let us make the induction step Let m be the minimal element of X, M be its maximal element Let
m + M
Since k+l−4k−2 = k+(l−1)−4
k−2 + (k−1)+l−4
(k−1)−2 , we can see that either
|Xm| ≥(k − 1) + l − 4
(k − 1) − 2
k − 2
+ 1
In the first case we apply the inductive assumption to Xm and either obtain a decreasing sequence of length l with the required properties (in this case the inductive step is made), or obtain an increasing sequence {xi}k−1i=1 ⊆
Xm of length k − 1 Then we note that the sequence {x1, x2, , xk−1, M } ⊆ X has length k and all the required properties
In the case |XM| ≥ k+(l−1)−4k−2 + 1 the inductive step is made in a similar way Thus the lemma is proved The reader may check that the number k+l−4k−2 + 1 cannot be smaller in the lemma
Problem 6 For every complex number z /∈ {0, 1} define
f (z) :=X(log z)−4, where the sum is over all branches of the complex logarithm
points]
b) Show that for all z ∈ C \ {0, 1}
f (z) = zz
2+ 4z + 1
Solution 1 It is clear that the left hand side is well defined and independent of the order of summation, because
we have a sum of the typeP n−4, and the branches of the logarithms do not matter because all branches are taken
It is easy to check that the convergence is locally uniform on C \ {0, 1}; therefore, f is a holomorphic function on the complex plane, except possibly for isolated singularities at 0 and 1 (We omit the detailed estimates here.) The function log has its only (simple) zero at z = 1, so f has a quadruple pole at z = 1
Now we investigate the behavior near infinity We have Re(log(z)) = log |z|, hence (with c := log |z|)
|X(log z)−4| ≤ X| log z|−4 =X(log |z| + 2πin)−4+ O(1)
=
−∞
(c + 2πix)−4dx + O(1)
−∞
(1 + 2πix/c)−4dx + O(1)
−∞
(1 + 2πit)−4dt + O(1)
≤ α(log |z|)−3 for a universal constant α Therefore, the infinite sum tends to 0 as |z| → ∞ In particular, the isolated singularity
at ∞ is not essential, but rather has (at least a single) zero at ∞
Trang 7The remaining singularity is at z = 0 It is readily verified that f (1/z) = f (z) (because log(1/z) = − log(z)); this implies that f has a zero at z = 0
We conclude that the infinite sum is holomorphic on C with at most one pole and without an essential singularity
at ∞, so it is a rational function, i.e we can write f (z) = P (z)/Q(z) for some polynomials P and Q which we may as well assume coprime This solves the first part
Since f has a quadruple pole at z = 1 and no other poles, we have Q(z) = (z − 1)4 up to a constant factor which we can as well set equal to 1, and this determines P uniquely Since f (z) → 0 as z → ∞, the degree of P
is at most 3, and since P (0) = 0, it follows that P (z) = z(az2+ bz + c) for yet undetermined complex constants
a, b, c
There are a number of ways to compute the coefficients a, b, c, which turn out to be a = c = 1/6, b = 2/3 Since f (z) = f (1/z), it follows easily that a = c Moreover, the fact lim
z→1(z − 1)4f (z) = 1 implies a + b + c = 1 (this fact follows from the observation that at z = 1, all summands cancel pairwise, except the principal branch which contributes a quadruple pole) Finally, we can calculate
nodd
n≥1odd
n−4= 2π−4
X
n≥1
n≥1even
n−4
48 . This implies a − b + c = −1/3 These three equations easily yield a, b, c
Moreover, the function f satisfies f (z) + f (−z) = 16f (z2): this follows because the branches of log(z2) = log((−z)2) are the numbers 2 log(z) and 2 log(−z) This observation supplies the two equations b = 4a and a = c, which can be used instead of some of the considerations above
(log z) 2 first In the same way, g(z) = (z−1)dz 2 The unknown coefficient d can be computed from lim
z→1(z − 1)2g(z) = 1; it is d = 1 Then the exponent 2 in the denominator can be increased
by taking derivatives (see Solution 2) Similarly, one can start with exponent 3 directly
A more straightforward, though tedious way to find the constants is computing the first four terms of the Laurent series of f around z = 1 For that branch of the logarithm which vanishes at 1, for all |w| < 12 we have
2
w3
w4
5);
after some computation, one can obtain
1 log(1 + w)4 = w−4+ 2w−2+7
6w
−2+1
6w
−1+ O(1)
The remaining branches of logarithm give a bounded function So
f (1 + w) = w−4+ 2w−2+7
6w
−2+ 1
6w
−1
(the remainder vanishes) and
f (z) = 1 + 2(z − 1) +
7
6(z − 1)2+ 16(z − 1)3
2+ 4z + 1) 6(z − 1)4 Solution 2 ¿From the well-known series for the cotangent function,
lim
N →∞
N
X
k=−N
1
w + 2πi · k =
i
2cot
iw 2 and
lim
N →∞
N
X
k=−N
1 log z + 2πi · k =
i
2cot
i log z
i
2 · i
e2i·i log z2 + 1
e2i·i log z2 − 1
1
z − 1. Taking derivatives we obtain
(log z)2 = −z · 1
1
z − 1
0
(z − 1)2,
(log z)3 = −z
2 ·
z (z − 1)2
0
2(z − 1)3
and
(log z)4 = −z
3·
z(z + 1) 2(z − 1)3
0
2+ 4z + 1) 2(z − 1)4
...48 . This implies a − b + c = −1/3 These three equations easily yield a, b, c
Moreover, the function f satisfies f (z) + f (−z) = 16f (z2): this follows because...
Hence i and j can be consecutive iff (i + 1, j + 1)-th entry of An is Denoting this entry by ai,j, the sum S(Ak−1
n ) =P2... data-page="4">
11th International Mathematical Competition for University Students
Skopje, 25–26 July 2004 Solutions for problems on Day
Problem Let S be an infinite set of real numbers such