POSSIBLE INFLUENCES OF RAILWAY CHARGES ON THE DEVELOPMENT OF HIGH SPEED
3.3. METHODOLOGY AND ASSUMPTIONS FOR THE CALCULATION OF THE RAIL INFRASTRUCTURE CHARGES APPLIED TO THE EUROPEAN
3.3.1. Sources of information and tools available for the calculation of rail infrastructure charges in European railway networks
MathProblems (4)1 (2014) Brendan W. Sullivan Emmanuel College, USA
Abstract: We have a rectangular chessboard 3× 20 made of square boxes 1 × 1.
We also have 30 dominoes of size 1× 2 or 2 × 1 that we want to use to cover the chessboard. In this note we address the question of counting the number of different ways to do that? (Problem proposed by Callegari Emanuele, Math. Dept.
“Tor Vergata” University, Rome, Italy.)
Solution: Let T (2n) be the number of ways to tile a 3×2n rectangular chessboard using dominoes. (Note: there are trivially no domino tilings of a 3×k board when k is odd.) This problem seeks a number for T (20). We solve this problem by proving the more general recursive formula
T (2n) = 4· T (2n − 2) − T (2n − 4)
for n≥ 2, with the additional convention that T (0) = 1. (To motivate this conven-tion, consider: How many ways can we tile an empty board? One way: I just did it!) This will lead us to the answer: T (20) = 413403.
First, observe that T (2) = 3:
Now, consider a 3× 2n board. We will describe all of the cases wherein one can construct a domino tiling of this board. These disjoint cases are based on the dominoes in the (2n−2)-th and (2n−1)-th columns of the board; more specifically,
these cases depend on how many domino tiles cross the boundary between those two columns.
3× 2n − 2 portion boundary
· · ·
(i) 0 crossings. This means we have a proper tiling of the final two columns, appended to a proper tiling of the first 2n− 2 columns. We know that there are 3 ways to tile the final two columns (since T (1) = 3), and there are T (2n−2) ways to tile the remaining portion. Thus, this case contributes the term 3· T (2n − 2) to the total number of tilings.
(ii) 1 crossing. By exhibiting three cases, we can show that this case actu-ally contributes no proper tilings: each possibility yields an irreconcilable situation in the 2n-th column.
Firstly, if the one and only crossing occurs on the top row, we can attempt to tile the surrounding squares in two ways:
· · · ?
· · ·
?
Secondly, by obvious symmetry, the situations where the one and only crossing occurs on the bottom row yield no tilings, as well.
Finally, when the one and only crossing occurs in the middle row, we have one possibility:
· · · ?
Thus, the case of exactly 1 crossing yields no tilings.
(iii) 2 crossings. This is the interesting case!
First, we observe that if the two crossings occur on the top and bottom row, then this yields no tilings:
· · ·
?
?
Next, we consider the more fruitful cases where the two crossings occur on the top two rows. (By symmetry, this will be analogous to the cases where the two crossings occur on the bottom two rows; thus, the terms we derive presently will then be doubled to account for all the tilings.) What we will observe is that such cases do, indeed, yield proper tilings.
Furthermore, each case can be extended arbitrarily far to the left, in blocks of two columns. Thus, we will generate tilings that correspond to this summation of terms: T (2n− 4) + T (2n − 6) + T (2n − 8) + · · · T (2) + T (0).
One possibility is as follows:
3× 2n − 4 portion
· · ·
This amounts to a particular proper tiling of the final 4 columns, which can be appended to any proper tiling of the first 2n− 4 columns. Thus, the number of such tilings is T (2n− 4).
We can extend this to the left by converting the leftmost vertical domino into two horizontal ones, like so:
· · ·
3× 2n − 6 portion
This amounts to a particular proper tiling of the final 6 columns, which can be appended to any proper tiling of the first 2n− 6 columns. Thus, the number of such tilings is T (2n− 6).
Observe that this extension process can be continued until we reach the leftmost side of the board. That final case amounts to a particular proper tiling of the entire board, “appended” to the only tiling of the empty board. Therefore, we include the terms of the summation mentioned above to account for all of these tilings. Also, as mentioned above, the cases where the two crossings of the boundary column occur on the bottom two rows are symmetrically analogous. In total, then, we include the terms 2·Pn−2
k=0T (2k).
(iv) 3 crossings. Observe that this case yields no proper tilings:
· · · ?
Overall, we have now deduced that
T (2n) = 3· T (2n − 2) + 2 ·
n−2X
k=0
T (2k)
provided n > 1 (otherwise the first part of the case involving 2 crossings above is invalid).
The next step is to simplify this recurrence to two terms. Consider the difference T (2n)− T (2n − 2), for n ≥ 1, and observe that we can cancel terms:
T (2n)− T (2n − 2) = 3T (2n − 2) + 2T (2n − 4) + 2T (2n − 6) + · · · + 2T (0)
− 3T (2n − 4) + 2T (2n − 6) + · · · + 2T (0)
= 3T (2n− 2) − T (2n − 4) Adding T (2n− 2) to both sides, we conclude that
T (2n) = 4T (2n− 2) − T (2n − 4) as claimed, originally.
Knowing that T (0) = 1 (by convention), or even just counting to observe that T (2) = 3 and T (4) = 11, we can plug numbers into this recurrence and deduce that
n 0 1 2 3 4 5 6 7 8 9 10
T (2n) 1 3 11 41 153 571 2131 7953 29681 110771 413403 Thus, the desired answer is that there are T (20) = 413403 ways to properly tile a 3× 20 rectangular chessboard using dominoes.
Finally, note that the above recurrence is linear, so one can solve for a closed form using standard techniques (i.e. generating functions). This yields
T (2n) = 1 +√ 3 2√
3
!
· (2 +√
3)n+ −1 +√ 3 2√
3
!
· (2 −√ 3)n =
&
(2 +√ 3)n+1 3 +√
3 '
. This can be used somewhat more efficiently for larger n. Plugging in n = 10 (i.e.
2n = 20) yields the same answer as above.
ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 3, Issue 1 (2013), Pages 118–139
Editors: Valmir Krasniqi, Jos´e Luis D´ıaz-Barrero, Paolo Perfetti, Armend Sh. Shabani, Valmir Bucaj, Mih´aly Bencze, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, J´ozsef S´andor, David R. Stone, Roberto Tauraso, Francisco Javier Garca Capit´an, Emanuele Callegari, Ercole Suppa, Cristinel Mortici.
PROBLEMS AND SOLUTIONS
Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction.
Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]
Solutions to the problems stated in this issue should arrive before June 15, 2013
Problems
58. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. 1) Let a be a real number such that −1 ≤ a ≤ 1. Prove that
∞
X
i=1
∞
X
j=1
(i − 1)!(j − 1)!
(i + j)! ai+j = 2Li2
a 2 − a
− 2Li2
− a 2 − a
− 2Li2(a),
where Li2(z) is the Dilogarithm function defined by Li2(z) = −Rz 0
ln(1 − t) t dt for all z ∈ {z ∈ C : |z| ≤ 1}.
2) Show that
∞
X
i=1
∞
X
j=1
(i − 1)!(j − 1)!
(i + j)! = ζ(2), where ζ(z) represents the Riemann zeta function.
c
2010 Mathproblems, Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e.
118
59. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). Let n be a positive integer. Prove that
LnL2n+2 Ln+3
+ Ln+1L2n+3 Ln+ Ln+2
+ (Ln+ Ln+2)2≥ 2√ 6 ·p
LnLn+1· Ln+2,
where Ln represents the nthLucas number defined by L0= 2, L1 = 1, and for all n ≥ 2, Ln= Ln−1+ Ln−2.
60. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. De-termine all functions f : R \ {−2, −1, 0, 1, 2} → R, which satisfy the relation
f (x) + f 7 + x 2 − x
= ax + b where a, b ∈ R.
61. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). If a ∈ R+, b, c, d, xk∈ R∗+, Xn=Pn
k=1xkand cXn> d max
1≤k≤nxk, then prove that
n
X
k=1
aXn+ bxk cXn− dxk
≥(an + b)n cn − d .
62. Proposed by Mih´aly Bencze, Bra¸sov, Romania Show that if ak, bk, ck ∈ R∗(k = 1, 2), such that a21+ b21+ c21= a22+ b22+ c22, then at least one of the equations
a1x2+ 2c2x + b1= 0; b1x2+ 2a2x + c1= 0; c1x2+ 2b2x + a1= 0 has real roots.
63. Proposed by Anastasios Kotronis, Athens, Greece. Let k ≥ 3 be an integer number. Show that
X
n≥0
1
(2n + 1)(3n + 2) · · · (kn + k − 1) = 1
k!
k
X
m=2
(−1)m−1 k m
(m − 1)mk−2 π 2cot π
m − ln m +
m−1
X
`=1
cos2`π m · ln
sin`π
m
! .
64. Proposed by Florin Stanescu, Serban School Cioculescu, Gaesti, Romania. Let f : [0, 1] → R be a continuous function. If
Z 1 0
x2f (x)dx = −2 Z 1
1/2
F (x)dx, where F (x) =Rx
0 f (t)dt, x ∈ [0, 1], then prove that Z 1
0
f2(x)dx ≥ 24
Z 1 0
f (x)dx
2 .
65.∗Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. Let Ωn
represents the volume of the unit ball in Rn, that is Ωn= πn/2
Γ(n/2 + 1), n = 1, 2 . . . , where Γ(x) =R∞
0 tx−1e−tdt, x > 0. For all natural numbers n, show that r n + 2α
n + 3 6 (Ωn+1)n+11
(Ωn)n1 6 r n + 2β n + 3 with the best possible constant factors α = 2 and β = ln(34)
ln(
√π
2 ) = 2.3818 · · ·
Solutions
No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.
50. Proposed by Anastasios Kotronis, Athens, Greece. Show that
+∞
X
k=0
(−1)k
ln(n + k) = 1
2 ln n+ O(n−1ln−2n) n → +∞.
Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. More generally we have :
Proposition. Let f : (a, ∞) → R be a continuously differentiable convex function such that limx→∞f (x) = 0. Then, for every x > a, the series P∞
k=0(−1)kf (x + k) does converge, and its sum satisfy the following inequality:
0 ≤ −1 2f (x) +
∞
X
k=0
(−1)kf (x + k) ≤ −1 2f0(x).
Proof. First, note that the conditions on f imply that f is nonnegative decreasing and that limx→∞f0(x) = 0. Indeed, if for some b ∈ (a, +∞) we have f0(b) > 0, then the convexity of f implies that f (x) ≥ f (b) + (x − b)f0(b) and this leads to the contradiction limx→∞f (x) = +∞. So, f0≤ 0 and f must be nonnegative and decreasing. On the other hand, since f0 is increasing and negative, there exists a real number ` ≤ 0 such that limx→∞f0(x) = `. If ` < 0, then the convexity of f implies that for y > b we have f (y) − f (b) ≤ f0(y)(y − b) ≤ `(y − b). Taking the limit as y tend to +∞ we obtain the contradiction −f (b) ≤ −∞. So, we must have
` = 0.
Now, for each x ∈ (a, +∞) the sequence {f (k + x)}k≥0 is decreasing to zero, and the alternating series test proves the convergence of the seriesP(−1)kf (x + k). So, let S(x) be defined, for x ∈ (a, +∞), by
S(x) =
∞
X
k=0
(−1)kf (x + k) =
∞
X
k=0
f (2k + x) − f (2k + 1 + x)
On the other hand
f (x) = f (x + 2m + 2) +
m
X
k=0
f (2k + x) − f (2k + 2 + x)
So, letting m tend to +∞, we obtain f (x) =
∞
X
k=0
f (2k + x) − f (2k + 2 + x)
Combining the preceding, we get
2S(x) − f (x) =
∞
X
k=0
f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x)
But using the convexity of f we have
f (2k+1+x)−f (2k+x) ≥ f0(2k+x) and f (2k+2+x)−f (2k+1+x) ≤ f0(2k+2+x) Thus
f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f0(2k + 2 + x) − f0(2k + x) On the other hand, again using the convexity of f , we have
f (2k + 1 + x) ≤1
2(f (2k + x) + f (2k + 2 + x)) It follows that
0 ≤
m
X
k=0
f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f0(2m + 2 + x) − f0(x)
So letting m tend to +∞ we obtain using (3) that 0 ≤ 2S(x) − f (x) ≤ −f0(x).
Applying the Lemma to the function f : (1, +∞) → R, f (x) = 1/ ln(x) we obtain
∀ x > 1, 1 2 ln x ≤
+∞
X
k=0
(−1)k
ln(x + k) ≤ 1
2 ln x + 1 2x ln2x from which the statement follows.
Also solved by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Moubinool Omarjee, Paris, France; and the proposer.
51. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). Let a ∈ (1, ∞) and b ∈ (0, ∞). Calculate:
n→∞lim n 2 − exp
n
X
k=1
(n + k)a−1 (n + k)a+ b
!!
Solution 1 by Moti Levy. The first step is to show that the limit does not dependent of a and b. Indeed, we have
n
X
k=1
(n + k)a−1 (n + k)a+ b = 1
n
n
X
k=1
1 1 +nk + b
n(n+k)a−1
If we put en,k =n(n+k)b a−1, 1 ≤ k ≤ n, then 0 < en,k≤ nba and By Taylor’s theorem,
exp (x + h) − exp (x) = exp (x) It follows from the preceding that
n→∞lim n 2 − exp
The second step is evaluation of limn→∞n So now we estimate ∆n. Since the function 1+x1 is convex then the area of the triangle formed by the secant is greater than ∆n. That is,
∆n≤ 1 Since the function 1+x1 is convex then the area of the triangle formed by the tangent is less than ∆n. Namely,
Solution 2 by Anastasios Kotronis, Athens, Greece.
We use the fact that
Hn= ln n + γ + 1
2n+ O(n−2) n → +∞. (1)
Note that for m > 1, it is
X
k≥n+1
1
km = O(n1−m), (2)
Applying Ces`aro–Stolz criterium, we have
n→+∞lim nm−1 X On account of the above
n (2 − exp Sa,b,n) = n Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Moubinool Omarjee, Paris, France;
Adrian Naco, Polytechnic University, Tirana, Albania; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer.
52. Proposed by Yuanzhe Zhou, The School of Physics and Technology (SPT) at Wuhan University. Let a, b, c > 0, prove that,
cos2
Solution by Moti Levy. Setting x = a
a + b + c, y = b
a + b + c, z = c a + b + c, then the inequality becomes
cos2
x 1 − x
+ cos2
y 1 − y
+ cos2
z 1 − z
>3
2, x + y + z = 1.
Using the well-known trigonometric identity cos2α = 12 +12cos 2α , and setting f (x) = cos
2x 1−x
, the inequality becomes
f (x) + f (y) + f (z) > 0, x + y + z = 1.
The function f (x) is concave for 0 < x < 12, so we can find a piecewise linear lower bound for f (x). Let g (x) be defined by
g (x) =
−2 (1 − cos 2) x + 1, 0 < x <12
−1, 12≤ x ≤ 1
then f (x) ≥ g (x) for 0 < x ≤ 1.
WLOG, we may assume that 0 < x ≤ y ≤ z ≤ 1. Then it will be suffice to consider the following two cases: 1) z < 12 and 2) z ≥12.
Case 1) z < 12 :
f (x) + f (y) + f (z) ≥ (−2 (1 − cos 2) x + 1) + (−2 (1 − cos 2) y + 1) + (−2 (1 − cos 2) z + 1)
= −2 (1 − cos 2) (x + y + z) + 3 = −2 (1 − cos 2) + 3 > 0 Case 2) z ≥ 12 : x + y < 12
f (x) + f (y) + f (z) ≥ −2 (1 − cos 2) x + 1 − 2 (1 − cos 2) y + 1 − 1
≥ −2 (1 − cos 2) (x + y) + 2 − 1
= 1 − (1 − cos 2) (x + y) ≥ 1 −1
2(1 − cos 2) > 0.
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; and the proposer.
53. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. The Stirling numbers of the first kind denoted by s(n, k), are the spe-cial numbers defined by the generating function z(z − 1)(z − 2) · · · (z − n + 1) = Pn
k=0s(n, k)zk. Let n and m be nonnegative integers with n > m − 1. Prove that Z 1
0
lnnx (1 − x)mdx =
((−1)n· n! · ζ(n + 1), m = 1,
(−1)n+m−1· (m−1)!n! ·Pm−1
i=1 (−1)i· s(m − 1, i) · ζ(n + 1 − i), m ≥ 2, where ζ denotes the Riemann zeta function.
Solution 1 by Anastasios Kotronis, Athens, Greece. By repeated integration by parts one can easily show that for any non negative integer k and any positive integer n, we have
Z
xklnnx dx = xk+1 lnnx
k + 1 −n lnn−1x
(k + 1)2 +n(n − 1) lnn−2x
(k + 1)3 − . . . + (−1)nn!
(k + 1)n+1
+c
Furthermore, we will use that Now, we have
Z 1 But for m ≥ 2, from the definition of Stirling numbers, we have
−k − 1 and we get what we wanted.
Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. First let us calculate the integralR1
0 xαlnβ(1/x) dx
Since all the terms are positive we can integrate on [0, 1] and exchange the order of
On the other hand, by definition we havePm−1
i=0 s(m−1, i)zi = z(z−1)(z−2) · · · (z−
Replacing in (1) we obtain 1
It follows that Z 1
Remark. The same proof shows that, more generally, for a real β and an integer m such that 1 ≤ m ≤ β we have Also solved by Moti Levi; and the proposer.
54. Proposed by Moubinool OMARJEE, Paris, France. Let f : [0, +∞[→ R be a measurable function such that g(t) = etf (t) ∈ L1(R+) ; the space of Lebesgue integrable functions. Find
n→+∞lim Z
R+
f (t)
4 sinh nt + 1 2
sinh nt − 1 2
n1 dt where sinh(x) = ex−e2−x
Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. We prove that
sinhnt + 1
2 sinhnt − 1 2 ≤ etn Indeed, putting etn= x, we have
√ e√
x − 1
√e√ x
√x
√e −
√e
√x
≤ x from which follows
e2− x(e2+ 1) ≤ 0 This holds true since x ≥ 1. Now define
fn(t) = f (t)
4 sinhnt + 1
2 sinhnt − 1 2
n1
|fn| ≤ et|f (t)| ∈ L1(R+)
by the Lebesgue integrability of etf (t). The dominated convergence theorem allows us to take the limit under integral getting
n→+∞lim Z
R+
f (t)
4 sinhnt + 1
2 sinhnt − 1 2
1n dt = Z
R+
f (t) lim
n→+∞
4 sinhnt + 1
2 sinhnt − 1 2
1n dt =
Z
R+
f (t)etdt
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Anastasios Kotronis, Athens, Greece;
Moti Levi; and the proposer.
55. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Let x, y, z, α be real positive numbers. Show that if
X
cycl
nx3+ (n + 1)x x2+ 1 = α then
X
cycl
1 x> 2α
3 + 27n3 9n2α + α3 where n is a natural number.
Solution by Moti Levi. Let x = y = z = 1 and α = 301.5; then n = 100 satisfies the condition.
X
cycl 1
x = 3 and 2α3 +9n27n2α+α3 3 = 2×301.53 +9×100227×100×301.5+301.53 3 = 201. 50.
Clearly 3 > 201. 50 is false, which implies that the claim in Problem 55 is not true.
Author’s Comment: I assume that either I did not understand the problem statement or there are typo errors.
56. Proposed byJos´e Luis D´ıaz–Barrero, BARCELONA TECH, Barcelona, Spain.
Let n ≥ 2 be a positive integer. Find all possible values of the number k such that Fn2(1 + Fn+12 Fn+22 )
Fn−1Fn+1 +Fn+22 (1 + Fn2Fn+12 )
FnFn+1 = k +Fn+12 (1 + Fn+22 Fn2) Fn−1Fn
Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let
An= Fn2 Fn−1Fn+1
+ Fn+22 FnFn+1
− Fn+12 Fn−1Fn
Bn
= Fn2Fn+12 Fn+22
1
Fn−1Fn+1+ 1
FnFn+1− 1 Fn−1Fn
so, that k = An+ Bn. Clearly, Bn= Fn2Fn+12 Fn+22
Fn−1FnFn+1
(Fn+ Fn−1− Fn+1) = 0 On the other hand
An= Fn3− Fn+13 + Fn+22 Fn−1 Fn−1FnFn+1 . Now
Fn3− Fn+13 + Fn+22 Fn−1= −Fn−1(Fn2+ Fn+12 + FnFn+1) + Fn+22 Fn−1
= Fn−1 Fn+22 − Fn2− Fn+12 − FnFn+1
= Fn−1 (Fn+1+ Fn)2− Fn2− Fn+12 − FnFn+1
= Fn−1(2FnFn+1− FnFn+1)
= Fn−1FnFn+1
Thus An= 1 and consequently k = 1.
Also solved by Moti Levi; and the proposer.
MATHCONTEST SECTION
This section of the Journal offers readers an opportunity to solve interesting and ele-gant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always wel-comed. The source of the proposals will appear when the solutions be published.
Proposals
41. Find a cubic polynomial which zeros are the square of the zeros of p(x) = x3+ 2x2+ 3x + 4.
42. Let n be an odd positive integer. A knight is placed at random in one of the n2 squares of a chessboard. It is possible that knight back to its initial position after passing once through each square?
43. Compute
sin 1◦+ sin 2◦+ . . . + sin 133◦+ sin 134◦ cos 1◦+ cos 2◦+ . . . + cos 133◦+ cos 134◦
44. Let n, m be positive integers. Find the value of the following expression 1
23n
n+m
X
k=0
7k X
i+j=k
n i
m j
45. Let AGB, BHC and CKA be equilateral triangles constructed on the sides of any triangle ABC. Prove that the centers of the circumcircles of those equilateral triangles D, E, F themselves form an equilateral triangle.
Solutions
36. Find all positive integers n such that 17n−1+ 19n−1divides 17n+ 19n. (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´anchez (student), CFIS, BARCELONA TECH, Barcelona, Spain. We have that 17(17n−1+ 19n−1) < 17n+ 19n <
19(17n−1+ 19n−1) as can be easily checked. Since 17n−1+ 19n−1divides 17n+ 19n, then 17n+19n= 18(17n−1+19n−1) = (17+1)17n−1+(19−1)19n−1= 17n+17n−1+ 19n− 19n−1 from which follows 17n−1= 19n−1.The preceding is only possible for n = 1 for which 17n−1+ 19n−1= 2 that divides 17n+ 19n = 36, and we are done.
2
Also solved by Bruno Salguerico, Viveiro, Spain ; Ioan Viorel Codeanu, Satulung, Rumania ; Jos´e Gibergnas-B´aguena, BARCELONA TECH, Barcelona, Spain ; Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
37. Let α, β and γ be three distinct complex numbers. Show that they are collinear if, and only if, Im(αβ + βγ + γα) = 0.
(Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. The equation of the line through γ = (c, d), β = (a, b), z = (x, y) is
x − c
a − c = y − d
b − d ⇐⇒ Im z − γ β − γ
= 0
Therefore,α, β, γ are collinear if, and only if, Im α − γ β − γ
= 0, or
Im (α − γ)(β − γ)
|β − γ|2
=Im αβ − γβ − αγ + |γ|2
|β − γ|2 = 0 which is equivalent to
Im αβ − γβ − αγ = 0
Since for any complex number z = x + iy is Im(−z) = −y = Im z, then
Im αβ − γβ − αγ = Im(αβ) + Im(−γβ + Im(−αγ) = Im(αβ + βγ + γα) = 0 and we are done.
2 Also solved by Jos´e Gibergnas-B´aguena, BARCELONA TECH, Barcelona, Spain, Daniel V´acaru, Pitesti, Rumania, Bruno Salguerico, Viveiro, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
38. Through the midpoint M of a chord P Q of a circle, any other chords AB and CD are drawn; chords AD and BC meet P Q at points X and Y respectively. Prove that M is the midpoint of XY.
(Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´e Gibergnas-B´aguena, BARCELONA TECH, Barcelona, Spain. We begin by dropping perpendiculars x1 = XX00 and y1 = Y Y00 from X and Y to AB, x2= XX0 and y2= Y Y0 from X and Y to CD.
y2
x2
y1 x1
Y'' x'' Y'
x'
Y X
M Q
P
D
C
B A
Figure 1. Butterfly’s Theorem
Letting a = P M = M Q, x = XM, y = M Y, and observing the pairs of similar tri-angles: 4M XX00∼ 4M Y Y00, 4M XX0∼ 4M Y Y0, 4AXX00∼ 4CY Y0,4DXX0∼ 4BY Y00, yields
x y = x1
y1
, x y =x2
y2
, x1 y2
=AX CY , x2
y1
= XD Y B from which follows
x2
y2 = x1x2 y1y2
= x1 y2
·x2 y1
=AX · XD
CY · Y B = P X · XQ P Y · Y Q
= (a − x)(a + x)
(a + y)(a − y) = a2− x2
a2− y2 =(a2− x2) + x2 (a2− y2) + y2 =a2
a2 = 1 and x = y, as we wanted to prove.
2 Also solved by Bruno Salguerico, Viveiro, Spain and Jos´e Luis D´ ıaz-Barrero, BARCELONA TECH, Barcelona, Spain and Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
39. The students of a University Course in Mathematics take their exams in Cal-culus, Algebra, Physics and Geometry. It is known that 73% passed Calculus exam, 82% passed Algebra, 77% passed Physics and 89% passed Geometry. At least, how many students have passed the exam of all four subjects?
(Training Sessions of Spanish Team for First Stage OME 2013)
Solution by Francesc Gispert S´anchez (student), CFIS, BARCELONA TECH, Barcelona, Spain. First, we calculate the number of students that passed Calculus and Algebra. That is,
N (C ∪ A) = N (C) + N (A) − N (C ∩ A) and
N (C ∩ A) = N (C) + N (A) − N (C ∪ A)
Since N (C) = 73, N (A) = 82, then N (C∩A) will be minimum when N (C∪A) = 100 that it is the maximum value that it may attain. So,
N (C ∩ A) = 73 + 82 − 100 = 55
Now, we calculate the minimum number of students that pass the exam of Calculus, Algebra and Physics. Let us denote by N (B) = N (C ∩ A) the number of student that passed the exam of Calculus and Algebra under the hypothesis that N (C∪A) = 100. Then
N (B ∪ P ) = N (B) + N (P ) − N (B ∩ P ) and
N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P )
The minimum value of N (B ∩ P ) will be got when N (B ∪ P ) = 100. So, N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P ) = 55 + 77 − 100 = 32
Finally, we compute the number of students that passed the four exams. Let us denote by B ∩ P = C ∩ A ∩ P = D. Then,
N (D ∪ G) = N (D) + N (G) − N (D ∩ G) and
N (D ∩ G) = N (D) + N (G) − N (B ∪ P )
The minimum value of N (D ∩G) corresponds to the maximum value of N (B ∪P ) = 100. So,
N (D ∩ G) = N (D) + N (G) − N (B ∪ P ) = 32 + 89 − 100 = 21
and we are done. 2
Also solved by Bruno Salguerico, Viveiro, Spain and Jos´e Luis D´ ıaz-Barrero BARCELONA TECH, Barcelona, Spain.
40. Let α, β, γ be the angles of an acute triangle ABC. Prove that 1
3 X
cyclic
tan2α tan β tan γ + 3
1
tan α + tan β + tan γ
2/3
≥ 2
(Mediterranean Mathematical Olympiad 2012)
Solution by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Using the well-known identity tan α tan β tan γ = tan α + tan β + tan γ and rearranging terms, we can write the inequality claimed as
tan3α + tan3β + tan3γ 3 tan α tan β tan γ + 3√3
tan α tan β tan γ tan α + tan β + tan γ ≥ 2 Now, applying two times AM-GM inequality, we have
tan3α + tan3β + tan3γ 3 tan α tan β tan γ + 3√3
tan α tan β tan γ tan α + tan β + tan γ
= tan3α + tan3β + tan3γ
3 tan α tan β tan γ +3 3√3
tan α tan β tan γ tan α + tan β + tan γ
−2 3√3
tan α tan β tan γ tan α + tan β + tan γ
≥ 4 4 s
9(tan3α + tan3β + tan3γ) (tan α + tan β + tan γ)3 − 2 So, it will be suffice to prove that
4
s
9(tan3α + tan3β + tan3γ) (tan α + tan β + tan γ)3 ≥ 1 The preceding is equivalent to see that
9(tan3α + tan3β + tan3γ) ≥ (tan α + tan β + tan γ)3 It trivially holds from
tan α + tan β + tan γ
3 ≤ 3
rtan3α + tan3β + tan3γ 3
Equality holds when tan α = tan β = tan γ. That is when 4ABC is equilateral and
we are done. 2
Also solved by Bruno Salguerico, Viveiro, Spain and Jos´e Gibergnas-B´aguena, BARCELONA TECH, Barcelona, Spain.