This section of the Journal offers readers an opportunity to solve interesting and el-egant 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
65. Let a, b, c, d be digits such that d > c > b > a≥ 0. How many numbers of the form 1a1b1c1d1 are multiples of 33?
66. Given trapezoid ABCD with parallel sides AB and CD, let E be a point on line BC outside segment BC, such that segment AE intersects segment CD. Assume that there exists a point F inside segment AD such that ∠EAD = ∠CBF . Denote by I the point of intersection of CD and EF , and by J the point of intersection of AB and EF . Let K be the midpoint of segment EF , and assume that K is different from I and J. Prove that K belongs to the circumcircle of 4ABI if and only if K belongs to the circumcircle of4CDJ
67. For all positive real numbers a, b, c, d prove the inequality pa4+ c4+p
a4+ d4+p
b4+ c4+p
b4+ d4≥ 2√
2(ad + bc)
68. Consider a sequence of equilateral triangles Tn as represented below:
T1 T2 T3 T4 T5
The length of the side of the smallest triangles is 1. A triangle is called a delta if its vertex is at the top; for example, there are 10 deltas in T3. A delta is said to be perfect if the length of its side is even. How many perfect deltas are there in T20?
69. Consider a chess board, with the numbers 1 through 64 placed in the squares as in the diagram below.
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64
Assume we have an infinite supply of knights. We place knights in the chess board squares such that no two knights attack one another and compute the sum of the numbers of the cells on which the knights are placed. What is the maximum sum that we can attain?
Note. For any 2× 3 or 3 × 2 rectangle that has the knight in its corner square, the knight can attack the square in the opposite corner.
Solutions
61. Find all real solutions of the following system of equations:
px2+ y2+ 6x + 9 +p
x2+ y2− 8y + 16 = 5, 9y2− 4x2 = 60.
(50th Catalonian Mathematical Olympiad) Solution 1 by Eloi Torrent Juste, AULA Escola Europea, Barcelona, Spain. First we observe that points (x, y) that satisfy the first equation are those that the sum of their distances to A(−3, 0) and B(0, 4) is equal to 5. Moreover, if a point P lies out of the segment AB then AP + P B > AB = 5. This let us to conclude that points (x, y) solution of the system must lie on AB. The equation of AB is y = 4
3x + 4 or
x,4
3x + 4
with −3 ≤ x ≤ 0. Substituting these values in the second equation, yields
9
4 3x + 4
2
− 4x2= 60⇔ x2+ 8x + 7 = 0
with roots x =−7 and x = −1. Since only the second lie in [−3, 0], then the unique solution of the given system is (−1, 8/3).
Solution 2 by Arkady Alt, San Jose, California, USA. Squaring both sides of the equationp
x2+ y2+ 6x + 9 +p
x2+ y2− 8y + 16 = 5 we have
px2+ y2+ 6x + 9 +p
x2+ y2− 8y + 162
= 25⇔ 4x − 3y + 12 = 0 Then, from
4x− 3y + 12 = 0 9y2− 4x2= 60
⇔ 3y = 4x + 12 (4x + 12)2− 4x2= 60
⇔ 3y = 4x + 12 12 (x + 7) (x + 1) = 0
we obtain
(x, y) =
−1,8 3
(x, y) =
−7,16 3
By substitution immediately follows that only (x, y) =
−1,8 3
satisfies the given
system and it is the desired solution.
Also solved by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain.
62. Let P be an interior point to an equilateral triangle ABC. Draw perpendiculars P X, P Y and P Z to the sides BC, CA and AB, respectively. Compute the value of
BX + CY + AZ P X + P Y + P Z
(First BARCELONATECH MATHCONTEST 2014)
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. First let us denote the side length of the triangle by a. The area of the triangle can be calculated in two ways and we get√
3
2 a2= a(P X + P Y + P Z), hence
P X + P Y + P Z =
√3
2 a (1)
On the other hand.
aBX =−−→BC·−−→BP , aCY =−→CA·−−→CP , aAZ =−−→AB·−→AP hence
a(BX + CY + AZ) =−−→BC·−−→BP +−→CA·−−→CP +−−→AB·−→AP
= (−−→BC +−→CA +−−→AB)
| {z }
~0
·−−→BP +−→CA·−−→CB +−−→AB·−−→AB
=−→CA·−−→CB +−−→AB·−−→AB
= a21 2 + a2 so
BX + CY + AZ = 3
2a (2)
Thus, from (1) and (2) we get
BX + CY + AZ P X + P Y + P Z =√
3.
Solution 2 by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Joining A, B, C with P we obtain three pairs of right triangles: AZP, AY P ; BZP, BXP and CXP, CY P. If a is the length of the side of 4ABC, then on account of Pithagoras theorem, we have
AZ2+ ZP2= (a− CY )2+ P Y2 BX2+ P X2= (a− AZ)2+ P Z2 CY2+ P Y2= (a− BX)2+ P X2 Developing and adding up, yields
BX + CY + AZ = 3a 2
On the other hand the sum of the areas of triangles AP B, BP C, CAP is the area of4ABC. That is,
a(P X + P Y + P Z)
2 = a2√
3
4 ⇔ P X + P Y + P Z = a√ 3 2 From the preceding immediately follows that
BX + CY + AZ P X + P Y + P Z =√
3
and we are done.
Solution 3 by Arkady Alt, San Jose, California, USA. Let a = BC = CA = AB, x = BX, y = CY, z = AZ, u = P X, v = P Y, w = P Z and h = a√
3
2 be height of the equilateral triangle ABC, then
[ABC] = [P BC] + [P CA] + [P AB]⇔ ah 2 = au
2 +av 2 +az
2 ⇐⇒ u + v + w = h
and BX + CY + AZ
P X + P Y + P Z = x + y + z
u + v + w = x + y + z
h = 2 (x + y + z) a√
3
Applying Pythagorean theorem to chain of right triangles4P XB, 4P BZ, 4P ZA, 4P AY, 4P XB, 4P Y C, 4P CX, we obtain
u2+ x2= w2+ (a− z)2 w2+ z2= v2+ (a− y)2 v2+ y2= u2+ (a− x)2 Adding all equations we get
X
cyc
u2+ x2
=X
cyc
w2+ (a− z)2
⇔ 3a2= 2a (x + y + z)
S0, x + y + z = 3a
2 and, therefore, BX + CY + AZ P X + P Y + P Z =√
3.
Also solved by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.
63. How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram, 5 of two grams, and 6 of five grams, respectively?
(Training Catalonian Team for OME 2014) Solution 1 by Jos´e Luis D´ıaz-Barrero BARCELONA TECH, Barcelona, Spain. The required number is the number of solutions of x + y + z = 31 with
x∈ {0, 1, 2, 3, 4, 5, 6, 7}, y ∈ {0, 2, 4, 6, 8, 10}, z ∈ {0, 5, 10, 15, 20, 25, 30}
We claim that the number of solutions of this equation equals the coefficient of x31 in the product
(1 + x + x2+ . . . + x7) (1 + x2+ x4+ . . . + x10) (1 + x5+ x10+ . . . + x30) Indeed, a term with x31is obtained by taking some term x from the first parentheses, some term y from the second, and z from the third, in such a way that x+y+z = 31.
Each such possible selection of x, y and z contributes 1 to the considered coefficient of x31in the product. Since,
(1 + x + x2+ . . . + x7) (1 + x2+ x4+ . . . + x10) (1 + x5+ x10+ . . . + x30)
= 1 + x + . . . + 10x30+ 10x31+ 10x32+ . . . + x46+ x47,
then the number of ways to obtain 31 grams is 10, and we are done. Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We are looking for the number of triplets (k, l, m) such that
k∈ {0, . . . , 7}, l ∈ {0, . . . , 5}, m ∈ {0, . . . , 6}, k + 2l + 5m = 31.
Since k + 2l≤ 17 we conclude that 5m ≥ 14, so m ∈ {3, 4, 5, 6}.
• m = 3, then k + 2l = 16 or k = 2(8 − l) ≥ 2(8 − 5) = 6 this gives the unique solution (k, l, m) = (6, 5, 3).
• m = 4, then k + 2l = 11 or k − 1 = 2(5 − l) ≤ 6. So every l ∈ {2, 3, 4, 5}
yields a solution, and we get four solutions:
(k, l, m)∈ {(7, 2, 4), (5, 3, 4), (3, 4, 4), (1, 5, 4)}
• m = 5, then k + 2l = 6 or k = 2(3 − l). So every l ∈ {0, 1, 2, 3} yields a solution, and we get four solutions:
(k, l, m)∈ {(6, 0, 5), (4, 1, 5), (2, 2, 5), (0, 3, 5)}
• m = 6, then k + 2l = 1, and this yields the unique solution (k, l, m) = (1, 0, 6).
So, the total number of ways to weight 31 grams is 10. Solution 3 by Arkady Alt, San Jose, California, USA. We have to compute the number of elements of the set
S :={(x, y, z) | x, y, z ∈ Z and x + 2y + 5z = 31, 0 ≤ x ≤ 7, 0 ≤ y ≤ 5, 0 ≤ z ≤ 6} Also solved by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.
64. Let A(x) be a polynomial with integer coefficients such that for 1≤ k ≤ n + 1, holds:
A(k) = 5k Find the value of A(n + 2).
(Training UPC Team for IMC 2014) Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
Remarks:
• If n ≥ 3 no such polynomial exists. Indeed, if a polynomial P has integer coefficients then for every distinct integers a and b, we have (b−a) | (P (b)−
P (a)). So, when n≥ 3, if there is a polynomial A with integer coefficients such that A(1) = 5 and A(4) = 54, then 3 must divide A(4)− A(1) = 620 which is absurd.
• When n ∈ {0, 1, 2}, (which are the only possible values left for n,) the polynomial A is not uniquely determined by the conditions that it has integer coefficients and that it satisfies A(k) = 5k for 1 ≤ k ≤ n + 1.
Indeed, when n = 0 the polynomial A(X) = λ(X− 1) + 5(2 − X), (for an arbitrary λ∈ Z,) satisfies A(1) = 5 and A(2) = λ. Similarly, when n = 1, the polynomial A(X) = 20X− 15 + λ(X − 1)(X − 2) satisfies A(1) = 5, A(2) = 25 and takes an arbitrary odd value at X = 3. A similar conclusion also holds when n = 2.
In view of the above, I propose a modified statement of the problem as follows:
Generalization of Proposal 64. Let An(X) be a polynomial of degree n such that A(k) = 5k for 1≤ k ≤ n + 1. Find the value of A(n + 2).
Solution. First, note that the existence and uniqueness, of An is guaranteed, from general theorems about Lagrange interpolation. Now, given An(X) for some n > 0 we consider
Qn(X) = 1
4(An(X + 1)− An(X))
Clearly deg Qn = n− 1, and Qn(k) = 5k for 1≤ k ≤ n. Hence Qn(X) = An−1(X).
This allows us to conclude that
An(X + 1)− An(X) = 4An−1(X).
Let bn= An(n + 2), from the above formula we see that bn= 5n+1+ 4bn−1 This is equivalent to
bn
4n+1−bn−1
4n =
5 4
n+1
. Adding these equalities and noting that b0= 5 we see that
bn
4n+1−5 4 =
Xn k=1
5 4
k+1
Thus,
bn = 4n+1 Xn k=0
5 4
k+1
= 5(5n+1− 4n+1).
which is the desired conclusion.
Solution 2 by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain. We observe that for all k≥ 1, holds:
5k = (1 + 4)k= Xk j=0
k j
4j=
k 0
+
k 1
4 + . . . +
k k
4k
Now, we consider the polynomial
Also solved by by Arkady Alt, San Jose, California, USA and Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain.
65. Let a0, a1, . . . , an and b0, b1,· · · , bn be complex numbers. If n≥ 2, then prove
(Training UPC Team for IMC 2014) Solution by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. We begin with the following claim:
Letα, a0, a1, . . . , an andb0, b1,· · · , bn be complex numbers, then it holds and the inequality claimed follows.
Now, we prove that for all n≥ 2, the following identityPn
k=0k2 nk2 holds. Indeed, applying two times the operator xdxd to the binomial identity (1 + x)n=Pn
Multiplying the first and the third and equating terms of degree n, we obtain
To prove the statement, we put into the claim α = k
n
After adding up the above expressions, we get n2n−1Re
on account that Xn
xk. From the preceding the inequality claimed immediately follows and the proof is complete. Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let un be defined by
un= 4n + 1
It is easy to check that un+1
un − 1 = (n + 2)(2n− 1)
2n(n + 1) − 1 = n− 2 2n(n + 1) ≥ 1
So, the sequence {un}n≥2 is increasing with u2 = 32 ≥ 1. Thus un ≥ 1 for every n≥ 2. Now let us come to our problem. Using the simple inequality X+Y ≥√
4XY we conclude that
1
where used the Cauchy-Schwarz inequality.
Also solved by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.