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Liceul Internaţional de Informatică Bucureşti
Societatea de Ştiinţe Matematice
Stelele Matematicii 2015, Juniori
Problema 1. Fie a, b, c trei numere reale pozitive, astfel ı̂ncât
ab + bc + ca + 2abc = 1.
Arătaţi că
√
a+
√
b+
√
c ≥ 2 şi determinaţi cazurile de egalitate.
Problema 2. Arătaţi că există o infinitate de numere naturale impare
m1 < m2 < · · · şi o infinitate de numere naturale n1 < n2 < · · · , astfel
ı̂ncât mk să fie relativ prim cu nk şi m4k − 2n4k să fie pătrat perfect, oricare
ar fi indicele k.
Problema 3. Fie ABCD un patrulater inscriptibil, fie γ cercul circumscris
acestui patrulater şi fie M mijlocul arcului AB al lui γ, care nu conţine
punctele C şi D. Dreapta care trece prin M şi prin punctul de intersecţie
a diagonalelor AC şi BD, intersectează a doua oară cercul γ ı̂n punctul N .
Fie P şi Q două puncte situate pe latura CD, astfel ı̂ncât ∠AQD = ∠DAP
şi ∠BP C = ∠CBQ. Arătaţi că cercul circumscris triunghiului N P Q este
tangent cercului γ.
Problema 4. Fie n un număr natural mai mare sau egal cu 5 şi fie
√
{a1 , a2 , . . . , an } = {1, 2, . . . , n}. Arătaţi că cel puţin b nc + 1 dintre resturile ı̂mpărţirii la n a numerelor a1 , a1 + a2 , . . ., a1 + a2 + · · · + an , sunt
√
√
distincte; b nc este cel mai mare număr ı̂ntreg, mai mic sau egal cu n.
Liceul Internaţional de Informatică Bucureşti
Societatea de Ştiinţe Matematice
Stars of Mathematics 2015, Junior Level — Solutions
Problem 1.
non-negative real numbers a, b, c such that ab + bc + ca + 2abc = 1, show
√ Given
√
√
that a + b + c ≥ 2 and determine the cases of equality.
Flavian Georgescu
Solution. The condition in the statement
√ is equivalent to 1/(1 + a) + 1/(1 + b) + 1/(1 + c) = 2.
If t is a non-negative real number, then t ≥ 2t/(1 + t) = 2 − 2/(1 + t), and equality holds if
and only if t is either 0 or 1. Consequently,
√
√
√
a + b + c ≥ 2 − 2/(1 + a) + 2 − 2/(1 + b) + 2 − 2/(1 + c)
= 6 − 2(1/(1 + a) + 1/(1 + b) + 1/(1 + c)) = 2.
By the preceding, equality holds if and only if one of the numbers a, b, c is 0, and the other
two are both equal to 1.
Remark.
P 1,√and a1 , . . ., an are non-negative real numbers such
P If n is an integer greater than
that ni=1 1/(1 + ai ) = n − 1, then ni=1 ai ≥ 2, and equality holds if and only if n − 2 of
the ai vanish, and the other two are both equal to 1. The proof goes along the same lines.
Alternatively, but equivalently, the condition on the ai reads
n
X
(k − 1)
k=1
X
ai1 · · · aik = 1.
1≤i1 <···<ik ≤n
Problem 2. Show that there are positive odd integers m1 < m2 < · · · and positive integers
n1 < n2 < · · · such that mk and nk are relatively prime, and m4k − 2n4k is a perfect square for
each index k.
Folklore
Solution. Let m and n be relatively prime positive integers such that m is odd and m4 − 2n4
is a perfect square, e.g., m = 3 and n = 2. Write `2 = m4 − 2n4 , so `4 = (m4 − 2n4 )2 =
(m4 + 2n4 )2 − 8m4 n4 , and `4 − 8m4 n4 − (m4 + 2n4 )2 = −16m4 n4 = −(2mn)4 . Multiply the
latter by `4 − 8m4 n4 + (m4 + 2n4 )2 = 2`4 to get
`4 − 8m4 n4 + (m4 + 2n4 )2 `4 − 8m4 n4 − (m4 + 2n4 )2 = −2 · (2`mn)4 ;
that is, (`4 − 8m4 n4 )2 − (m4 + 2n4 )4 = −2 · (2`mn)4 . Letting m0 = m4 + 2n4 and n0 = 2`mn,
clearly m0 > m, m0 is odd, n0 > n, the difference m04 − 2n04 is a perfect square, and it is readily
checked that m0 and n0 are relatively prime. The conclusion follows.
Problem 3. Let ABCD be a cyclic quadrangle, let γ be its circumcircle, and let M be the
midpoint of the arc AB of γ, not containing the vertices C and D. The line through M and
the point where the diagonals AC and BD cross one another, crosses γ again at N . Let P
and Q be points on the side CD such that ∠AQD = ∠DAP and ∠BP C = ∠CBQ. Show
that the circles N P Q and γ are tangent to one another.
Flavian Georgescu
Solution. Since the lines AC, BD and M N are concurrent, (AM/M B)(BC/CN )(N D/DA) =
1, by Ceva’s theorem in trigonometric form along with the sine law in the triangle ABN ; and
since M A = M B, it follows that CN/DN = BC/AD.
Next, the triangles ADP and QDA are similar, so AD2 = DP · DQ. Similarly, BC 2 =
CP · CQ, so, by the preceding, (CN/DN )2 = (CP/DP )(CQ/DQ).
By a well-known theorem of Steiner, the lines N P and N Q are isogonal with respect to
the lines N C and N D; that is, the angles CN P and DN Q are congruent.
Finally, if X is a point on the tangent t at N to γ, other than N , then a standard angle
chase shows the angle QN X congruent to one of the angles CP N , N P Q, depending on which
side of N the point X lies on t, so t is also tangent at N to the circle N P Q. The conclusion
follows.
Problem 4. Given an integer n ≥ 3 and a permutation a1 , a2 , . . ., an of the first n positive
√
integers, show that at least n distinct residue classes modulo n occur in the list a1 , a1 + a2 ,
. . ., a1 + a2 + · · · + an .
Amer. Math. Monthly
Solution. Let rk be the remainder of the partial sum a1 + a2 + · · · + ak upon division by
n, and let m be the number of distinct residues in the list r1 , r2 , . . ., rn . Since there are
m distinct integers in the list, the number of ordered pairs with different entries that can
be formed from these m integers does not exceed m(m − 1). On the other hand, of the
n − 1 distinct ordered pairs (rk , rk+1 ), k = 1, 2, . . . , n − 1, at√most one has equal entries, so
√
m(m − 1) ≥ (n − 1) − 1 = n − 2. Consequently, m ≥ (1 + 4n − 7)/2 ≥ n if n ≥ 4, the
latter inequality being strict if n ≥ 5. If n = 4, and a1 = 2,√a2 = 4, a3 = 3 and a4 = 1, then
√
the corresponding list of residues is 2, 2, 1, 2, so m = 2 = √
4 = n. Finally, if n = 3, then
every permutation produces a list consisting of√exactly 2 > 3 distinct residues. (If n = 2,
√
the identity produces the list 1, 1, so m = 1 < 2 = n.)
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