Liceul International de Informatica Bucuresti

Societatea de Stiinte Matematice

Stelele Matematicii 2015, Juniori

Problema 1. Fie a, b, c trei numere reale pozitive, astfel ıncat

ab+ bc+ ca+ 2abc = 1.

Aratati ca√a+√b+√c ≥ 2 si determinati cazurile de egalitate.

Problema 2. Aratati ca exista o infinitate de numere naturale imparem1 < m2 < · · · si o infinitate de numere naturale n1 < n2 < · · · , astfelıncat mk sa fie relativ prim cu nk si m4

k − 2n4k sa fie patrat perfect, oricarear fi indicele k.

Problema 3. Fie ABCD un patrulater inscriptibil, fie γ cercul circumscrisacestui patrulater si fie M mijlocul arcului AB al lui γ, care nu continepunctele C si D. Dreapta care trece prin M si prin punctul de intersectiea diagonalelor AC si BD, intersecteaza a doua oara cercul γ ın punctul N .Fie P si Q doua puncte situate pe latura CD, astfel ıncat ∠AQD = ∠DAPsi ∠BPC = ∠CBQ. Aratati ca cercul circumscris triunghiului NPQ estetangent cercului γ.

Problema 4. Fie n un numar natural mai mare sau egal cu 5 si fie{a1, a2, . . . , an} = {1, 2, . . . , n}. Aratati ca cel putin b

√nc + 1 dintre res-

turile ımpartirii la n a numerelor a1, a1 + a2, . . ., a1 + a2 + · · · + an, suntdistincte; b

√nc este cel mai mare numar ıntreg, mai mic sau egal cu

√n.

Liceul International de Informatica Bucuresti

Societatea de Stiinte Matematice

Stars of Mathematics 2015, Junior Level — Solutions

Problem 1. Given non-negative real numbers a, b, c such that ab+ bc+ ca+ 2abc = 1, showthat

√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 othertwo are both equal to 1.

Remark. If n is an integer greater than 1, and a1, . . ., an are non-negative real numbers suchthat

∑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∑k=1

(k − 1)∑

1≤i1<···<ik≤nai1 · · · aik = 1.

Problem 2. Show that there are positive odd integers m1 < m2 < · · · and positive integersn1 < n2 < · · · such that mk and nk are relatively prime, and m4

k − 2n4k is a perfect square foreach 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 − 8m4n4, and `4 − 8m4n4 − (m4 + 2n4)2 = −16m4n4 = −(2mn)4. Multiply thelatter by `4 − 8m4n4 + (m4 + 2n4)2 = 2`4 to get(

`4 − 8m4n4 + (m4 + 2n4)2) (`4 − 8m4n4 − (m4 + 2n4)2

)= −2 · (2`mn)4;

that is, (`4 − 8m4n4)2 − (m4 + 2n4)4 = −2 · (2`mn)4. Letting m′ = m4 + 2n4 and n′ = 2`mn,clearly m′ > m, m′ is odd, n′ > n, the difference m′4−2n′4 is a perfect square, and it is readilychecked that m′ and n′ are relatively prime. The conclusion follows.

Problem 3. Let ABCD be a cyclic quadrangle, let γ be its circumcircle, and let M be themidpoint of the arc AB of γ, not containing the vertices C and D. The line through M andthe point where the diagonals AC and BD cross one another, crosses γ again at N . Let Pand Q be points on the side CD such that ∠AQD = ∠DAP and ∠BPC = ∠CBQ. Showthat the circles NPQ and γ are tangent to one another.

Flavian Georgescu

Solution. Since the linesAC, BD andMN are concurrent, (AM/MB)(BC/CN)(ND/DA) =1, by Ceva’s theorem in trigonometric form along with the sine law in the triangle ABN ; andsince MA = MB, it follows that CN/DN = BC/AD.

Next, the triangles ADP and QDA are similar, so AD2 = DP · DQ. Similarly, BC2 =CP · CQ, so, by the preceding, (CN/DN)2 = (CP/DP )(CQ/DQ).

By a well-known theorem of Steiner, the lines NP and NQ are isogonal with respect tothe lines NC and ND; that is, the angles CNP and DNQ are congruent.

Finally, if X is a point on the tangent t at N to γ, other than N , then a standard anglechase shows the angle QNX congruent to one of the angles CPN , NPQ, depending on whichside of N the point X lies on t, so t is also tangent at N to the circle NPQ. The conclusionfollows.

Problem 4. Given an integer n ≥ 3 and a permutation a1, a2, . . ., an of the first n positiveintegers, 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 byn, and let m be the number of distinct residues in the list r1, r2, . . ., rn. Since there arem distinct integers in the list, the number of ordered pairs with different entries that canbe formed from these m integers does not exceed m(m − 1). On the other hand, of then − 1 distinct ordered pairs (rk, rk+1), k = 1, 2, . . . , n − 1, at most one has equal entries, som(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, thenthe 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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