Romanian Problem Book

8/14/2019 Romanian Problem Book

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l Romanian Problem Book File: b75-77 Version 01975"t;i

75. \ (A 1). The positive numbers aI, a2, . . . , an, bl bz . . . b.; satisfy the inequali, itiesi ... ,l Prove that va; + va2 + ... + Vb; + Vb; + ... + A.L

Solution. The solution uses Abel's summation formula. The AM-GM inequalityyields

for each k = 1,2, . . . ,11., so we havefi l l + va;- + ... + [ ( + ~ ... + ~ + ( ~ + ... + Jb:)].In view of this, it suffices to prove that

If we set(k=l ,2 , .. . ,n),

then Sk Tk by hypothesis, so

al a2 an 1 (1 1) (1 1)+ - + ... + - = -Sn + - - 5n - l + ... + - - - 51


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Romanian Problem Book File: b75-77 Version 0Solution. Let ak = + Ck, where 0 :s Ck :s ~ k = 1 , ~ , .. , ,11.. Then

n (1 ) (1 ) 11. 1 n 1 n n"'b n = L "2 + Ck "2 - Ck+l = 4" + 2" LCk - 2" L C k + l - LC kC k+ l .k=l k=1 k=1 k=l

The equality an+l = a l implies Cn+l = Cn, hence l : ~ = 1 Ck = l : ~ = 1 Ck+l' and, thereforen

bn = - L Ck Ck+lk=1

Since Ck 2: 0, k = 1,2, . .. ,11., we obtain bn :::; 11./4. The equality holds, for instance,for Ck = 0, k = 1,2, ... ,n, that is, ak = 1/2, k = 1,2, . . . ,11.. So the maximum value of bnis n]4. '75.3 (A3). Consider the sets of real numbers A = {al,a2,"'} and B = {bl , b2, ... }

such that al = bl and bn + l = bn (1 - an + l ) + (1 - bn )an + lfor any positive integer n. Prove that E A if and only if E B.

Solution. Assume EA. Then either = al (= bd, so E B, or there exists apositive integer, 11., such that an+l = ~ in which casej1i

If E B, then = bl (= ad, so E A, or there exists a positive integer, 11., suchthat b+1 = ~ Let -no be the least su;h positive integer. Then

which implies (2an o+ l - 1)(2bn o - 1) = O. Since the choice of no gives bn o =1= ~ it followsthat ano+l = ~ hence E A.

75.4 (SG 1) . Given a cone, consider the set 51 formed by the planes which are parallelto it s base and divide the cone into n solids of equal volumes, and the, set 52 formed by th eplanes which are also parallel to the cone's base but divide it s lateral surface into 11. partsI.'

If of equal areas. Prove that 51 n 52 =1= 0 if and only if n contains at least one factor whichl i t is a power, with an exponent greater than 2, of a prime.Solution. Let hI, h2 , .. . ,h n - 1 be the altitudes of the cones formed by the planes in 51

and iI, i z, . . . , i n - I those determined by the elements of 52. The cone with the altitude h kis similar to the original one; the ratio of similitude is hkl H, where H is the altitude ofthe given cone.

Since the ratio of their volumes is kin, one obtains

k=I,2,oo. ,n-1.

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Romanian Problem Book File: b75-77 Version 0Similarly,

= 3fT... H V;" 1= 1,2, . . . ,12-1.I f 51 n 82 =I- (/) then there exist k, I E {I , 2, ... ,12-1} such that hk = it . It follows thatyJk/n = vrrn, and so 12k2 = 13. Clearly, I i= k (otherwise 12 = I, which is absurd). Let pbe a prime factor of I which is not .a factor of k. Then I = nip ~ n gcd(k, p) = 1, hence12k!. = 1n 3]} implies that 12 contains ] i as a factor.

Conversely, if 12 contains p3 as a factor then 12 = 12lp3. Considering k = 11,1 and I = p12l,we get nk 2 = [3 , which is equivalent to hk = i/ and 51 n 82 i= 0.

75. 5 (NT 1). Let p and q be relatively prime odd integers, greater than 1. Prove that

Solution. Let 5 be th e set of all integers of the form kp - Iq, where 1 ::; k ::; (q - 1)/2and 1 ::; l ::; (p - 1)/2 . All its elements are nonzero and pairwise distinct. Indeed, if. weassume that kp - lq = 0 then p divides lq, and since gcd'(p,q) = 1, P II , absurd since p > I.And if we set kIP -/ lq = k2P -/2q for k l i= k2, it follows that (k l - k2)p = (l 1 -l2)q,hence p I(It - l2 )q, so since gcd(p, q) = 1, p IIt -lz, absurd since 1 ::; it, 12 ::; (p - 1)/2and 11 =1= h imply 0 < III -1 2 1 < p. Therefore 5 contains P;1 . q ; l elements. For a fixed k;lk(; Jof them are positive. Indeed, kp - lq > 0 is equivalent to I < 7, so I can be any of thenumbers 1,2, . . . , l ~ J. Hence, 5 contains 2 : . 1 ~ / ) / 2 l J positive elements, and similarly

- I ) / 2 l / iI J . Tl lU S2:. (P - negative ones.[= 1 p ,p -1 q -1= - - .- 2 2

75.6 (PG 1). Prove that for any 12 greater than 3 there exists a convex polygon with 12sides, not all equal, such that the sum of th e distances from any interior point to its sidesis a constant. '

Solution. Note first that the only triangle with the property that the su m of thedistances from any interior point to its sides is a constant is th e equilateral one, whenceth e condition 12 > 3.

Take a convex n-gon with the given property. Choose a direction not parallel to any ofits sides and, drawing two parallels to this direction, cut off two vertices together with partsof the sides having them as endpoints, as shown in the figure. We obtain a convex polygonwith 12+2 sides, and one can assume that th e two newly formed sides have different lengths(clearly the cut can be performed in a convenient way to ensure this). The new polygonalso has the stated properties. Indeed, not all of its sides are equal. Furthermore, the sumof the distances from any interior point to its sides is equal to the sum 51 of the distances

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Romanian Problem Book File: b75-77 Version 0from the same point to the sides of the old polygon plus the sum 52 of the distances fromthe point to the two parallels. But 51 is a constant according to the induction hypothesis,and is also a constant, being equal to the distance between two fixed parallel lines.Hence 5 = 51 + 52 is a constant, too.

The above argument shows that the property in question can be transmitted from n ton + 2. In order to prove it by induction, we just need to check the base cases n = 4 and11 = 5. For 11 = 4 it suffices to take any rectangle. The case n = 5 is, however, not obviousif considered directly. But we can start with an equilateral triangle which is known tosatisfy our conditions, to perform the construction described above and obtain the desiredpentagon.

75.7 (SG2). For a tetrahedron, 1', let 5(1') be the set which consists of the midpointsof the edges, the feet of the altitudes of each face, and the midpoints of the segments whichconnect, on each face, the orthocenter with the vertices. Prove that if 5(1') contains asubset, 5'(1'), which lies on a sphere, such that two faces of T contain each at least threepoints of 5'(1') but 5'(1') is not a subset of the union of the two faces then:

a) 5(1') contains at most 24 elements;b) the opposite edges of T are perpendicular;c) 5 (1'J lies on a sphere.Solution. Let T = ABCD. One knows that the points considered on each face areconcyclic (they lie on the Euler's of the face). Assume that ABC and ABD are the faces

that each contain at least three points of 5'(1'). Since the sphere on which 5'(1') liescontains three points on the face ABC, it contains the entire Euler's circle of ABC, inparticular the midpoints of AC and BC. Similarly, the above sphere contains all pointsof 5(1') from ABD, in particular the midpoints of AD an d BD. Since 5'(1') is notcontained in the union of the two faces, there exists a point, say on the face BCD, whichdoes not belong to ABC and ABD, but is situated on 5'(1'), and hence on the spherecontaining 5'(1'). But this sphere contains, also, the midpoints of BC and BD, so itcontains the entire Euler's circle of BCD, in particular the midpoint of CD. Since themidpoint of AC and AD also belong to this sphere, the Euler's circle of ADC is situatedon the sphere. Hence 5(1') is on the sphere, and c) is proven.

Consider the edge AB. It contains its midpoint, P, and the feet of the altitudes from Cand D, q and R, respectively. These three points are on a sphere, and AB is a line, sothey cannot be all distinct. We distinguish the following cases:

.. .1. P = Q = R. Then AB 1- CD and AB 1- DR, so AB 1- CD.2. P:j:. Q = R. Analogously, AB 1- CD.3. Q:j:. P = R.4. Q = P f R.L The cases 3 and 4 are eliminated by the following lemma: In an isosceles triangle, theEuler's circle is tangent to the unequal side. The proof is obvious.In the case 3, for example, the triangle B DA is isosceles (D A = DB), hence AB ISl tangent to the sphere on which 5(1') lies, so P = Q = R, absurd. Similarly for case 4.

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Romanian Problem Book File: b75-77 Version 0Reiterating the above reasoning, it follows that the opposite edges are pairwise perpen-

dicular, hence b) is proven. In case 2, we obtain the maximum number of points: 6 edgestimes 2J)()ints plus 4 faces times 3 points, 24 points in all, so a) is also proven.

75.8 (SG3). Given four parallel planes, construct a cube having four noncoplanarvertices situated one in each of the given planes. How many such cubes of different sizescan be built?

Solution. Any four vertices of a cube which are not situated in the same plane deter-mine a tetrahedron. There are four such possible cases of nonsimilar tetrahedra.

There are figures missing here in the original.

Given four parallel planes, PI, P2,P3 ,P4 in that order, and a tetrahedron ABCD, thereexists a unique tetrahedron A'B'C'D' similar to ABCD with A', B', C', D' in PI, P2,P3,P4,respectively.

Indeed, consider E and F on AD such that AE, EF, F D are proportional to thedistances between PI and p- , P2 and P3, P3 and P4, respectively. Construct also EG II FCand F H II EB. G E AC, H E B D. The planes BEG and H FC are parallel. Consideringthe planes' through A. and D parallel to these, we get four parallel planes whose mutualdistances are proportional to the distances between the original ones. Constructing, witha convenient ratio of proportionality, the figure similar to the figure above, we obtain thedesired cube.Uniqueness follows immediately (start from the final construction and go back throughsimilarity to the original one).

Considering another tetrahedron WXYZ nonsimilar to ABCD whose vertices are situ-ated in the four planes, we obtain, in general, other sizes of the tetrahedron's edges. Let,then, VI V2 V3V4 be a tetrahedron and define for the set of permutations of {I, 2, 3, 4} thebinary relation byv

~ This is an equivalence relation, and the number of equivalence classes gives the number oftetrahedra similar to the given one and whose vertices are situated on the four planes. Forthe first case there exists a single such class .(the tetrahedron is regular). In the secondII.

11- ' case, there exist 4 cla.sses, and for each of the cases 3 and 4 there exist 12 classes. Thus.. the answer to the second part of the problem is 29.75.9 (A4). Solve the system

sin a:l cos X2 = sin X2 cos X3 = ... = sin Xn-l cos X n = sin X n cos X l,twhere :1:1 + .r 2 + ... + l.:n = Tr and 0 :S Xk :S -' k = 1,2, . . . ,n.

Solution. We consider three cases.Case 1: None of the numbers sin Xk, cos Xk is zero.

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Romanian Problem Book File: b75-77 Version 0I f X l = X2 then sin X l = sin X2 and, from the first equation, we obtain cos X2 = cos X3,

so X2 = X3 . Similarly, X3 = X4, and so on; thus Xk = ;, k = 1,2, . . . , n.If X =f X2, assume that X l < X2 . Then sin X l < sin X2, hence cos X2 > cos X3 , and so

X2 < X3 . Repeating this argument, we get X l < X2 < ... < X n < X l, absurd.Case 2: There exists an m E {1,2, .. . ,n} such that sinx m = O.Then X m = 0 and sin Xm - l cos Xm = sinz..; cos Xm+ l = 0, hence sin Xm- l = O. Thus

: rm- l = O. Continuing this process, one obtains Xk = 0, k = 1, 2 , . ~ . ,n,which violates thecondition X l + X2 + ... + X n = t t ,

Case 3: There exists an mE {1, 2, . . . ,n} such that cos Xm = O.Then X m = I and

o= sin 1: m - l cos Xm = sin X m cos Xm+ l = 0,hence cos XII/+J = O. Therefore Xm+ l = T' Repeating this reasoning, we obtain Xk = T,k = 1,2, . . . , n . which impliesn = 2.

In conclusion, the system has the unique solution X l = X2 = . . . = X n = ~ 75. 10 (A 5) . Determine all functions f: Q Q which satisfy the equation

f(x + y) + f(x - y) = 2(f(x) + f(y) + 1)for every X, y E Q.

Solution. For X = Y = 0 we obtain f(O) = -1 , and X = 0 gives f(y) = f( -y) for every y E Q, so f is even. Let X = ky, where k is a positive integer. Then

f((k + l)y) - f(ky) = f(ky) - f((k - 1)y) +2(f(y) +1).Replacing k: by 1,2, . . . ,q and adding up the resulting relations, we obtain

f(qy) - f((q - 1)y) = (2q - 1)(f(y) + 1).t.iiiLetting q be, successively, 1,2, . .. , n., and adding up the relations obtain, we get

f(ny) + 1 = n 2( f (y ) + 1} (1)01 ' 1f(y) + 1 = 2(f(ny) + 1)n \for any y E Q. Then

f ( ~ + 1 = -;(f(1) + 1), (2)n nBy (1) and (2),

2f c::) + 1 = f (m . ~ + 1 = m( f ( ~ + 1) = : : (f(1) + 1).~ ' fk.'.. 6.. .k.ill


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Romanian Problem Book File: b75-77 Version 0Since / is even, we conclude that the general form of the solution is Ie (x) = cx2 - 1, wherec = /(1) + 1 is a generic rational number.

7 5 . ~ 1 (A6). Solve, in complex numbers, the system(X l + X 2 +Xa ) X 4 = a(X l + X 2 +X4 ) X a = a(X l + Xa + X4 )X 2 = a(X 2 +Xa + X4)XI = a,

where a E R.Solution. Set 8 = :1:1 + X2 + X3 + X4. The system becomes

xi - SX l + a = 0(s - Xa)X3 = a(s - .L:4):l:4 = a

x - SX 2 + a = 0or(8 - X2)X2 = a x - sXa + a = 0(8 - XI )X l = a, x -,- 'SX4 + a = 0.

2Hence X l, X2, :r3, .1:4 are zeros of the equation x - SX + a =ith roots !(s vs 2 - 4a),\Ve distinguish the following cases.

Case 1: .1;1 = .1:2 =.1:3 = .1:4. From Xl + X2 + Xa + X4 = S one gets 2s 2vs2 - 4a = s,so s = 4 ~ . This yields the solutions

(va!3,V03,va!3,va j 3) and (-va!3,-va!3,-va!3,-va!3).Case 2: X l = X2 = X3 == ~ ( - V82 - 4a) and X4 = !(s + vs 2 - 4a) or vice versa. It

then follows that 28 vs 2 - 4a = s, and hence a = O. Thus, for a =I 0, the system has nosolutions in case 2; for a = 0, it has the solutions

(O,O,O,S), (0,0,8,0), (O,s,O,O), (8,0,0,0).Case 3: :1:1 = ,l:2 = ~ ( - V82 - 4a) and X3 = X4 = ~ ( + vs 2 - 4a). In this case, .s =

and Xl = X2 = - J-a, X3 = X4 = V ~ a . Onegets the solution ( -Fa, - ~ , ~ , ~ ) and the five other permutations of it. . ,In conclusion, for a =I 0, the system -admits the eight solutions d ~ c r i b e d in cases 1

and 3, and for a = 0, the four described in case 2 (all the others become (0,0,0,0), whichis included in case 2).

Note. In a similar manner, one can solve the systemk = 1,2, . . . ,n

for 'It = 3 or n 5.7


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Romanian Problem Book File: b75-77 Version 075.12 (A 7). Determine the realnumbers a,'b,c,d for which

m ~ x ( a , b+ c+ d) = max(b, a + c + d) = max(c, a + b+d) = max(d, a + b+ c) = 1.State an d solve the analogous problem for n numbers.

Solution. Clearly, none of th e four numbers is greater than 1.If at least three of the numbers, say a, b, c, are equal to 1 then n:rax(d, a + b+ c) = 3 =I=- 1.i f two of these numbers are equal to 1, for instance a, b then c an d d are less than 1, and

a + b+ d = a + b+ c = 1,hence c = d = -1. We obtain the solution (11, -1, -1) and it s other five permutations.I f exactly one of the numbers is 1, say a, then b, c, dare all less than 1, and

a + c + d = a + b + d = a + b + c = 1,.so b = c = d = O. This yields the solution (1,0,0,0) an d it s other three permutations.

I f all the numbers are less than 1 thenb+c+d=a+c+d=a+b+d=a+b+c= 1.

Adding these equations, we obtain 3(a +b+ c + d) = 4, an d so ( ~ ~ ~ ~ ) Extension. For n 3, determine the real numbers aI, a2, . . . ,an such that

for each k = 1,2, . . . , n,75.13 (NT2). Find a polynomial, P(n), with integer coefficients, such that P(n) + 4n

is divisible by 33 for any positive integer n.(iL Solution. 'Write

for any positive integer '11.Let Q(n) = -1 - 3n - g n ( r ~ - l ) . Then Q(1'i) + 4n = 0 (mod 33 ) . Considering th e polynomial P( /I,) = 28Q( n), we obtain a polynomial with integer coefficients such that

P(n) + 4 11 = 27Q(n) + [Q(n) + 4H ] == 0 (mod 33 ) ,for any positive integer ri ,

75.14 (C 1). I f a, b, c are the 'lengths of th e sides of a triangle with perimeter 6and A, B, C the measures, in radians, of its angles, prove that

27r ::; aA + bB + cC < 37r.8

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Romanian Problem Book File: b75-77 Version 0

Solution. More generally, we will prove that1r aA + bB + cC t t- <


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Romanian Problem Book File: b75-77 Version 075. 16 (A 8) . Let n be an integer greater' than 1. Prove that there is no irrational

number a such that :Ja+ va 2 - 1 + Va - Va 2 - 1is rational.Solution. We will prove by induction that if x + l /x is rational then xn + l /x nis rational for any positive integer n. This clearly holds for n = 1. Assuming it truefor 1 k n - 1 and applying th e identity

xn + = (x + 2:.) (x n-1+_1 ) _ (xn-2 +_1 ),x n X x n - 1 Xn-2we deduce that the statement is also true for k = n. This completes the induction.

Assume now that there exists an irrational number a such that2Va +Ja - 1 + Va - va 2 - 1

is rational. Setting .7: = Va + Ja2 - 1, we obtain from above thatn 2x + = (a + Va 2 - 1) + (a - Ja - 1) = 2a:r n

is rational, a contradiction.75.17 (C 2). Let 5 be a set with S elements and M a set with the largest number of

4-element subsets of 5, such that , if A an d B are distinct parts of M, then An B has atmost 2 elements. How many elements does M have?r Solution. Let JII! = {51, 52, . . . ,5n} and Sf, 52"'" S'4n be the 3-element subsetsL of 51 ,52, .. . ,5n. By hypothesis, 5f , 52" .. ,5 '4n are pairwise distinct. Their number cannot exceed ( ~ = 56, the number of 3-element subsets of M. Hence 4n 56, i. e. n 14.For 5 = {1,2, . . . ,S}, the set consisting of {1,2,3,4},.{1,2,5,6}, {1,2, 7,8}, {3,4,5,6},{3,4, 7,8}, {5,6, 7,8}, {1,3,5, 7}, {1,3,6,8}, {1,4,5,8}, {1,4,6, 7}, {2,4,6,8}, {2,4,5, 7},{2, 5, 6, I} , {2. :3, 5. 8} has the desired properties.

75. 18 (C 3). Let ,1:1, 2 , " . ,X n E {-I , 1} be such thatX1X2 + X2X3 +... +Xn-1Xn + XnXl = O.

Prove that '/1 is divisible by 4. \Solution. Each of the n terms on the left-hand side of the given equality is either 1

or -1 . Let k be the number of -1 'so The remaining n - k terms are 1's, and since thesum of all terms is 0, we must have k = n - k, so n = 2k. Moreover,

so k has to be even. Therefore n, which is twice k, is divisible by 4.10


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Romanian Problem Book File: b75-77 Version 075 . 19 (P G 3) . Consi del' a circle of radius 1 and the points Al , A2 , ... ,An in its plane.Prove that the circle contains a point, P, such that

...

PAl +PA2 +...+ PAn n, -Solution. Consider any point Po on the given circle, different from AI, A 2 , ,An.If POA1 + PoA 2 + ... + PoAn n, we are done. If not, one has(1)

Call PI the opposite diametral point of Po on the circle. From the triangle inequality,

so n nL POAk + L PIAk 2n.k= l k= l

Hence, by (1), PI Al + PI .42 +... + PI An n, and, therefore, PI has the required property.75.20 (PG4). Give an example of two noncongruent quadrilaterals which have the

angles, respectively, congruent, the diagonals, respectively, congruent, and equal perimeters.

Solution. Consider an isosceles trapezoid, ABCD, whose diagonals have length 1,and whose angles at the base are 30. Let A' and B' be the feet of the altitudes from Aand B, respectively, and set BB' = x. We get B'C' = xV3, BC = 2x, DB' = )1- x2 , soAB = A'B' = )1 - x2 - xV3. The perimeter of the trapezoid is p(x) = 4x + 2)1 - x2 .We will show that there exist XI,X2 E (0,1), Xl i= X2, such that p(xd = P(X2), i. e. that pis not one-to-one in (0, 1).

Set Xl = sina, a E (0, f), and X2 = sinb, bE (0, ~ ) b =1= a. Then p(xd = P(X2) is equivalent to 4 sin a + 2 cos a = 4 sin b + 2 cos b, which, after standard trig manipulations, trans lates into tan arb = 2. Notice that 2 tan- 1 2 E ( ~ 7l"). Considering ,~ ....Ii .r} = sina and X2 = sin(2tan-1 2 - a),

for some a E (0, I) ' we obtain two quadrilaterals (even trapezoids) which satisfy the statedconclitions.75.21 (C4). Prove that any polynomial with real coefficients can be written as the

difference of some two increasing polynomial functions (over the reals).Solution. Let P(x) = ao + alX + ... +anxn. Then

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Romanian Problem Book File: b75-77 Version 0where M > max(/all, 2Ia21, nlanl). Note that 'if Ixl 1 then 1 + Ixl + ' " + !xn-Il n,and if Ixl > 1 then 1 + Ixl + + Ixn-Il < nx 2n. Hence, for any real number x,

IP'(x) < Mn(x + x2n)j.Consider the polynomials

2n+1),Pdx) = P(x) +Nfn(x + XVve have P(.r) = PI (1') - P2(.t'). Also, PI and P2 are increasing functions, because

2n) > 0;(x) = P'(x) +Nfn(l + (2n + 1)x2n) 2:: P'(x) + Mn(l + xand

P ~ ( x ) = 1\([n(1 + (2n + 1)x2n) > 0for any x E R.

75.22 (PG5). Prove that among any nine points situated in the interior of a squareof side 1, there are three which form a triangle of area no more than l/S.

Solution. Consider the line segment joining the midpoints of the two pairs of oppositesides of the square. They divide it into four congruent smaller squares. By the pigeonholeprinciple, one of these contains at least three of he given points, say A, B, C. One of theparallels through A, B, C to the sides of the smaller square will intersect the opposi te sideof 6ABC. Assume this is the parallel through A. Let it meet BC at A', and let B', C'be the projections of B, C onto AA', respectively. Then

, ,1, ( , C') 1 1 1 1ABC = ABA + ACA = -AA BB + C < - . - . - = 2 .: - 2 2 2 S'and the claim follows.

75.23 (C 5). Prove that the equation

admits integer solutions if and only if n = 4.Solution. For n = 4, the equation admits integer solutions, for example (1,1,0,0). One

easily checks that there are no solutions for n = 1,2,3, so le t us assumethat n 2:: 5. Writethe equation as follows:

') ? ( )2 ( )2 ( )2( ; f l - : r 2 t+ ( : r l - ;T3 t+ + XI-X n + X2-x3 + ... + xn - l - x n =n. (1)Each unknown occurs n - 1 times in the left-hand side of (1), and out of the G) terms,at most n are nonzero. Hence at least G) - n terms are 0, and since ( ~ - n > (n;2) for n 2:: 5, there are at least n - 1 unknowns appearing in zero terms.

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Romanian Problem Book File: b75-77 Version 0On the other hand, assume that each of th e numbers Xl, X2 , . . . ,X n appears in a zero

term. Then ;1:1 = X2 = ... = :Tn , and (X l ,X2 , ' " , x n) is no t a solution of (1). We infer f r o ~ p here that all the unknowns except one are equal. If Xl = X2 = . . . = Xn - l = Xand X n i= X, (1) reduces to (11, - l ) ( x n - x)2 = 11,. Hence IX n - xl = Vn ~ l ' which is absurd (because 1 < Vn ~ < 2). In conclusion, the given equation admits integer solutionsif an d only if 11, = 4. .75.24 (A 9). Let a1, a2, . . . ,an be real numbers. Determine th e real numbers Xl, X2 , . . . ,xn,1Xl :::; X2 :::; . . . :::; Xn, such that

max la k - xk ll $k$nis minimized. Solution. Consider any nondecreasing real number sequence Xl, X2 , . . . ,Xk an d set.M = max- $/.;$11 la/.; - Xk I. Then ak - NI :::; Xk :::; ak + M for k = 1,2, . . . ,11" and if i < i,

~ i we obtain a, - 111 :::; Xi :::; Xj :::; a j + M. Thus M a i ~ a j whenever 1 :::; i < j ::; 11" so a' - a 'M > max J = Mo. (1)- l $ i < j $n 2'''.Te will show that .Mo is the required minimum. All that remains is to find real num

a,> bers X'l, X2 , . . . X:n such that Xl :::; X2 ::; . . ::; Xn and maXl$k$n la k - Xk I ::; Mo. To doIi this, observe that the definition of kIo implies a i - a j ::; 2Mo for any i, j , i < j. Thenmax{aI , a2 , . . . , ak } - ak :::; 21v[0 for each k = 1,2, . . . ,11" which can be written asL

fl.:. This means thatItl Thus. setting Xk = -lvIo+ max{al,a2,'" ,ad, we have Xl : : ; X2:: : ; ' " :::;,x n and

The proof is complete.

75 . 25 (SG4). On the edges of a tetrahedron, A IA 2 A 3A 4 , considervsorne points, B i j ,B i j E AjA j , i ,j E {I, 2, 3, 4}. Prove that he spheres circumscribed about the four tetrahedra AIB12B13B14, A2B12B23B24, A3B13B23B34' A4Bl4B24B34 have a common point.Solution. This is a three-dimensional variant of the following theorem from plane

geometry:Let AI. Bs ; Cl be points on the sides BC, CA, AB of the triangle ABC, respectively. Then the circumcircles of the triangles ABCI, BCA I, CAB I have a common

point.13


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Romanian Problem Book File: b75-77 Version 0Indeed, let the circles ABlC l and BClA I meet again at P. For definiteness, assumethat P is interior to 6ABC. From the cyclic quadrilaterals ABIPC l and BAlPC l wehave LfbPCl = 1800 - LA, LClPA l = 1800 - LB, so

LAIPB, = 360 0 - (LBlPC l + LCIPAd = LA + LB = 1800 - LC.The latter means that AlCB 1P is also cyclic, and the proof is complete.-o prove the main result, denote by S, the sphere through Ai, Bi], Bik' Bil (assuming{i, j, k, I} = {1, 2, 3, 4}). Let Cijk be the intersection of 51 with the plane AiAjAk. Consideralso {Di j} = 5i n 5j, {Md = Cjkl n Cn, n Cu, and le t I be an inversion with center AI.The images I(C1j) are lines, dij, and since Al Dli ' the images I(DIi) = Wi are circles.The inversion is a bijection, therefore dij and Wi lie in PI; also, {I(B l i)} = dij n dik nWiand {I(A1il} = dj k n Wj nWk, {i,), k} = {2,3, 4}. By the statement proved above, we conclude that the circles wz, W3, W4 have a common point P. Hence D 12 n D13 n D 14 = Q,where Q = I - I (P). Thus (51 n 52) n (51 n 53) n (51 n 54) = Q, i. e. the spheres 51, 52,S3, S4 have a point in common.

75. 26 (C 6). Prove that among .any ten consecutive positive integers, there exists atleast one which is relatively prime with any of the other nine.

Solution. Our array of ten consecutive positive integers contains five odd numbers.Among them, there are at most two which are divisible by 3, one multiple of 5, and atmost one multiple of 7. Hence, at least one of these five odds is note divisible by 3,5,7(and, of course, by 2). Call this number m, and suppose it is not relatively prime with anyof the other nine. Then the array contains a number n such that gcd( m, n) = d > 1. Thedifference m - n is then divisible by d, and since 1m - nl ::; 9, its divisor d can only be 2,3, 5 or 7. None of these can occur since, on the other hand, d divides m -a contradiction., 75.27 (C 7). Consider a partition of th e plane into' congruent equilateral triangles.i1 Does there exist a square whose vertices are all vertices of these triangles?Solution. Assume that the equilateral triangles are of side length 1. Consider acoordinate system whose x-axis lies on the line containing the side of a certain triangle.Let 1V[ and N be any vertices of the equilateral triangles of the partition. Their z- and y-coordinates differ by numbers of the form and If for some integers k and e. Then thedistance between any two vertices of the grid has the form! vu 2 + 3v 2 with u,v integers.

Now, if there is a square satisfying the hypothesis then certain integers a, b, c, d shouldsatisfy the equality

(1)We will prove that this is impossible by observing that 2 enters the prime factorization ofany number uz + 3v 2 with an even power. (This implies that (1) cannot hold.) Indeed,let u = 2'cp , v = 2Yq , where p, q are odd. If x =f:. y then u 2 +3v 2 is the product of 2m in (x ,y)and an odd number, so we are done. And if x = y, we obtain

22xu2 + 3v2 = [p 2 + 3q 2 ].The square of each odd number is 1 modulo 8, so pZ + 3q2 = 4 (mod 8). This implies thatu 2 + 3v2 is divisible by 22 x +2 but not by 22x +3 , and theclaim is proven.

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Romanian Problem Book File: b75-77 Version 0Thus, the answer to the question is negative. ' ,75.28 (C8). Prove that if the binomial coefficient G) is divisible by p'"; where p is aprime, ~ e 'It ;::: pili.Solution. The exponent of a prime p in the f a c t o r i z a t i o ~ of n! into primes is L:i2::1 l; J,therefore the exponent of p in the factorization of ( ~ is '

E(p) = (l;J-l:, J-ln ; . kJ) .If G) is divisible by pm then E(p) ;::: m. Since the sum in E(p) has a finite number of

terms, let the rthe term be the last nonzero one. Then

l;:J > l;J+ ln ; kJ'so n ;:::p".I t is known thatfor any real z , y. So


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Romanian Problem Book File: b75-77 Version 0and the sum of these equal numbers is O. Then bi = b2 = . . . = bk , an d this is againimpossible.

75."'30 (C9). A sloppy tailor uses a machine to cu t 120 square patches of side 1 froma 25 x 20 rectangle sheet. Prove that, no matter how he performs the cutting, he can stillcu t a circular patch of diameter 1 from the remaining material,

Solution. For each square, Sk, consider the "rounded square", Rk, obtained from Skby adding four 1 x rectangles on its sides and four quarter-circles of radius at thecorners.

The area of R ~ is 1+ 4 + i = 3 + i, k = 1,2, . . . ,120. Note that120 (3 + ~ < 360 + 30 x 3.2 = 24 x 19.

It then follows from the pigeonhole principle that the rounded rectangles Rk do no t coverthe 24 x 19 rectangle, whose sides are parallel to those of the original one, at a distanceof from them.

Any disk of radius ~ centered at one of the uncovered points of the 24 x 19 rectangle,does not have any point in common with the squares Sk. Hence it is contained in th er e m a i n i n ~ material.

75.31 (A 11). Given that the equation ax 2 + bx + c = 0, where a, b, c are integersand a > O. has two distinct zeros in (0,1), prove that a;::: 5 and give an example of suchan equation for a = 5.

Solution. Let P(x) = a;12 + bx + c. The zeros Xl',:C2 of P are in (0,1), so the integers P(O) and P(l) are nonzero, and hence IP(O)P(I)I;::: 1. Since P(x) = a(x - XI)(X - X2)and a > 0, this can be written as

(1)Now, by the AM-GM inequality, we have x(l - x) ::; i for x E (0,1), with equality if

and only if x = ~ ThenI 1L x2(1 - X2) ::; 4' (2)

therefore(3)\Note that the inequality is strict since, by hypothesis, Xl and X2 cannot be both equal

to ~ 2Combined with (1), (3) yields a > 16, and since a is a positive integer, we obtain a;::: 5.

Since 1(0)f(2) 2:: :2 implies a2 > 32 and a ;::: 6, the only suitable example for a = 5 is th eequation 5a:(:c - 1) + 1 = O.

75.32 (NT3). Let a2 = :2 and {(1I+1 = (11. + l)U n for n, = 2,3, . .. ,8 . Find the last twodigits of a9.

16


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Romanian Problem Book File: b75-77 Version 06asSolution. Note first that a6 = is even since as > 1. Also, 7 = -1 (mod 4) and7a 6a7 = == (-lt 6 == 1 (mod 4).A t ~ v i a l induction shows that 84k+1 =8 (mod 10) for each nonnegative k. Then

8a 7a8 = == 8 (mod 10), so as = lOR+ 8 for some nonnegative integer Z, By the binomialtheorem,gas = (10 - I ts == -10(10l +8) + 1 (mod 100),...so ag == -79 == 21 (mod 100). Therefore, the last two digits of ag are 21.

75.33 (A 12). Prove that for any Xk E [1,2], k = 1,2, .. . ,17"

Are the conditions XI.: :::; :2 essential?Solution. The condition :rk E [1,2] is equivalent to(Xk

standard manipulation. gives- l)(xk - 2) :::; 0, which, after a

,L Adding up over k =')

;Lh, + :::; 3,X J . ~

1,2, . .. ,17" we obtaink = 1,2, . . . ,17,.

(1)

l On the other hand, from the AM-GM inequality,(2)

Clearly (1) and (2) yield the desired result.The conditions :r/.: :::; 2 essential. Indeed, set Xl = X2 = ... = Xn - l = 1 and X n = X > 2.

Then

'" )('" 1)2 3 .l 'll(.r) = { ; q. {; .q . - 11. = x: [(x - 2)17,2 - (x - 1)(2x -\Vn + (x - 1)2].(For .r > 2, we have lim .n. -+ 00 In (x) = 00 , and therefore j n(x) > afor 17, sufficiently large;hence the original inequality is violated.75.34 (SG5). Consider a pyramid, OABCD, where ABCD is a square, and theprojection of 0 on ABCD is its center. Determine the locus of centers of spheres whichpass through V and intersect the edges OA, OB, OC, OD at four coplanar points.17


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rL

..

Romanian Problem Book File: b75-77 Version 0Solution. Suppose a sphere with center S passes through 0 an d meets the edges OA,DB, OC, OD in the coplanar points A', B', C', D', respectively, We start by proving thatA' B' C D' is an isosceles trapezoid.Indeed, le t P be the center of ABCD, and le t P' = A' C' n B'D'. The lines A'C'

and B'D' are contained in the planes 0 A'C' an d 0 B'D', so P' is contained in theirintersection, the line OP. Now imagine rotating 60AC at 90 about OP, so that it landson 60BD; we then have th e following diagram.

I have some difflculties with both the diagram and theargument following it; it is probably Titu who will help.

Going back to the main solution, we observe that the locus in question splits into twoparts, corresponding to whether A'B' II C'D' or A'D' II B'C'. Suppose that A'B' II C'D'.

Let 0 1 , 0:3, O ~ 0(3 be the circumcenters of 0 AB, OCD, 0 A' B', OC'D', respectively.5 must lie 011 the perpendicular bisector of the segments A' B' an d C' D', which is theplane 00 10:3. In fact, 5 is the intersection in the plane 00103 of th e perpendicularsto 00 1 at O and to 00 3 at 0; . . Now since O ~ 0; can lie anywhere along the linesegments 00 1 and 00 3 , the locus in this case is th e rhombus, whose sides are formed bythe perpendiculars to 00 1 at 0 and 0 1 and to 00 3 at 0 an d 0 3 .The desired locus is the union of this rhombus with' th e rhombus obtained from theother case, which is simply the same figure, rotated about it s axis of symmetry OP by 90.

75.35 (A 13). Prove that every infinite arithmetic progression of positive integers:a) contains an infinite geometric progression;b) contains, for any odd n, a geometric progression of length n whose sum of terms is

equal to the sum of some n consecutive terms of the arithmetic progression.Solution. a) Let th e first term of the arithmetic progression be a and its common

difference d. Consider th e set {a(1 + d ) n } ~ = o ' It is an infinite geometric progression, andits elements are terms of the given progression, because

is divisible by d ..b) Consider the geometric progression {a(1 + nd)i}:01. Its terms are all in the arith

metic progression, because. nn )a( 1 + nd) l - a = a L (. {nd)l

j= l Jis a multiple of d. The sum of this progression

a(1 + nd)n - a = - . t ( 1 ~ ) (nd)j = at ( ~ (nd)j-1nd nd . J . 1 JJ= l J=18


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Romanian Problem Book

is clearly divisible by 17" andFile: b75-77 Version 0

So ( l ( l + ~ ~ ) n _ ( t = n( a + kcl), wherek = t ( ~ (17,d)j-2

)=2 JThus the sum equa.ls

17,; 1 for all n.

and the proof is complete.

\

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Romanian Problem Book