First selection test, May 1 st, 2008

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1 First selectio test, May st, 2008 Problem. Let p be a prime umber, p 3, ad let a, b be iteger umbers so that p a + b ad p 2 a 3 + b 3. Show that p 2 a + b or p 3 a 3 + b 3. Problem 2. Prove that for ay positive iteger there exists a multiple of with the sum of its digits equal to. Mihai Băluă Problem 3. Let ABC be a acute-agled triagle. Cosider the equilateral triagle A UV, with A (BC), U (AC), V (AB) such that UV BC. Similarly, defie poits B (AC) ad C (AB). Show that the lies AA, BB ad CC are cocurret. Vasile Pop Problem 4. Let ABC be a triagle ad D the midpoit of BC. O the sides AB ad AC there are poits M, N respectively, other tha the midpoits of these segmets, so that AM 2 + AN 2 = BM 2 + CN 2 ad MDN = BAC. Prove that A = 90. Fracisc Bozga Problem 5. Let N, 2. Cosider the itegers a, a 2,... a with 0 < a k k, for all k =, 2,...,. Give that a + a a is a eve umber, prove that oe ca choose the sigs + ad so that a ± a 2 ±... ± a = 0. Secod selectio test, Jue 5 th, 2008 Problem 6. Cosider a acute-agled triagle ABC, the height AD ad the poit E where the diameter from A of the circumcircle itersects the lie BC. Let M, N be the mirror images of D across the lies AC ad AB. Show that EMC = BNE. Diu Şerbăescu Problem 7. A sequece of itegers a, a 2,..., a is give so that a k is the umber of all multiples of k i this sequece, for ay k =, 2,...,. Fid all possible values for. Cristia Magra Problem 8. Let N ad let a, a 2,..., a be positive real umbers so that a + a a = a 2 + a a 2. Prove that for ay m =, 2...,, there exist m umbers amog the give oes with the sum ot less tha m. Adrei Ciupa & Flavia Georgescu Problem 9. Let a, b be real umbers with the property that the iteger part of a + b is a eve umber, for all N. Show that a is a eve iteger. Diu Şerbăescu

2 Third selectio test, Jue 6 th, 2008 Problem 0. Te umbers are chose at radom from the set,2,3,...,37. Show that oe ca select four distict umbers from the chose oes so that the sum of two of them is equal to the sum of the other two. Vasile Pop Problem. Let a, b, c be positive real umbers with + bc + ca = 3. Prove that + a 2 (b + c) + + b 2 (c + a) + + c 2 (a + b) c. Vlad Matei Problem 2. Fid all primes p, q satisfyig the equatio 2p q q p = 7. Fracisc Bozga Problem 3. Let d be a lie ad let M, N be two poits o d. Circles α, β, γ, δ cetered at A, B, C, D are taget to d i such a maer that circles α, β are exterally taget at M, while circles γ, δ are exterally taget at N. Moreover, poits A ad C lies o the same side of lie d. Prove that if there exists a circle taget to all circles α, β, γ, δ, cotaiig all of them i the iterior, the lies AC, BD ad d are cocurret or parallel. Flavia Georgescu Fourth selectio test, Jue 9 th, 2008 Problem 4. Let ABCD be a quadrilateral with o two opposite sides parallel. The parallel from A to BD meets the lie CD at poit F ad the parallel from D at AC meet the lie AB at poit E. Cosider the midpoits M, N, P, Q of the segmets AC, BD, AF, DE respectively. Show that lies MN, P Q ad AD are cocurret. Diu Şerbăescu Problem 5. Let m, N ad A = {, 2,..., }, B = {, 2,..., m}. A subset S of the set product A B has the property that for ay pairs (a, b), (x, y) S, the (a x)(b y) 0. Show that S has at most m + elemets. Diu Şerbăescu Problem 6. Fid all pairs of itegers (m, ),, m > so that m divides 3. Fracisc Bozga Problem 7. Determie the maximum value of the real umber k such that ( (a + b + c) a + b + b + c + ) a + c k k, for all real umbers a, b, c 0 with a + b + c = + bc + ca. Adrei Ciupa 2

3 SOLUTIONS Problem. Suppose that p 2 a + b. It suffices to prove that p 3 a 3 + b 3. Ideed, if p 2 (a + b) 3 3(a + b), we have p 3. As p 3 is prime, it follows that p a or p b. Sice p a + b, we get p a ad p b. As a cosequece, p 3 a 3 ad p 3 b 3, implyig p 3 a 3 + b 3. Problem 2. Let ad let 0 k, k N. Cosider all the remaiders of the umbers 0 k at the divisio by. Sice there are oly fiitely may residues, there exists a = 0,,..., so that 0 m a (mod ) for ifiitely may values of m N. Amog them select oly, amely 0 m, 0 m2,..., 0 m, with m > m 2 >... > m. The umber A = 0 m + 0 m m has digits equal to ad the rest equal to 0, has the sum of all digits is. Moreover, A a 0 (mod ), so A, which cocludes the proof. Problem 3. Cosider the equilateral triagle BCA, costructed i the exterior of the triagle ABC. The poits A, A, A are colliear, through the homothety of ceter A which map poits U, V i B, C, respectively. Sice AA, BB, CC cocur i the Fermat-Torricelli poit of the triagle ABC, the claim is proved. Problem 4. Let E ad F be the midpoit of the sides AC ad AB ad let P be the mirror image of D across E. The relatio AM 2 + AN 2 = BM 2 + CN 2 gives ( c 2 + F M)2 + ( b 2 NE)2 = ( c 2 F M)2 + ( b 2 + NE)2, hece c F M = b NE. The F M AC = NE AB, so F M F D = NE EP. Sice MF D = NEP, we get MF D NEP, which implies MDF = NP E. O the other had MDN = BAC = F DE, so MDF = NDE. Now the triagle NP D is isosceles ad NE is a media i this triagle, so NE DP, i other words A = 90. Problem 5. Cosider A = a a. Sice a ad a, we have A. If A = 0, that is a = a, the a + a a 2 is eve ad the claim reduces to the case of 2 umbers. If A > 0, the a + a a 2 + A is eve ad the claim reduces to the case of umbers. Problem 6. Observe that AD = AN = AM ad AND = AMC = 90, due to the reflectios across AB ad AC. It is kow that AD ad AE are isogoal cevias, that is BAD = EAC. The N AE = N AB + BAE = BAD + BAE = EAC + DAC = EAC + CAM = EAM ad cosequetly NAE EAM. It follows that ENA = EMA, so BNE = 90 ENA = 90 EMA = EMC, as desired. Problem 7. Notice that a =, as divides ay a j, ad a j, for ay j =, 2,...,. Cosider a array with rows correspodig to,2,..., ad colums correspodig to the umbers a, a 2,..., a. Defie the etries as follows: o the positio (i, a j ) put if i divides a j ad 0 otherwise. Now observe that the sum of the umbers placed of the i th row is equal to a i, as we see the etries for 3

4 each multiple of i i the sequece a, a 2,..., a. Hece the sum of all the etries i the array is a j. O the other had, o the a th j colum we get ad etry as log as we cout the divisors of a j, so the sum of the umbers o the j th colums is the umber of divisors of a j. This implies that the sum of all the etries i the array is the sum of all divisors of the umbers a j. Usig this double-coutig, together with the obvious fact that the umber of divisor of a is less tha a - uless a is or 2, show that = or = 2. Alterative solutio. Recall that a = ad a i, for all i =, 2,...,. Assume that > 3. As a, the exists a multiple of, where >, i the give sequece; let a k, k > be such a multiple. The coditio a i shows that a k =, i other words there are multiples of k i the sequece. As ad are coprime, k does ot divide a =, so k divides a 2,..., a. But k 2 ad k a, therefore a >. Thus must occur at least twice i the sequece, so, beside a we have a j =, j >. Hece k, a cotradictio. As before, = or = 2 are the oly possible values. Problem 8. It is clear that we eed to prove that a +a a. Let us otice that this is eough: let m =, 2,..., ad assume that ay selectio of m umbers from the give oes has the sum less tha. The add the iequalities a + a a m < m, a 2 + a a m+ < m,. a + a a m < m, to get (a + a a ) < m, which is a cotradictio. Back to the top, let g be the geometrical mea of the umbers a, a 2,..., a ad suppose that a + a a <. By AM-GM iequality, we have g a + a a <, while >, which gives g >, a cotradic- g2 a + a a = tio a 2 a 2 2 a 2 Problem 9. Let [a + b] = 2x, for all itegers > 0. The 2x a + b < 2x +, () 2x + a( + ) + b < 2x + +. (2) Subtractig () from (2) we get 2(x + x ) < a < 2(x + x ) +, for all > 0. As 2(x + x ) is a odd iteger, it follows that all umbers 2(x + x ) must be equal, as otherwise a belogs to two ope itervals of legths 2 havig the left margis at a differece of at least 2, which is impossible. Hece 2(x + x ) = 2s, s Z, so x + x = s ad the x = s + p, p Z, > 0. Pluggig i () we get 2p b (a 2s) < 2p b+, > 0, so a = 2s, sice otherwise the set of the positive itegers will have a upper margi. Observe that s is a iteger, so a is a eve iteger, as eeded. Problem 0. Cosider all positive differeces a b amog all 0 umbers. Sice there are C 2 0 = 45 positive differeces ad all belog to the set,2,3,...,36, 4

5 at least two of them are equal. Let them be a b ad c d, with a > c. If a, b, c, d are all distict, we are doe; if ot, the b = c, so b = c is oe of the 8 umbers which are either the lowest or the greatest umber from the iitial oes. Now observe that we have 45 positive differeces ad 36 possible values for them, so either 3 positive differeces are equal or there are 9 pairs of equal positive differeces. The first case gives a b = c d = e f, with a > c > e. Sice we caot have b = c, b = e ad d = e, we are doe. The secod case gives at least oe pair of positive differeces i which case b = c is excluded, as oly 8 cadidates for b = c exist, so we are doe. Problem. Usig the AM-GM iequality we derive 3 (c)2. As + bc + ca = 3, the c. Now + bc + ca 3 + a 2 (b + c) = + a( + ac) = + a(3 bc) = 3a + ( c) + bc + ca = =, as claimed. 3a 3c c Problem 2. It is easy to observe that p is odd ad p q, i other words p 3 ad (p, q) =. If q = 2, the 2 p+ = 7+p 2. The oly solutio is p = 3, as 2 + > 7+ 2, for all 4. For q 3, by Little Fermat s Theorem we get p 2q 7 ad q p + 7. Set p + 7 = kq, k N. If 2q 7 0, we have q = 3 ad p, false. If 2q 7 > 0, the 2q 7 p, so 2q p + 7 kq, therefore k = or k = 2. For k = we obtai p + 7 = q, so p 2p + 7. This implies p = 7 ad the q = 4, false. Hece k = 2 ad p + 7 = 2q. Suppose p > q; as p, q 3 we get q p q p ad the 7 = 2q p p q q p 3 3 = 27, a cotradictio. Thus q > p ad the p + 7 = 2q > 2p, which yields p = 3 or p = 5. For p = 3 we have q = 5, while p = 5 gives q 2, with o solutio. To coclude, the solutios are (p, q) = (3, 2), (3, 5). Problem 3. Let a, b, c, d be the radii of the circles α, β, γ, δ. It suffices to prove that a b = c d, i other words the ratio a b is costat while poit M varies o lie d. Let R ad S be the midpoits of the arcs determied by d o the fifth circle K, the oe taget simultaeously to α, β, γ, δ, ad let N be o the same side of d as A. Deote by A ad B the tagecy poit of α ad β to K, respectively. Observe that poits A, M, R are colliear - via the homothety which maps circle α oto circle K - ad similarly poits B, M, S are colliear. Sice RS is a diameter of K, agles RA S ad SB R are right. If lies B R ad A S meet as poit V, the M is the orthoceter of the triagle V RS. Notice that d RS, hece V d; deote by O the itersectio poit of d ad RS. Lies A S ad B R itersect the circles α ad β at poits U ad Z respectively. Sice RA S = SB R = 90, the segmets UM ad ZM are diameters i circles α, β, so a b = UM ZM = RO SO. The latter ratio is costat, as claimed. Problem 4. Let O be the midpoit of AD, R be the itersectio poit of lies AC ad BD ad S be the itersectio poit of lies AF ad DE. Sice N ad Q are the midpoits of the sides DB ad DE of the triagle DBE, we 5

6 have O NQ ad similarly O MP. Moreover, as DRAS is a parallelogram, the diagoal RS passes through the midpoit O of the other diagoal, AD. Now apply Desargues Theorem for triagles NRM ad SP Q, give that O lies simultaeously o lies NQ, MP, RS ad we are doe. Problem 5. Cosider a set S which satisfies all requiremets. For each i A = {, 2,..., }, defie B i B the set of all elemets j B for which the pair (i, j) belogs to the set S - otice that some subsets B i ca be empty. Coutig all pairs i S over all secod elemet i each pair, we have S = B + B B. The mai idea is to observe the chai of iequalities B B 2 B, where by X Y we mea that x y, for ay x X ad y Y, X, Y beig sets of itegers. (This defiitio allows the empty set to occupy ay positio i this chai). Sice B B 2... B = B = {, 2,..., m} ad ay two cosecutive subsets B i share i commo at most oe elemet, we get - by sieve thorem - that S m +, as claimed. Problem 6. The solutios are (k, k 2 ) ad (k 2, k), with k >. We have m ( 3 )m 2 (m ) = 2 m. O the other had, m m( 2 m) (m ) = m 2. If > m 2, the m m 2, so m, false. If = m 2, the obviously m 3 m 6, so all pairs (m, m 2 ), m > are solutios. If < m 2, from m 3 we derive that < m 2. The m m 2 < m 2, so < m. If 2 m > 0, we obtai m 2 m < 2, so m <, a cotradictio. Hece = m 2, which holds, sice m 3 m 3, so all pairs ( 2, ), > are also solutios. Problem 7. Observe that the umbers a = b = 2, c = 0 fulfill the coditio ( + bc + ca = a + b + c. Pluggig ito the give iequality, we derive that ) 2 k k, hece k. We claim that the iequality holds for k =, provig that the maximum value of k is. For this, rewrite the iequality as ( ( + bc + ca) a + b + b + c + + bc + ca a + b ( ) a + b + c + bc + ca + ) a + c + bc + ca + a + b. Notice that a + b, sice a, b, c 0. Summig over a cyclic a + b + c permutatio of a, b, c we get as edeed. a + b + bc + ca = =, a + b + c a + b + c 6

7 Alterative solutio. The iequality is equivalet to the followig: S = a + b + c ( a + b + c + a + b + b + c + ) k. a + c Usig the give coditio, we get hece a + b + b + c + a + c = a2 + b 2 + c 2 + 3( + bc + ca) (a + b)(b + c)(c + a) = a2 + b 2 + c 2 + 2( + bc + ca) + a + b + c (a + b + c)( + bc + ca) c (a + b + c)(a + b + c + ) = (a + b + c) 2, c S = (a + b + c) 2 (a + b + c) 2 c. Now it is clear that S, ad the equality occurs for c = 0. Therefore k = is the maximum value. 7

Solutions for May. 3 x + 7 = 4 x x +

Solutions for May. 3 x + 7 = 4 x x + Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits