The First TST for the JBMO Satu Mare, April 6, 2018


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1 The First TST for the JBMO Satu Mare, April 6, 08 Problem. Prove that the equation x +y +z = x+y +z + has no rational solutions. Solution. The equation can be written equivalently (x ) + (y ) + (z ) = 7. If (x, y, z) would be a solution of this equation with rational components, denoting x = a b, y = a b, z = a 3 b 3, one would have integers a, b, a, b, a 3, b 3 satisfying the equality (a b b 3 ) + (b a b 3 ) + (b b a 3 ) = 7(b b b 3 ). This would lead to the existence of four integers a, b, c, d such that a + b + c = 7d. If the greatest common divisor of a, b, c, d is k, dividing by k one would obtain a solution (a, b, c, d) Z of the equation a + b + c = 7d with g.c.d.(a, b, c, d) =. Then a, b, c, d can not all be even. But a perfect square is congruent to either 0,, or modulo 8, so the left hand side can only be,, 3, 5 or 6 modulo 8, while the right hand side, 7d, is 0, or 7 modulo 8. In conclusion, the equality can not hold. Problem. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that When does the equality hold? Solution. The inequality can be written a + 3 b + 5 c a + 3b + c. a + 3 b + 5 c + b + c (a + b + c ) or a + b + c + b + b + c + c. This follows from the following inequalities by using the AMGM inequality: a + b + c abc Marius Stănean because On the other hand, 3 = a + b + c 3 3 a b c = abc. b + b = b + b + b 3 3 b b b = 3 c + c = c + c + c 3 3 b b b = 6.
2 The equality holds if a = b = c =. Remark. One can apply directly the AMGM inequality for the following numbers: a, b, b, b, c, c, c, c, c, b, c, c. Solution. As x 3 3x + 0, x 0 (equivalent to (x ) (x + ) 0; alternatively, the previous inequality follows from AMGM: x x 3 = 3x; the equality holds whence x = ). We deduce that x 3 x. Writing this for a, b, c and multiplying these inequalities by, 3, and 5, respectively, we obtain by summing a + 3 b + 5 c 7 a + 3b 3 + 5c. Using 7 = 9(a + b + c ) one obtains the conclusion. The equality holds if and only if a = b = c =. Problem 3. Let ABC be a triangle with AB > AC. Point P (AB) is such that ACP = ABC. Let D be the reflexion of P into the line AC and let E be the point in which the circumcircle of BCD meets again the line AC. Prove that AE = AC. Vlad Robu Solution. Let Q be the point in which the circumcircle of BCD meets again the line AB. Then QEA = QBC = ECP, hence EQ P C. Moreover, ECP = ECD implies QD is parallel to EC, hence EQ = CD = CP. It follows that EQCP is a parallelogram, which leads to the conclusion. Solution. (given in the contest) If {O} = BD AC, it is easy to prove that BA and BO are isogonal in the angle EBC, therefore, in order to show that BA is the median, one has to prove that BO is the simedian. This follows readily by computation, using Steiner s Theorem. Indeed, EBA = BAC BEA = 80 ABC ACB BDC = 80 ECD
3 ECB BDC = CBD. Triangles OEB and ODC are similar, hence EO OD = EB CD. () Triangles OCB and ODE are also similar, therefore OC OD = BC ED. () We have ADC = AP C = P BC + P CB = ACB = EDB, which shows that the EB AD CD AC triangles ACD and EBD are similar. Hence ED = and, as AD = AC BC EB CD (from ACB AP C ADC), it follows that ED = BC. From () and () we obtain EO ( ) EB ED EB = OC CD BC =, whih shows that BO is indeed BC the simedian. Solution 3. (given in the contest by Andrei Mărginean) Let a = EBA, x = ABC. We have that DCE = ACP = x and EBC = a + x. As B, C, D, E are concyclic, it follows that CDE = 80 a x, DEC = 80 EDC DCE = a = EBA. But AE is the perpendicular bisector of [P D], hence P EA = DEC = EBA, which shows that the triangles AP E and AEB are similar. It follows that AE = AB AP. But triangles ACP and ABC are also similar, hence AC = AP AB = AE, and the conclusion follows immediately.. Problem. What is the maximum number of rooks one can place on a chessboard such that any rook attacks exactly two other rooks? (We say that two rooks attack each other if they are on the same line or on the same column and between them there are no other rooks.) Alexandru Mihalcu Solution. We say that there can be two types of rooks on the chessboard: a rook of type T is a rook that is attacked from two perpendicular lines; a rook is of type T if it is attacked by two rooks situated on the same line, but opposite directions. Suppose one can place m rooks of type T and n of type T, with m + n = x. Each rook of type T determines two lines from which it is attacked (out of the 6 of the chessboard: 8 horizontal and 8 vertical ones), while exactly one other rook of type T is situated on each of these two lines. Notice that a rook of type T can not be attacked from any other line than the one it is already attacked from, hence, for each of these m rooks, there are m lines on which there can be no other rooks. In total, we could have at most n + m = n + m lines. Hence 6 m + n = x. To prove that 6 is indeed the desired maximum, it is sufficient to exhibit an example of 6 rooks that attack exactly two other ones. 3
4 ele sunt atacate din direcţii perpendiculare, respectiv de tipul T dacă ele sunt atacate din aceeaşi direcţie, dar sens contrar. Presupunem că putem plasa maxim x ture, m de tipul T şi n de tipul T. Observăm că orice tură de tipul T determină linii de pe care este atacată, iar exact o altă tură de tipul T se află pe oricare dintre aceste două linii. Pentru turele de tipul T, putem observa că ele nu pot fi atacate din altă direcţie decât cea din care este atacată deja, deci pentru fiecare din cele m ture, există m linii care nu mai pot fi ocupate de alte ture. În total am putea avea n + m = n + m linii. Deci 6 m + n = x. Problema se termină când găsim un exemplu cu 6 ture. 8 RSRSRSRS S0Z0Z0ZR a b c d e f g h Remark. Other elegant examples can be obtained by putting rooks on the two diagonals or, another example, regrouping squares into a square and placing rooks in each unit square of such squares that cover one of the diagonals. Solution. (given in the contest by Andrei Mărginean) We prove by induction after n that on an n n board one can place at most n rooks with the above restrictions. For n = the statement is obvious. For the inductive step, assume the statement above to hold for an arbitrary n and let us prove it for n +. Assume that on an (n + ) (n + ) one could place at least n + 3 rooks such that each of them attacks exactly two other ones. From the Pigeonhole Principle it follows that there exists at least one horizontal line containing at least 3 rooks and, also, a vertical line with 3 or more rooks. On a horizontal line with 3 or more rooks there exists at least one rook that is attacked by two rooks situated on the same horizontal line with it. Therefore this rook must stand alone on the vertical line it occupies. Similarly, there must be a horizontal line containing exactly one rook. Eliminating these two lines, we obtain an n n board with n + rooks that still satisfies the condition that each rook attacks exactly two other rooks. This contradicts our inductive hypothesis for n. Again, an example with 6 rooks finishes the proof. (Andrei gave the one from the first solution.) Remark. By induction after n + m one can prove that the maximum number of rooks one can place on an m n board such that each rook attacks exactly two other rooks is m + n. Solution 3. (given in the contest by Iustinian Constantinescu) Let us consider a configuration of rooks attacking each other. A rook can attack on rays, by opposite. On exactly of these rays one must have rooks, on each of the other two rays the rook... attacks one of the 3 segments that constitute the margins of the chessboard. Each of these 3 segments is attacked by a different rook, and each rook attacks two such segments, therefore one can have at most 6 rooks on the board. An example with 6 rooks on the board finishes the proof. (Iustinian placed the rooks on the two diagonals.)
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