| {"year": "2015", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "We say that a finite set \\(S\\) of points in the plane is balanced if, for any two different points \\(A\\) and \\(B\\) in \\(S\\) , there is a point \\(C\\) in \\(S\\) such that \\(AC = BC\\) . We say that \\(S\\) is center-free if for any three different points \\(A\\) , \\(B\\) and \\(C\\) in \\(S\\) , there are no points \\(P\\) in \\(S\\) such that \\(PA = PB = PC\\) . \n\n(a) Show that for all integers \\(n \\geq 3\\) , there exists a balanced set consisting of \\(n\\) points. \n(b) Determine all integers \\(n \\geq 3\\) for which there exists a balanced center-free set consisting of \\(n\\) points.", "solution": "We \n\nFor part (a), take a circle centered at a point \\(O\\) , and add \\(n - 1\\) additional points by adding pairs of points separated by an arc of \\(60^{\\circ}\\) or similar triples. An example for \\(n = 6\\) is shown below. \n\n\n \n\nFor part (b), the answer is odd \\(n\\) , achieved by taking a regular \\(n\\) - gon. To show even \\(n\\) fail, note that some point is on the perpendicular bisector of \n\n\\[\\left\\lceil \\frac{1}{n} {\\binom{n}{2}}\\right\\rceil = \\frac{n}{2}\\] \n\npairs of points, which is enough. (This is a standard double- counting argument.) \n\nRemark. As an aside, there is a funny joke about this problem. There are two types of people in the world: \n\n- Those who solve (b) quickly and then take forever to solve (a), \n\n- those who solve (a) quickly and then can't solve (b) at all. \n\n(Empirically true when the Taiwan IMO 2014 team was working on it.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2015/1, proposed by Merlijn Staps (NLD) \n"}} |
| {"year": "2015", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Find all positive integers \\(a\\) , \\(b\\) , \\(c\\) such that each of \\(ab - c\\) , \\(bc - a\\) , \\(ca - b\\) is a power of 2 (possibly including \\(2^0 = 1\\) ).", "solution": "Fi \n\nHere is the solution of Telv Cohl, which is the shortest solution I am aware of. We will prove the only solutions are \\((2,2,2)\\) , \\((2,2,3)\\) , \\((2,6,11)\\) and \\((3,5,7)\\) and permutations. \n\nWLOG assume \\(a \\geq b \\geq c > 1\\) , so \\(ab - c \\geq ca - b \\geq bc - a\\) . We consider the following cases: \n\n- If \\(a\\) is even, then \n\n\\[ca - b = \\gcd (ab - c,ca - b)\\leq \\gcd (ab - c,a(ca - b) + ab - c)\\] \\[\\qquad = \\gcd \\left(ab - c,c(a^{2} - 1)\\right).\\] \n\nAs \\(a^{2} - 1\\) is odd, we conclude \\(ca - b \\leq c\\) . This implies \\(a = b = c = 2\\) . \n\n- If \\(a\\) , \\(b\\) , \\(c\\) are all odd, then \\(a > b > c > 1\\) follows. Then as before \n\n\\[ca - b\\leq \\gcd (ab - c,c(a^{2} - 1))\\leq 2^{a_{2}(a^{2} - 1)}\\leq 2a + 2\\leq 3a - b\\] \n\nso \\(c = 3\\) and \\(a = b + 2\\) . As \\(3a - b = ca - b \\geq 2(bc - a) = 6b - 2a\\) we then conclude \\(a = 7\\) and \\(b = 5\\) . \n\n- If \\(a\\) is odd and \\(b\\) , \\(c\\) are even, then \\(bc - a = 1\\) and hence \\(bc^{2} - b - c = ca - b\\) . Then from the miraculous identity \n\n\\[c^{3} - b - c = (1 - c^{2})(ab - c) + a(\\underbrace{bc^{2} - b - c}_{= ca - b}) + (ca - b)\\] \n\nso we conclude \\(\\gcd (ab - c,ca - b) = \\gcd (ab - c,c^{3} - b - c)\\) , in other words \n\n\\[bc^{2} - b - c = ca - b = \\gcd (ab - c,ca - b) = \\gcd (ab - c,c^{3} - b - c).\\] \n\nWe thus consider two more cases: \n\n- If \\(c^{3} - b - c \\neq 0\\) then the above implies \\(|c^{3} - b - c| \\geq bc^{2} - b - c\\) . As \\(b \\geq c > 1\\) , we must actually have \\(b = c\\) , thus \\(a = c^{2} - 1\\) . Finally \\(ab - c = c(c^{2} - 2)\\) is a power of 2, hence \\(b = c = 2\\) , so \\(a = 3\\) . \n\n- In the second case, assume \\(c^{3} - b - c = 0\\) , hence \\(c^{3} - c\\) . From \\(bc - a = 1\\) we obtain \\(a = c^{4} - c^{2} - 1\\) , hence \n\n\\[ca - b = c^{5} - 2c^{3} = c^{3}(c^{2} - 2)\\] \n\nis a power of 2, hence again \\(c = 2\\) . Thus \\(a = 11\\) and \\(b = 6\\) . \n\nThis finishes all cases, so the proof is done.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2015/2, proposed by Dušan Djukić (SRB) \n"}} |
| {"year": "2015", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABC\\) be an acute triangle with \\(AB > AC\\) . Let \\(\\Gamma\\) be its circumcircle, \\(H\\) its orthocenter, and \\(F\\) the foot of the altitude from \\(A\\) . Let \\(M\\) be the midpoint of \\(\\overline{BC}\\) . Let \\(Q\\) be the point on \\(\\Gamma\\) such that \\(\\angle HQA = 90^\\circ\\) and let \\(K\\) be the point on \\(\\Gamma\\) such that \\(\\angle HKQ = 90^\\circ\\) . Assume that the points \\(A\\) , \\(B\\) , \\(C\\) , \\(K\\) and \\(Q\\) are all different and lie on \\(\\Gamma\\) in this order. Prove that the circumcircles of triangles \\(KQH\\) and \\(FKM\\) are tangent to each other.", "solution": "Letr. \n\nLet \\(L\\) be on the nine- point circle with \\(\\angle HML = 90^{\\circ}\\) . The negative inversion at \\(H\\) swapping \\(\\Gamma\\) and nine- point circle maps \n\n\\[A\\longleftrightarrow F,\\quad Q\\longleftrightarrow M,\\quad K\\longleftrightarrow L.\\] \n\nIn the inverted statement, we want line \\(ML\\) to be tangent to \\((AQL)\\) . \n\n\n \n\nClaim — \\(\\overline{LM} \\parallel \\overline{AQ}\\) . \n\nProof. Both are perpendicular to \\(\\overline{MHQ}\\) . \n\nClaim — \\(LA = LQ\\) .\n\n\n\nProof. Let \\(N\\) and \\(T\\) be the midpoints of \\(\\overline{HQ}\\) and \\(\\overline{AH}\\) , and \\(O\\) the circumcenter. As \\(\\overline{MT}\\) is a diameter, we know \\(LTNM\\) is a rectangle, so \\(\\overline{LT}\\) passes through \\(O\\) . Since \\(\\overline{LOT} \\perp \\overline{AQ}\\) and \\(OA = OQ\\) , the proof is complete. \\(\\square\\) \n\nTogether these two claims solve the problem.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "3. ", "solution_match": "## §1.3 IMO 2015/3, proposed by Danylo Khilko and Mykhailo Plotnikov (UKR) \n"}} |
| {"year": "2015", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Triangle \\(ABC\\) has circumcircle \\(\\Omega\\) and circumcenter \\(O\\) . A circle \\(\\Gamma\\) with center \\(A\\) intersects the segment \\(BC\\) at points \\(D\\) and \\(E\\) , such that \\(B\\) , \\(D\\) , \\(E\\) , and \\(C\\) are all different and lie on line \\(BC\\) in this order. Let \\(F\\) and \\(G\\) be the points of intersection of \\(\\Gamma\\) and \\(\\Omega\\) , such that \\(A\\) , \\(F\\) , \\(B\\) , \\(C\\) , and \\(G\\) lie on \\(\\Omega\\) in this order. Let \\(K = (BDF) \\cap \\overline{AB} \\neq B\\) and \\(L = (CGE) \\cap \\overline{AC} \\neq C\\) and assume these points do not lie on line \\(FG\\) . Define \\(X = \\overline{FK} \\cap \\overline{GL}\\) . Prove that \\(X\\) lies on the line \\(AO\\) .", "solution": "Since \\(\\overline{AO} \\perp \\overline{FG}\\) for obvious reasons, we will only need to show that \\(XF = XG\\) , or that \\(\\angle KFG = \\angle LGF\\) . \n\nLet line \\(FG\\) meet \\((BDF)\\) and \\((CGE)\\) again at \\(F_{2}\\) and \\(G_{2}\\) . \n\n\n \n\nClaim — Quadrilaterals \\(FBDF_{2}\\) and \\(G_{2}ECG\\) are similar, actually homothetic through \\(\\overline{FG} \\cap \\overline{BC}\\) . \n\nProof. This is essentially a repeated application of being \"anti- parallel\" through \\(\\angle (FG, BC)\\) . Note the four angle relations \n\n\\[\\angle (FD,FG) = \\angle (BC,GE) = \\angle (G_{2}C,FG)\\Rightarrow \\overline{FD}\\parallel \\overline{G_{2}C\\] \\[\\angle (F_{2}B,FG) = \\angle (BC,FD) = \\angle (GE,FG)\\Rightarrow \\overline{F_{2}B}\\parallel \\overline{GE\\] \\[\\angle (FB,FG) = \\angle (BC,GC) = \\angle (G_{2}E,FG)\\Rightarrow \\overline{FB}\\parallel \\overline{G_{2}E\\]\n\n\n\n\\[\\angle (F_{2}D,F G) = \\angle (B C,F B) = \\angle (G C,F G)\\Rightarrow \\overline{{F_{2}D}}\\parallel \\overline{{G C}}.\\] \n\nThis gives the desired homotheties. \n\nTo finish the angle chase, \n\n\\[\\angle G F K = \\angle F_{2}B K = \\angle F_{2}B F - \\angle A B F = \\angle F_{2}D F - \\angle A B F\\] \\[\\qquad = \\angle F_{2}D F - \\angle G C A = \\angle G C G_{2} - \\angle G C A\\] \\[\\qquad = \\angle L C G_{2} = \\angle L G F\\] \n\nas needed. (Here \\(\\angle A B F = \\angle G C A\\) since \\(A F = A G\\) .)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2015/4, proposed by Silouanos Brazitikos and Evangelos Psychas (HEL) \n"}} |
| {"year": "2015", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Solve the functional equation \n\n\\[f(x + f(x + y)) + f(xy) = x + f(x + y) + yf(x)\\] \n\nfor \\(f: \\mathbb{R} \\to \\mathbb{R}\\) .", "solution": "\\) \n\nThe answers are \\(f(x)\\equiv x\\) and \\(f(x)\\equiv 2 - x\\) . Obviously, both of them work. \n\nLet \\(P(x,y)\\) be the given assertion. We also will let \\(S = \\{t\\mid f(t) = t\\}\\) be the set of fixed points of \\(f\\) \n\nFrom \\(P(0,0)\\) we get \\(f(f(0)) = 0\\) \n\nFrom \\(P(0,f(0))\\) we get \\(2f(0) = f(0)^{2}\\) and hence \\(f(0)\\in \\{0,2\\}\\) \n\nFrom \\(P(x,1)\\) we find that \\(x + f(x + 1)\\in S\\) for all \\(x\\) \n\nWe now solve the case \\(f(0) = 2\\) \n\nClaim If \\(f(0) = 2\\) then \\(f(x)\\equiv 2 - x\\) \n\nProof. Let \\(t\\in S\\) be any fixed point. Then \\(P(0,t)\\) gives \\(2 = 2t\\) or \\(t = 1\\) ; so \\(S = \\{1\\}\\) . But we also saw \\(x + f(x + 1)\\in S\\) , which implies \\(f(x)\\equiv 2 - x\\) . \\(\\square\\) \n\nHenceforth, assume \\(f(0) = 0\\) \n\nClaim If \\(f(0) = 0\\) then \\(f\\) is odd. \n\nProof. Note that \\(P(1, - 1)\\Rightarrow f(1) + f(- 1) = 1 - f(1)\\) and \\(P(- 1,1)\\Rightarrow f(- 1)+\\) \\(f(- 1) = - 1 + f(1)\\) , together giving \\(f(1) = 1\\) and \\(f(- 1) = - 1\\) . To prove \\(f\\) odd we now obtain more fixed points: \n\nFrom \\(P(x,0)\\) we find that \\(x + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFrom \\(P(x - 1,1)\\) we find that \\(x - 1 + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFrom \\(P(1,f(x) + x - 1)\\) we find \\(x + 1 + f(x)\\in S\\) for all \\(x\\in \\mathbb{R}\\) \n\nFinally \\(P(x, - 1)\\) gives \\(f\\) odd. \n\nTo finish from \\(f\\) odd, notice that \n\n\\[P(x, - x)\\Rightarrow f(x) + f(- x^{2}) = x - xf(x)\\] \\[P(- x,x)\\Rightarrow f(- x) + f(- x^{2}) = - x + xf(- x)\\] \n\nwhich upon subtracting gives \\(f(x)\\equiv x\\)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2015/5, proposed by Dorlir Ahmeti (ALB) \n"}} |
| {"year": "2015", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "The sequence \\(a_1, a_2, \\ldots\\) of integers satisfies the conditions: \n\n(i) \\(1 \\leq a_j \\leq 2015\\) for all \\(j \\geq 1\\) , \n\n(ii) \\(k + a_k \\neq \\ell + a_\\ell\\) for all \\(1 \\leq k < \\ell\\) . \n\nProve that there exist two positive integers \\(b\\) and \\(N\\) for which \n\n\\[\\left|\\sum_{j = m + 1}^{n}(a_{j} - b)\\right| \\leq 1007^{2}\\] \n\nfor all integers \\(m\\) and \\(n\\) such that \\(n > m \\geq N\\) .", "solution": "give two equivalent solutions with different presentations, one with \"arrows\" and the other by \"juggling\". \n\n\\(\\P\\) First solution (arrows). Consider the map \n\n\\[f\\colon k\\mapsto k + a_{k}.\\] \n\nThis map is injective, so if we draw all arrows of the form \\(k\\mapsto f(k)\\) we get a partition of \\(\\mathbb{N}\\) into one or more ascending chains (which skip by at most 2015). \n\nThere are at most 2015 such chains, since among any 2015 consecutive points in \\(\\mathbb{N}\\) every chain must have an element. \n\nWe claim we may take \\(b\\) to be the number of such chains, and \\(N\\) to be the largest of the start- points of all the chains. \n\nConsider an interval \\(I = [m + 1,n]\\) . We have that \n\n\\[\\sum_{m< j\\leq n}a_{j} = \\sum_{\\mathrm{chain~}c}[\\min \\{x > n,x\\in c\\} -\\min \\{x > m,x\\in c\\} ].\\] \n\nThus the upper bound is proved by the calculation \n\n\\[\\sum_{m< j\\leq n}(a_{j} - b) = \\sum_{\\mathrm{chain~}c}[(\\min \\{x > n,x\\in c\\} -n) - (\\min \\{x > m,x\\in c\\} -m)]\\] \\[\\qquad = \\sum_{\\mathrm{chain~}c}[(\\min \\{x > n,x\\in c\\} -n)] - \\sum_{\\mathrm{chain~}c}[\\min \\{x > m,x\\in c\\} -m]\\] \\[\\qquad \\leq (1 + 2015 + 2014 + \\dots +(2015 - (b - 2))) - (1 + 2 + \\dots +b)\\] \\[\\qquad = (b - 1)(2015 - b)\\] \n\nfrom above (noting that \\(n + 1\\) has to belong to some chain). The lower bound is similar. \n\n\\(\\P\\) Second solution (juggling). This solution is essentially the same, but phrased as a juggling problem. Here is a solution in this interpretation: we will consider several balls thrown in the air, which may be at heights 0, 1, 2, ..., 2014. The process is as follows:\n\n\n\n- Initially, at time \\(t = 0\\) , there are no balls in the air. \n\n- Then at each integer time \\(t\\) thereafter, if there is a ball at height 0, it is caught; otherwise a ball is added to the juggler's hand. This ball (either caught or added) is then thrown to a height of \\(a_{t}\\) . \n\n- Immediately afterwards, all balls have their height decreased by one. \n\nThe condition \\(a_{k} + k \\neq \\ell + a_{\\ell}\\) thus ensures that no two balls are ever at the same height. In particular, there will never be more than 2016 balls, since there are only 2015 possible heights. \n\nWe claim we may set. \n\n\\(b =\\) number of balls in entire process \n\n\\(N =\\) last moment in time at which a ball is added. \n\nIndeed, the key fact is that if we let \\(S_{t}\\) denote the sum of the height of all the balls just after time \\(t + \\frac{1}{2}\\) , then \n\n\\[S_{t + 1} - S_{t} = a_{t + 1} - b\\] \n\nAfter all, at each time step \\(t\\) , the caught ball is thrown to height \\(a_{t}\\) , and then all balls have their height decreased by 1, from which the conclusion follows. Hence the quantity in the problem is exactly equal to \n\n\\[\\left|\\sum_{j = m + 1}^{n}(a_{j} - b)\\right| = |S_{m} - S_{n}|.\\] \n\nFor a fixed \\(b\\) , we easily have the inequalities \\(0 + 1 + \\dots +(b - 1) \\leq S_{t} \\leq 2014 + 2013 + \\dots +(2015 - b)\\) . Hence \\(|S_{m} - S_{n}| \\leq (b - 1)(2015 - b) \\leq 1007^{2}\\) as desired.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2015-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2015/6, proposed by Ross Atkins and Ivan Guo (AUS) \n"}} |
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