# New Zealand Mathematical Olympiad Committee # NZMO Round One 2019 — Solutions 1. How many positive integers less than 2019 are divisible by either 18 or 21, but not both? Solution: For any positive integer \(n\) , the number of multiples of \(n\) less than or equal to 2019 is given by \[\left\lfloor \frac{2019}{n}\right\rfloor .\] So there are \(\left\lfloor \frac{2019}{18}\right\rfloor = 112\) multiples of 18, and \(\left\lfloor \frac{2019}{21}\right\rfloor = 96\) multiples of 21. Moreover, since \(lcm(18,21) = 126\) there are \(\left\lfloor \frac{2019}{126}\right\rfloor = 16\) positive integers less than 2019 which are a multiple of both 18 and 21. Therefore the final answer is \[\left\lfloor \frac{2019}{18}\right\rfloor +\left\lfloor \frac{2019}{21}\right\rfloor -2\left\lfloor \frac{2019}{126}\right\rfloor = 112 + 96 - 2\times 16 = 176.\] 2. Find all real solutions to the equation \[(x^{2} + 3x + 1)^{x^{2} - x - 6} = 1.\] Solution: Let \(a = x^{2} + 3x + 1\) and let \(b = x^{2} - x - 6\) . The only way to have \(a^{b} = 1\) , is if \(a = \pm 1\) or \(b = 0\) . - If \(b = 0\) , then we solve the quadratic \(x^{2} - x - 6 = 0\) which has solutions \(x = -2,3\) (we would also have to check that \(a \neq 0\) in this case) - If \(a = 1\) , then we solve the quadratic \(x^{2} + 3x + 1 = 1\) which has solutions \(x = 0, -3\) . - If \(a = -1\) , then we solve the quadratic \(x^{2} + 3x + 1 = -1\) which has solutions \(x = -1, -2\) . (we also have to check that \(b\) is an even integer in this case) Therefore there are a total of 5 candidate solutions: \(x = - 3, - 2, - 1,0,3\) . Remark: In order to receive full marks, a student would have to demonstrate that \(x = - 3, - 2, - 1,0,3\) are actually all solutions, by substituting each of these values into the expression, and verify that the result is indeed 1. 3. In triangle \(ABC\) , points \(D\) and \(E\) lie on the interior of segments \(AB\) and \(AC\) , respectively, such that \(AD = 1\) , \(DB = 2\) , \(BC = 4\) , \(CE = 2\) and \(EA = 3\) . Let \(DE\) intersect \(BC\) at \(F\) . Determine the length of \(CF\) . Solution: First notice that the sidelengths of \(\triangle ABC\) are 3, 4 and 5. By Pythagoras this implies that triangle \(ABC\) is right- angled at \(B\) . Now we can put the diagram on coordinate axes such that \(B = (0,0)\) and \(A = (0,3)\) and \(C = (4,0)\) . Furthermore we get \(D = (0,2)\) and since \(E\) divides \(CA\) into the ratio \(2:3\) we get \(E = (2.4,1.2)\) , as shown in the diagram. ![md5:938072234333898a25a67f943dd7a9a6](938072234333898a25a67f943dd7a9a6.jpeg) Now we can calculate the slope of the line \(D E\) to be \(\frac{- 0.8}{2.4} = - \frac{1}{3}\) . This means that the equation of line \(D E\) is given by \(y = - \frac{x}{3} +2\) . Therefore the \(x\) - intercept of this line is the solution to \(0 = - \frac{x}{3} +2\) . The solution is when \(x = 6\) , and thus \(F = (6,0)\) . Hence \(C F = 2\) . \(\square\) 4. Show that the number \(122^{n} - 102^{n} - 21^{n}\) is always one less than a multiple of 2020, for any positive integer \(n\) . Solution: Let \(f(n) = 122^{n} - 102^{n} - 21^{n}\) . We consider \(f(n)\) in mod 101 and in mod 20 separately. Consider \(f(n)\) mod 101. \[f(n) = 122^{n} - 102^{n} - 21^{n\] \[\equiv 21^{n} - 1^{n} - 21^{n} \pmod {101\] \[= -1\] Consider \(f(n)\) mod 20. \[f(n) = 122^{n} - 102^{n} - 21^{n\] \[\equiv 2^{n} - 2^{n} - 1^{n} \pmod {20\] \[= -1\] Therefore \(f(n) \equiv - 1\) both in mod 20 and in mod 101. Since 20 and 101 are relatively prime, this means \(f(n) \equiv - 1\) (mod 2020). As required. \(\square\) 5. Find all positive integers \(n\) such that \(n^{4} - n^{3} + 3n^{2} + 5\) is a perfect square. Solution: Let \(f(n) = 4n^{4} - 4n^{3} + 12n^{2} + 20 = 4(n^{4} - n^{3} + 3n^{2} + 5)\) and note that \((n^{4} - n^{3} + 3n^{2} + 5)\) is a perfect square if and only if \(f(n)\) is. First note that: \[(2n^{2} - n + 5)^{2} - f(n) = 9n^{2} - 10n + 5 = 4n^{2} + 5(n - 1)^{2} > 0.\] Also note that \[f(n) - (2n^{2} - n + 2)^{2} = 3n^{2} + 4n + 16 = 2n^{2} + (n + 2)^{2} + 12 > 0.\] Therefore \((2n^{2} - n + 2)^{2}< f(n)< (2n^{2} - n + 5)^{2}\) , so the only way \(f(n)\) could be a perfect square is if it is \((2n^{2} - n + 3)^{2}\) or \((2n^{2} - n + 4)^{2}\) . Solving \(f(n) = (2n^{2} - n + 3)^{2}\) gives us the quadratic \(n^{2} - 6n - 11 = 0\) which has no integer solutions. Solving \(f(n) = (2n^{2} - n + 4)^{2}\) gives us \(5n^{2} - 8n - 4 = (5n + 2)(n - 2) = 0\) . which has only one integer solution \(n = 2\) . Checking \[(2)^{4} - (2)^{3} + 3(2)^{2} + 5 = 25\] which is a perfect square. Therefore the only solution is \(n = 2\) . \(\square\) 6. Let \(V\) be the set of vertices of a regular 21-gon. Given a non-empty subset \(U\) of \(V\) , let \(m(U)\) be the number of distinct lengths that occur between two distinct vertices in \(U\) . What is the maximum value of \(\frac{m(U)}{|U|}\) as \(U\) varies over all non-empty subsets of \(V\) ? Solution: To simplify notation, we will let \(m\) be \(m(U)\) and let \(n\) be \(|U|\) . First note that there are 10 different diagonal-lengths in a regular 21- gon. Now consider the following set of 5 vertices. ![md5:cdf1ef8261596f1ada833898c5a5bf7b](cdf1ef8261596f1ada833898c5a5bf7b.jpeg) Note that each of the 10 different diagonal- lengths appear (exactly once each). So for this set of 5 vertices we have \(\frac{m}{n} = \frac{10}{5} = 2\) . We will now show that this is the maximum possible value for \(\frac{m}{n}\) . If \(U\) is an arbitrary non- empty set of vertices, then there are two cases: - Case 1: \(n < 5\) . The total number of pairs of vertices in \(U\) is given by \(\frac{1}{2} n(n - 1)\) . Since \(n - 1 < 4\) this gives us the bound: \[m \leq \frac{n(n - 1)}{2} < \frac{n \times 4}{2} = 2n.\] Thus \(\frac{m}{n} < 2\) in this case. - Case 1: \(n \geq 5\) . The total number of distances in \(U\) is at most 10 because there are only 10 different diagonal lengths in the 21-gon. Therefore \[\frac{m}{n} \leq \frac{10}{n} \leq \frac{10}{5} = 2\] as required. Remark: The construction given is unique up to rotations and reflections. I.e. all sets that achieve the value \(\frac{m}{n} = 2\) are congruent to the example given here. \(\square\) 7. Let \(ABCDEF\) be a convex hexagon containing a point \(P\) in its interior such that \(PABC\) and \(PDEF\) are congruent rectangles with \(PA = BC = PD = EF\) (and \(AB = PC = DE = PF\) ). Let \(\ell\) be the line through the midpoint of \(AF\) and the circumcentre of \(PCD\) . Prove that \(\ell\) passes through \(P\) . Solution: Let \(M\) be the midpoint of \(AF\) and let \(O\) be the circumcentre of triangle \(CPD\) . Now construct \(Q\) to be the point such that \(CPDQ\) is a parallelogram, and let \(R\) be the centre of this parallelogram (i.e. \(R\) is the intersection of \(PQ\) with \(CD\) , and also \(R\) is the midpoint of \(PQ\) ). ![md5:7f7e1f9db688bdbb4b80cb850a19bf58](7f7e1f9db688bdbb4b80cb850a19bf58.jpeg) Note that \(Q D = C P = F P\) and \(D P = P A\) and \(\angle Q D P = 180^{\circ} - \angle D P C = \angle F P A\) Therefore (by SAS) we have a pair of congruent triangles: \[\triangle Q D P\cong \triangle F P A.\] Therefore \(\angle M A P = \angle R P D\) and \(A F = P Q\) . Thus \(A M = \frac{1}{2} A F = \frac{1}{2} P Q = P R\) . Therefore (by SAS) we have another pair of congruent triangles: \[\triangle M A P\cong \triangle R P D.\] Therefore \(\angle A P M = \angle R D P\) . Let \(x = \angle A P M\) so that \(\angle C D P = \angle R D P = x\) also. ![md5:459d4f57d9a2fcdc07adf11012a15437](459d4f57d9a2fcdc07adf11012a15437.jpeg) Since the angle subtended at the circumcentre is double the angle subtended at the circumference, we get \(\angle C O P = 2x\) (recall that \(O\) is the circumcentre of \(\triangle P C D\) ). Finally we get \(\angle O P C = 90^{\circ} - x\) because \(\triangle C O P\) is isosceles. Putting this all together, we get \[\angle O P M = \angle O P C + \angle C P A + \angle A P M = (90^{\circ} - x) + 90^{\circ} + x = 180^{\circ}.\] Therefore \(\angle O P M\) is a straight line. 8. Suppose that \(x_{1},x_{2},x_{3},\ldots x_{n}\) are real numbers between 0 and 1 with sum \(s\) . Prove that \[\sum_{i = 1}^{n}{\frac{x_{i}}{s + 1 - x_{i}}} + \prod_{i = 1}^{n}(1 - x_{i})\leq 1.\] Solution: Let \(i\) be arbitrary and consider the set \(A = \{a_{1},a_{2},\ldots ,a_{n}\}\) defined by \(a_{i} = s + 1 - x_{i}\) and let \(a_{j} = 1 - x_{j}\) for all \(j\neq i\) . For example, if \(i = 2\) then \(A\) would be \(\{1 - x_{1},s + 1 - x_{2},1 - x_{3},\ldots ,1 - x_{n}\}\) . The AM- GM inequality on \(A\) tells us \[1 = \frac{(s + 1 - x_{i}) + \sum_{j\neq i}(1 - x_{j})}{n}\geq \left((1 + s - x_{i})\prod_{j\neq i}(1 - x_{j})\right)^{\frac{1}{n}}\] Which rearranges to give us \[1 - (s + 1 - x_{i})\prod_{j\neq i}(1 - x_{j})\geq 0.\] From here we can multiply both sides by \((1 - x_{i})\) , then add \(s\) to both sides and factorise the LHS to get: \[(s + 1 - x_{i})\left(1 - \prod_{j = 1}^{n}(1 - x_{j})\right)\geq s.\] Now multipy both sides by \(\frac{x_{i}}{s(s + 1 - x_{i})}\) to get the following equation. \[\left(1 - \prod_{j = 1}^{n}(1 - x_{j})\right)\frac{x_{i}}{s}\geq \frac{x_{i}}{s + 1 - x_{i}} \quad (1)\] Note that this equation holds for all \(i\) . Now consider the sum of Equation 1 over all \(1\leq i\leq n\) . Since \((1 - \prod (1 - x_{j}))\) is constant and \(\sum \frac{x_{i}}{s} = 1\) , the sum of all the LHS equals \(\left(1 - \prod (1 - x_{j})\right)\) . So we get \[1 - \prod_{j = 1}^{n}(1 - x_{j})\geq \sum_{i = 1}^{n}\frac{x_{i}}{s + 1 - x_{i}}\] as required.