# New Zealand Mathematical Olympiad Committee # NZMO Round Two 2020 — Solutions 1. Problem: Let \(P(x) = x^{3} - 2x + 1\) and let \(Q(x) = x^{3} - 4x^{2} + 4x - 1\) . Show that if \(P(r) = 0\) then \(Q(r^{2}) = 0\) Solution: Notice that \(P(x) = x^{3} - 2x + 1 = (x - 1)(x^{2} + x - 1)\) , and the roots of \(x^{2} + x - 1\) are \(\frac{- 1\pm\sqrt{5}}{2}\) . Therefore the roots of \(P(x)\) are \[1, \frac{\sqrt{5} - 1}{2} \text{and} \frac{-1 - \sqrt{5}}{2}.\] Notice also that \(Q(x) = x^{3} - 4x^{2} + 4x - 1 = (x - 1)(x^{2} - 3x + 1)\) , and the roots of \(x^{2} - 3x + 1\) are \(\frac{3\pm\sqrt{5}}{2}\) . Therefore the roots of \(Q(x)\) are \[1, \frac{3 + \sqrt{5}}{2} \text{and} \frac{3 - \sqrt{5}}{2}.\] So it suffices to check that \(1^{2} = 1\) and that \(\left(\frac{- 1\pm\sqrt{5}}{2}\right)^{2} = \frac{1\pm 2\sqrt{5} + 5}{4} = \frac{3\pm\sqrt{5}}{2}\) . ## Alternative Solution: First notice that \[(x^{3} - 2x - 1)P(x) = (x^{3} - 2x - 1)(x^{3} - 2x + 1)\] \[\qquad = x^{6} - 4x^{4} + 4x^{2} - 1\] \[\qquad = Q(x^{2}).\] If \(r\) is any root of \(P(x)\) then \(P(r) = 0\) . This implies that \[Q(r^{2}) = (r^{3} - 2r + 1)P(r) = (r^{3} - 2r + 1)\times 0 = 0.\] Hence \(r^{2}\) is a root of \(Q(x)\) . 2. Problem: Find the smallest positive integer \(N\) satisfying the following three properties. - \(N\) leaves a remainder of 5 when divided by 7. - \(N\) leaves a remainder of 6 when divided by 8. - \(N\) leaves a remainder of 7 when divided by 9. Solution: We notice that \(\{5,6,7\}\) are each 2 less than \(\{7,8,9\}\) respectively. Therefore \(N + 2\) must be a multiple of 7, 8 and 9. Since 7, 8 and 9 are pairwise coprime, this means that \[(N + 2) \text{is a multiple of} 7 \times 8 \times 9 = 504.\] Therefore the smallest positive possibility for \(N + 2\) is 504. Thus \(N = 502\) . 3. Problem: There are 13 marked points on the circumference of a circle with radius 13. Prove that we can choose three of the marked points which form a triangle with area less than 13. Solution: Divide the circle into 6 equal ( \(60^{\circ}\) ) arcs. By the pigeon-hole principle (since \(13 > 2 \times 6\) ) there exists at least one arc which contains at least three of the marked points. Let this \(60^{\circ}\) arc be \(AB\) , and let the three marked points be \(X\) , \(Y\) and \(Z\) in that order. ![md5:99abfa9e89b6df2b8bb12a8f787f605a](99abfa9e89b6df2b8bb12a8f787f605a.jpeg) Let \(M\) be the midpoint of arc \(AB\) . Points \(X\) and \(Z\) both lie within the minor arc \(AB\) so the base of triangle \(\triangle XYZ\) is less than or equal to the base of triangle \(\triangle AMB\) . Also the distance from \(Y\) to \(XZ\) is less than or equal to the distance from \(Y\) to \(AB\) , which is at most the distance from \(M\) to \(AB\) . Hence: the base and height of triangle \(\triangle XYZ\) are less than or equal to the base and height of triangle \(\triangle AMB\) respectively. Therefore \[\mathrm{area}(\triangle XYZ)\leq \mathrm{area}(\triangle AMB).\] So it suffices to show that \(\mathrm{area}(\triangle AMB)< 13\) . Let \(O\) be the centre of the circle and let \(D\) be the foot of the altitude from \(M\) to \(AB\) . Note that \(\triangle ABO\) is equilateral and \(D\) is the midpoint of \(AB\) so \(AD = \frac{13}{2}\) . Furthermore, by Pythagoras in \(\triangle ADO\) we get \[OD = \sqrt{OA^{2} - AD^{2}} = \sqrt{13^{2} - \left(\frac{13}{2}\right)^{2}} = \frac{13}{2}\sqrt{3}.\] \[\Rightarrow MD = MO - OD = 13 - \frac{13}{2}\sqrt{3} = 13\left(1 - \frac{\sqrt{3}}{2}\right).\] Now we can calculate the area of triangle \(\triangle AMB\) . The base is \(AB = 13\) because \(ABO\) is equilateral. The height is \(MD = 13\left(1 - \frac{\sqrt{3}}{2}\right)\) . Therefore \[\mathrm{area}(\triangle AMB) = \frac{13 \times 13\left(1 - \frac{\sqrt{3}}{2}\right)}{2} = \frac{169(2 - \sqrt{3})}{4}.\] And we can confirm that \(\frac{169(2 - \sqrt{3})}{4} < 13\) because \[\frac{169(2 - \sqrt{3})}{4} < 13\] \[\frac{13(2 - \sqrt{3})< 4}{26 - 13\sqrt{3} < 4\] \[\frac{22< 13\sqrt{3}}{22^{2}< 13^{2}\times 3\] \[484< 507.\] as required. 4. Problem: Let \(\Gamma_{1}\) and \(\Gamma_{2}\) be circles internally tangent at point \(A\) , with \(\Gamma_{1}\) inside \(\Gamma_{2}\) . Let \(BC\) be a chord of \(\Gamma_{2}\) which is tangent to \(\Gamma_{1}\) at point \(D\) . Prove that line \(AD\) is the angle bisector of \(\angle BAC\) . Solution: Let \(\lambda\) be the common tangent of \(\Gamma_{1}\) and \(\Gamma_{2}\) at point \(A\) . Let \(P\) be a point on \(\lambda\) such that \(P\) and \(C\) are on opposite sides of line \(AB\) . Let \(Q\) and \(R\) be the points of intersection of \(\Gamma_{1}\) with \(AB\) and \(AC\) respectively. ![md5:ec506ddeedd78c0f7630939e755f3f41](ec506ddeedd78c0f7630939e755f3f41.jpeg) \[\angle BCA = \angle BAP\] \[\qquad = \angle QAP\] \[\qquad = \angle QRA\] Therefore lines \(BC\) and \(QR\) are parallel. Now consider \(\angle BAD\) . \[\angle BAD = \angle QAD\] \[\qquad = \angle QRD\] \[\qquad = \angle CDR\] \[\qquad = \angle DAR\] \[\qquad = \angle DAC.\] Since \(\angle BAD = \angle DAC\) , we are done. ## Alternative Solution: Consider the dilation centered at \(A\) which sends \(\Gamma_{1}\) to \(\Gamma_{2}\) . This dilation sends point \(D\) to the point \(D^{\prime} \in \Gamma_{2}\) such that points \(A\) , \(D\) and \(D^{\prime}\) are colinear. This dilation also sends line \(BC\) to the line tangent to \(\Gamma_{2}\) at point \(D^{\prime}\) . Therefore the tangent to \(\Gamma_{2}\) at point \(D^{\prime}\) is parallel to chord \(BC\) . Hence \(D^{\prime}\) is the midpoint of arc \(BC\) . I.e. \(BD^{\prime}\) and \(D^{\prime}C\) have the same arc- length. Since equal arcs subtend equal angles, we deduce that \(\angle BAD^{\prime} = \angle D^{\prime}AC\) as required. 5. Problem: A sequence of \(A s\) and \(B s\) is called antipalindromic if writing it backwards, then turning all the \(A s\) into \(B s\) and vice versa, produces the original sequence. For example \(A B B A A B\) is antipalindromic. For any sequence of \(A s\) and \(B s\) we define the cost of the sequence to be the product of the positions of the \(A s\) . For example, the string \(A B B A A B\) has cost \(1\cdot 4\cdot 5 = 20\) . Find the sum of the costs of all antipalindromic sequences of length 2020. Solution: For each integer \(0\leq k\leq 1009\) define a \(k\) - pal to be any sequence of 2020 \(A s\) and \(B s\) , where the first \(k\) terms are \(B\) , the last \(k\) terms are \(B\) , and the middle \((2020 - 2k)\) terms form an antipalindromic sequence. Now for any \(k\) , define \(f(k)\) to be sum of the costs of all \(k\) - pals. Note that any \(k\) - pal can be created from a \((k + 1)\) - pal by either - (A) replacing the \(B\) in position \((k + 1)\) with an \(A\) , or - (B) replacing the \(B\) in position \((2021 - k)\) with an \(A\) . Therefore the sum of the costs of all \(k\) - pals formed using operation (A) is \((k + 1)\times f(k + 1)\) . Similarly the sum of the costs of all \(k\) - pals formed using operation (B) is \((2021 - k)\times f(k + 1)\) . Hence \[f(k) = (k + 1)f(k + 1) + (2020 - k)f(k + 1) = ((k + 1) + (2020 - k))f(k + 1) = 2021f(k + 1).\] Now we note that there are two different 1009- pals, with costs equal to 1010 and 1011 respectively. So \[f(1009) = 1010 + 1011 = 2021.\] Now if we use the formula \(f(k) = 2021f(k + 1)\) iteratively, we get \(f(1010 - i) = 2021^{i}\) for each \(i = 1,2,3,\ldots\) . Therefore \[f(0) = 2021^{1010}\] which is our final answer. ## Alternative Solution: Let \(n\) be a positive integer. We will find an expression (in terms of \(n\) ) for the sum of the costs of all antipalindromes of length \(2n\) . Note that a string of \(A s\) and \(B s\) of length \(2n\) is an antipalindrome if and only if for each \(i\) , exactly one of the \(i^{\mathrm{th}}\) and \((2n + 1 - i)^{\mathrm{th}}\) letters is an \(A\) (and the other is a \(B\) ). Let \(x_{1},x_{2},\ldots ,x_{2n}\) be variables. For any \(1\leq a(1)< a(2)< \dots < a(k)\leq 2n\) , consider the string of \(A s\) and \(B s\) of length \(2n\) , such that the \(a(j)^{\mathrm{th}}\) letter is \(A\) for all \(j\) (and all the other letters are \(B\) ). Let this string correspond to the term \(t = x_{a(1)}x_{a(2)}x_{a(3)}\dots x_{a(k)}\) . If \(x_{i} = i\) for all \(i\) then the value of \(t\) is equal to the cost of it's corresponding string. Now consider the expression \[y = (x_{1} + x_{2n})(x_{2} + x_{2n - 1})\cdot \cdot \cdot (x_{n} + x_{n + 1}) = \prod_{j = 1}^{n}(x_{j} + x_{2n + 1 - j}).\] If we expand the brackets then we get \(2^{n}\) terms, each in the form \(t = x_{a(1)}x_{a(2)}x_{a(3)}\dots x_{a(n)}\) such that for each \(j = 1,2,\ldots ,n\) either \(a(j) = j\) or \(a(j) = 2n + 1 - j\) . Therefore \(y\) is the sum of all terms that correspond to antipalindromes. Hence if we substitute \(x_{i} = i\) for all \(i\) , then the value of \(y\) would be the sum of the costs of all antipalindromes. So the final answer is: \[\prod_{j = 1}^{n}(j + (2n + 1 - j)) = \prod_{j = 1}^{n}(2n + 1) = (2n + 1)^{n}.\] ## Alternative Solution 2: Let \(n\) be a positive integer. We will find an expression (in terms of \(n\) ) for the sum of the costs of all antipalindromes of length \(2n\) . Let \(\mathcal{P}\) denote the set of all antipalindromes of length \(2n\) , and let \(P\) be an antipalindrome chosen uniformly from \(\mathcal{P}\) . Note that for each \(j = 1,2,\ldots ,n\) the \(j^{\mathrm{th}}\) and \((2n + 1 - j)^{\mathrm{th}}\) must be an \(A\) and a \(B\) in some order. Let \(X_{j}\) be the random variable defined by: \(\cdot X_{j} = j\) if the \(j^{\mathrm{th}}\) letter of \(P\) is an \(A\) and the \((2n + 1 - j)^{\mathrm{th}}\) letter is a \(B\) \(\cdot X_{j} = 2n + 1 - j\) if the \(j^{\mathrm{th}}\) letter of \(P\) is a \(B\) and the \((2n + 1 - j)^{\mathrm{th}}\) letter is an \(A\) Notice that the cost of \(P\) is given by the product \(X_{1}X_{2}X_{3}\cdot \cdot \cdot X_{n}\) . Now consider \(f_{j}:\mathcal{P}\to\) \(\mathcal{P}\) to be the function which swaps the \(j^{\mathrm{th}}\) and \((2n + 1 - j)^{\mathrm{th}}\) letters of the string. Notice that \(f_{j}\) is a bijection that toggles the value of \(X_{j}\) . This means that \(X_{j}\) is equal to \(j\) or \((2n + 1 - j)\) with equal probabilities. Therefore \[\mathbb{P}(X_{j} = j) = \mathbb{P}(X_{j} = 2n + 1 - j) = \frac{1}{2}.\] Furthermore \(f_{j}\) preserves the value of \(X_{i}\) for all \(i\neq j\) . Therefore the variables \(X_{i}\) and \(X_{j}\) are independent. Therefore the expected value of the cost of \(P\) is given by: \[\mathbb{E}[\mathrm{cost}(P)] = \mathbb{E}\left[\prod_{i = 1}^{n}X_{i}\right]\] \[\qquad = \prod_{i = 1}^{n}\mathbb{E}\left[X_{i}\right]\] \[\qquad = \prod_{i = 1}^{n}\left(\frac{1}{2} (i) + \frac{1}{2} (2n + 1 - i)\right)\] \[\qquad = \prod_{i = 1}^{n}\frac{2n + 1}{2}\] \[\qquad = \left(\frac{2n + 1}{2}\right)^{n}.\] Now the number of antipalindromes of length \(2n\) is simply \(2^{n}\) (one for each choice of the variables \(X_{j}\) ). Therefore the sum of the costs of all antipalindromes of length \(2n\) is simply \(2^{n}\) multiplied by the expected value of the cost of \(P\) . This is \[2^{n}\times \left(\frac{2n + 1}{2}\right)^{n} = (2n + 1)^{n}.\]