# New Zealand Mathematical Olympiad Committee # 2019 NZMO Round 2 # Solutions 1. A positive integer is called sparkly if it has exactly 9 digits, and for any \(n\) between 1 and 9 (inclusive), the \(n^{\text{th}}\) digit is a positive multiple of \(n\) . How many positive integers are sparkly? Solution: For each \(n = 1,2,\ldots ,9\) there are \(\lfloor 9 / n\rfloor\) different possibilities for the \(n^{\mathrm{th}}\) digit. For example there are \(\lfloor 9 / 2\rfloor = 4\) possible choices for the second digit (these being 2, 4, 6 and 8). Therefore the answer is \[\left|\frac{9}{1}\right|\times \left|\frac{9}{2}\right|\times \left|\frac{9}{3}\right|\times \left|\frac{9}{4}\right|\times \left|\frac{9}{5}\right|\times \left|\frac{9}{6}\right|\times \left|\frac{9}{7}\right|\times \left|\frac{9}{8}\right|\times \left|\frac{9}{9}\right|\] which equals \(9\times 4\times 3\times 2\times 1\times 1\times 1\times 1\times 1 = 216\) 2. Let \(X\) be the intersection of the diagonals \(AC\) and \(BD\) of convex quadrilateral \(ABCD\) . Let \(P\) be the intersection of lines \(AB\) and \(CD\) , and let \(Q\) be the intersection of lines \(PX\) and \(AD\) . Suppose that \(\angle ABX = \angle XCD = 90^{\circ}\) . Prove that \(QP\) is the angle bisector of \(\angle BQC\) . Solution: First note that quadrilateral \(ABCD\) is cyclic because \(\angle ABD = \angle ACD = 90^{\circ}\) . Also, since \(AP \perp DX\) and \(DP \perp AX\) , we see that \(X\) is the orthocentre of triangle \(APD\) . Hence \(PX \perp AD\) . Therefore quadrilaterals \(ABXQ\) and \(QXCD\) are cyclic (opposite angles are supplementary). Now we perform a simple angle chase \[\angle XQB = \angle XAB = \angle CAB = \angle CDB = \angle CDX = \angle CQX.\] Since \(\angle XQB = \angle CQX\) , it follows that \(QX\) is the angle bisector of \(\angle CQB\) as required. 3. Let \(a\) , \(b\) and \(c\) be positive real numbers such that \(a + b + c = 3\) . Prove that \[a^{a} + b^{b} + c^{c}\geq 3.\] Solution: We start with a general fact about any positive real number \(x\) . There are two cases: either \(x \geq 1\) or \(x < 1\) . - If \(x \geq 1\) then \(x^{p} \geq x^{q}\) for any \(p > q\) . Substituting \(p = x\) and \(q = 1\) gives us \(x^{x} \geq x^{1} = x\) . - If \(x < 1\) then \(x^{p} < x^{q}\) for any \(p > q\) . Substituting \(p = 1\) and \(q = x\) gives us \(x = x^{1} < x^{x}\) . In either case we get \(x^{x} \geq x\) for all positive real numbers \(x\) . Applying this fact for \(a\) , \(b\) and \(c\) gives us \[a^{a} + b^{b} + c^{c} \geq a + b + c = 3\] as required. 4. Show that for all positive integers \(k\) , there exists a positive integer \(n\) such that \(n2^{k} - 7\) is a perfect square. Solution: Proof by induction on \(k\) . For the base cases ( \(k \leq 3\) ) we can simply choose \(n = 2^{3 - k}\) to get \(n2^{k} - 7 = 2^{3} - 7 = 1^{2}\) . For the inductive step let \(k \geq 3\) and assume there exist integers \(a\) and \(n\) such that \[a^{2} = n2^{k} - 7.\] We will now endeavour to find integers \(b\) and \(m\) such that \(b^{2} = m2^{k + 1} - 7\) . To do this we have two cases: - If \(n\) is even then choose \(b = a\) and \(m = n / 2\) . Thus \(b^{2} = (2m)2^{k} - 7 = m2^{k + 1} - 7\) , as required. - If \(n\) is odd, then note that \(a\) must also be odd. Let \(n = 2x + 1\) and let \(a = 2y + 1\) . Now consider \((a + 2^{k - 1})^{2}\) . \[(a + 2^{k - 1})^{2} = a^{2} + 2^{k}a + 2^{2k - 2}\] \[\qquad = \left(n2^{k} - 7\right) + 2^{k}a + 2^{2k - 2}\] \[\qquad = {\bigl (}(2x + 1)2^{k} - 7{\bigr)} + 2^{k}\left(2y + 1\right) + 2^{2k - 2}\] \[\qquad = \left(x + y + 1 + 2^{k - 3}\right)2^{k + 1} - 7.\] So in this case we can simply choose \(b = a + 2^{k - 1}\) and \(m = x + y + 1 + 2^{k - 3}\) . Note here that this inductive step only works when \(k \geq 3\) (otherwise \(m = x + y + 1 + 2^{k - 3}\) is not an integer). 5. An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size. Solution: For the purpose of this proof, we will consider a vertex to be any point which is a corner of at least one of the triangular pieces. Define an edge to be any line segment between two vertices, which is part of a side of a triangular piece but does not pass through any other vertex. Note that each vertex must be one of the following types: - Type \(A\) : incident with only 2 edges at \(60^{\circ}\) .- Type \(B\) : incident with exactly 4 edges at angles \(60^{\circ}, 60^{\circ}, 60^{\circ}, 180^{\circ}\) .- Type \(C\) : incident with exactly 6 edges forming six \(60^{\circ}\) angles. There can be no other types of vertex, because all the angles must be \(60^{\circ}\) or \(180^{\circ}\) (or \(300^{\circ}\) in the corners of the original large triangle). We will now colour each edge- end either green or blue, as shown in the diagram (the green edge- ends are also a bit thicker). ![md5:5bbd2bd546501f0b6ffceaf0aa5606a4](5bbd2bd546501f0b6ffceaf0aa5606a4.jpeg) ![md5:018509b36f48f8170b65bdbc076f8f1c](018509b36f48f8170b65bdbc076f8f1c.jpeg) ![md5:f62fe60f7215cc7178e7522106b9480f](f62fe60f7215cc7178e7522106b9480f.jpeg) An edge- end is coloured green if it touches a Type \(B\) vertex along the \(180^{\circ}\) angle, otherwise it is coloured blue. Now observe that there must be exactly 3 vertices of type \(A\) (the three corners of the original large triangle before it was partitioned). If there are \(x\) vertices of type \(B\) and \(y\) vertices of type \(C\) then this makes a total of \((6 + 2x + 6y)\) blue ends, but only \((2x)\) green ends. Note also that the 6 edges with an endpoint at a type \(A\) vertex must all have their other end being green. There are more blue edge- ends than green edge- ends, so it can't be the case that every blue edge- end is connected to a green edge- end. There must therefore be an edge such that both of its ends are blue. Since neither endpoint of such an edge is a type \(A\) vertex, we can conclude that all four of the angles at the endpoints of this double blue- ended edge must be \(60^{\circ}\) . ![md5:d48c633d9f0fd9e595a6cc1b371d7452](d48c633d9f0fd9e595a6cc1b371d7452.jpeg) Therefore this edge is a shared side of two triangular pieces. These two triangular pieces must therefore be the same size. \(\square\) September 2019 www.mathsolympiad.org.nz