# NZMO Round Two 2023 — Solutions 1. Problem: For any positive integer \(n\) let \(n! = 1 \times 2 \times 3 \times \dots \times n\) . Do there exist infinitely many triples \((p, q, r)\) , of positive integers with \(p > q > r > 1\) such that the product \[p!\cdot q!\cdot r!\] is a perfect square? Solution A: (James Xu) Yes. Let \(t\) be an arbitrary positive integer and consider the following perfect square: \[(t!)^{2} = (t!)\cdot (t! - 1)!\cdot t!.\] So if we consider \((p,q,r) = (t!,t! - 1,t)\) then \(p!q!r! = (t!)^{2}\) which is a perfect square. Since the choice of \(t\) is arbitrary, there must be infinitely many such triples. Solution B: (Josie Smith) Yes. Consider the substitution \((p,q,r) = (6t^{2},6t^{2} - 1,3)\) \[p!\cdot q!\cdot r! = (6t^{2})!\cdot (6t^{2} - 1)!\cdot 3! = ((6t^{2} - 1)!\times 6t)^{2}\] which is a perfect square. Since the choice of \(t\) is arbitrary, there must be infinitely many such triples. Solution C: (Michael Albert) This solution shows a stronger result. For any positive integer \(k\) , there exist infinitely many \(q\) such that \((q + 1)!\cdot q!\cdot k\) is a perfect square. This can be seen by setting \(q + 1 = kx^{2}\) for any \(x\) . This of course implies the required result. Comment: There are many more triples \((p,q,r)\) that work. For example \((p,q,r) = (10,7,6)\) . But this does not matter because the problem only asked us to determine whether or not there are infinitely many triples that work. 2. Problem: Let \(a\) , \(b\) and \(c\) be positive real numbers such that \(a + b + c = abc\) . Prove that at least one of \(a\) , \(b\) or \(c\) is greater than \(\frac{17}{10}\) . Solution A: (Ross Atkins) Wlog assume \(a\geq b\geq c\) . Therefore \(a + b + c\geq 3c\) . Now for a proof by contradiction, assume \(a\leq \frac{17}{10}\) and \(b\leq \frac{17}{10}\) . Since \(\left(\frac{17}{10}\right)^2 = \frac{289}{100} < 3\) , it follows that \(\frac{17}{10} < \sqrt{3}\) and therefore \(ab\leq 3\) . However this implies: \[3c\leq a + b + c = abc< 3c\] which is a contradiction. Solution B: (Michael Albert) From \(abc = a + b + c\) we get \[1 = \frac{a + b + c}{abc} = \frac{1}{bc} +\frac{1}{ca} +\frac{1}{ab}\] As all three terms are positive, at least one must be less than or equal to \(1 / 3\) and, without loss of generality, we can assume \(1 / bc\leq 1 / 3\) . But then \(bc\geq 3\) and at least one of \(b\) or \(c\) must be greater than \(\sqrt{3}\) . Since \(\sqrt{3} >17 / 10\) we're done. Solution C: (Ross Atkins) By the AM- GM inequality we have \(\frac{a + b + c}{3} \geq \sqrt[3]{abc}\) and thus \(abc \geq 3\sqrt[3]{abc}\) , which rearranges to give \(\sqrt[3]{abc} \geq \sqrt{3}\) . Now wlog \(a \geq b \geq c\) and so \[a = \sqrt[3]{a^3} \geq \sqrt[3]{abc} \geq \sqrt{3} > \frac{17}{10}.\] 3. Problem: Let \(ABCD\) be a square (vertices labelled in clockwise order). Let \(Z\) be any point on diagonal \(AC\) between \(A\) and \(C\) such that \(AZ > ZC\) . Points \(X\) and \(Y\) exist such that \(AXYZ\) is a square (vertices labelled in clockwise order) and point \(B\) lies inside \(AXYZ\) . Let \(M\) be the point of intersection of lines \(BX\) and \(DZ\) (extended if necessary). Prove that \(C\) , \(M\) and \(Y\) are collinear. Solution: (Kevin Shen) Since \(Z\) lies on diagonal \(AC\) , we have \(\angle DAZ = 45^{\circ}\) and \(\angle ZAB = 45^{\circ}\) . Therefore \(B\) lies on diagonal \(AY\) of square \(AXYZ\) and \(\angle BAX = 45^{\circ}\) . ![md5:4202f96cc9b60e6fbaca4c97d9e55390](4202f96cc9b60e6fbaca4c97d9e55390.jpeg) Since \(AB = AD\) and \(AX = AZ\) and \(\angle BAX = 45^{\circ} = \angle DAZ\) , we have congruent triangles \[\triangle DAZ\equiv \triangle BAX\qquad (SAS).\] Therefore let \(x = \angle ZDA = \angle XBA\) and \(y = \angle AZD = \angle AXB\) . By the angle sum in triangle \(DAZ\) we have \(\angle DAZ + \angle AZD + \angle ZDA = 180^{\circ}\) . Therefore \(x + y = 135^{\circ}\) . Now by the angle sum in quadrilateral \(DAXM\) we get \(\angle DAX + \angle AXM + \angle XMD + \angle MDA = 360^{\circ}\) . Therefore \[\angle BMD = 90^{\circ}.\] Hence \(ABMCD\) is cyclic (the circle with diameter \(BD\) ). Therefore \[\angle DMC = \angle DAC = 45^{\circ}.\] Also \(AXYMZ\) is cyclic (the circle with diameter \(XZ\) ). Therefore \[\angle YMX = \angle YAX = 45^{\circ}.\] Hence \(\angle YMC = \angle YMX + \angle YMD + \angle DMC = 45^{\circ} + 90^{\circ} + 45^{\circ} = 180^{\circ}\) as required. 4. Problem: For any positive integer \(n\) , let \(f(n)\) be the number of subsets of \(\{1,2,\ldots ,n\}\) whose sum is equal to \(n\) . Does there exist infinitely many positive integers \(m\) such that \(f(m) = f(m + 1)\) ? (Note that each element in a subset must be distinct.) Solution: (Michael Albert) Let \(S(n)\) be the set of such subsets. Consider the map from \(S(n)\) to \(S(n + 1)\) that adds one to the largest element of each \(A \in S(n)\) . This map is an injection (needs proof but easy) and not a surjection provided that \(S(n + 1)\) contains a set whose largest and second largest elements differ by one. For even \(n = 2k \geqslant 2\) this is true since we can take \(\{k, k + 1\} \in S(n + 1)\) and for odd \(n = 2k + 1 \geqslant 5\) this is true since we can take \(\{1, k, k + 1\}\) . So for \(n \geqslant 5\) , we must have \(f(n) < f(n + 1)\) and there do not exist infinitely many such pairs. 5. Problem: Let \(x\) , \(y\) and \(z\) be real numbers such that: \(x^{2} = y + 2\) , and \(y^{2} = z + 2\) , and \(z^{2} = x + 2\) . Prove that \(x + y + z\) is an integer. Solution A: (Ross Atkins) First we exclude \(- 1\) and 2: \(x = 2\) implies \(y = 2\) implies \(z = 2\) implies \(x = 2\) \(x = - 1\) implies \(y = - 1\) implies \(z = - 1\) implies \(x = - 1\) In both these cases we have \(x + y + z\) being an integer. So henceforth we assume none of \(x,y,z\) are 2 nor \(- 1\) . Now let \(x,y,z\) be the roots of the following cubic equation. \[(\lambda -x)(\lambda -y)(\lambda -z) = \lambda^{3} - A\lambda^{2} + B\lambda -C.\] Applying Viete's formula to this cubic gives us \(A = x + y + z\) , \(B = xy + yz + zx\) and \(C = xyz\) . This means that \(x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2(xy + yz + zx) = A^{2} - 2B\) . Now sum the three given equations ( \(x^{2} = y + 2\) and \(y^{2} = z + 2\) and \(z^{2} = x + 2\) ) to get \[A^{2} - 2B = x^{2} + y^{2} + z^{2} = (y + 2) + (z + 2) + (x + 2) = A + 6.\] \[A^{2} - A - 2B = 6. \quad (1)\] Next rearrange the equations to be \(x^{2} - 1 = y + 1\) and \(y^{2} - 1 = z + 1\) and \(z^{2} - 1 = x + 1\) . These can then be multiplied to get \[(x^{2} - 1)(y^{2} - 1)(z^{2} - 1) = (y + 1)(z + 1)(x + 1)\] \[(x - 1)(x + 1)(y - 1)(y + 1)(z - 1)(z + 1) = (y + 1)(z + 1)(x + 1)\] \[(x - 1)(y - 1)(z - 1) = 1\] \[xyz - (xy + yz + zx) + (x + y + z) - 1 = 1\] \[B - A + 2 = C \quad (2)\] In the above algebraic manipulation, we are allowed to cancel the \((x + 1)(y + 1)(z + 1)\) factor because none of \(x,y,z\) are equal to \(- 1\) . Finally rearrange the equations to be \(x^{2} - 4 = y - 2\) and \(y^{2} - 4 = z - 2\) and \(z^{2} - 4 = x - 2\) . These can then be multiplied to get \[(x^{2} - 4)(y^{2} - 4)(z^{2} - 4) = (y - 2)(z - 2)(x - 2)\] \[(x - 2)(x + 2)(y - 2)(y + 2)(z - 2)(z + 2) = (y - 2)(z - 2)(x - 2)\] \[(x + 2)(y + 2)(z + 2) = 1\] \[xyz + 2(xy + yz + zx) + 4(x + y + z) + 8 = 1\] \[C + 2B + 4A = -7\] \[C = -4A - 2B - 7 \quad (3)\] In the above algebraic manipulation, we are allowed to cancel the \((x - 2)(y - 2)(z - 2)\) factor because none of \(x,y,z\) are equal to 2. Combining equations (2) and (3) gives us \(B - A + 2 = C = - 4A - 2B - 7\) which rearranges to give us \(B = - A - 3\) . Substituting this into 1 gives us. \[A^{2} - A - 2B = 6\] \[A^{2} - A - 2(-A - 3) = 6\] \[A(A + 1) = 0.\] Therefore \(A = 0\) or \(A = - 1\) both of which are integers. Since \(A = x + y + z\) we are done. Solution B: (Ross Atkins) Consider the polynomial \(P\) defined by \[P(\lambda) = \lambda^{8} - 8\lambda^{6} + 20\lambda^{4} - 16\lambda^{2} - \lambda +2\] \[\qquad = (\lambda +1)(\lambda -2)(\lambda^{3} - 3\lambda +1)(\lambda^{3} + \lambda^{2} - 2\lambda -1).\] If we substitute \(z = y^{2} - 2\) into \(z^{2} = x + 2\) gives us \((y^{2} - 2)^{2} = x + 2\) . Then substitute \(y = x^{2} - 2\) to get \(\left((x^{2} - 2)^{2} - 2\right)^{2} = x + 2\) . Expanding gives \[x^{8} - 8x^{6} + 20x^{4} - 16x^{2} - x + 2 = 0.\] Therefore \(x\) is a root of the polynomial \(P\) . By symmetry we must have all of \(x,y,z\) being roots of \(P\) . Now we consider cases: Case 1: at least one of \(x,y,z\) is equal to \(- 1\) . Wlog assume \(x = - 1\) . Using \(y = x^{2} - 2\) we get \(y = - 1\) . Then using \(z = y^{2} - 2\) we get \(z = - 1\) . In this case we get \((x,y,z) = (- 1, - 1, - 1)\) which has sum \(- 3\) which is an integer. Case 2: at least one of \(x,y,z\) is equal to 2. Wlog assume \(x = 2\) . Using \(y = x^{2} - 2\) we get \(y = 2\) . Then using \(z = y^{2} - 2\) we get \(z = 2\) . In this case we get \((x,y,z) = (2,2,2)\) which has sum 6 which is an integer. Case 3: at least one of \(x,y,z\) is a root of \((\lambda^{3} - 3\lambda +1)\) . Wlog assume \(x^{3} - 3x + 1 = 0\) . Note that \(z = y^{2} - 2 = (x^{2} - 2)^{2} - 2 = x^{4} - 4x + 2\) . Now consider the sum of \(x\) and \(y = x^{2} - 2\) and \(z = x^{4} - 4x + 2\) , \[x + y + z = x + (x^{2} - 2) + (x^{4} - 4x^{2} + 2) = x^{4} - 3x^{2} + x = (x^{3} - 3x + 1)x.\] But since \(x^{3} - 3x + 1 = 0\) this means \(x + y + z = 0\) in this case. Case 4: some two of \(x,y,z\) are the same. Wlog assume \(x = y\) therefore \(x = y = x^{2} - 2\) . Hence \[0 = x^{2} - x - 2 = (x - 2)(x + 1).\] and so \(x = - 1\) or \(x = 2\) , and this was covered in cases 1 and 2. Case 5: Since \(x,y,z\) are all roots of \(P\) , the only remaining possibility is that \(x,y\) and \(z\) are distinct roots of \(\lambda^{3} + \lambda^{2} - 2\lambda - 1\) . By Viete's formula this means that the sum of the roots is \(- 1\) in this case. In all cases we conclude that \(x + y + z\) is an integer. Comment: Each of the eight roots of \(P\) lead to genuine solutions. For example the roots of \(\lambda^{3} - 3\lambda + 1\) are approximately \(0.3473\ldots , - 1.8794\ldots\) and \(1.5321\ldots\) , and so \[(x,y,z) = (0.3473\ldots , - 1.8794\ldots ,1.5321\ldots)\] is a valid solution. Similarly, the roots of \(\lambda^{3} + \lambda^{2} - 2\lambda - 1\) are approximately \(1.2470\ldots , - 0.44504\ldots\) and \(- 1.8019\ldots\) , and so \[(x,y,z) = (1.2470\ldots , - 0.44504\ldots , - 1.8019\ldots)\] is also a valid solution.