NINETEENTH IRISH MATHEMATICAL OLYMPIAD. Saturday, 6 May a.m. 1 p.m. First Paper. 1. Are there integers x, y, and z which satisfy the equation

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1 NINETEENTH IRISH MATHEMATICAL OLYMPIAD Saturday, 6 May a.m. 1 p.m. First Paper 1. Are there integers x, y, and z which satisfy the equation when (a) n = 2006 (b) n = 2007? z 2 = (x 2 + 1)(y 2 1) + n 2. P and Q are points on the equal sides AB and AC respectively of an isosceles triangle ABC such that AP = CQ. Moreover, neither P nor Q is a vertex of ABC. Prove that the circumcircle of the triangle AP Q passes through the circumcentre of the triangle ABC. 3. Prove that a square of side 2.1 units can be completely covered by seven squares of side 1 unit. 4. Find the greatest value and the least value of x + y, where x and y are real numbers, with x 2, y 3, and x 2 x + 2 = 2 y + 3 y. 5. Determine, with proof, all functions f : R R such that f(1) = 1, and f(xy + f(x)) = xf(y) + f(x) for all x, y R. Notation: R denotes the set of real numbers.

2 Solutions and Marking Schemes 1. Proposed by Richard Watson. Solution: (a) When n = 2006 the equation may be rewritten as x 2 y 2 + z 2 x 2 y 2 = Consider x 2 y 2 = Both sides factor to give (x + y)(x y) = 5.401, so one possibility is x + y = 401, x y = 5. Solving these simultaneous equations yields x = 203, y = 198, and if we set z = xy = , then we obtain integers which satisfy the given equation. (b) Suppose that there are integers x, y, z which satisfy z 2 = (x 2 + 1)(y 2 1) It is straightforward to check that z 2 0, 1 or 4 (mod 8), while the right hand side is congruent to 2, 5, 6 or 7 (mod 8). This is a contradiction, so no such integers exist. Marking scheme: 5 marks for each part with 1 mark in each part for a reasonable but unsuccessful attempt to use modular arithmetic to show the nonexistence of a solution. 2. Proposed by Jim Leahy. Solution: Draw the perpendicular bisectors of P Q and AC and let them meet at O. OP = OQ, OA = OC and AP = CQ. Thus AP O CQO. As then CQO = AP O, AP OQ is a cyclic quadrilateral. Also OCQ = OAP = OQP = OP Q = OAQ Hence OA is the angle bisector of BAC which is the perpendicular bisector of BC as ABC is isosceles. O lies on the perpendicular bisector of AC and BC and is thus the circumcentre of ABC. Marking scheme: There are other ways to go about this problem. Complete solution: 10 marks. Partial solution: Finding congruent triangles. (3 marks). Finding equal angles in AP OQ or other properties that indicate that AP OQ is cyclic.(3 marks). A partial solution should earn no more than 6 marks. An alternate solution. Denote by O the circumcentre of ABC. Since AO = BO = CO, the triangles AOB, BOC, COA are isosceles. Since ABC is also isosceles, BAO = OBA = OCA. Hence AP O, OCQ are similar. Therefore, OP A = OQC = π OQA. In other words, AP OQ is cyclic, i.e., O lies on the circumcircle of the triangle AP Q.

3 3. Proposed by Gordon Lessells. Solution and marking scheme: A complete solution (10 marks) must indicate clearly the positions and orientations of the seven squares used and a proof that all the square of side 2.1 is covered. Inverting the problem to one of arranging 7 squares of side 1 to include a square in the interior is worth 3 marks. An arrangement which works but with no calculations should earn at most 5 marks. The two most obvious solutions are A: Squares with sides parallel to the axes with centres at (0.5, 0.5), (0.5, 1.5), (0.5, 2.5), (1.5, 1), (1.5, 2), ( 0.5, 1), ( 0.5, 2). If O = (0, 0), P = (1.5, 1.5), Q = (0, 3), R = ( 1.5, 1.5),then OP QR is a square of side 3 2 > B: Arrange 5 squares with sides parallel to the axes with centres at (0.5, 0.5), (1.5, 0.5), (0.5, 1.5), (1.5, 1.5), (1.6, 1.6) This covers all of the square of side 2.1 apart from two rectangles of dimension Each of these can be covered by a square of side 1 at a forty five degree angle. This needs to be checked by means of Pythagoras Theorem. 4. Proposed by Gordon Lessells. Solution: Rewriting the equation we obtain x + y = 2 x y + 3 Thus there is a solution of the system of equations x + y = t 2 x y + 3 = t We need to find the values of t for which this system of equations has a solution. Let u = x + 2 and v = y + 3. Then x = u 2 2 and y = v 2 3. Hence u 2 + v 2 = 5 + x + y = 5 + t. Now 2uv = (u + v) 2 (u 2 + v 2 ) = (t/2) 2 (t + 5). u and v are roots of the quadratic z 2 (t/2)z + t2 8 t = 0 Noting that these roots are real and positive we obtain the following three conditions. t/2 0 (1) t 2 8 t (2) (t/2) 2 4( t2 8 t ) 0 (3)

4 (2) is equivalent to (t 2) (3) is equivalent to (t 4) Combining the three conditions we see that t For the range of t values given our original system of equations has a solution. Thus max value of x+y is and the minimum value of x+y is Marking scheme: Complete solution 10 marks. Complete solution to either the max or min 5 marks. Partial solutions: a) Any attempt in which completion of the square is used or b) Any attempt in which square roots are cleared in a sensible fashion or c) Any attempt in which there is an attempt to find the range of values of, say, x 2 x + 2 or 2 y + 3 y or d) Any attempt in which change of variable is used to get rid of the square roots 3 marks. Discovery of values of x and y which attain the max and min without proof 2 marks each. An alternate solution. Let u = x + 2, v = y + 3. Then u, v 0, and satisfy the equation u 2 + v 2 2u 2v 5 = 0, (u 1) 2 + (v 1) 2 = 7. Thus the point P (u, v) lies on that part of the circle centred at (1, 1) of radius 7 that is contained in the first quadrant of the (u, v)-plane. Also, x + y = u 2 +v 2 5, and u 2 + v 2 is the distance from P to the origin. Hence, effectively, the problem is to find the points in the first quadrant on the circle that are nearest to and furthest from the origin. The nearest points are when uv = 0, giving (0, 1 + 6) and (1 + 6, 0). The furthest away point is given by u = v = (2 + 14)/2. Hence min x+y = (1+ 6) 2 5 = 2+2 6, max x+y = 2( ) 2 5 = Proposed by Prithwijit De and Finbarr Holland. Solution and marking scheme: We begin by showing that f(0) = 0. To see this, let y = 0. Then f(f(x)) = xf(0) + f(x), x R. In particular, f(f(0)) = f(0). Now set y = 0, x = f(0) and deduce that f(0) = f(f(0)) = f(f(0).0 + f(f(0))) = (f(0)) 2 + f(f(0)) = (f(0)) 2 + f(0), whence f(0) = 0. [2 marks for showing this.]

5 One consequence of this is seen by setting y = 0, namely, f(f(x)) = f(x), x R. (4) [1 marks for deducing this.] Setting x = 1 in the equation we get that [1 mark for noting this.] f(y + 1) = f(y) + 1, y R. (5) Replacing y by f(x) in (5) and using (4), we infer that f(1 + f(x)) = f(f(x) + 1) = f(x) + 1, x R. (6) [2 marks for establishing this.] Returning to the functional equation, let x 0 and put y = 1/x. Then, using (6), 1 + f(x) = f(1 + f(x)) = f(xy + f(x) = xf(y) + f(x), i.e., 1 = xf(y) = xf( 1 x ), f( 1 x ) = 1 x. Thus f(x) = x, x 0. But f(0) = 0. Hence f(x) = x, x. [4 marks for completing the argument.]

1. Suppose that a, b, c and d are four different integers. Explain why. (a b)(a c)(a d)(b c)(b d)(c d) a 2 + ab b = 2018.

New Zealand Mathematical Olympiad Committee Camp Selection Problems 2018 Solutions Due: 28th September 2018 1. Suppose that a, b, c and d are four different integers. Explain why must be a multiple of