Solutions and Marking Schemes

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1 TWENTY NINTH IRISH MTHEMTICL OLYMPID 1. Proposed by Gordon Lessells. Saturday, 23 pril 2016 First Paper s and Marking Schemes If the three-digit number C is divisible by 27, prove that the three-digit numbers C and C are also divisible by 27. If 27 divides C then 3 divides + +C. C C = C = 9(11 10 C) = 9( C 21 12C) = 99(+ +C) 27(7 +4C) Therefore, if 27 divides C then 27 divides C. Now, applying the result just proved to C, we deduce that 27 also divides C. Full solution: 10 marks. by enumeration of all multiples of 27 less than 1000 earns 10 marks provided these 37 numbers are grouped into 12 sets of three together with the number 999. Partial solution: 1 mark for observing that for multiples of three, the sum of the digits is divisible by 3. 2 marks for observing that for multiples of 9, the sum of the digits is divisible by 9. These 2 marks are not additive with the previous mark. 2 marks for obtaining an expression similar to C C = C, and a further 1 mark for taking out the factor 9 from this expression. 1

2 2. Proposed by Jim Leahy. In triangle C we have C. The bisectors of C and C meet C and at E and F, respectively, and intersect at I. If EI = FI find the measure of C. Let α = C, β = C and γ = C, then α+β +γ = 180. Step 1: Let F be the reflection of F in the line I. Step 2: ecause I is the angle bisector of C, F is on the line C. Moreover, because F I = FI and F = F, the two triangles FI and F I are congruent. Note that F I = EI by assumption. Step 3: We first show that F E. For a proof by contradiction we assume that F = E. Then FI and EI are congruent, hence FI = EI. These two angles are external angles of the triangles FC and EC, respectively. Therefore, we obtain β + 1γ = γ + 1 β, that is β = γ. This shows that 2 2 F = E can only happen in a triangle with = C, which was excluded. C C E F I F E I F F Step 4: One of the two angles F I and EI is a base angle of the isosceles triangle EIF and the other of these two angles is an external angle at the base. Therefore, the sum of these two angles is equal to 180. Step 5: On the other hand, F I = FI is equal to β + 1γ (external 2 angle of triangle FC) and EI = γ+ 1 β (external angle of triangle EC). 2 Therefore, 180 = 3(β +γ), and so β +γ = This implies that C = α = = mark for constructing F (Step 1). 1 mark for showing that triangles FI and F I are congruent (Step 2). 3 marks for showing that F E (Step 3). 2 marks for showing that F I + EI = 180 (Step 4). 3 marks for completing the proof (Step 5). 2

3 3. Proposed by Mark Flanagan. DothereexistfourpolynomialsP 1 (x),p 2 (x),p 3 (x),p 4 (x)withrealcoefficients, such that the sum of any three of them always has a real root, but the sum of any two of them has no real root? There do not exist four such polynomials. We show this as follows. Suppose that there do exist four such polynomials. If a polynomial has no real roots, it is either positive for all real x, or else it is negative for all real x. Consider the complete graph with the four polynomials as vertices. Colour the edgep i P j whiteifp i (x)+p j (x) > 0forallrealx, andblackifp i (x)+p j (x) < 0 for all real x. We cannot have a triangle P i P j P k of the same colour (if we did, we could deduce that P i (x) + P j (x) + P k (x) has a constant sign for all real x, and therefore this sum of three polynomials would not have a real root). y the Pigeonhole Principle, at least three of the edges must have the same colour let s say this colour is black. If they have a common vertex, then in order to avoid forming a black triangle, the other three edges would need to be white, thus forming a white triangle a contradiction. So, without loss of generality, we can consider the case where P 1 P 2, P 2 P 3 and P 3 P 4 are black. Then P 1 P 3 and P 2 P 4 must be white and therefore we have (P 1 (x)+p 3 (x))+ (P 2 (x) + P 4 (x)) > 0 for all real x. ut since P 1 P 2 and P 3 P 4 must be black, we have (P 1 (x)+p 2 (x))+(p 3 (x)+p 4 (x)) < 0 for all real x. The case where P 1 P 2, P 2 P 3 and P 3 P 4 are white yields a similar contradiction. We conclude that there do not exist four such polynomials. (a) 2 marks for proposing a representation via a graph, with vertices representing polynomials, and edges indicating a positive or negative polynomial sum. (b) 2 marks for proving that we cannot have a triangle of the same colour. (c) 1 mark for proving that there must be three edges of the same colour. (d) 2 marks for proving that these three edges do not share a common vertex. (e) 3 marks for finishing the proof. ny partial solution which is presented without the graph theory framework and terminology should be marked in such a way that the same marks are allocated to the achievement of the same milestones. The initial 2 marks allocated to (a) can be awarded as: an additional mark once (c) is achieved; and another additional mark once (d) is achieved. 3

4 4. Proposed by ernd Kreussler. Let C be a triangle with C C. Let P and Q be the intersection points of the line with the internal and external angle bisectors at C, so that P is between and. Prove that if M is any point on the circle with diameter PQ, then MP = MP. The internal and external angle bisectors divide the segment internally and externally in the same ratio C : C. This can be seen, for example, with the aid of the Sine Rule, applied to the triangles CP and CP for the internal bisector and to the triangles CQ and CQ for the external angle bisector. In particular, we obtain P P = Q Q. C P Q Let X and Y be points on the line M be such that X MQ and Y MP. Y M X P Q ecause X MQ, we have M / XM = Q / Q and because Y MP, we have M / YM = P / P. This implies M / XM = M / YM, hence XM = YM. 4

5 ecause M is on the circle with diameter PQ, PMQ = 90 and so also YX = 90. Therefore, is on the circle with diameter XY. s M was shown to be the midpoint of XY, we obtain M = XM = YM. This implies now P P = M M. If P is the intersection point of the angle bisector of M and, then P P = M M = P P. This implies that P = P, and so MP is the angle bisector of M. Remark. The circle with diameter PQ is the circle of pollonius which consists of all points M for which the ratio M : M of the distances to the two points and is a given constant. In our case, this constant is equal to the ratio C : C. The two pairs of points, and P,Q form a so-called harmonic range. 8 marks for proving the equality P P = M M. If this is not achieved, partial marks could be awarded as follows: 2 marks for finding the equality P = Q P Q. No mark should be given if only P P = C C is mentioned. More marks can be awarded for some progress towards the equality P = M. P M The remaining 2 marks for completing the solution. s that differ from the one presented here should be marked accordingly. 5

6 5. Proposed by Gordon Lessells. Let a 1,a 2,...,a m be positive integers, none of which is equal to 10, such that a 1 +a 2 + +a m = 10m. Prove that (a 1 a 2 a 3 a m ) 1/m ssume m = 2k and that a 1 a 2 a 3... a m. Let 1 = a 1 +a 2 + +a k k, 2 = a k+1 +a k+2 + +a 2k k. Clearly 1 2 and = 10. Hence 1 = 10 δ and 2 = 10 + δ for some δ 0. If δ < 1 then 1 > 9 and 2 < 11. The first of these inequalities implies that a k > 9. This in turn implies that a k 11, which means that a k+1,a k+2,...,a 2k 11. ut then we have 2 11, which is a contradiction. Therefore we must have δ 1. Nowapply them-gminequality separately tothenumbersa 1,a 2,...,a k and to the numbers a k+1,a k+2,...,a 2k. We obtain since δ 1. m a i ( 1 2 ) k = (100 δ 2 ) k 99 k = 9 k 11 k i=1 Taking mth roots completes the proof for the case m = 2k. To deal with the case where m = 2k+1, let a m+1 = a 1, a m+2 = a 2,..., a 2m = a m. pplying the result just proved to the numbers a 1,a 2,...,a 2m yields the result in this case. 10 marks for complete solution. 7 marks for proof of the case where m is even. Partial solutions: 1 mark for proving the case m = 2. 1 mark for indication of knowledge of the M-GM inequality. 1 mark for putting the numbers a 1,a 2,...,a m in order. Up to 3 marks for splitting the set into two, and making observations about both subsets. 6