Paper 1

EIGHTEENTH IRISH MATHEMATICAL OLYMPIADSaturday, 7 May 200510 a.m.–1 p.m.First Paper1. Prove that 2005 2005 is a sum of two perfect squares, but not the sum of twoperfect cubes.2. Let ABC be a triangle and let D, E and F , respectively, be points on the sidesBC, CA and AB, respectively—none of which coincides with a vertex of thetriangle—such that AD, BE and CF meet at a point G. Suppose the trianglesAGE, CGD and BGF have equal area. Prove that G is the centroid of ABC.3. Prove that the sum of the lengths of the medians of a triangle is at least threequarters of the sum of the lengths of the sides.4. Determine the number of different arrangements a 1 , a 2 , . . . , a 10 of the integers1, 2, . . . , 10 such thata i > a 2i for 1 ≤ i ≤ 5,anda i > a 2i+1 for 1 ≤ i ≤ 4.5. Suppose a, b and c are non-negative real numbers. Prove that13 [(a−b)2 +(b−c) 2 +(c−a) 2 ] ≤ a 2 +b 2 +c 2 −3 3√ a 2 b 2 c 2 ≤ (a−b) 2 +(b−c) 2 +(c−a) 2 .OUTLINE SOLUTIONS1. Prove that 2005 2005 is a sum of two perfect squares, but not the sum of twoperfect cubes. [GL]Solution. 2005 = 5 × 401. 5 = 2 2 + 1 2 and 401 = 20 2 + 1 2 . Using the factthat the modulus of a product of two complex numbers is the product of theirmoduli, or the formula(a 2 + b 2 )(c 2 + d 2 ) = (ac − bd) 2 + (bc + ad) 2 ,

we find 2005 = 41 2 + 18 2 = 39 2 + 22 2 .Then2005 2005 = 2005 × 2005 2004 = (41 2 + 18 2 )(2005 1002 ) 2 = a 2 + b 2where a = 41 × 2005 1002 and b = 18 × 2005 1002 .Alternatively, we apply the formula repeatedly to express 2005 2005 as a sum oftwo squares.(Since the set of sums of squares {a 2 +b 2 : a, b ∈ N} is closed under multiplication(?), 2005 n belongs to this set for any natural number n. Also, as FiachraKnox observed, it is a standard fact (due to Euler (?)) that an integer belongsto this set iff if it is of the form xy 2 , where x, y ∈ N and x has no prime factorof the form 4k + 3. Thus every prime congruent to 1 modulo 4 belongs to theset, a fact that was proved/conjectured (?) by Fermat.)Cubes are 0, 1 or −1 modulo 7. Thus a sum of two cubes is 0, 1, 2, −1, −2modulo 7. 2005 = 3mod7. Hence 2005 2005 = 3 2005 mod7 = 3 6.334 × 3 = 3mod7.Hence 2005 2005 cannot be a sum of two cubes. A mod 13 argument will alsowork.2. Let ABC be a triangle and let D, E and F , respectively, be points on the sidesBC, CA and AB, respectively—none of which coincides with a vertex of thetriangle—such that AD, BE and CF meet at a point G. Suppose the trianglesAGE, CGD and BGF have equal area. Prove that G is the centroid of ABC.[JL]Solution. Let △P QR stand for the area of a triangle P QR. In this notation,leta = △AGE, x = △EGC, b = △CGD, y = △DGB, c = △BGF, z = △F GA.Theni.e.,or, inverting this,Hence△AGB△ADB = AGAD = △AGC△ADC ,z + cz + c + y = AGAD =yz + c =ba + x .a + xa + x + b ,ay + xy = bz + bc. (1)

AFGEBDCSimilarly,cx + zx = ay + ab, bz + yz = cx + ca. (2)Adding these equations we deduce thatCeva’s Theorem impliesab + bc + ca = xy + yz + zx.abcxyz= 1, abc = xyz (3)These facts are independent of the hypotheses. If we now bring these into play,we can assume that a = b = c = 1 in which case xyz = 1 and xy +yz +zx = 3.Thus the geometric mean and the arithmetic mean of xy, yz, zx are equal.Hence x = y = z = 1 and the result follows.Alternatively, by (1) and (2), z +1 = y(x+1), y +1 = x(z +1) and eliminatingx and y, we obtain z 4 + 3z 3 − z 2 − 4z + 1 = 0. Factorising the quartic weobtain (z − 1)(z 3 + 4z 2 + 3z − 1) = 0. Hence z = 1 or z 3 + 4z 2 + 3z − 1 = 0.

The cubic has a single real root between 0 and 1. x and y satisfy the sameequation. But x, y, z > 0 and xyz = 1. This is only possible if x = y = z = 1.D, E, F are then the mid points of BC, CA and CB. Hence G is the centroid.3. Prove that the sum of the lengths of the medians of a triangle is at least threequarters of the sum of the lengths of the sides. [TL]Solution. Let m a , m b , m c denote the lengths of the medians of a triangle ABCwith centroid G. Then m a = 3|AG|/2, m b = 3|BG|/2, m c = 3|CG|/2, andtheir sum is given bym a + m b + m c = 3 [|AG| + |BG| + |CG|]2= 3 ((|AG| + |BG|) + (|BG| + |CG|) + (|CG| + |AG|))4> 3 (|AB| + |BC| + |CA|),4by three applications of the triangle inequality.Remark.|AG| + |BG| + |CG| ≥min [|AP | + |BP | + |CP |] = |AF | + |BF | + |CF |,P ∈ABCwhere F is the Fermat-Torricelli point. Equality holds here iff the triangle isequilateral.Alternatively, denoting the vertices of the triangle by vectors a, b, c, then, bydefinition,Hencem a = | b + c2m a + m b + m c = 3 2− a|, m b = | c + a2− b|, m c = | a + b2− c|.a + b + c[|m − a| + |m − b| + |m − c|] (m = )3= 3 [(|m − a| + |m − b|) + (|m − b| + |m − c|)4+ (|m − c| + |m − a|)]> 3 [|a − b| + |b − c| + |c − a|].4Another approach would be to determine expressions for the medians in termsof the side lengths. Two applications of the Cosine Rule tell us that4m 2 a = 2(b 2 + c 2 ) − a 2 = b 2 + c 2 + 2bc cos A = |b + ce iA | 2 ,

with similar formulae for m b , m c . Thus4 ∑ m 2 a = 3 ∑ a 2 , 2 ∑ m a = ∑ √ 2(b 2 + c 2 ) − a 2 .While it’s a short step to see from the first identity that∑ma > 1 (a + b + c),2this appears to be the best that one can deduce from it. And working withthe second seems a hopeless task!Note too that the constant 3/4 is sharp. To see this, let ABC be a trianglewith b = c = 1, a = 2ɛ, 0 < ɛ < 1 so thatsince2(m a + m b + m c )a + b + c== 0.(1 + ɛ)2(1 + ɛ) 2It follows from this that there is no absolute constant ν > 3/4 such thatm a + m b + m c ≥ ν(a + b + c),for all triangles.There are also direct and successful approaches using the cosine and sine rules.Here’s something else that you might like to try your hands at:Problem 1 Prove thatm a + m b + m c < a + b + c.Prove also that this is sharp.Problem 2 Prove that m a , m b , m c are the side lengths of a triangle—called themedian dual triangle of ABC—the lengths of whose medians are 3a/4, 3b/4, 3c/4.

Problem 3 Prove thatm a + m b + m c ≥ a sin A + b sin B + c sin C,with equality iff the triangle is equilateral.Problem 4 Prove that the sum of the lengths of the medians of a triangle is atleast three times the sum of the lengths of the perpendiculars from the centroidto the sides. Prove also that there is equality iff the triangle is equilateral.4. Determine the number of different arrangements a 1 , a 2 , . . . , a 10 of the integers1, 2, . . . , 10 such thata i > a 2i for 1 ≤ i ≤ 5,anda i > a 2i+1for 1 ≤ i ≤ 4. [DH]Solution. Suppose {a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 , a 8 , a 9 , a 10 } is such an arrangement.Then the inequalities that hold can be represented by the following treediagramwhere a i → a j means a i > a j : Clearly, the tree has two main branches,a_1a_8a_4a_2a_5a_6a_3a_7a_9a_10one bifurcating at a 2 and the other at a 3 . Since a 1 is the largest, a 1 = 10.So, to count the number of different arrangements subdivide the numbers1, 2, . . . , 9 into two sets {a 2 , a 4 , a 5 , a 8 , a 9 , a 10 } and {a 3 , a 6 , a 7 }. We can carryout this subdivision in ( 93)= 84 ways. For each subdivision, a2 and a 3 areuniquely determined as the largest elements in their subsets. For the choice a 6and a 7 we have two possibilities, and for the subsets {a 4 , a 8 , a 9 } and {a 5 , a 10 }we have ( 52)= 10 ways to do it. For each of these choices, a4 , a 5 , a 10 are already

determined. There are again two possibilities for a 8 and a 9 . Thus the numberof possibilities to arrange the full set as required is( ( 9 5× 2 × × 2 = 84 × 2 × 10 × 2 = 33603)2)ways.5. Suppose a, b and c are non-negative real numbers. Prove that13 [(a−b)2 +(b−c) 2 +(c−a) 2 ] ≤ a 2 +b 2 +c 2 −3 3√ a 2 b 2 c 2 ≤ (a−b) 2 +(b−c) 2 +(c−a) 2 . [FH]Solution. Noting that(a − b) 2 + (b − c) 2 + (c − a) 2 = 3(a 2 + b 2 + c 2 ) − (a + b + c) 2 ,the left-hand inequality follows from the fact that3 3√ ( ) ( )3√ 2 2 a + b + ca 2 b 2 c 2 = 3 abc ≤ 3 =3To establish the right-hand inequality we must show that2(ab + bc + ca) ≤ a 2 + b 2 + c 2 + 3 3√ a 2 b 2 c 2 .(a + b + c)2.3Since the expressions are symmetric we can assume without loss of generalitythat a is the minimum of {a, b, c}. With this assumption,a 2 + b 2 + c 2 + 3 3√ a 2 b 2 c 2 = a 2 + 3 3√ a 2 b 2 c 2 + b 2 + c 2sincewith equality iff a = 0.≥ 4 4√ a 2 (a 2 b 2 c 2 ) + b 2 + c 2= 4a √ bc + (b − c) 2 + 2bc= 2a(2 √ bc − (b + c)) + (b − c) 2 + 2(ab + bc + ca)= −2( √ c − √ b) 2 a + (c − b) 2 + 2(ab + bc + ca)= ( √ c − √ b)(( √ 2 c + √ )b) 2 − 2a + 2(ab + bc + ca)≥2(ab + bc + ca),( √ c + √ b) 2 − 2a ≥ 4a − 2a = 2a ≥ 0,As Fiachra Knox observed—and Stephen Ryan wondered about—Schur’s inequalityto the effect that if r ≥ 0 and x, y, z are positive numbers, thenx r (x − y)(x − z) + y r (y − z)(y − x) + z r (z − x)(z − y) > 0,

unless x = y = z can be used to establish the right-hand side. [Hint: Take r =1, expand and rearrange this inequality and set x = a 2/3 , y = b 2/3 , z = c 2/3 .]Also, as Richard Ellard essentially noted, the left-hand side can be strengthened:Problem 5 Prove that12 [(a − b)2 + (b − c) 2 + (c − a) 2 ] ≤ a 2 + b 2 + c 2 − 3 3√ a 2 b 2 c 2 .Problem 6 Prove Schur’s inequality.