The 42^nd IMO 
(official site : The 42nd International Mathematical Olympiad )
(official scoring site : Scores of the 42nd IMO)

Problems

Day one

1. Let ABC be an acute-angled triangle which satisfies the following property: ACB ³ 30 + ABC. Let P be the foot of the perpendicular from A to BC. Prove that COP + BAC < 90.

2. Let a, b, c be positive real numbers. Prove that :

3. 21 boys and 21 girls take part in a mathematical competition where several problems are proposed. It is known that each boy and each girl solves at most 6 of the proposed problems. Furthermore, for any girl and any boy there is at least one problem solved by both of them. Prove that there is a problem solved by at least 3 boys and at least 3 girls.

4. Let n be an odd integer greater than 1 and k₁, k₂, ..., k_n be given integers. For each of the n! permutations of {1, 2, ..., n} let S(a) = k₁a(1) + k₂a(2) + ... + k_na(n). Prove that there are two distinct permutations so that n! divides S(b) - S(c).

5. In a triangle ABC let P bisect angle BAC (P on BC) and BQ bisect angle ABC (Q on AC). It is known that BAC = 60 and AB + BP = AQ + QB. What are the angles of triangle ABC.

6. Let a, b, c, d be integers with a > b > c > d > 0. Suppose that ac + bd = (b + d + a - c)(b + d - a + c). Prove that ab + cd is not a prime.

1. (My solution) I use the classical notations for the elements of the triangle. Clearly OCB = 90 – A. From the sine law in triangle OPC we have PC/sin POC = OC/sin(180 – POC – (90-A)) = R/sin(POC + 90 – A) = R/cos(A - POC).
If POC = x we have PC/sin x = R/cos(A-x).
PC = b cosC = 2R sinB cosC. Therefore 2sinB cosC = f(x) = sin x/cos(A – x). But f’(x) = cosA/(cos(A-x))² > 0. So the function f is decreasing.
Assume x ³90-A. That means that 2sinB cosC = f(x) £ f(90-x) = sin(90 – A)/cos(2A – 90) = cosA/sin2A = 1/(2sinA). So we have 4sinBcosCsinA ³ 1.
sinA = sin(B+C) = sinBcosC + cosBsinC. So we have 4sin2Bcos2C + sin2Bsin2C ³ 1. Add -2sin2B
and get 2sin2B(2cos2C – 1) + sin2Bsin2C ³ 1-2sin2B.
So 2sin2Bcos2C + sin2Bsin2C ³ 1-2sin2B. Add –cos2C
and get –cos2Bcos2C + sin2Bsin2C ³ 1-2sin2B – cos2C = cos2B – cos2C.
So we have –cos(2B+2C) £ cos2B – cos2C. i.e. cos2A + cos2B £ cos2C.
cos2A £ cos2C – cos2B = 2sin(B-C)sin(B+C) = 2sinAsin(B-C) £ -sinA since C ³ B+30.
cos2A = 1-2sin²A £ -sinA. Solving the inequality 2x² – x – 1 ³ 0 we get either x £  - 1/2 or x ³ 1. Since ABC is acute-angled we have x is in the interval (0,1). Therefore our assumption false so x < 90-A.
 

2. I didn't solve this problem but here is the easiest solution: using the Cauchy Schwarz inequality twice we get , since (a + b + c)³ = a³ + b³ + c³ + 3(a+b)(b+c)(c+a). Clearly this ends the proof.
 

3. (My solution of the fact that #1 + #2 is at least 22, Mihai Manea's solution of the #1 + #2 is at most 15).
Place the problems solved by each contestant in a table with columns representing problems and rows representing contestants. Mark a 1 if the contestant solved the problem and a 0 otherwise. The rows are g₁, ..., g₂₁ and b₁, ..., b₂₁ and the columns 1, 2, ..., s. For each column i let u_i and v_i be the number of problems solved by the girls and boys respectively. Since each solved at most 6 problems and since each boy and girl have a problem in common we have the following three inequalities:
Su_i £ 6*21
Sv_i £ 6*21
Su_iv_i³ 21²
If one of the numbers u_ior v_i is 0 then that problem can be discarded from the set of problems solved by that particular boy or girl and the problem rests equivalent. So we may assume that these numbers u_i and v_i are at least 1. Assume no problem is solved by three boys and three girls. Then divide the set of problems into three groups:
the problems solved by at most 2 girls and at least 3 boys, the problems solved by at most 2 boys and at least 3 girls and the problems solved by at most 2 girls and at most 2 boys. Denote the sets by 1, 2 and 3.

S_1,2(u_i - 2)(v_i - 2) + S₃(u_i - 2)(v_i - 2) = Su_iv_i - 2(Su_i+ Sv_i) + 4(#1,2,3) ³ 21*21 - 24*21 + 4(#1,2,3). Where 4(#1,2,3) is the number of elements of the sets 1, 2 and 3. On the left side, the first sum is negative and the second sum is positive, due to the intial definition of the sets. Let A and B denote these two sets. B = S₃(u_i - 2)(v_i - 2) £(#3), since if u and v are either 1 or 2 then (u-2)(v-2) £ 1. So #3 ³ A + (#3) ³ A + B ³ -63 + 4(#1,2) + 4(#3) since A is not positive. Therefore 4(#1,2) + 3(#3) £ 63 which means that #1 + #2 £ 15.
Due to the Dirichlet principle, for each boy or girl there is a problem (among the ones he or she solved) which is solved by at least 4 other contestants of the oposite sex. So in 1 and 3, since at most 2 girls solve these problems, we must have theboys solving at least 4 problems. Clearly this may occur only for set 1.
In each column of 1 we have at most 2 ones next to the girls and some columns have at least 4 ones next to the boys. Each girl must have a one in a column that has 4 ones next to the boys. In each column we have at most 2 girls and therefore the number of columns having 4 ones next to the boys is at least 21/2 so at least 11. Therefore #1 ³ 11. Similarly #2 ³ 11. Then 15 ³  #1 + #2 ³ 22, which is false. So the problem is solved.

4. (My solution) This is an easy problem. If no two of S(a) have the same remainder mod n! then the set S_n has all the remainders 1, 2, ..., n!-1 mod n!. Take the sum of all S(a) on the set S_n. Denote this sum by A. Then mod n! we have A = 1 + 2 + ... + n!-1 = n!(n!-1)/2. Then A = S_aS_ik_ia(i) = S_ik_i S_aa(i) = S_ik_i S_aa(1) = (n-1)!(S_ik_i)(1+2+...+n) since S_aa(i) = S_aa(1) = (n-1)!(1+2+...+n). This is obvious. But then we have mod n!, A = n!(n+1)(S_ik_i)/2 = 0 = n!(n!-1)/2. But n!-1 is odd and prime with n!. Therefore n!/2 = 0 mod n!. This is clearly impossible. Therefore the conclusion follows.

5. (Mihai Manea's solution) More generally I shall prove that if BAC is at least 60 and if AB + BP = AQ + QB then B=2C. Rotate BP around B to BU, B on segment AU. Rotate QB around Q until QV so that C is on segment AV. We have AU = AV and AP is the bisector of UAV. So UP = PV and PVQ = PUB = BPU = B/2 = QBP. Then BQ = QC and QBP = PVQ therefore QP bisects angle BQC. Clearly triangle BUP is isoscleles (using angles) so BU = BP = PV = PU which means that BUP is equilateral. So B = 120, but this means that A < 60. Therefore V must conincide with C. Then B/2 = QBC = QVP = QCP = C. So B = 2C.
If A=60 then 3B/2 + 60 = 180 which means that B = 80 and C = 40.

6. (My solution).
The given equation can be rewritten as b² + bd + d² = a²- ac + c², so (b-wd)(b-w²d) = (a+wc)(a+w²c) = k. Decompose this integer in primes in Z[w] then we must get x, x', ..., z, z'. For b-wd and a+wc, if we place x in b-wd (or a+wc) then we must place x' in the conjugate of b-wd (or a+wc). (x' is the conjugate of x). Similarly for all primes in the decomposition of k. Then there is u and v so that b-wd = uv and a+wc = uv'. (u consists of all primes that appear either simply or conjugately in both b-wd and a+wc, and v consists of all the other primes).
Calculating a, b, c and d from the two equations and their conjugates we get:
a = (u'v - wuv')/(1-w)
b = (u'v' - wuv)/(1-w)
c = (uv' - u'v)/w(1-w)
d = (u'v' - uv)/w(1-w)
so, using w³ = 1, we get ab + cd = (u'v - wuv')(u'v' - wuv)/(1-w)²+ w(uv' - u'v)(u'v' - uv)/(1-w)²= vv'(u'2 - wu2)/(1-w). If ab + cd were a prime in the Z[w] prime decomposition it would have at most two prime factors. So it is enough to prove that it has at least three prime factors and the problem would be finished.
Since w⁴ = w we have (u'² - wu²) = (u' - w²u)(u' + w²u). Assume that v is a unit. Then since v'(b-wd) = v(a+wc) we get the equations: -d = c, or d = -c, or b = c, neither of which is possible. So v is not a unit. Assume v equals (is equivalent to in Z[w]) to 1-w. Then we get the system of equations 2b + d = a + c and b - d = 2c - a (or another similar system). This leads to b = c which is not possible.
Assume (u' - w²u)(u' + w²u) is a unit. Then u' - w²u and u' + w²u are both units in Z[w]. Conjugating we get that u' - wu and u' + wu are units. Let x and y be the values of the two units. Then u' = (x+y)/2 and u=w(y-x)/2. Since the two expressions are conjugates we get that x'+y' = w(y-x). Conjugating the equality and adding it to its conjugate times w we get: wx + x' = 0. This, however, is not satisfied for any unit x.
Assume that (u' - w²u)(u' + w²u) is equivalent to 1-w. Then one of the possibilities is u' - w²u = 1 and u' + w²u = 1-w. Then we have 2u' = 2-w which is not possible, since 2 is prime in Z[w]. Similarly for u' - w²u = -1. If u' - w²u = w then u' + w²u = w²-1. Then u' = u = -1. This, however, does not satisfy the first equation.
Then we find the following conclusions:
- v is not a unit nor equivalent to 1-w.
- (u'²- wu²) is not a unit nor equivalent to 1-w.
- ab + cd has three primes coming from:
2 from vv', and one from (u'² - wu²)/(1-w), since 1-w is a prime (has norm 3) and (u'² - wu²) is not equivalent to 1-w. From here the conclusion follows.

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