EUROPEAN MATHEMATICAL OLYMPIAD

Consider a pair of distinct, nonconstant repetitive words with periods ga and gb, where a, b = 1 and a, b > 1, the first word is made up of kb repetitions of the subword of length ga and

European Girls’ Mathematical Olympiad

EGMO | 2012

Thursday, April 12, 2012 Problem 1 Let ABC be a triangle with circumcentre O The points D, E and F lie in the interiors of the sides BC, CA and AB respectively, such that DE is perpendicular to CO and DF is perpendicular to BO (By interior we mean, for example, that the point D lies on the line BC and D is between B and C on that line.) Let K be the circumcentre of triangle AF E Prove that the lines DK and BC are perpendicular.

Problem 2 Let n be a positive integer Find the greatest possible integer m, in terms of n, with the following property: a table with m rows and n columns can be lled with real numbers in such a manner that for any two dierent rows [a1, a2, , an] and [b1, b2, , bn] the following holds:

max( |a 1 − b 1 |, |a 2 − b 2 |, , |a n − b n |) = 1.

Problem 3 Find all functions f : R → R such that

f yf (x + y) + f (x)

= 4x + 2yf (x + y) for all x, y ∈ R.

Problem 4 A set A of integers is called sum-full if A ⊆ A + A, i.e each element a ∈ A is the sum of some pair of (not necessarily dierent) elements b, c ∈ A A set A of integers is said to be zero-sum-free if 0 is the only integer that cannot be expressed as the sum of the elements of a nite nonempty subset of A.

Does there exist a sum-full zero-sum-free set of integers?

Language: English

Day: 1

European Girls’ Mathematical Olympiad 2012—Day 1 Solutions

Problem 1 Let ABC be a triangle with circumcentre O The points D, E and F lie in the interiors of the sides BC, CA and AB respectively, such that DE is perpendicular to CO and DF is perpendicular to BO (By interior we mean, for example, that the point D lies on the line BC and D is between B and C on that line.)

Let K be the circumcentre of triangle AF E Prove that the lines DK and BC are perpendicular.

Origin Netherlands (Merlijn Staps).

K

Solution 1 (submitter) Let `C be the tangent at C to the circumcircle of 4ABC As CO ⊥ `C, the lines

DE and `C are parallel Now we find that

∠CDE = ∠(BC, ` C) = ∠BAC, hence the quadrilateral BDEA is cyclic Analogously, we find that the quadrilateral CDF A is cyclic As we now have ∠CDE = ∠A = ∠F DB, we conclude that the line BC is the external angle bisector of ∠EDF Furthermore, ∠EDF = 180◦− 2∠A Since K is the circumcentre of 4AEF , ∠F KE = 2∠F AE = 2∠A So

∠F KE + ∠EDF = 180◦, hence K lies on the circumcircle of 4DEF As |KE| = |KF |, we have that K is the midpoint of the arc EF of this circumcircle It is well known that this point lies on the internal angle bisector

of ∠EDF We conclude that DK is the internal angle bisector of ∠EDF Together with the fact that BC is the external angle bisector of ∠EDF , this yields that DK ⊥ BC, as desired.

Solution 2 (submitter) As in the previous solution, we show that the quadrilaterals BDEA and CDF A are both cyclic Denote by M and L respectively the circumcentres of these quadrilaterals We will show that the quadrilateral KLOM is a parallelogram The lines KL and M O are the perpendicular bisectors of the line segments AF and AB, respectively Hence both KL and M O are perpendicular to AB, which yields KL k M O.

In the same way we can show that the lines KM and LO are both perpendicular to AC and hence parallel

as well We conclude that KLOM is indeed a parallelogram Now, let K 0 , L 0 , O 0 and M 0 be the respective projections of K, L, O and M to BC We have to show that K0 = D As L lies on the perpendicular bisector

of CD, we have that L0 is the midpoint of CD Similarly, M0 is the midpoint of BD and O0 is the midpoint

of BC Now we are going to use directed lengths Since KLOM is a parallelogram, M0K0= O0L0 As

O0L0= O0C − L0C =12· (BC − DC) = 1

2 · BD = M0D,

we find that M0K0 = M0D, hence K0= D, as desired.

Solution 3 (submitter) Denote by `A, `B and `C the tangents at A, B and C to the circumcircle of 4ABC Let A0 be the point of intersection of `B and `C and define B0 and C0 analogously As in the first solution,

we find that DE k `C and DF k `B Now, let Q be the point of intersection of DE and `A and let R be the point of intersection of DF and `A We easily find 4AQE ∼ 4AB 0 C As |B 0 A| = |B 0 C|, we must have

|QA| = |QE|, hence 4AQE is isosceles Therefore the perpendicular bisector of AE is the internal angle bisector

of ∠EQA = ∠DQR Analogously, the perpendicular bisector of AF is the internal angle bisector of ∠DRQ.

We conclude that K is the incentre of 4DQR, thus DK is the angle bisector of ∠QDR Because the sides of the triangles 4QDR and 4B0A0C0 are pairwise parallel, the angle bisector DK of ∠QDR is parallel to the angle bisector of ∠B0A0C0 Finally, as the angle bisector of ∠B0A0C0 is easily seen to be perpendicular to BC (as it is the perpendicular bisector of this segment), we find that DK ⊥ BC, as desired.

Remark (submitter) The fact that the quadrilateral BDEA is cyclic (which is an essential part of the first two solutions) can be proven in various ways Another possibility is as follows Let P be the midpoint of BC Then, as ∠CP O = 90◦, we have ∠P OC = 90◦−∠OCP Let X be the point of intersection of DE and CO, then

we have that ∠CDE = ∠CDX = 90◦− ∠XCD = 90 ◦ − ∠OCP Hence ∠CDE = ∠P OC = 1

2 ∠BOC = ∠BAC From this we can conclude that BDEA is cyclic.

Solution 4 (PSC) This is a simplified variant of Solution 1 ∠COB = 2∠A (angle at centre of circle ABC) and OB = OC so ∠OBC = ∠BCO = 90◦− ∠A Likewise ∠EKF = 2∠A and ∠KF E = ∠F EK = 90◦− ∠A Now because DE ⊥ CO, ∠EDC = 90◦− ∠DCO = 90 ◦ − ∠BCO = ∠A and similarly ∠BDF = ∠A, so

∠F DE = 180◦− 2∠A So quadrilateral KF DE is cyclic (opposite angles), so (same segment) ∠KDE =

∠KF E = 90◦− ∠A, so ∠KDC = 90 ◦ and DK is perpendicular to BC.

Problem 2 Let n be a positive integer Find the greatest possible integer m, in terms of n, with the following property: a table with m rows and n columns can be filled with real numbers in such a manner that for any two different rows [a1, a2, , an] and [b1, b2, , bn] the following holds:

max(|a1 − b1|, |a2 − b2|, , |an − bn|) = 1.

Origin Poland (Tomasz Kobos).

Solution 1 (submitter) The largest possible m is equal to 2 n

In order to see that the value 2 n can be indeed achieved, consider all binary vectors of length n as rows of the table We now proceed with proving that this is the maximum value.

(2) For every i 6= j, ij ∈ E(Gk) for some k This follows directly from the problem statement.

For every graph Gk fix some bipartition (Ak, Bk) of {1, 2, , m}, i.e., a partition of {1, 2, , m} into two disjoint sets Ak, Bk such that the edges of Gk traverse only between Ak and Bk If m > 2 n , then there are two distinct indices i, j such that they belong to exactly the same parts Ak, Bk, that is, i ∈ Ak if and only if j ∈ Akfor all k = 1, 2, , n However, this means that the edge ij cannot be present in any of the graphs G1, G2, , Gn, which contradicts (2) Therefore, m ≤ 2 n

Solution 2 (PSC) In any table with the given property, the least and greatest values in a column cannot differ by more than 1 Thus, if each value that is neither least nor greatest in its column is changed to be equal

to either the least or the greatest value in its column (arbitrarily), this does not affect any |ai − bi| = 1, nor does it increase any difference above 1, so the table still has that given property But after such a change, for any choice of what the least and greatest values in each column are, there are only two possible choices for each entry in the table (either the least or the greatest value in its column); that is, only 2n possible distinct rows, and the given property implies that all rows must be distinct As in the previous solution, we see that this number can be achieved.

Solution 3 (Coordinators) We prove by induction on n that m ≤ 2

First suppose n = 1 If real numbers x and y have |x − y| = 1 then bxc and byc have opposite parities and hence it is impossible to find three real numbers with all differences 1 Thus m ≤ 2.

Suppose instead n > 1 Let a be the smallest number appearing in the first column of the table; then every entry in the first column of the table lies in the interval [a, a + 1] Let A be the collection of rows with first entry a and B be the collection of rows with first entry in (a, a + 1] No two rows in A differ by 1 in their first entries, so if we list the rows in A and delete their first entries we obtain a table satisfying the conditions of the problem with n replaced by n − 1; thus, by the induction hypothesis, there are at most 2 n−1 rows in A Similarly, there are at most 2 n−1 rows in B Hence m ≤ 2 n−1 + 2 n−1 = 2 n As before, this number can be achieved.

Solution 4 (Coordinators) Consider the rows of the table as points of R n As the values in each column differ by at most 1, these points must lie in some n-dimensional unit cube C Consider the unit cubes centred

on each of the m points The conditions of the problem imply that the interiors of these unit cubes are pairwise disjoint But now C has volume 1, and each of these cubes intersects C in volume at least 2−n: indeed, if the unit cube centred on a point of C is divided into 2ncubes of equal size then one of these cubes must lie entirely within C Hence m ≤ 2n As before, this number can be achieved.

Solution 5 (Coordinators) Again consider the rows of the table as points of R n The conditions of the problem imply that these points must all lie in some n-dimensional unit cube C, but no two of the points lie in any smaller cube Thus if C is divided into 2 n equally-sized subcubes, each of these subcubes contains at most one row of the table, giving m ≤ 2 n As before, this number can be achieved.

Problem 3 Find all functions f : R → R such that

f yf (x + y) + f (x)
= 4x + 2yf (x + y) for all x, y ∈ R.

Origin Netherlands (Birgit van Dalen).

Solution 1 (submitter) Setting y = 0 yields

from which we derive that f is a bijective function Also, we find that

f (0) = f (4 · 0) = f (f (f (0))) = 4f (0), hence f (0) = 0 Now set x = 0 and y = 1 in the given equation and use (1) again:

Problem 4 A set A of integers is called sum-full if A ⊆ A + A, i.e each element a ∈ A is the sum of some pair of (not necessarily different) elements b, c ∈ A A set A of integers is said to be zero-sum-free if 0 is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of A.

Does there exist a sum-full zero-sum-free set of integers?

Origin Romania (Dan Schwarz).

Remark The original formulation of this problem had a weaker definition of zero-sum-free that did not require all nonzero integers to be sums of finite nonempty subsets of A.

Solution (submitter, adapted) The set A = {F2n : n = 1, 2, } ∪ {−F2n+1 : n = 1, 2, }, where Fk is the kth Fibonacci number (F1 = 1, F2 = 1, Fk+2 = Fk+1 + Fk for k ≥ 1) qualifies for an example We then have F2n = F2n+2 + (−F2n+1) and −F2n+1 = (−F2n+3) + F2n+2 for all n ≥ 1, so A is sum-full (and even with unique representations) On the other hand, we can never have

0 =

s X i=1

F2ni−

t X j=1

F2nj+1,

owing to the fact that Zeckendorf representations are known to be unique.

It remains to be shown that all nonzero values can be represented as sums of distinct numbers 1, −2, 3,

−5, 8, −13, 21, This may be done using a greedy algorithm: when representing n, the number largest

in magnitude that is used is the element m = ±Fk of A that is closest to 0 subject to having the same sign

as n and |m| ≥ |n| That this algorithm terminates without using any member of A twice is a straightforward induction on k; the base case is k = 2 (m = 1) and the induction hypothesis is that for all n for which the above algorithm starts with ±F` with ` ≤ k, it terminates without having used any member of A twice and without having used any ±Fj with j > `.

Remark (James Aaronson and Adam P Goucher) Let n be a positive integer, and write u = 2n; we claim that the set

{1, 2, 4, , 2 n−1 , −u, u + 1, −(2u + 1), 3u + 2, −(5u + 3), 8u + 5, }

is a sum-full zero-sum-free set The proof is similar to that used for the standard examples.

European Girls’ Mathematical Olympiad

EGMO | 2012

Friday, April 13, 2012 Problem 5 The numbers p and q are prime and satisfy

Problem 6 There are innitely many people registered on the social network Mugbook Some pairs of (dierent) users are registered as friends, but each person has only nitely many friends Every user has at least one friend (Friendship is symmetric; that is, if A is a friend of B, then B is a friend of A.)

Each person is required to designate one of their friends as their best friend If A designates B as her best friend, then (unfortunately) it does not follow that B necessarily designates A as her best friend Someone designated as a best friend is called a 1-best friend More generally, if n > 1 is a positive integer, then a user

is an n-best friend provided that they have been designated the best friend of someone who is an (n − 1)-best friend Someone who is a k-best friend for every positive integer k is called popular.

(a) Prove that every popular person is the best friend of a popular person.

(b) Show that if people can have innitely many friends, then it is possible that a popular person is not the best friend of a popular person.

Problem 7 Let ABC be an acute-angled triangle with circumcircle Γ and orthocentre H Let K be a point

of Γ on the other side of BC from A Let L be the reection of K in the line AB, and let M be the reection

of K in the line BC Let E be the second point of intersection of Γ with the circumcircle of triangle BLM Show that the lines KH, EM and BC are concurrent (The orthocentre of a triangle is the point on all three

of its altitudes.)

Problem 8 A word is a nite sequence of letters from some alphabet A word is repetitive if it is a catenation of at least two identical subwords (for example, ababab and abcabc are repetitive, but ababa and aabb are not) Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical (Note that one may swap two adjacent identical letters, leaving a word unchanged.)

con-Language: English

Day: 2

European Girls’ Mathematical Olympiad 2012—Day 2 Solutions

Problem 5 The numbers p and q are prime and satisfy

Origin Luxembourg (Pierre Haas).

Solution 1 (submitter) Rearranging the equation, 2qn(p + 1) = (n + 2)(2pq + p + q + 1) The left hand side is even, so either n + 2 or p + q + 1 is even, so either p = 2 or q = 2 since p and q are prime, or n is even.

If p = 2, 6qn = (n + 2)(5q + 3), so (q − 3)(n − 10) = 36 Considering the divisors of 36 for which q is prime, we find the possible solutions (p, q, n) in this case are (2, 5, 28) and (2, 7, 19) (both of which satisfy the equation).

If q = 2, 4n(p + 1) = (n + 2)(5p + 3), so n = pn + 10p + 6, a contradiction since n < pn, so there is no solution with q = 2.

Finally, suppose that n = 2k is even We may suppose also that p and q are odd primes The equation becomes 2kq(p + 1) = (k + 1)(2pq + p + q + 1) The left hand side is even and 2pq + p + q + 1 is odd, so k + 1

is even, so k = 2` + 1 is odd We now have

`(p + 1) = (` + 1)(p + r), a contradiction since ` < ` + 1 and p + 1 ≤ p + r.

Thus the possible values of q − p are 2, 3 and 5.

Solution 2 (PSC) Subtracting 2 and multiplying by −1, the condition is equivalent to

q − p − 1 = 4(p + 1)q

n + 2 .The expression on the right is a positive integer, and q must cancel into n + 2 else q would divide p + 1 < q Let (n + 2)/q = u a positive integer.

Now

q − p − 1 = 4(p + 1)

u so

Solution 3 (Coordinators) Subtract 2 from both sides to get

q − p − 1 q(p + 1) =

4

n + 2.Now (q, q − p − 1) = (q, p + 1) = 1 and (p + 1, q − p − 1) = (p + 1, q) = 1 so the fraction on the left is in lowest terms Therefore the numerator must divide the numerator on the right, which is 4 Since q − p − 1 is positive,

it must be 1, 2 or 4, so that q − p must be 2, 3 or 5 All of these can be attained, by (p, q, n) = (3, 5, 78), (2, 5, 28) and (2, 7, 19) respectively.

Problem 6 There are infinitely many people registered on the social network Mugbook Some pairs of (different) users are registered as friends, but each person has only finitely many friends Every user has at least one friend (Friendship is symmetric; that is, if A is a friend of B, then B is a friend of A.)

Each person is required to designate one of their friends as their best friend If A designates B as her best friend, then (unfortunately) it does not follow that B necessarily designates A as her best friend Someone designated as a best friend is called a 1-best friend More generally, if n > 1 is a positive integer, then a user

is an n-best friend provided that they have been designated the best friend of someone who is an (n − 1)-best friend Someone who is a k-best friend for every positive integer k is called popular.

(a) Prove that every popular person is the best friend of a popular person.

(b) Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person.

Origin Romania (Dan Schwarz) (rephrasing by Geoff Smith).

Remark The original formulation of this problem was:

Given a function f : X → X, let us use the notations f0(X) := X, fn+1(X) := f (fn(X)) for n ≥ 0, and also

fω(X) := \

n≥0

fn(X) Let us now impose on f that all its fibres f−1(y) := {x ∈ X | f (x) = y}, for y ∈ f (X), are finite Prove that f (fω(X)) = fω(X).

Solution 1 (submitter, adapted) For any person A, let f0(x) = x, let f (A) be A’s best friend, and define

f k+1 (A) = f (f k (A)), so any person who is a k-best friend is f k (A) for some person A; clearly a k-best friend

is also an `-best friend for all ` < k Let X be a popular person For each positive integer k, let xk be a person with f k (xk) = X Because X only has finitely many friends, infinitely many of the f k−1 (xk) (all of whom designated X as best friend) must be the same person, who must be popular.

If people can have infinitely many friends, consider people Xifor positive integers i and Pi,jfor i < j positive integers Xi designates Xi+1 as her best friend; Pi,i designates X1 as her best friend; Pi,j designates Pi+1,j as her best friend if i < j Then all Xi are popular, but X1 is not the best friend of a popular person.

Solution 2 (submitter, adapted) For any set S of people, let f−1(S) be the set of people who designated someone in S as their best friend Since each person has only finitely many friends, if S is finite then f−1(S) is finite.

Let X be a popular person and put V0 = {X} and Vk = f−1(Vk−1) All Vi are finite and (since X is popular) nonempty.

If any two sets Vi, Vj, with 0 ≤ i < j are not disjoint, define fi(x) for positive integers i as in Solution 1.

It follows ∅ 6= f i (Vi∩ Vj ) ⊆ f i (Vi) ∩ f i (Vj) ⊆ V0∩ Vj−i , thus X ∈ Vj−i But this means that f j−i (X) = X, therefore f n(j−i) (X) = X Furthermore, if Y = f j−i−1 (X), then f (Y ) = X and f n(j−i) (Y ) = Y , so X is the best friend of Y , who is popular.

If all sets Vn are disjoint, by K¨ onig’s infinity lemma there exists an infinite sequence of (distinct) xi, i ≥ 0, with xi∈ Vi and xi= f (xi+1) for all i Now x1 is popular and her best friend is x0= X.

If people can have infinitely many friends, proceed as in Solution 1.

Problem 7 Let ABC be an acute-angled triangle with circumcircle Γ and orthocentre H Let K be a point

of Γ on the other side of BC from A Let L be the reflection of K in the line AB, and let M be the reflection

of K in the line BC Let E be the second point of intersection of Γ with the circumcircle of triangle BLM Show that the lines KH, EM and BC are concurrent (The orthocentre of a triangle is the point on all three

of its altitudes.)

Origin Luxembourg (Pierre Haas).

Solution 1 (submitter) Since the quadrilateral BM EL is cyclic, we have ∠BEM = ∠BLM By tion, |BK| = |BL| = |BM |, and so (using directed angles)

construc-∠BLM = 90◦− 1

2 ∠M BL = 90◦− 180◦− 1

2∠LBK −12 ∠KBM

= 12∠LBK +12 ∠KBM
− 90 ◦ = (180◦− ∠B) − 90◦= 90◦− B.

We see also that ∠BEM = ∠BAH, and so the point N of intersection of EM and AH lies on Γ.

Let X be the point of intersection of KH and BC, and let N0 be the point of intersection of M X and AH Since BC bisects the segment KM by construction, the triangle KXM is isosceles; as AHkM K, HXN0 is isosceles Since AH ⊥ BC, N0 is the reflection of H in the line BC It is well known that this reflection lies on Γ, and so N0 = N Thus E, M , N and M , X, N0 all lie on the same line M N ; that is, EM passes through X.

M E

N = N0X

Remark (submitter) The condition that K lies on the circumcircle of ABC is not necessary; indeed, the solution above does not use it However, together with the fact that the triangle ABC is acute-angled, this condition implies that M is in the interior of Γ, which is necessary to avoid dealing with different configurations including coincident points or the point of concurrence being at infinity.

Solution 2 (PSC) We work with directed angles Let HK meet BC at X Let M X meet AH at HA on Γ (where HA is the reflection of H in BC) Define E 0 to be where HAM meets Γ (again) Our task is to show that ∠M E0B = ∠M LB.

Observe that

∠M E0B = ∠HA AB (angles in same segment)

= BcNow

= ∠BKHC (reflecting in the line AB)

= ∠BCH C (angles in the same segment)

= Bc.

Problem 8 A word is a finite sequence of letters from some alphabet A word is repetitive if it is a catenation of at least two identical subwords (for example, ababab and abcabc are repetitive, but ababa and aabb are not) Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical (Note that one may swap two adjacent identical letters, leaving a word unchanged.)

con-Origin Romania (Dan Schwarz).

Solution 1 (submitter) In this and the subsequent solutions we refer to a word with all letters identical as constant.

Let us consider a nonconstant word W , of length |W | = w, and reach a contradiction Since the word

W must contain two distinct adjacent letters, be it W = AabB with a 6= b, we may assume B = cC to be non-empty, and so W = AabcC By the proper transpositions we get the repetitive words W0 = AbacC = Pw/p,

of a period P of length p | w, 1 < p < w, and W00= AacbC = Qw/q, of a period Q of length q | w, 1 < q < w However, if a word U V is repetitive, then the word V U is also repetitive, of a same period length; therefore we can work in the sequel with the repetitive words W00 = CAbac, of a period P0 of length p, and W000= CAacb,

of a period Q 0 of length q The main idea now is that the common prefix of two repetitive words cannot be too long.

Now, if a word a1a2 aw= T w/t is repetitive, of a period T of length t | w, 1 ≤ t < w, then the word (and any subword of it) is t-periodic, i.e ak= ak+t, for all 1 ≤ k ≤ w − t Therefore the word CA is both p-periodic and q-periodic.

We now use the following classical result:

Wilf-Fine Theorem Let p, q be positive integers, and let N be a word of length n, which is both p-periodic and q-periodic If n ≥ p + q − gcd(p, q) then the word N is gcd(p, q)-periodic (but this need not be the case if instead n ≤ p + q − gcd(p, q) − 1).

By this we need |CA| ≤ p + q − gcd(p, q) − 1 ≤ p + q − 2, hence w ≤ p + q + 1, otherwise W00 and W000would

be identical, absurd Since p | w and 1 < p < w, we have 2p ≤ w ≤ p + q + 1, and so p ≤ q + 1; similarly we have q ≤ p + 1.

If p = q, then |CA| ≤ p + p − gcd(p, p) − 1 = p − 1, so 2p ≤ w ≤ p + 2, implying p ≤ 2 But the three-letter suffix acb is not periodic (not even for c = a or c = b), thus must be contained in Q0, forcing q ≥ 3, contradiction.

If p 6= q, then max(p, q) = min(p, q) + 1, so 3 min(p, q) ≤ w ≤ 2 min(p, q) + 2, hence min(p, q) ≤ 2, forcing min(p, q) = 2 and max(p, q) = 3; by an above observation, we may even say q = 3 and p = 2, leading to c = b It follows 6 = 3 min(p, q) ≤ w ≤ 2 min(p, q) + 2 = 6, forcing w = 6 This leads to CA = aba = abb, contradiction.

Solution 2 (submitter) We will take over from the solution above, just before invoking the Wilf-Fine Theorem, by replacing it with a weaker lemma, also built upon a seminal result of combinatorics on words.

Lemma Let p, q be positive integers, and let N be a word of length n, which is both p-periodic and periodic If n ≥ p + q then the word N is gcd(p, q)-periodic.

q-Proof Let us first prove that two not-null words U , V commute, i.e U V = V U , if and only if there exists

a word W with |W | = gcd(|U |, |V |), such that U = W|U |/|W |, V = W|V |/|W | The “if” part being trivial, we will prove the “only if” part, by strong induction on |U | + |V | Indeed, for the base step |U | + |V | = 2 we have |U | = |V | = 1, and so clearly we can take W = U = V Now, for |U | + |V | > 2, if |U | = |V | it follows

U = V , and so we can again take W = U = V If not, assume without loss of generality |U | < |V |; then

V = U V0, so U U V0 = U V0U , whence U V0 = V0U Since |V0| < |V |, it follows 2 ≤ |U | + |V0| < |U | + |V |,

so by the induction hypothesis there exists a suitable word W such that U = W|U |/|W |, V0 = W|V0|/|W |, so

By this we need |CA| ≤ p + q − 1, hence w ≤ p + q + 2, otherwise by the previous lemma W00 and W000would

be identical, absurd Since p | w and 1 < p < w, we have 2p ≤ w ≤ p + q + 2, and so p ≤ q + 2; similarly we have

q ≤ p + 2 That implies max(p, q) ≤ min(p, q) + 2 Now, from k max(p, q) = w ≤ p + q + 2 ≤ 2 max(p, q) + 2

we will have (k − 2) max(p, q) ≤ 2; but max(p, q) ≤ 2 is impossible, since the three-letter suffix acb is not periodic (not even for c = a or c = b), thus must be contained in Q0, forcing q ≥ 3 Therefore k = 2, and so

w = 2 max(p, q).

If max(p, q) = min(p, q), then w = 2p = 2q, for a quick contradiction.

If max(p, q) = min(p, q) + 1, it follows 3 min(p, q) ≤ w = 2 max(p, q) = 2 min(p, q) + 2, hence min(p, q) ≤ 2, forcing min(p, q) = 2 and max(p, q) = 3; by an above observation, we may even say q = 3 and p = 2, leading to

c = b It follows w = 2 max(p, q) = 6, leading to CA = aba = abb, contradiction.

If max(p, q) = min(p, q) + 2, it follows 3 min(p, q) ≤ w = 2 max(p, q) = 2 min(p, q) + 4, hence min(p, q) ≤ 4 From min(p, q) | w = 2 max(p, q) then follows either min(p, q) = 2 and max(p, q) = 4, thus w = 8, clearly contradictory, or else min(p, q) = 4 and max(p, q) = 6, thus w = 12, which also leads to contradiction, by just

a little deeper analysis.

Solution 3 (PSC) We define the distance between two words of the same length to be the number of positions

in which those two words have different letters Any two words related by a transposition have distance 0 or 2; any two words related by a sequence of two transpositions have distance 0, 2, 3 or 4.

Say the period of a repetitive word is the least k such that the word is the concatenation of two or more identical subwords of length k We use the following lemma on distances between repetitive words.

Lemma Consider a pair of distinct, nonconstant repetitive words with periods ga and gb, where (a, b) = 1 and a, b > 1, the first word is made up of kb repetitions of the subword of length ga and the second word is made up of ka repetitions of the subword of length gb These two words have distance at least max(ka, kb).

Proof We may assume k = 1, since the distance between the words is k times the distance between their initial subwords of length gab Without loss of generality suppose b > a.

For each positive integer m, look at the subsequence in each word of letters in positions congruent to m (mod g) Those subsequences (of length ab) have periods dividing a and b respectively If they are equal, then they are constant (since each letter is equal to those a and b before and after it, mod ab, and (a, b) = 1) Because a > 1, there is some m for which the first subsequence is not constant, and so is unequal to the second subsequence Restrict attention to those subsequences.

We now have two distinct repetitive words, one (nonconstant) made up of b repetitions of a subword of length a and one made up of a repetitions of a subword of length b Looking at the first of those words, for any

1 ≤ t ≤ b consider the letters in positions t, t + b, , t + (a − 1)b These letters cover every position (mod a); since the first word is not constant, the letters are not all equal, but the letters in the corresponding positions

in the second word are all equal At least one of these letters in the first word must change to make them all equal to those in the corresponding positions in the second word; repeating for each t, at least b letters must

In the original problem, consider all the words (which we suppose to be repetitive) obtained by a transposition

of two adjacent letters from the original nonconstant word; say that word has length n Suppose those words include two distinct words with periods n/a and n/b; those words have distance at most 4 If a > 4 or b > 4,

we have a contradiction unless a | b or b | a If a > 4 is the greatest number of repetitions in any of the words (n/a is the smallest period), then unless all the numbers of repetitions divide each other there must be words with 2 or 4 repetitions, words with 3 repetitions and all larger numbers of repetitions must divide each other and be divisible by 6.

We now divide into three cases: all the numbers of repetitions may divide either other; or there may be words with (multiples of) 2, 3 and 6 repetitions; or all words may have at most 4 repetitions, with at least one word having 3 repetitions and at least one having 2 or 4 repetitions.

Case 1 Suppose all the numbers of repetitions divide each other Let k be the least number of repetitions Consider the word as being divided into k blocks, each of ` letters; any transposition of two adjacent letters leaves those blocks identical If any two adjacent letters within a block are the same, then this means all the blocks are already identical; since the word is not constant, the letters in the first block are not all identical,

so there are two distinct adjacent letters in the first block, and transposing them leaves it distinct from the other blocks, a contradiction Otherwise, all pairs of adjacent letters within each block are distinct; transposing any adjacent pair within the first block leaves it identical to the second block If the first block has more than two letters, this is impossible since transposing the first two letters has a different result from transposing the second two So the blocks all have length 2; similarly, there are just two blocks, the arrangement is abba but transposing the adjacent letters bb does not leave the word repetitive.

Case 2 Suppose some word resulting from a transposition is made of (a multiple of) 6 repetitions, some

of 3 repetitions and some of 2 repetitions (or 4 repetitions, counted as 2) Consider it as a sequence of 6 blocks, each of length ` If the six blocks are already identical, then as the word is not constant, there are some

pattern BAAAAA, which cannot have two, three or six repetitions So the six blocks are not already identical.

If a transposition within a block results in them being identical, the blocks form a pattern (without loss of generality) BAAAAA, ABAAAA or AABAAA In any of these cases, apply the same transposition (that converts between A and B) to an A block adjacent to the B block, and the result cannot have two, three or six repetitions Finally, consider the case where some transposition between two adjacent blocks results in all six blocks being identical The patterns are BCAAAA, ABCAAA and AABCAA (and considering the letters at the start and end of each block shows B 6= C) In all cases, transposing two adjacent distinct letters within an

A block produces a result that cannot have two, three or six repetitions.

Case 3 In the remaining case, all words have at most 4 repetitions, at least one has 3 repetitions and

at least one has 2 or 4 repetitions For the purposes of this case we will think of 4-repetition words as being 2-repetition words The number of each letter is a multiple of 6, so n ≥ 12; consider the word as made of six blocks of length `.

If the word is already repetitive with 2 repetitions, pattern ABCABC, any transposition between two distinct letters leaves it no longer repetitive with two repetitions, so it must instead have three repetitions after the transposition If AB is not all one letter, transposing two adjacent letters within AB implies that

CA = BC, so A = B = C, the word has pattern AAAAAA but transposing within the initial AA means it no longer has 3 repetitions This implies that AB is all one letter, but similarly BC must also be all one letter and

so the word is constant, a contradiction.

If the word is already repetitive with 3 repetitions, it has pattern ABABAB and any transposition leaves it

no longer having 3 repetitions, so having 2 repetitions instead ABA is not made all of one letter (since the word

is not constant) and any transposition between two adjacent distinct letters therein turns it into BAB; such a transposition affects at most two of the blocks, so A = B, the word has pattern AAAAAA and transposing two adjacent distinct letters within the first half cannot leave it with two repetitions.

So the word is not already repetitive, and so no two adjacent letters are the same; all transpositions give distinct strings Consider transpositions of adjacent letters within the first four letters; three different words result, of which at most one is periodic with two repetitions (it must be made of two copies of the second half

of the word) and at most one is periodic with three repetitions, a contradiction.

European Girls’ Mathematical Olympiad 2012—Problem Selection

and Coordination

Problem Selection Committee

Dr Geoff Smith MBE (Chair)

Dr Joseph Myers

The following countries submitted a total of 33 problems for consideration for EGMO 2012: Belarus, embourg, Netherlands, Poland, Romania, Turkey, Ukraine, USA From these problems, two proposed papers were selected with an alternative to each question also being available The Jury identified two of the questions

Lux-on the proposed papers as being already known, so those questiLux-ons were replaced by the associated alternative questions.

Coordinators

Prof Imre Leader (Chief Coordinator)

Dr Gerry Leversha (Problem 1, Senior Coordinator for Problems 1 and 7)

James Gazet (Problem 1)

Dr Paul Russell (Problem 2, Senior Coordinator for Problems 2 and 8)

Jonathan Lee (Problem 2)

Prof Dan Crisan (Problem 3, Senior Coordinator for Problems 3 and 5)

Alexander Betts (Problem 3)

Dr Tony Gardiner (Problem 4, Senior Coordinator for Problems 4 and 6)

Dr Joseph Myers (Problem 4)

Jack Shotton (Problem 5)

Aled Walker (Problem 5)

Dr James Cranch (Problem 6)

David Mestel (Problem 6)

Dr Vesna Kadelburg (Problem 7)

Tom Lovering (Problem 7)

Sean Moss (Problem 8)

Craig Newbold (Problem 8)

Administrative assistance with coordination was provided by Beverley Detoeuf and Hannah Roberts Coordination started at 08:35 on Saturday 14 April The first scores, for Poland on Problem 8, were entered into the scoring system at 08:39:04; the last scores, for Switzerland on Problem 2, were entered at 20:02:18 The entry of scores for each question finished at the following times: Problem 1 at 14:49:03, Problem 2 at 20:02:18, Problem 3 at 13:56:23, Problem 4 at 17:12:53, Problem 5 at 16:39:17, Problem 6 at 17:05:09, Problem 7 at 18:14:27, Problem 8 at 16:55:55 The medal boundaries were entered at 21:46:28.

Note on the solutions

The solutions given have been edited based on solutions from the original submitters, those from the Problem Selection Committee and those found during EGMO by various people Earlier-listed solutions are not always those that most illuminate the nature of the problem and the issues involved in solving it We are aware of further solutions to some of the problems that are not yet included in these documents and hope to expand the documents in future to include these further solutions.

EGMO 2013European Girls' Mathematical Olympiad

Wednesday, April 10, 2013

Problem 1 The side BC of the triangle ABC is extended beyond C to D so that CD = BC The

side CA is extended beyond A to E so that AE = 2CA.

Prove that, if AD = BE, then the triangle ABC is right-angled.

Problem 2 Determine all integers m for which the m×m square can be dissected into five rectangles,

the side lengths of which are the integers 1, 2, 3, , 10 in some order.

Problem 3 Let n be a positive integer.

(a) Prove that there exists a set S of 6n pairwise different positive integers, such that the least common multiple of any two elements of S is no larger than 32n2

(b) Prove that every set T of 6n pairwise different positive integers contains two elements the least common multiple of which is larger than 9n2

Language: English Time: 4 hours and 30 minutes

Language: English

Day: 1

EGMO 2013European Girls' Mathematical Olympiad

takes integer values

Problem 5 Let Ω be the circumcircle of the triangle ABC The circle ω is tangent to the sides

AC and BC, and it is internally tangent to the circle Ω at the point P A line parallel to AB and intersecting the interior of triangle ABC is tangent to ω at Q.

Prove that ∠ACP = ∠QCB.

Problem 6 Snow White and the Seven Dwarves are living in their house in the forest On each of

16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarvescollected berries in the forest No dwarf performed both types of work on the same day On any twodifferent (not necessarily consecutive) days, at least three dwarves each performed both types of work.Further, on the first day, all seven dwarves worked in the diamond mine

Prove that, on one of these 16 days, all seven dwarves were collecting berries

Language: English

Day: 2

EGMO 2013

Problems with Solutions

Problem Selection Committee:

Charles Leytem (chair), Pierre Haas, Jingran Lin,

Christian Reiher, Gerhard Woeginger.

The Problem Selection Committee gratefully acknowledges the receipt

of 38 problems proposals from 9 countries:

Belarus,

Bulgaria,

Finland,

the Netherlands, Poland, Romania,

Slovenia, Turkey, the United Kingdom.

Problem 1. (Proposed by David Monk, United Kingdom)

The side BC of the triangle ABC is extended beyond C to D so that CD = BC The side CA is extended beyond A to E so that AE = 2CA.

Prove that if AD = BE, then the triangle ABC is right-angled.

Solution 1: Define F so that ABF D is a parallelogram Then E, A, C, F are collinear (as diagonals of a parallelogram bisect each other) and BF = AD = BE Further, A is the midpoint of EF , since AF = 2AC, and thus AB is an altitude of the isosceles triangle

EBF with apex B Therefore AB ⊥ AC.

A

D E

F

A Variant. Let P be the midpoint of [AE], so that AP = AB because AE = 2AB.

of the isosceles triangle CP Q In other words, P A ⊥ AB, which completes the proof.

[ACD] = [ABC] = [EBP ].

But AD = BE, and, as mentioned previously, AC = EP , so this implies that

i.e ∠BEP = ∠CAD It follows that triangles BEP and DAC are congruent, and thus

∠BP A = ∠ACB But AP = AC, so BA is a median of the isosceles triangle BCP Thus

AB ⊥ P C, completing the proof.

B

C A

z2 = x2+ y2+ 2xy cos β in triangle ACD;

z2 = 9x2+ y2− 6xy cos β in triangle BCE.

Hence 4z2 = 12x2+ 4y2 or z2− y2 = 3x2 Let H be the foot of the perpendicular through

B to AC, and write h = BH Then

Hence z2− y2 = EH2 − CH2 Substituting from the above,

Thus H = A, and hence the triangle ABC is right-angled at A.

Remark. It is possible to conclude directly from z2 − y2 = 3x2 = (2x)2 − x2 usingCarnot’s theorem

Solution 5: Writing a = BC, b = CA, c = AB, we have

a2 = b2+ c2− 2bc cos ∠A

c2 = a2+ b2− 2ab cos ∠C

)

in triangle ABC;

and thus AC ⊥ AB, whence triangle ABC is right-angled at A.

Remark. It is perhaps more natural to introduce −→

Solution 7: Let a, b, c, d, e denote the complex co-ordinates of the points A, B, C, D,

E and take the unit circle to be the circumcircle of ABC We have

Thus b − e = (d − a) + 2(b − a), and hence

BE = AD ⇐⇒ (b − e)(b − e) = (d − a)(d − a)

⇐⇒ 2(d − a)(b − a) + 2(d − a)(b − a) + 4(b − a)(b − a) = 0

⇐⇒ 2(d − a)(a − b) + 2(d − a)(b − a)ab + 4(b − a)(a − b) = 0

⇐⇒ (d − a) − (d − a)ab + 2(b − a) = 0

⇐⇒ 2c − b − a − 2cab + a + b + 2(b − a) = 0

⇐⇒ (b + c)(c − a) = 0, implying c = −b and that triangle ABC is right-angled at A.

Solution 8: We use areal co-ordinates with reference to the triangle ABC Recall that if (x1, y1, z1) and (x2, y2, z2) are points in the plane, then the square of the distance between

these two points is −a2vw − b2wu − c2uv, where (u, v, w) = (x1− x2, y1− y2, z1− z2)

In our case A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1), so E = (3, 0, 2) and, introducing point F as in the first solution, F = (−1, 0, 2) Then

and thus a2 = b2 + c2 Therefore ∠BAC is a right angle by the converse of the theorem

of Pythagoras

Problem 2. (Proposed by Matti Lehtinen, Finland)

Determine all integers m for which the m × m square can be dissected into five rectangles, the side lengths of which are the integers 1, 2, 3, , 10 in some

order.

Solution: The solution naturally divides into three different parts: we first obtain some

bounds on m We then describe the structure of possible dissections, and finally, we deal

with the few remaining cases

In the first part of the solution, we get rid of the cases with m 6 10 or m > 14 Let `1, , `5 and w1, , w5 be the lengths and widths of the five rectangles Then therearrangement inequality yields the lower bound

m > 10, each side of the square has at least two adjacent rectangles Hence each side of

the square has precisely two adjacent rectangles, and thus the only way of partitionningthe square into five rectangles is to have a single inner rectangle and four outer rectangleseach covering of the four corners of the square, as claimed

Let us now show that a square of size 12 × 12 cannot be dissected in the desired

boundary of the square) If an outer rectangle has a side of length s, then some adjacent outer rectangle must have a side of length 12 − s Therefore, neither of s = 1 or s = 6

can be sidelengths of an outer rectangle, so the inner rectangle must have dimensions

1 × 6 One of the outer rectangles (say R1) must have dimensions 10 × x, and an adjacent

(12 − y) × z, and rectangle R4 has dimensions (12 − z) × (12 − x) Note that exactly one

of the three numbers x, y, z is even (and equals 4 or 8), while the other two numbers are

odd Now, the total area of all five rectangles is

144 = 6 + 10x + 2y + (12 − y) z + (12 − z)(12 − x),

which simplifies to (y − x)(z − 2) = 6 As exactly one of the three numbers x, y, z is even, the factors y − x and z − 2 are either both even or both odd, so their product cannot equal 6, and thus there is no solution with m = 12.

Finally, we handle the cases m = 11 and m = 13, which indeed are solutions The corresponding rectangle sets are 10 × 5, 1 × 9, 8 × 2, 7 × 4 and 3 × 6 for m = 11, and

10 × 5, 9 × 8, 4 × 6, 3 × 7 and 1 × 2 for m = 13 These sets can be found by trial and

error The corresponding partitions are shown in the figure below

10×5

1×9

8×2 7×4

Remark. The configurations for m = 11 and m = 13 given above are not unique.

A Variant for Obtaining Bounds. We first exclude the cases m 6 9 by the

observa-tion that one of the small rectangles has a side of length 10 and must fit into the square;

hence m > 10.

To exclude the cases m > 14, we work via the perimeter: as every rectangle has at most two sides on the boundary of the m × m square, the perimeter 4m of the square is bounded by 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55; hence m 6 13.

We are left to deal with the case m = 10: clearly, the rectangle with side length 10

must have one its sides of length 10 along the boundary of the square The remaining

rectangle R of dimensions 10 × s, say, would have to be divided into four rectangles with

different sidelengths strictly less than 10 If there were at least two rectangles adjacent

to one of the sides of length s of R, removing these two rectangles from R would leave

a polygon with at least six vertices (since the sidelengths of the rectangles partitioning

R are strictly less than 10) It is clearly impossible to partition such a polygon into no

more than two rectangles with different sidelengths Hence, given a side of length s of R,

there is only one rectangle adjacent to that side, so the rectangles adjacent to the sides

of length s of R would have to have the same length s, a contradiction.

Remark. Note that the argument of the second part of the main solution cannot be

directly applied to the case m = 10.

A Variant for Dealing with m = 12. As in the previous solution, we show that theinner rectangle must have dimensions 1 × 6 Since the area of the square and the area ofthe inner rectangle are even, the areas of the four outer rectangles must sum to an evennumber Now the four sides of the square are divided into segments of lengths 2 and 10,

3 and 9, 4 and 8, and 5 and 7 Hence the sides with adjacent segments of lengths 3 and

9, and 5 and 7 must be opposite sides of the square (otherwise, exactly one of the outerrectangles would have odd area) However, the difference of two rectangle side lengths onopposite sides of the square must be 1 or 6 (in order to accomodate the inner rectangle)

This is not the case, so there is no solution with m = 12.

Remark. In the case m = 12, having shown that the inner rectangle must have

dimen-sions 1 × 6, this case can also be dealt with by listing the remaining configurations one

by one

Problem 3. (Proposed by Dan Schwarz, Romania)

Let n be a positive integer.

(a) Prove that there exists a set S of 6n pairwise different positive integers, such that the least common multiple of any two elements of S is no larger than 32n2

(b) Prove that every set T of 6n pairwise different positive integers contains two elements the least common multiple of which is larger than 9n2

Solution: (a) Let the set A consist of the 4n integers 1, 2, , 4n and let the set B consist of the 2n even integers 4n + 2, 4n + 4, , 8n We claim that the 6n-element set

S = A ∪ B has the desired property.

Indeed, the least common multiple of two (even) elements of B is no larger than

element of A ∪ B is at most their product, which is at most 4n · 8n = 32n2

(b) We prove the following lemma: “If a set U contains m + 1 integers, where m > 2,

that are all not less than m, then some two of its elements have least common multiple

Let the elements of U be u1 > u2 > · · · > u m+1 > m Note that 1/u1 6 1/u i 6 1/m

length By the pigeonhole principle, there exist indices i, j with 1 6 i < j 6 m + 1 such that 1/u i and 1/u j belong to the same subinterval Hence

Now 1/u j −1/u i is a positive fraction with denominator lcm(u i , u j) The above thus yields

the lower bound lcm(u i , u j ) > m2, completing the proof of the lemma

Applying the lemma with m = 3n to the 3n + 1 largest elements of T , which are all not less than 3n, we arrive at the desired statement.

A Variant. Alternatively, for part (b), we prove the following lemma: “If a set U

contains m > 2 integers that all are greater than m, then some two of its elements have

Let u1 > u2 > · · · > u m be the elements of U Since u m > m = m2/m, there exists

a smallest index k such that u k > m2/k If k = 1, then u1 > m2, and the least common

multiple of u1 and u2 is strictly larger than m2 So let us suppose k > 1 from now on, so that we have u k > m2/k and u k−1 6 m2/(k − 1) The greatest common divisor d of u k−1 and u k satisfies

This implies m2/(dk) > k − 1 and u k /d > k − 1, and hence u k /d > k But then the least

common multiple of u k−1 and u k equals

and the proof of the lemma is complete

If we remove the 3n smallest elements from set T and apply the lemma with m = 3n

to the remaining elements, we arrive at the desired statement

Problem 4. (Proposed by Vesna Irˇsiˇc, Slovenia)

Find all positive integers a and b for which there are three consecutive integers

at which the polynomial

P (n) = n

5 + a

b

takes integer values.

Solution 1: Denote the three consecutive integers by x − 1, x, and x + 1, so that

By computing the differences of the equations in (1) we get

For b = 1, it is trivial to see that a polynomial of the form P (n) = n5+ a, with a any

positive integer, has the desired property

For b = 11, we note that

Hence a polynomial of the form P (n) = (n5+ a)/11 has the desired property if and only

if a ≡ ±1 (mod 11) This completes the proof.

A Variant. We start by following the first solution up to equation (4) We note that

have different parity, b must be odd As B in (3) is a multiple of b, we conclude that (i)

(4) is divisible by b, we altogether derive

Together with (2) this implies that

Hence b = 11 is the only remaining candidate, and it is handled as in the first solution.

Solution 2: Let p be a prime such that p divides b For some integer x, we have

(x − 1)5 ≡ x5 ≡ (x + 1)5 (mod p).

Now, there is a primitive root g modulo p, so there exist u, v, w such that

The condition of the problem is thus

either one is the other’s inverse, or one is the other’s square In the first case, we have

or

Thus, since r, s 6≡ 1 (mod p), we have r ≡ −2 (mod p) or s ≡ −2 (mod p), and thus

p ≡ 1 (mod 5), it follows that p = 11, i.e b is a power of 11.

Examining the fifth powers modulo 11, we see that b = 11 is indeed a solution with

a ≡ ±1 (mod 11) and, correspondingly, x ≡ ±4 (mod 11) Now suppose, for the sake of

contradiction, that 112 divides b Then, for some integer m, we must have

contra-Finally, we conclude that the positive integers satisfying the original condition are

b = 11, with a ≡ ±1 (mod 11), and b = 1, for any positive integer a.

Solution 3: Denote the three consecutive integers by x−1, x, and x+1 as in Solution 1.

By computing the differences in (1), we find

By determining the polynomial greatest divisor of F (x) and G(x) using the Euclidean

algorithm, we find that

Problem 5. (Proposed by Waldemar Pompe, Poland)

Let Ω be the circumcircle of the triangle ABC The circle ω is tangent to the sides AC and BC, and it is internally tangent to Ω at the point P A line parallel to AB and intersecting the interior of triangle ABC is tangent to ω

at Q.

Prove that ∠ACP = ∠QCB.

Solution 1: Assume that ω is tangent to AC and BC at E and F , respectively and let P E, P F , P Q meet Ω at K, L, M , respectively Let I and O denote the respective

the arc CA Similarly, L is the midpoint of the arc BC and M is the midpoint of the arc

BA It follows that arcs LM and CK are equal, because

Thus arcs F Q and DE of ω are equal, too, where D is the intersection of CP with ω Since

CE and CF are tangents to ω, this implies that ∠DEC = ∠CF Q Further, CE = CF ,

and thus triangles CED and CF Q are congruent In particular, ∠ECD = ∠QCF , as

required

C

E

F ω

P

Ω

Q K

L M

D

I

A Variant. As above, we show that arcs F Q and DE of ω are equal, which implies that DEF Q is an isoceles trapezoid, and so we have ∠F ED = ∠QF E Together with

|F Q| = |DE|, this implies that, since E and F are images of each other under reflection

in the angle bisector CI of ∠C, so are the segments [EQ] and [F D], and, in particular,

D and Q In turn, this yields ∠ECD = ∠QCF , as required.

Remark. Let J denote the incentre of ABC By Sawayama’s theorem, J is the midpoint

of [EF ], i.e P J is a median of P F E Since C is the intersection of the tangents AC and

BC to the circumcircle of P F E at E and F , respectively, P C is a symmedian of P F E.

Thus ∠CP E = ∠F P J But, since the arcs F Q and DE of ω are equal, ∠CP E = ∠F P Q This shows that J lies on the line P Q.

Another Variant. We show that arcs QE and F D are equal, and then finish as in the

common tangent to ω and Ω, and tangential angles.) Now, by power of a point,

Now DZ k CB implies BZ/BP = CD/CP , and so, dividing the two previous tions by each other, and taking square roots, BF/CF = BP/CP Hence P F bissects angle ∠BP C Now let ∠BP F = ∠F P C = β By tangential angles, it follows that

equa-∠CF D = β Further, ∠BAC = ∠BP C = 2β Let the tangent to ω through Q and parallel to AB meet AC at X Then ∠QXC = 2β, so, since XQ = XE by tangency,

∠QEX = β By tangential angles, it follows that arcs F D and QE are equal, as claimed.

C

E

F ω

Solution 2: Let I and O denote the respective centres of ω and Ω Observe that

CI is the angle bisector of angle ∠C, because ω is tangent to AC and BC Consider the

IQ ⊥ AB, so OM ⊥ AB Thus the diameter OM of Ω passes through the midpoint of

the arc AB of Ω, which also lies on the angle bisector CI This implies that ∠ICM = 90◦

We next show that P, I, Q, C lie on a circle Notice that

Hence P, I, Q, C lie on a circle But P I = IQ, so CI is the angle bisector of ∠P CQ Since

CI is also the angle bisector of angle ∠C, it follows that ∠ACP = ∠QCB, as required.

C

ω

P

O Ω

Q M

I

T

A Variant. We show that P IQC is cyclic by chasing angles Define α = ∠BAC,

β = ∠CBA and γ = ∠ACP For convenience, we consider the configuration where A

and P lie one the same side of the angle bisector CI of ∠C In this configuration,

Let AB and P O intersect at T Then

But QI ⊥ AB by construction, and thus

AC at S Hence ID ⊥ ` By construction, P , I, O lie one a line, and hence the isosceles

triangles P ID and P OC are similar In particular, it follows that OC ⊥ `, so C is the midpoint of the arc of Ω defined by the points of intersection of ` with Ω It is easy to

see that this implies that

∠DSC = ∠ABC.

Under reflection in the angle bisector CI of ∠C, ` is thus mapped to a tangent to ω parallel to AB and intersecting the interior of ABC, since ω is mapped to itself under this reflection In particular, D is mapped to Q, and thus ∠QCB = ∠ACD, as required.

C

ω

P

O Ω

I

D S

`

Remark. Conceptually, this solution is similar to Solution 1, but here, we proceedmore directly via the reflectional symmetry Therefore, this solution links Solution 1 toSolution 4, in which we use an inversion

Solution 4: Let the tangent to ω at Q meet AC and BC at X and Y , respectively Then

circle Γ Under I ,

In particular, ω0, the image of ω, is a circle tangent to AC, BC and A0B0, so it is either

the excircle of CA0B0 opposite C, or the incircle of CA0B0 Let ω be tangent to BC at F ,

Solution 5: Let r be the radius such that r2 = AC ·BC LetJ denote the composition

of the inversionI in the circle of centre C and radius r, followed by the reflection in the

angle bisector of ∠C Under J ,

A 7−→ B, B 7→ A;

Ω 7−→ the line AB;

AB, so it is either the incircle of ABC, or the excircle opposite vertex C Observe that

r > min {AC, BC}, so the image of the points of tangency of ω must lie outside ABC,

and thus ω0 cannot be the incircle Thus ω0 is the excircle opposite vertex C as claimed Further, the point of tangency P is mapped to Q0

CQ0 underJ , CP and CQ0 are images of each other under reflection in the angle bisector

onto the excircle ω0 This completes the proof

Now, a point Z on the circle Ω satisfies

|z − o|2 = (o + 1)2 ⇐⇒ zz∗− o(z + z∗) − 2o − 1 = 0.

The triangle ABC is defined by the points E and F on ω, the intersection C of the corresponding tangents lying on Ω Thus c = 2ef.(e + f ), and further

and this is the equality defining o The points A and B are the second intersection points

of Ω with the tangents to ω at E and F respectively A point Z on the tangent through



2e − e2z∗z∗− o2e − e2z∗+ z∗− 2o − 1 = 0

⇐⇒ −e2z∗2+2e + oe2− oz∗−2eo + 2o + 1= 0

⇐⇒ z∗2−2e∗+ o − oe∗2z∗+2e∗o + 2oe∗2 + e∗2= 0, since |e| = 1 Thus

using p = −1 to obtain the final equality.

Now, it suffices to show that (i) DQ k EF ⊥ CI and (ii) the midpoint of [DQ] is on

CI The desired equality then follows by symmetry with respect to the angle bisector of

the angle ∠ACB Notice that (i) is equivalent with

I, and the third equality follows from the expression for c But we have just shown that

dq = ef This proves (ii), justifies the choice of sign for q a posteriori, and thus completes

the solution of the problem

Problem 6. (Proposed by Emil Kolev, Bulgaria)

Snow White and the Seven Dwarves are living in their house in the forest.

On each of 16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarves collected berries in the forest No dwarf performed both types of work on the same day On any two different (not necessarily consecutive) days, at least three dwarves each performed both types of work Further, on the first day, all seven dwarves worked in the diamond mine.

Prove that, on one of these 16 days, all seven dwarves were collecting berries.

Solution 1: We define V as the set of all 128 vectors of length 7 with entries in {0, 1} Every such vector encodes the work schedule of a single day: if the i-th entry is 0 then the i-th dwarf works in the mine, and if this entry is 1 then the i-th dwarf collects berries The 16 working days correspond to 16 vectors d1, , d16 in V , which we will call day-

vectors The condition imposed on any pair of distinct days means that any two distinct

day-vectors d i and d j differ in at least three positions

We say that a vector x ∈ V covers some vector y ∈ V , if x and y differ in at most one position; note that every vector in V covers exactly eight vectors For each of the 16 day-vectors d i we define B i ⊂ V as the set of the eight vectors that are covered by d i As,

for i 6= j, the day-vectors d i and d j differ in at least three positions, their corresponding

sets B i and B j are disjoint As the sets B1, , B16 together contain 16 · 8 = 128 = |V | distinct elements, they form a partition of V ; in other words, every vector in V is covered

by precisely one day-vector

day-vectors d1, , d16 by their weights, and let us discuss how the vectors in V are covered

by them

1 As all seven dwarves work in the diamond mine on the first day, the first day-vector is

d1 = (0000000) This day-vector covers all vectors in V with weight 0 or 1.

2 No day-vector can have weight 2, as otherwise it would differ from d1 in at most twopositions Hence each of the 72 = 21 vectors of weight 2 must be covered by someday-vector of weight 3 As every vector of weight 3 covers three vectors of weight 2,

exactly 21/3 = 7 day-vectors have weight 3.

3 How are the73= 35 vectors of weight 3 covered by the day-vectors? Seven of them areday-vectors, and the remaining 28 ones must be covered by day-vectors of weight 4 As

every vector of weight 4 covers four vectors of weight 3, exactly 28/4 = 7 day-vectors

To summarize, one day-vector has weight 0, seven have weight 3, and seven have weight 4.None of these 15 day-vectors covers any vector of weight 6 or 7, so that the eight heavy-

weight vectors in V must be covered by the only remaining day-vector; and this remaining

vector must be (1111111) On the day corresponding to (1111111) all seven dwarves arecollecting berries, and that is what we wanted to show

Solution 2: If a dwarf X performs the same type of work on three days D1, D2, D3,

then we say that this triple of days is monotonous for X We claim that the following configuration cannot occur: There are three dwarves X1, X2, X3 and three days D1, D2,

D3, such that the triple (D1, D2, D3) is monotonous for each of the dwarves X1, X2, X3.(Proof: Suppose that such a configuration would occur Then among the remaining

dwarves there exist three dwarves Y1, Y2, Y3 that performed both types of work on day

D1 and on day D2; without loss of generality these three dwarves worked in the mine on

day D1 and collected berries on day D2 On day D3, two of Y1, Y2, Y3 performed the same

type of work, and without loss of generality Y1 and Y2 worked in the mine But then on

days D1 and D3, each of the five dwarves X1, X2, X3, Y1, Y2 performed only one type ofwork; this is in contradiction with the problem statement.)

Next we consider some fixed triple X1, X2, X3 of dwarves There are eight possible

working schedules for X1, X2, X3 (like mine-mine-mine, mine-mine-berries, mine, etc) As the above forbidden configuration does not occur, each of these eightworking schedules must occur on exactly two of the sixteen days In particular thisimplies that every dwarf worked exactly eight times in the mine and exactly eight times

mine-berries-in the forest

For 0 6 k 6 7 we denote by d(k) the number of days on which exactly k dwarves

were collecting berries Since on the first day all seven dwarves were in the mine, on each

of the remaining days at least three dwarves collected berries This yields d(0) = 1 and

d(1) = d(2) = 0 We assume, for the sake of contradiction, that d(7) = 0 and hence

As every dwarf collected berries exactly eight times, we get that, further,

Next, let us count the number q of quadruples (X1, X2, X3, D) for which X1, X2, X3

are three pairwise distinct dwarves that all collected berries on day D As there are

7 · 6 · 5 = 210 triples of pairwise distinct dwarves, and as every working schedule for three

fixed dwarves occurs on exactly two days, we get q = 420 As every day on which k dwarves collect berries contributes k(k − 1)(k − 2) such quadruples, we also have

3 · 2 · 1 · d(3) + 4 · 3 · 2 · d(4) + 5 · 4 · 3 · d(5) + 6 · 5 · 4 · d(6) = q = 420,

which simplifies to

Finally, we count the number r of quadruples (X1, X2, X3, D) for which X1, X2, X3 are

three pairwise distinct dwarves that all worked in the mine on day D Similarly as above

we see that r = 420 and that

7 · 6 · 5 · d(0) + 4 · 3 · 2 · d(3) + 3 · 2 · 1 · d(4) = r = 420,

which simplifies to

Multiplying (1) by −40, multiplying (2) by 10, multiplying (3) by −1, multiplying (4) by

4, and then adding up the four resulting equations yields 5d(3) = 30 and hence d(3) = 6 Then (4) yields d(4) = 11 As d(3) + d(4) = 17, the total number of days cannot be 16.

We have reached the desired contradiction

A Variant. We follow the second solution up to equation (3) Multiplying (1) by 8,multiplying (2) by −3, and adding the two resulting equations to (3) yields

As d(5) and d(6) are positive integers, (5) implies 0 6 d(6) 6 2 Only the case d(6) = 1 yields an integral value d(5) = 4 The equations (1) and (2) then yield d(3) = 10 and

d(4) = 0.

Now let us look at the d(3) = 10 special days on which exactly three dwarves were

collecting berries One of the dwarves collected berries on at least five special days (if every

dwarf collected berries on at most four special days, this would allow at most 7 · 4/3 < 10 special days); we call this dwarf X On at least two out of these five special days, some dwarf Y must have collected berries together with X Then these two days contradict

the problem statement We have reached the desired contradiction

Comment. Up to permutations of the dwarves, there exists a unique set of day-vectors(as introduced in the first solution) that satisfies the conditions of the problem statement: