mosp 2004 solution Solutions to the Mathematics Olympiad Test 1

mosp 2004 solution Solutions to the Mathematics Olympiad Test 1

MOP 2004 - 2005Solutions to the Mathematics Olympiad Test 1

1.1 Let n be a positive integer There are n wood blocks, numbered

1, 2, , n placed on a circle in clockwise order Zachary is playingthe following game At each step, he is allowed to pick fourconsecutive blocks and reverse their order For n = 20, determine

if it is possible for Zachary to obtain blocks 6, 1, 2, 3, 4, 5, 7, 8, , 20 in clockwise order in a finite number of steps

(Query: What if n = 19?)

Solution: The answer is yes We use letters a, b, , t instead

of numbers 1, 2, , 20 Applying the sequence of moves

bcde F → bFedc → deFbc → dcbFe → F bcde

repeatedly, one can obtain

a bcde F ghi jklm nopq rst → a F bcde ghi jklm nopq rst

→ a bcde ghi jklm nopq F rst →

→ abcd eghi F jklm nopq rst →

→ F abcd eghi jklm nopq rst.Starting with 1, 2, , 20, one ends up exactly with the desiredsequence 6, 1, 2, 3, 4, 5, 7, 8, , 20

Note: If you think the bcdeF → Fbcde thing is tricky, you’rewrong It’s entirely obvious Given five blocks on a line, onlytwo moves are possible: flip the first four, or flip the last four.Making the same choice twice in a row amounts to doing nothing.Therefore, the only sensible thing to do is to alternate betweenthe two

In the case n = 19, one cannot obtain 6, 1, 2, 3, 4, 5, 7, 8, , 19.The operation abcd → dcba is an even permutation, and so isthe operation of cyclically permuting 19 blocks Because thoseare the only two types of operations allowed, no odd permutation,such as the one desired, can be obtained

1.2 Let ABCD be a convex quadrilateral with AB not parallel to

CD, and let X be a point inside ABCD such that ∠ADX =

∠BCX < 90◦and ∠DAX = ∠CBX < 90◦ If the perpendicularbisectors of segments AB and CD intersect at Y , prove that

∠AY B = 2∠ADX

First Solution:

C D

X

W Z

Figure 1.1.

Construct point Z inside ABCD such that AZ = ZB and

∠AZB = 2∠ADX It suffices to show that DZ = CZ Construct

W within the non-reflex angle AZB so that triangle AZW issimilar to triangle ADX or, equivalently, triangle BZW is similar

to triangle BCX Then triangle ADZ is similar to triangle AXWand triangle BCZ is similar to triangle BXW , implying that

Second Solution: (By Tony Zhang) Construct point M on theside of line AB opposite X such that triangle M AB is similar

to triangle Y DC Construct point N on the side of line CDopposite X such that triangle N CD is similar to triangle Y BA.Construct point X1 on the same side of line AD as B such thattriangle X1AD is similar to triangle CY N , and construct point

X2 on the same side of line BC as D such that triangle X2BC issimilar to triangle DY N It suffices to show X1= X2, because X

is the unique point with the properties described in the problem.Note that the triangles AY M , BY M , DY N , CY N , AX D and

BX2C are all similar Now use complex numbers; each lowercaseletter will denote the complex number coordinate of the pointnamed by the corresponding uppercase letter Set y = 0 and

r = d/n Then

d = rn c = rn a = m − rm b = m − rmand

Note: We start with the following lemma

Lemma 1 For any odd positive integers a, b,

22b−a(2a+ 1) ≡ 22b−a(2a+ 1) − (2b+ 1)(2b− 1)

≡ 22b−a+ 1 (mod 2b+ 1)

In either case,

gcd(2a+ 1, 2b+ 1) = gcd(2|a−2b|+ 1, 2b+ 1),

contradicting the assumption of the minimality of a + b

Please also compare this lemma to lemma 2 in the next session

In this problem, the full generality of lemma 1 is not necessary

The following proofs rely only on the weaker statement that, forany relatively prime odd positive integers a, b not divisible by

3, gcd(2a + 1, 2b+ 1) = 3 The proof of this is a bit simpler.Suppose that p is prime and 2a ≡ 2b≡ −1 (mod p) Then if c isthe minimum positive integer such that 2c ≡ 1 (mod p), then cmust divide both 2a and 2b, so c is either 1 or 2, which implies

p = 3 It remains to note that 9 divides neither 2a+ 1 nor 2b+ 1.First Solution: Call an integer “tenebrous” if it is odd, square-free, not divisible by 3, and at least 5 For any integer m, letψ(m) denote the number of distinct prime factors of m, and letd(m) denote the number of factors of m We wish to prove thatd(2a+ 1) ≥ 4ψ(a) for all tenebrous integers a

Induct on ψ(a) For the base case ψ(a) = 1, 2a+ 1 is divisible

by 3 exactly once and is greater than 3, so ψ(2a + 1) ≥ 2 andd(2a+ 1) ≥ 4

Now let a, b be relatively prime tenebrous integers such thatthe claim holds for both a and b Clearly 2ab+ 1 is divisible byboth 2a+ 1 and 2b+ 1, so we can write

2ab+ 1 = C · lcm[2a+ 1, 2b+ 1]

Because ab − 2a − 2b − 4 = (a − 2)(b − 2) − 8 > 0,

2ab+ 1 > 22a+2b+4> (2a+ 1)2(2b+ 1)2> lcm[2a+ 1, 2b+ 1]2,

so C ≥ lcm[2a+ 1, 2b+ 1] From the comment, we have gcd(2a+

1, 2b+ 1) = 3, so as 3 divides each of 2a+ 1 and 2b+ 1 exactlyonce,

d(lcm[2a+ 1, 2b+ 1]) = d(2

a+ 1)d(2b+ 1)

2 ≥ 22ψ(a)+2ψ(b)−1.For every divisor m of lcm[2a+ 1, 2b+ 1], both m and Cm aredivisors of 2ab+ 1 As C > lcm[2a+ 1, 2b+ 1],

d(2ab+ 1) ≥ 2 · d(lcm[2a+ 1, 2b+ 1]) ≥ 4ψ(a)+ψ(b),completing the induction

Second Solution: Following the notation of the first solution,

a stronger claim is that, for any tenebrous integer a,

ψ(2a+ 1) ≥ 2ψ(a)

We proceed by induction on ψ(a) The base case is the same as

in the first solution

Now let a, b be coprime tenebrous integers We claim thatψ(2ab+ 1) ≥ ψ(2a+ 1) + ψ(2b+ 1)

¶(−2a− 1)i−1

≡ b −

µb2

¶(2a+ 1) (mod (2a+ 1)2),

so if a prime p divides 2a+ 1 exactly k ≥ 1 times, then p divides

2ab+ 1 either k times (if p doesn’t divide b) or k + 1 times (if pdivides b) In any case p divides 2ab+ 1 at most twice as manytimes as p divides 2a+ 1 The same is true for prime factors of

2b+ 1

As in the first solution, 2ab+ 1 > (2a+ 1)2(2b+ 1)2, so, in light

of the above, 2ab+ 1 must have a prime factor dividing neither

2a+ 1 nor 2b+ 1

Clearly 2ab+1 is divisible by lcm[2a+1, 2b+1] Because 2ab+1has a prime factor not dividing lcm[2a+ 1, 2b+ 1], we haveψ(2ab+ 1) ≥ ψ(lcm[2a+ 1, 2b+ 1]) + 1

= ψ(2a+ 1) + ψ(2b+ 1) − ψ(gcd(2a+ 1, 2b+ 1)) + 1

= ψ(2a+ 1) + ψ(2b+ 1) − ψ(3) + 1

= ψ(2a+ 1) + ψ(2b+ 1),

completing the induction

1.4 Let n be an integer greater than 1 Let a1, a2, , an, b1, b2, , bn

be nonnegative real numbers with a1a2· · · an = b1b2· · · bn, b1+

Determine the maximum value of a1+ a2+ · · · + an

Solution: The answer is n−1, which can be obtained by setting

a1 = 0 and a2 = · · · an = 1, and b1 = b2 = · · · = bn−1 = 0 and

bn = 0 Now we show that

establishing the inequality (∗)

If a1 > 0 and n = 2 Then the given inequality becomes

a2−a1≤ b2−b1 It follows that (a1+ a2)2= 4a1a2+ (a1−a2)2≤4a1a2+(b2−b1)2= 4b1b2+(b2−b1)2= (b2+1b1)2= 1, establishingthe inequality (∗)

Now we assume that a1> 0 and n > 2 We approach indirectly

by assuming that the inequality (∗) is false; that is

³

n 2(n−1)

´n−1

(n − 1)! <

n − 1

n ,implying that Pn

i=1ai ≤ nan < n − 1, contradicts to ourassumption (∗′) Thus our assumption was wrong and the desiredinequality (∗) holds

Note: One of the key to this solution is always considering theequality case of the inequality (∗) by taking out terms ai− ajfor

1 < i < j ≤ n

Some Important Ideas You Should Know

F1.1 To clip a convex n-gon means to choose a pair of consecutive sides

AB, BC and to replace them by the three segments AM, M N ,and N C, where M is the midpoint of AB and N is the midpoint

of BC In other words, one cuts off the triangle M BN to obtain

a convex (n + 1)-gon A regular hexagon P6 of area 1 is clipped

to obtain a heptagon P7 Then P7is clipped (in one of the sevenpossible ways) to obtain an octagon P8, and so on Prove that

no matter how the clippings are done, the area of Pn is greaterthan 1/3, for all n ≥ 6

Solution: The key observation is that for any side S of of P6,there is some sub-segment of S that is a side of Pn (This is easilyproved by induction on n.) Thus Pn has a vertex on each side of

P6 Since Pn is convex, it contains a hexagon Q with (at least)one vertex on each side of P6 (The hexagon may be degenerate,

as some of its vertices may coincide.)

Let P6= A1A2A3A4A5A6, and let Q = B1B2B3B4B5B6, with

Bion AiAi+1(indices are considered modulo 6) The side BiBi+1

of Q is entirely contained in triangle AiAi+1Ai+2, so Q enclosesthe smaller regular hexagon R (shaded in the diagram below)whose sides are the central thirds of the segments AiAi+2, 1 ≤ i ≤

6 The area of R is 1/3, as can be seen from the fact that its sidelength is 1/√

3 times the side length of P6, or from a dissectionargument (count the small equilateral triangles and halves thereof

in the diagram below) Thus [Pn] ≥ [Q] ≥ [R] = 1/3 We obtainstrict inequality by observing that mathcalPn is strictly largerthat Q: if n = 6, this is obvious; if n > 6, then Pn cannot equal

Q because Pn has more sides

Note: With a little more work, one could improve 1/3 to 1/2.The minimal area of a hexagon Q with one vertex on each side

of P6 is in fact 1/2, attained when the vertices of Q coincide inpairs at every other vertex of P6, so the hexagon Q degeneratesinto an equilateral triangle If the conditions of the problem werechanged so that the cut-points could be anywhere within adjacentsegments instead of just at the midpoints, then the best possiblebound would be 1/2

F1.2 Find all integers n for which a regular hexagon can be dividedinto n parallelograms of equal area.

F1.3 Let n be a positive integer greater than or equal to three.(a) Let C be a circle with center O and radius 1, and let

S1, S2, Sn be n semicircles with center O and radius 1.Circle C is covered by semicircles S1, S2, , and Sn Provethat one can find indices i1, i2, and i3 such that sphere C iscovered by Si 1, Si 2, and Si 3

(b) Let S be a sphere with center O and radius 1, and let

H1, H2, Hn+1 be n + 1 hemispheres with center O andradius 1 Sphere S is covered by hemispheres H1, H2, ,and Hn Prove that one can find indices i1, i2, i3, i4such thatsphere S is covered by Hi 1, Hi 2, Hi 3, Hi 4

F1.4 Let P be a 1000-sided regular polygon Some of its diagonals weredrawn to obtain a triangulation the polygon P (The region inside

P is cut into triangular regions, and the diagonals drawn onlyintersect at the vertices of P ) Let n be the number of differentlengths of the drawn diagonals Determine the minimum value ofn

(By Tiankai Liu) The answer is 10

This can be achieved by using diagonals corresponding to arclengths of 400, 200, 120, 80, 40, 24, 16, 8, 4, and 2 First

we can draw a regular pentagon with sides subtending arcs oflength 200; the interior of this pentagon can be triangulatedwith diagonals subtending arcs of length 400 Then we divideeach the remaining five congruent 201-gons into five congruent41-gons using diagonals subtending arcs of length 40; the interiors

of the hexagons thus formed can be triangulated by diagonalssubtending arcs of lengths 80 and 120 Proceeding in a sim-ilar manner, We divide each of the 25 congruent 41-gons intofive nonagons using diagonals subtending arcs of length 8; theinteriors of the hexagons thus formed can be triangulated bydiagonals subtending arcs of lengths 16 and 24 Each of theremaining nonagons can be triangulated in a straightforward waywith diagonals subtending arcs of lengths 8, 4, and 2

Suppose, for the sake of contradiction, that the answer is lessthan 10 Let the arc lengths corresponding to the different

diagonals be a1, , an, where we always choose the arc length to

be at most 500, such that 1 < a1< a2< · · · < an≤ 500 Set also

a0= 1; this corresponds to the edges of the original 1000-gon.Note that either an = 500 or there must exist some trianglecontaining the center of the 1000-gon, in which case there exist

i1, i2, i3such that ai 1+ ai 2+ ai 3 = 1000 In particular, an ≥ 334.Note also that for each 1 ≤ i ≤ n, ai≤ 2ai−1 Thus, if n ≤ 8,then an ≤ 28= 256 < 334, a contradiction, so n = 9

Now consider the number w of i, 1 ≤ i ≤ n, for which ai <2ai−1 If w = 0, then a9 = 29 = 512 > 500, a contradiction If

w ≥ 2, then it is not difficult to see that a9≤ 27·5/2 = 320 < 334,also a contradiction Therefore w = 1 Let k be the uniquenumber such that ak < 2ak−1 Note that a9≤ 28· 3/2 = 384 <

500 Then each of the numbers a0, a1, , ak−1 contains one 1 inits binary representation Each of the numbers ak, , a9containstwo 1s in its binary representation But because a9< 500, thereexist i1, i2, i3 for which ai 1+ ai 2+ ai 3 = 1000, and the number

1000 has six 1s in its binary representation There must be nocarrying in the addition ai 1+ ai 2+ ai 3, and so at least one of ai 1,

ai 2, ai 3 is at least as big as the place value of the highest 1 in thebinary representation of 1000 But ai 1, ai 2, ai 3 ≤ a9≤ 384 < 512,

a contradiction

Therefore 10 is indeed the answer

MOP 2004 - 2005Solutions to the Mathematics Olympiad Test 2

2.1 Let m be a fixed positive integer with m ≥ 3 A set S is good ifthere are elements s1, s2, , sm−1, sm (not necessarily distinct)

in S such that s1+s2+· · ·+sm−1= sm Determine the minimumpositive integer f (m) such that for any partition A and B of theset {1, 2, , f(m)}, at least one of the sets A and B is good

Solution: We show that f (m) = m2− m − 1 Indeed, we maypartition {1, 2, , m2− m − 2} into the following sets A, B thatare not good:

A = {1, 2, , m − 2, (m − 1)2, (m − 1)2+ 1, , m2− m − 2}

B = {m − 1, m, , m2− 2m}

Any sum of m−1 elements of A that all belong to {1, 2, , m−2}

is between m − 1 and (m − 2)(m − 1) inclusive, hence is notcontained in A; all other sums are at least (m − 1)2+ (m − 2) >

m2− m − 2, so A is not good Also, B is not good because theminimal sum of m − 1 elements of B is (m − 1)2> m2− 2m Bytaking subsets of A and B, we may obtain for any k ≤ m2−m−2partitions of {1, 2, , k} that are not good

It remains to show that any partition {1, 2, , m2− m − 1} =

A ∪ B contains at least one good set Assume the contrary, andassume without loss of generality that 1 ∈ A; then m−1 ∈ B and(m−1)2∈ A Also, m2−m−1 = (m−1)2+(m−2) = (m−1)2+1+1+· · ·+1 ∈ B, and m+m+· · ·+m+1 = (m−2)m+1 = (m−1)2, so

m ∈ B But then m+m+· · ·+m+(m−1) = (m−2)m+(m−1) =

m2− m − 1 is a sum of m − 1 elements of B that is itself in B, acontradiction

2.2 Each square of a (2n− 1) × (2n− 1) board contains either +1 or

−1 Such an arrangement is called successful if each number isthe product of its neighbors (squares sharing a common side withthe given square) Find the number of successful arrangements.First Solution: We show that there are two successful arrange-ments for n = 1 and only one arrangement for each n > 1 In

all cases, the matrix containing all +1’s is successful, and in thecase n = 1 the matrix containing a single −1 is also successful.

We induct on n to show that these are the only successfularrangements The base case n = 2 can be verified easily Assumenow that there is only one successful (2n− 1) × (2n− 1) matrix,and suppose for a contradiction that there exists a successful(2n+1− 1) × (2n+1− 1) matrix (aij) that contains at least one

−1 We will call a matrix containing at least one −1 nontrivial.Also, we let N = 2n+1

We claim that then there exists a nontrivial successful matrixthat has both vertical and horizontal lines of symmetry Indeed,

if (aij) does not have a horizontal line of symmetry, then thematrix (bij) given by

Consider now the middle row and middle column of (cij) Bysymmetry, the vertical neighbors of the center entry are equal,

as are the horizontal neighbors; hence, the center entry must be+1 If the horizontally adjacent entries are +1, then using thevertical symmetry repeatedly, we see that all entries of the centerrow are +1 Otherwise, the entries of the center row, startingfrom the center entry and working toward either side, must be+1, −1, −1, +1, −1, −1, +1, , −1, −1, +1, −1, −1 But then thewidth of the matrix is 2n+1− 1 ≡ 2 (mod 3), which is impossible

It follows that the center row must contain only +1’s, and likewisefor the center column

We may now observe that each of the four remaining (2n−1)×(2n− 1) squares obtained by deleting the center row and columnare successful Applying the induction hypothesis, we see thatthe trivial matrix is the only successful matrix in the case n + 1.Second Solution: We outline an approach using linear algebra

We may rephrase the problem additively, by substituting 0 for +1and 1 for −1, and replacing multiplication with addition (mod 2)

Then the condition that a matrix (aij) be successful becomes asystem of linear equations (mod 2, or more precisely, over thefield F2), of the form

ai−1,j+ ai,j−1+ ai,j+ ai+1,j+ ai,j+1= 0

for non-edge entries aij; when aij is an edge entry there willonly be four summands, and corner entries will give equationswith three summands, in the case n > 1 Thus, showing thatthe trivial arrangement is the only successful matrix amounts

to proving that this system of (2n− 1)2 equations in (2n− 1)2

unknowns has a unique solution

To do so, it suffices to show that the determinant of the system

is nonzero (mod 2), or in other words, is odd Recall that thedeterminant can be expanded as a sum of products of terms,one from each row and one from each column, with sign chosenappropriately In this case, the sign does not matter since we areworking mod 2 Hence, it suffices to show that the number ofnonzero terms is odd

Now the (2n−1)2×(2n−1)2matrix corresponding to our system

of equations is sparse; the row corresponding to the equation for

aij contains all 0’s except for 1’s as the coefficients of aij and itsneighbors Hence, the nonzero terms in the determinant, whicharise from choosing a 1 from each row and column, correspond

to bijections from {aij} to itself such that each aij is mapped

to a neighboring square We may think of such a bijection as adiagram of arrows pointing from the center of each square to thesquare it is mapped to, with dots (degenerate arrows) in squaresthat are mapped to themselves We wish to show that the number

of such “arrow diagrams” is odd

Note now that an arrow diagram that is not symmetric zontally can be paired with its reflection; likewise for diagramsthat are not symmetric vertically Thus, it suffices to show thatthere is an odd number of arrow diagrams with both horizontaland vertical symmetries Now, such diagrams must contain dots

hori-in all entries of the center row and center column But then thediagram is determined by its top-left (2n−1 − 1) × (2n−1 − 1)corner, which must be a legitimate diagram for the case n − 1.The result follows by induction

2.3 Let ABC be a triangle, and let D be a point on side BC suchthat ∠BAD = ∠CAD Line AD meet the common tangent lines

to the circumcircles of triangles ABD and ACD at points P and

Q Prove that P Q2= AB · AC

Note: Let ω1and ω2denote the circumcirlces of triangles ABDand ACD, respectively Let Y U and XZ be the common tangentlines to ω1 and ω2 If Y U k XZ, then it is not difficult to seethat AB = AC, ∠ADB = ∠ADC = 90◦, and P Q = AB = AC.Hence the desired result is trivial

In the following solutions, we assume that lines Y U and XZmeet at T Without loss of generality, we label the points asshown in Figure 2.1

Z

Figure 2.1.

First Solution: Consider the dilation T taking ω1 to ω2 It

is clear that T is the center of T Assume that line DT meet ω1

and ω2 again (other than D) at B1 and C1, respectively ThenT(B1) = D, T(D) = C1, and T([B1D) = [DC1, implying that

∠B1AD = ∠DAC1 On the other hand, consider a point B′

moving along omega1 from D to A, line B′D meet ω2 again C′

As B′moving from D towards A, C′is moving from A towards Dalong ω2; that is, ∠DAB′ is increasing and ∠C′AD is decreasing.Hence there is an unique point B such that ∠BAD = ∠CAD

We conclude that B1= B, C1= C; that lines BC, XZ, and Y Uare concurrent at T

Because T(B) = D and T(X) = Z, it follows that BX k DZ

Likewise, XD k ZC Because BX k DZ and ZT is tangent

to ω2 at Z, ∠BXT = ∠DZT = ∠ZCD, or BCZX is cyclic,and let ω3 denote the circumcircle of BCZX Therefore, lines

P Q, CZ, and BX are the radical axes of ω1and ω2, ω2 and ω3,and ω3 and ω1 Hence lines P Q, CZ, and BX are concurrent,and let W denote the point of concurrence Then DXW Z is aparallelogram, and so DQ = QW and P Q = AW It suffices toshow that AW2= AB · AC

By cyclic quadrilateral ACZD and parallelogram DXW Z,

∠ACW = ∠ACZ = ∠ZDW = ∠DW X = ∠AW B Likewise,

∠BAW = ∠AW C Hence triangle ABW is similar to triangle

Z

B C

K L

M

Figure 2.2.

As shown in Figure 2.2, let K and L be the circumcenters oftriangles ABD and ACD, respectively Because segment XZ issymmetric to segment Y U across line KL, XY U Z is an isoscelestrapezoid Since P, Q are the midpoints of segments Y U and XZ,respectively, P Q k XY k UZ and 2P Q = XY + UZ It suffices

to show that (U Z + XY )2= 4P Q2= 4AB · AC

Let r and r be the radius of ω and ω , respectively It is clear

that LZ k KX and LU k KY , implying that isosceles triangles

U LZ and Y KX are similar with ratio r 1

U Z2r2

∠M KL Note that M L = r2− r1 = AL − AK, which impliesthat, in right triangle KLM , sin ∠MKL =M L

KL andcos ∠MKL =p

K

L

2g

g 2g

a a

∠BKA Because ∠CLA = ∠XKY and both CAL and BAKare isosceles triangles, they are similar; that is, there is a spiralsimilaritycentered at A sending triangle ALC to triangle AKB.Consequently, there is a spiral similarity centered at A sendingtriangle ALK to triangle ACB, implying that triangle KAL issimilar to triangle BAC The equation (∗′) becomes

s

1 −

µ

AC − ABBC

p

BC2− AC2− AB2+ 2AC · AB

=p

2AC · AB(1 − cos 2α) = 2 sin α√AC · AB

It remains to show that

ABsin γ +

ACsin γ =

BDsin α +

CDsin α,which is evident, because the first (second) summand on theleft-hand side of the above equation is equal to the first (second)summand on the right-hand side, by applying the law of sines

to triangle ABD (ACD)

2.4 Let p be a prime number Prove that there exist infinitely manyprime numbers q such that for every integer n, the number np− p

is not divisible by q

First Solution: We will need the following fact

Lemma 2 If a, m, n ∈ Z, a > 1, m > 0, n > 0, then

gcd(am− 1, an− 1) = agcd(m,n)− 1

Proof: Induct on m + n ≥ 2 For m = n the statement istrivial Suppose that the statement holds for any m′, n′ ∈ N withsum m′+ n′< m + n Without loss of generality, we assume that

For each prime r 6= p let F (r) = (p(p prp−1)(p−1)(p−1)r −1) By lemma 2gcd(pr− 1, pp− 1) = p − 1 Since both pr− 1 and pp− 1 divide

prp− 1, their product divides (prp− 1)(p − 1) Therefore, F (r) is

an integer

Lemma 3 For any distinct prime numbers r, s 6= p

gcd(F (r), F (s)) = 1Proof: By lemma 2 gcd(prp− 1, psp− 1) = pp− 1 Therefore,gcd(pprpp −1−1,ppspp −1−1) = 1 Since F (r) | pprpp −1−1 and F (s) | pprpp −1−1,gcd(F (r), F (s)) = 1

Lemma 4 For any prime r 6= p there is a prime divisor q of

F (r) such that p | q − 1 but p26 |q − 1

Proof: Clearly, F (r) > 1 First we will show that any primedivisor q of F (r) is equivalent to 1 modulo p We know that

q | prp−1 Therefore, q 6= p and q | pq−1−1 by Fermat’s theorem.Applying lemma 2, we have q | pgcd(q−1,rp)− 1 This means thateither p | q − 1 or q | pr− 1 However,

This means, q cannot divide pr− 1 and p | q − 1

Now we know that all prime divisors of F (r) are equivalent to 1modulo p We need to find one that is not equivalent to 1 modulo

p2 If we assume that all prime divisors of F (r) are 1 modulo

p2, then F (r) ≡ 1 (mod p2) as their product On the other hand

F (r) = (p(prpp −1)(p−1)(p−1)r −1), where prp− 1 ≡ pp− 1 ≡ pr− 1 (mod p2).Thus, F (r) ≡ 1 − p (mod p2) This contradiction finishes theproof

Lemma 5 If r 6= p is a prime number and q is a prime divisor

of F (r) such that p | q − 1 and p26 |q − 1, then for any integer n

np− p is not divisible by q

Proof: Let q = kp+1, where p 6 |k Suppose np ≡ p (mod q) forsome integer n Then npk ≡ pk(mod q) Because clearly q 6= p,

npk = nq−1 ≡ 1 (mod q) by Fermat’s theorem This implies

q | pk − 1 On the other hand, q | prp− 1 Since p 6 |k, lemma

2 implies q | pr− 1 On the other hand,

This means, q cannot divide pr−1, which leads to a contradiction

By lemma 4 for each prime r 6= p we can pick such primenumber qr | F (r) that p | qr− 1 and p2 6 |qr− 1 Then lemma

5 shows that each qr satisfies the desired condition and lemma 3shows that all qrare different since they divide pairwise relativelyprime numbers F (r)

Second Solution: We start with two lemmas The first is wellknown, and the second is problem 6 of IMO 2003

Lemma 6 Let q be a prime, and let n be a positive integerrelatively prime toq Denote by dn the order ofn modulo q, that

is,dn is the smallest positive integer such thatnd n≡ 1 (mod q).Then for any positive integer m such that nm ≡ 1 (mod q), dn

A prime q is called good if and only if q satisfies the properties

of the problem

Lemma 7 There exists a good prime

Proof: We approach indirectly by assuming that such q doesnot exist Then for any fixed prime q, there is a positive integer

n such that np− p is divisible by q, that is

If q divides n, then q divides p, and so q = p We further assumethat q 6= p Hence q does not divide n

By Fermat’s little theorem, nq−1 ≡ 1 (mod q) Thus, by 6, dn

divides q −1 For the positive integer n, because np≡ p (mod q),

we have npd p ≡ pd p≡ 1 (mod q) Thus, by lemma 6, dn dividesboth q − 1 and pdp, implying that dn divides gcd(q − 1, pdp).Now we pick a prime q such that (a) q divides pp−1p−1 = 1 + p +

· · · + pp−1, and (b) p2 does not divide q − 1 First we show thatsuch a q does exist Note that 1 + p + · · · + pp−1 ≡ 1 + p 6≡ 1(mod p2) Hence there is a prime divisor of 1 + p + · · ·+pp−1 that

is not congruent to 1 modulo p2, and we can choose that prime

Thus, in any case, we have p ≡ 1 (mod q) But then by (a),

0 ≡ 1 + p + · · · + pp−1 ≡ p (mod q), implying that p = q, which

is a contradiction Therefore our original assumption was wrong,and there is a good prime q

Now we prove the desired result By 7, there exists one suchprime q We approach indirectly by assuming that there are onlyfinitely many good primes q1, q2, , and qk We set

a = pq1pq2p· · · qpk.Then

A = a

p− 1

a − 1 = a

p−1+ ap−2+ · · · + a2+ a + 1,

implying that A ≡ a + 1 6≡ 1 (mod p2) Thus, there exists prime

q divides A and p2does not divide q −1 Because q divides ap−1,

q is different from p, q1, q2, , and qk

We claim that q is good Assume not, there there exist positiveinteger n such that np≡ p (mod q) Set s = nq1q2· · · qk Thus

sp 2

≡ (npqp1qp2· · · qkp)p≡ (pq1pqp2· · · qkp)p≡ ap (mod q).Because q divides A, which divides ap− 1, it follows that ap≡ 1(mod q) Hence sp2 ≡ 1 (mod q) By Fermat’s little theorem,

sq−1 ≡ 1 (mod q) By 6, ds divides both q − 1 and p2 Because

p2 does not divide q − 1, ds divides p, implying that sp ≡ 1(mod q) Consequently,

a ≡ pqp1q2p· · · qpk≡ npqp1qp2· · · qkp≡ sp≡ 1 (mod q)

We obtain that p ≡ ap−1+ ap−2+ · · · + a + 1 ≡ A ≡ 0 (mod q);that is, q = p, which contradicts the fact that q is different from

p Hence our assumption was wrong, and q is good

Thus we find a new good prime, which contradicts the fact that

q1, q2, , qk are all the good primes Hence we assumption waswrong, and there are infinitely many good primes

Note: The proof can be shortened by starting directly with thedefinition of q as in the second half of the above proof But wethink the argument in the first part provides motivation for thechoice of this particular q

Many students were able to apply Fermat’s little theorem torealize that npd p ≡ pd p ≡ 1 (mod q) It is also not difficult tosee that there are integers n such that nd p6= 1 (mod q), because

of the existence of primitive roots modulo q By the minimality

of dp, we conclude that dp = pk, where k is some divisor of dp.Consequently, we have pk | (q − 1), implying that q ≡ 1 (mod p).This led people to think about various applications of Dirich-let’s theorem, which is an very popular but fatal approach tothis problem However, a solution with advanced mathematicsbackground is available It involves a powerful prime densitytheorem The prime q satisfies the required condition if andonly if q remains a prime in the field k = Q(√pp) By applyingChebotarev’s density theorem to the Galois closure of k, we canshow that the set of such q has density 1, implying that there are

infinitely many q satisfying the required condition Of course, thisapproach is far beyond the knowledge of most IMO participants.

Some Important Ideas You Should Know

F2.1 Suppose each vertex v of a cube is assigned a real number f (v)

so that |f(v) − f(w)| < 1 whenever v and w are adjacent vertices.Show that there must exist a vertex v such that |f(v)−f(v′)| < 1,where v′denotes the vertex of the icosahedron opposite v What

if it is an icosahedron instead of a cube? (An icosahedron is

a regular polyhedron whose faces are 20 congruent equilateraltriangles.)

Solution: Let u be a vertex on the icosahedron, and let u′

denote the vertex of the icosahedron opposite ui Take a path

P along the edges from u to u′ Assume that P travels throughvertices u1, u2, , uk with u1= u and uk = u′ For each i with

1 ≤ i ≤ k, define g(i) = f(ui) − f(u′

i) We keep track the values

of g(i) as i various from 1 to k The value at i = k is the negativevalue of the value at i = 1, and with each step the value changes

by less than 2 because of the inequalities |f(ui)−f(ui+1)| < 1 and

|f(u′

i) − f(u′

i+1)| < 1 Therefore for some 1 ≤ i ≤ k, the value

of g(i) must be in the interval (−1, 1), establishing the desiredresult

F2.2 Suppose the sequence of nonnegative integers a1, a2, , a1997

satisfies

ai+ aj≤ ai+j ≤ ai+ aj+ 1for all i, j ≥ 1 with i + j ≤ 1997 Show that there exists a realnumber x such that an= ⌊nx⌋ (the greatest integer ≤ nx) for all

¶, n = 1, 2, , 1997will have the desired property Let

or, equivalently,

for all m and n ranging from 1 to 1997 Then L < U , since L ≥ Uimplies that (∗) is violated when n = p and m = q Any point x

in [L, U ) has the desired property

We prove (∗) for all m and n ranging from 1 to 1997 by stronginduction The base case m = n = 1 is trivial The induction stepsplits into three cases If m = n, then (∗) certainly holds If m >

n, then the induction hypothesis gives (m − n)an< n(am−n+ 1),and adding n(am−n+ an) ≤ nam yields (∗) If m < n, thenthe induction hypothesis yields man−m< (n − m)(am+ 1), andadding man≤ m(am+ an−m+ 1) gives (∗)

F2.3 One the following two inequalities is always true

(sin x)sin x< (cos x)cos x or (sin x)sin x> (cos x)cos x

for all real number x such that 0 < x < π/4 Identify thatinequality and prove your result

First Solution: The first inequality is true Observe thatthe logarithm function is concave down We apply Jensen’sinequalityto the points sin x < cos x < sin x+cos x with weights

λ1 = tan x and λ2= 1 − tan x (because 0 < x < π/4, λ1 and λ2

are positive) to obtain

log(cos x) = log[tan x sin x + (1 − tan x)(sin x + cos x)]

> tan x log(sin x) + (1 − tan x) log(sin x + cos x).Since sin x + cos x = √

2 sin¡

x +π 4

¢

> 1 and tan x < 1 in thespecified interval, the second term is positive and we may drop it

to obtain

log(cos x) > tan x log(sin x)

Multiplying by cos x and exponentiating gives the required equality

in-Second Solution: We recall Bernoulli’s inequality: for realnumbers x and a with x > −1 and a > 1 ,

(1 + x)a ≥ 1 + ax

Hence for all real number x with 0 < x < π/4, we have

(cos2x)cosxsin x = (1 − sin x)cossinxx(1 + sin x)cosxsin x

≥ (1 − cos x)(1 + cos x) = sin2x,implying that (cos x)cos x> (sin x)sin x

F2.4 A sequence {an}∞

n=1 of positive integers contains each positiveinteger exactly once Moreover, for every pair of distinct positiveintegers m and n,

1

1998 <

|an− am|

|n − m| < 1998.

Show that |an− n| < 2000000 for all n

Solution: Consider the positive integers as blocks in a sidewalkfrom left to right, and consider a1, a2, as the sequence of blocks

we visit; we must visit every block exactly once

First let n = m+1 in the given inequality From the right-handside inequality, we have |am+1− am| < 1998, i.e., we can move atmost 1997 blocks on each move

Then let an= am+ 1 in the given inequality From the hand side inequality, we have |n − m| < 1998 i.e., we can take

right-at most 1997 moves from one block to an adjacent block

We claim that if we are on block ai, every unvisited number

to our left is greater than ai − 2000000 Assume the oppositefor the sake of contradiction, and say we have not visited m ≤

ai− 2000000 Let aj be the rightmost block with j ≥ i that

we visit before we visit block m From aj to m, we make atleast 2000000/1997 > 1000 moves By the definition of j, afterreaching m we still must visit aj+ 1 This takes at least another

1000 moves So, it takes at least 2000 moves to go from aj to

aj+ 1, a contradiction Thus our assumption is wrong and ourclaim is correct

Now, suppose that an− n ≥ 2000000 When we visit an, we’veonly visited n blocks So, there is a block between 1 and n that

we still must visit But this is at least 2000000 blocks to the left

of an, a contradiction

On the other hand, suppose that n−an ≥ 2000000 By the time

we visit a , there must be a block a > n we already visited But