mosp 2004 solution Solutions to the Mathematics Olympiad Test 1

17
MOP 2004 - 2005
Solutions 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 playing
the following game. At each step, he is allowed to pick four
consecutive 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 desired
sequence 6, 1, 2, 3, 4, 5, 7, 8, . . . , 20.
Note: If you think the bcdeF → Fbcde thing is tricky, you’re
wrong. It’s entirely obvious. Given five blocks on a line, only
two 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 between
the 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 is
the operation of cyclically permuting 19 blocks. Because those
are the only two types of operations allowed, no odd permutation,
such as the one desired, can be obtained.

18
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 perpendicular
bisectors of segments AB and CD intersect at Y , prove that
∠AY B = 2∠ADX.
First Solution:
A B
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 is
similar to triangle ADX or, equivalently, triangle BZW is similar
to triangle BCX. Then triangle ADZ is similar to triangle AXW
and triangle BCZ is similar to triangle BXW , implying that
DZ
XW
=
AD
AX
=

BC
BX
=
CZ
W X
,
and DZ = CZ, as desired.
Second Solution: (By Tony Zhang) Construct point M on the
side of line AB opposite X such that triangle MAB is similar
to triangle Y DC. Construct point N on the side of line CD
opposite X such that triangle NCD is similar to triangle Y BA.
Construct point X
1
on the same side of line AD as B such that
triangle X
1
AD is similar to triangle CY N, and construct point
X
2
on the same side of line BC as D such that triangle X
2
BC is
similar to triangle DY N. It suffices to show X
1
= X
2
, 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
1

D and
19
BX
2
C are all similar. Now use complex numbers; each lowercase
letter will denote the complex number coordinate of the point
named by the corresponding uppercase letter. Set y = 0 and
r = d/n. Then
d = rn c =
rn a = m −rm b = m −rm
and
x
1
= a +
r(d −a) = m −rm + r(rn − m + rm)
= m −
rm + r(rn − m + rm) = b + r(c −b) = x
2
,
as desired.
1.3. Let n be a positive integer, and let p
1
, p
2
, . . . , p
n
be distinct
primes greater than 3. Prove that 2
p
1

p
2
···p
n
+ 1 has at least 4
n
divisors.
Note: We start with the following lemma.
Lemma 1. For any odd positive integers a, b,
gcd(2
a
+ 1, 2
b
+ 1) = 2
gcd(a,b)
+ 1.
Proof: Suppose this were false; choose a counterexample
where a + b is minimum. Obviously we must have a = b; assume
a > b. If a −2b ≥ 0,
2
a
+ 1 ≡ 2
a
+ 1 −(2
b
+ 1)(2
b
− 1)(2
a−2b
+ 1)

≡ 2
a
+ 1 −(2
2b
− 1)(2
a−2b
+ 1)
≡ 2
a−2b
+ 1 (mod 2
b
+ 1),
and if a −2b ≤ 0,
2
2b−a
(2
a
+ 1) ≡ 2
2b−a
(2
a
+ 1) −(2
b
+ 1)(2
b
− 1)
≡ 2
2b−a
+ 1 (mod 2
b

+ 1).
In either case,
gcd(2
a
+ 1, 2
b
+ 1) = gcd(2
|a−2b|
+ 1, 2
b
+ 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.
20
The following proofs rely only on the weaker statement that, for
any relatively prime odd positive integers a, b not divisible by
3, gcd(2
a
+ 1, 2
b
+ 1) = 3. The proof of this is a bit simpler.
Suppose that p is prime and 2
a
≡ 2
b
≡ −1 (mod p). Then if c is
the minimum positive integer such that 2
c
≡ 1 (mod p), then c

must divide both 2a and 2b, so c is either 1 or 2, which implies
p = 3. It remains to note that 9 divides neither 2
a
+ 1 nor 2
b
+ 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, let
ψ(m) denote the number of distinct prime factors of m, and let
d(m) denote the number of factors of m. We wish to prove that
d(2
a
+ 1) ≥ 4
ψ(a)
for all tenebrous integers a.
Induct on ψ(a). For the base case ψ(a) = 1, 2
a
+ 1 is divisible
by 3 exactly once and is greater than 3, so ψ(2
a
+ 1) ≥ 2 and
d(2
a
+ 1) ≥ 4.
Now let a, b be relatively prime tenebrous integers such that
the claim holds for both a and b. Clearly 2
ab
+ 1 is divisible by
both 2
a

+ 1 and 2
b
+ 1, so we can write
2
ab
+ 1 = C ·lcm[2
a
+ 1, 2
b
+ 1].
Because ab −2a −2b −4 = (a −2)(b −2) − 8 > 0,
2
ab
+ 1 > 2
2a+2b+4
> (2
a
+ 1)
2
(2
b
+ 1)
2
> lcm[2
a
+ 1, 2
b
+ 1]
2
,

so C ≥ lcm[2
a
+ 1, 2
b
+ 1]. From the comment, we have gcd(2
a
+
1, 2
b
+ 1) = 3, so as 3 divides each of 2
a
+ 1 and 2
b
+ 1 exactly
once,
d(lcm[2
a
+ 1, 2
b
+ 1]) =
d(2
a
+ 1)d(2
b
+ 1)
2
≥ 2
2ψ(a)+2ψ(b)−1
.
For every divisor m of lcm[2

a
+ 1, 2
b
+ 1], both m and Cm are
divisors of 2
ab
+ 1. As C > lcm[2
a
+ 1, 2
b
+ 1],
d(2
ab
+ 1) ≥ 2 ·d(lcm[2
a
+ 1, 2
b
+ 1]) ≥ 4
ψ(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(2
a
+ 1) ≥ 2ψ(a).
21
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 that

ψ(2
ab
+ 1) ≥ ψ(2
a
+ 1) + ψ(2
b
+ 1).
Note that
2
ab
+ 1
2
a
+ 1
=
b

i=1

b
i

(−2
a
− 1)
i−1
≡ b −

b
2


(2
a
+ 1) (mod (2
a
+ 1)
2
),
so if a prime p divides 2
a
+ 1 exactly k ≥ 1 times, then p divides
2
ab
+ 1 either k times (if p doesn’t divide b) or k + 1 times (if p
divides b). In any case p divides 2
ab
+ 1 at most twice as many
times as p divides 2
a
+ 1. The same is true for prime factors of
2
b
+ 1.
As in the first solution, 2
ab
+ 1 > (2
a
+ 1)
2
(2

b
+ 1)
2
, so, in light
of the above, 2
ab
+ 1 must have a prime factor dividing neither
2
a
+ 1 nor 2
b
+ 1.
Clearly 2
ab
+1 is divisible by lcm[2
a
+1, 2
b
+1]. Because 2
ab
+1
has a prime factor not dividing lcm[2
a
+ 1, 2
b
+ 1], we have
ψ(2
ab
+ 1) ≥ ψ(lcm[2
a

+ 1, 2
b
+ 1]) + 1
= ψ(2
a
+ 1) + ψ(2
b
+ 1) −ψ(gcd(2
a
+ 1, 2
b
+ 1)) + 1
= ψ(2
a
+ 1) + ψ(2
b
+ 1) −ψ(3) + 1
= ψ(2
a
+ 1) + ψ(2
b
+ 1),
completing the induction.
1.4. Let n be an integer greater than 1. Let a
1
, a
2
, . . . , a
n
, b

1
, b
2
, . . . , b
n
be nonnegative real numbers with a
1
a
2
···a
n
= b
1
b
2
···b
n
, b
1
+
b
2
+ ···+ b
n
= 1, and

1≤i<j≤n
|a
i
− a

j
| ≤

1≤i<j≤n
|b
i
− b
j
|.
Determine the maximum value of a
1
+ a
2
+ ···+ a
n
.
Solution: The answer is n−1, which can be obtained by setting
a
1
= 0 and a
2
= ···a
n
= 1, and b
1
= b
2
= ··· = b
n−1
= 0 and

22
b
n
= 0. Now we show that
n

i=1
a
i
= a
1
+ a
2
+ ···+ a
n
≤ n −1. (∗)
Without loss of generality, we assume that a
1
≤ a
2
≤ ··· ≤ a
n
and b
1
≤ b
2
≤ ··· ≤ b
n
.
If a

1
= 0, then
n

i=1
a
i
≤

1≤i<j≤n
|a
i
− a
j
| ≤

1≤i<j≤n
|b
i
− b
j
|
=
n

i=1
(2i −n − 1)b
i
≤ (n −1)
n


i=1
b
i
= n −1,
establishing the inequality (∗).
If a
1
> 0 and n = 2. Then the given inequality becomes
a
2
−a
1
≤ b
2
−b
1
. It follows that (a
1
+a
2
)
2
= 4a
1
a
2
+(a
1
−a

2
)
2
≤
4a
1
a
2
+(b
2
−b
1
)
2
= 4b
1
b
2
+(b
2
−b
1
)
2
= (b
2
+1b
1
)
2

= 1, establishing
the inequality (∗).
Now we assume that a
1
> 0 and n > 2. We approach indirectly
by assuming that the inequality (∗) is false; that is
n

i=1
= a
1
+ a
2
+ ···+ a
n
> n −1. (∗
′
)
We have

1≤i<j≤n
|a
i
− a
j
| =
n

1=1
(2i −n − 1)a

i
=
n

i=1
a
i
+
n

1=1
(2i −n − 2)a
i
> n −1 +
n

i=1
(2i −n − 2)a
i
.
23
and

1≤i<j≤n
|b
i
− b
j
| =
n


1=1
(2i −n − 1)b
i
= (n −1)
n

i=1
b
i
− 2
n

1=1
(n −i)b
i
= n −1 − 2
n−1

i=1
(n −i)b
i
.
Since

1≤i<j≤n
|a
i
−a
j

| =

1≤i<j≤n
|b
i
−b
j
|, combining the last
two inequalities gives
2
n−1

i=1
(n −i)b
i
<
n

i=1
(n + 2 − 2i)a
i
= na
1
+ (n −2)a
2
+ ···+ (2 − n)a
n
= na
1
− (n −2)(a

n
− a
2
) −(n −4)(a
n−1
− a
3
) −···
≤ na
1
.
Consequently, by AM-GM inequality, we obtain
na
1
> 2
n−1

i=1
(n −i)b
i
≥ 2(n −1)
n−1

(n −1)!b
1
b
2
···b
n−1
= 2(n −1)

n−1

(n −1)!a
1
a
2
···a
n
b
n
> 2(n −1)
n−1

(n −1)!a
1
a
2
···a
n
≥ 2(n −1)a
1
n−1

(n −1)!a
n
.
Therefore, by noting n ≥ 3,
a
n
<


n
2(n−1)

n−1
(n −1)!
<
n −1
n
,
implying that

n
i=1
a
i
≤ na
n
< n − 1, contradicts to our
assumption (∗
′
). Thus our assumption was wrong and the desired
inequality (∗) holds.
24
Note: One of the key to this solution is always considering the
equality case of the inequality (∗) by taking out terms a
i
−a
j
for

1 < i < j ≤ n.
25
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, MN ,
and NC, where M is the midpoint of AB and N is the midpoint
of BC. In other words, one cuts off the triangle MBN to obtain
a convex (n + 1)-gon. A regular hexagon P
6
of area 1 is clipped
to obtain a heptagon P
7
. Then P
7
is clipped (in one of the seven
possible ways) to obtain an octagon P
8
, and so on. Prove that
no matter how the clippings are done, the area of P
n
is greater
than 1/3, for all n ≥ 6.
Solution: The key observation is that for any side S of of P
6
,
there is some sub-segment of S that is a side of P
n
. (This is easily
proved by induction on n.) Thus P
n

has a vertex on each side of
P
6
. Since P
n
is convex, it contains a hexagon Q with (at least)
one vertex on each side of P
6
. (The hexagon may be degenerate,
as some of its vertices may coincide.)
Let P
6
= A
1
A
2
A
3
A
4
A
5
A
6
, and let Q = B
1
B
2
B
3

B
4
B
5
B
6
, with
B
i
on A
i
A
i+1
(indices are considered modulo 6). The side B
i
B
i+1
of Q is entirely contained in triangle A
i
A
i+1
A
i+2
, so Q encloses
the smaller regular hexagon R (shaded in the diagram below)
whose sides are the central thirds of the segments A
i
A
i+2
, 1 ≤ i ≤

6. The area of R is 1/3, as can be seen from the fact that its side
length is 1/
√
3 times the side length of P
6
, or from a dissection
argument (count the small equilateral triangles and halves thereof
in the diagram below). Thus [P
n
] ≥ [Q] ≥ [R] = 1/3. We obtain
strict inequality by observing that mathcalP
n
is strictly larger
that Q: if n = 6, this is obvious; if n > 6, then P
n
cannot equal
Q because P
n
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 P
6
is in fact 1/2, attained when the vertices of Q coincide in
pairs at every other vertex of P
6
, so the hexagon Q degenerates
into an equilateral triangle. If the conditions of the problem were
changed so that the cut-points could be anywhere within adjacent
segments instead of just at the midpoints, then the best possible

bound would be 1/2.
26
F1.2. Find all integers n for which a regular hexagon can be divided
into 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
S
1
, S
2
, . . . S
n
be n semicircles with center O and radius 1.
Circle C is covered by semicircles S
1
, S
2
, . . . , and S
n
. Prove
that one can find indices i
1
, i
2
, and i
3
such that sphere C is
covered by S
i
1

, S
i
2
, and S
i
3
.
(b) Let S be a sphere with center O and radius 1, and let
H
1
, H
2
, . . . H
n+1
be n + 1 hemispheres with center O and
radius 1. Sphere S is covered by hemispheres H
1
, H
2
, . . . ,
and H
n
. Prove that one can find indices i
1
, i
2
, i
3
, i
4

such that
sphere S is covered by H
i
1
, H
i
2
, H
i
3
, H
i
4
.
F1.4. Let P be a 1000-sided regular polygon. Some of its diagonals were
drawn to obtain a triangulation the polygon P . (The region inside
P is cut into triangular regions, and the diagonals drawn only
intersect at the vertices of P .) Let n be the number of different
lengths of the drawn diagonals. Determine the minimum value of
n.
(By Tiankai Liu) The answer is 10.
This can be achieved by using diagonals corresponding to arc
lengths of 400, 200, 120, 80, 40, 24, 16, 8, 4, and 2. First
we can draw a regular pentagon with sides subtending arcs of
length 200; the interior of this pentagon can be triangulated
with diagonals subtending arcs of length 400. Then we divide
each the remaining five congruent 201-gons into five congruent
41-gons using diagonals subtending arcs of length 40; the interiors
of the hexagons thus formed can be triangulated by diagonals
subtending arcs of lengths 80 and 120. Proceeding in a sim-

ilar manner, We divide each of the 25 congruent 41-gons into
five nonagons using diagonals subtending arcs of length 8; the
interiors of the hexagons thus formed can be triangulated by
diagonals subtending arcs of lengths 16 and 24. Each of the
remaining nonagons can be triangulated in a straightforward way
with diagonals subtending arcs of lengths 8, 4, and 2.
Suppose, for the sake of contradiction, that the answer is less
than 10. Let the arc lengths corresponding to the different
27
diagonals be a
1
, . . . , a
n
, where we always choose the arc length to
be at most 500, such that 1 < a
1
< a
2
< ··· < a
n
≤ 500. Set also
a
0
= 1; this corresponds to the edges of the original 1000-gon.
Note that either a
n
= 500 or there must exist some triangle
containing the center of the 1000-gon, in which case there exist
i
1

, i
2
, i
3
such that a
i
1
+ a
i
2
+ a
i
3
= 1000. In particular, a
n
≥ 334.
Note also that for each 1 ≤ i ≤ n, a
i
≤ 2a
i−1
. Thus, if n ≤ 8,
then a
n
≤ 2
8
= 256 < 334, a contradiction, so n = 9.
Now consider the number w of i, 1 ≤ i ≤ n, for which a
i
<
2a

i−1
. If w = 0, then a
9
= 2
9
= 512 > 500, a contradiction. If
w ≥ 2, then it is not difficult to see that a
9
≤ 2
7
·5/2 = 320 < 334,
also a contradiction. Therefore w = 1. Let k be the unique
number such that a
k
< 2a
k−1
. Note that a
9
≤ 2
8
· 3/2 = 384 <
500. Then each of the numbers a
0
, a
1
, . . . , a
k−1
contains one 1 in
its binary representation. Each of the numbers a
k

, . . . , a
9
contains
two 1s in its binary representation. But because a
9
< 500, there
exist i
1
, i
2
, i
3
for which a
i
1
+ a
i
2
+ a
i
3
= 1000, and the number
1000 has six 1s in its binary representation. There must be no
carrying in the addition a
i
1
+ a
i
2
+ a

i
3
, and so at least one of a
i
1
,
a
i
2
, a
i
3
is at least as big as the place value of the highest 1 in the
binary representation of 1000. But a
i
1
, a
i
2
, a
i
3
≤ a
9
≤ 384 < 512,
a contradiction.
Therefore 10 is indeed the answer.
28
MOP 2004 - 2005
Solutions to the Mathematics Olympiad Test 2

2.1. Let m be a fixed positive integer with m ≥ 3. A set S is good if
there are elements s
1
, s
2
, . . . , s
m−1
, s
m
(not necessarily distinct)
in S such that s
1
+s
2
+···+s
m−1
= s
m
. Determine the minimum
positive integer f(m) such that for any partition A and B of the
set {1, 2, . . . , f(m)}, at least one of the sets A and B is good.
Solution: We show that f(m) = m
2
− m −1. Indeed, we may
partition {1, 2, . . . , m
2
−m −2} into the following sets A, B that
are not good:
A = {1, 2, . . . , m − 2, (m −1)
2

, (m −1)
2
+ 1, . . . , m
2
− m −2}
B = {m −1, m, . . . , m
2
− 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 not
contained in A; all other sums are at least (m −1)
2
+ (m −2) >
m
2
− m − 2, so A is not good. Also, B is not good because the
minimal sum of m −1 elements of B is (m − 1)
2
> m
2
−2m. By
taking subsets of A and B, we may obtain for any k ≤ m
2
−m−2
partitions of {1, 2, . . . , k} that are not good.
It remains to show that any partition {1, 2, . . . , m
2
−m −1} =
A ∪ B contains at least one good set. Assume the contrary, and
assume without loss of generality that 1 ∈ A; then m−1 ∈ B and

(m−1)
2
∈ A. Also, m
2
−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) =
m
2
−m −1 is a sum of m −1 elements of B that is itself in B, a
contradiction.
2.2. Each square of a (2
n
−1) ×(2
n
−1) board contains either +1 or
−1. Such an arrangement is called successful if each number is
the product of its neighbors (squares sharing a common side with
the 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
29
all cases, the matrix containing all +1’s is successful, and in the
case n = 1 the matrix containing a single −1 is also successful.

We induct on n to show that these are the only successful
arrangements. The base case n = 2 can be verified easily. Assume
now that there is only one successful (2
n
− 1) ×(2
n
− 1) matrix,
and suppose for a contradiction that there exists a successful
(2
n+1
− 1) × (2
n+1
− 1) matrix (a
ij
) that contains at least one
−1. We will call a matrix containing at least one −1 nontrivial.
Also, we let N = 2
n+1
.
We claim that then there exists a nontrivial successful matrix
that has both vertical and horizontal lines of symmetry. Indeed,
if (a
ij
) does not have a horizontal line of symmetry, then the
matrix (b
ij
) given by
b
ij
= a

ij
· a
N−i,j
is symmetric about its center row and is still successful. Further-
more, by the assumption that (a
ij
) was not symmetric, there exist
i, j such that a
ij
= a
N−i,j
, so (b
ij
) is nontrivial. By applying the
same argument, we may obtain a matrix (c
ij
) from (b
ij
) that has
both symmetries.
Consider now the middle row and middle column of (c
ij
). By
symmetry, 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 the
vertical symmetry repeatedly, we see that all entries of the center
row are +1. Otherwise, the entries of the center row, starting
from the center entry and working toward either side, must be
+1, −1, −1, +1, −1, −1, +1, . . . , −1, −1, +1, −1, −1. But then the

width of the matrix is 2
n+1
−1 ≡ 2 (mod 3), which is impossible.
It follows that the center row must contain only +1’s, and likewise
for the center column.
We may now observe that each of the four remaining (2
n
−1) ×
(2
n
−1) squares obtained by deleting the center row and column
are successful. Applying the induction hypothesis, we see that
the 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 +1
and 1 for −1, and replacing multiplication with addition (mod 2).
30
Then the condition that a matrix (a
ij
) be successful becomes a
system of linear equations (mod 2, or more precisely, over the
field F
2
), of the form
a
i−1,j
+ a
i,j−1
+ a
i,j

+ a
i+1,j
+ a
i,j+1
= 0
for non-edge entries a
ij
; when a
ij
is an edge entry there will
only be four summands, and corner entries will give equations
with three summands, in the case n > 1. Thus, showing that
the trivial arrangement is the only successful matrix amounts
to proving that this system of (2
n
− 1)
2
equations in (2
n
− 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 the
determinant can be expanded as a sum of products of terms,
one from each row and one from each column, with sign chosen
appropriately. In this case, the sign does not matter since we are
working mod 2. Hence, it suffices to show that the number of
nonzero terms is odd.
Now the (2

n
−1)
2
×(2
n
−1)
2
matrix corresponding to our system
of equations is sparse; the row corresponding to the equation for
a
ij
contains all 0’s except for 1’s as the coefficients of a
ij
and its
neighbors. Hence, the nonzero terms in the determinant, which
arise from choosing a 1 from each row and column, correspond
to bijections from {a
ij
} to itself such that each a
ij
is mapped
to a neighboring square. We may think of such a bijection as a
diagram of arrows pointing from the center of each square to the
square it is mapped to, with dots (degenerate arrows) in squares
that 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 hori-
zontally can be paired with its reflection; likewise for diagrams
that are not symmetric vertically. Thus, it suffices to show that
there is an odd number of arrow diagrams with both horizontal

and vertical symmetries. Now, such diagrams must contain dots
in all entries of the center row and center column. But then the
diagram is determined by its top-left (2
n−1
− 1) × (2
n−1
− 1)
corner, which must be a legitimate diagram for the case n − 1.
The result follows by induction.
31
2.3. Let ABC be a triangle, and let D be a point on side BC such
that ∠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 Q
2
= AB · AC.
Note: Let ω
1
and ω
2
denote the circumcirlces of triangles ABD
and ACD, respectively. Let Y U and XZ be the common tangent
lines to ω
1
and ω
2
. If Y U  XZ, then it is not difficult to see
that 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 XZ
meet at T . Without loss of generality, we label the points as
shown in Figure 2.1.
A
B B( )
1
C C( )1
D
P
Q
X
T
W
Y
U
Z
Figure 2.1.
First Solution: Consider the dilation T taking ω
1
to ω
2
. It
is clear that T is the center of T . Assume that line DT meet ω
1
and ω
2
again (other than D) at B
1
and C

1
, respectively. Then
T(B
1
) = D, T(D) = C
1
, and T(

B
1
D) =

DC
1
, implying that
∠B
1
AD = ∠DAC
1
. On the other hand, consider a point B
′
moving along omega
1
from D to A, line B
′
D meet ω
2
again C
′
.

As B
′
moving from D towards A, C
′
is moving from A towards D
along ω
2
; that is, ∠DAB
′
is increasing and ∠C
′
AD is decreasing.
Hence there is an unique point B such that ∠BAD = ∠CAD.
We conclude that B
1
= B, C
1
= C; that lines BC, XZ, and Y U
are concurrent at T .
Because T(B) = D and T(X) = Z, it follows that BX  DZ.
32
Likewise, XD  ZC. Because BX  DZ and ZT is tangent
to ω
2
at Z, ∠BXT = ∠DZT = ∠ZCD, or BCZX is cyclic,
and let ω
3
denote the circumcircle of BCZX. Therefore, lines
P Q, CZ, and BX are the radical axes of ω
1

and ω
2
, ω
2
and ω
3
,
and ω
3
and ω
1
. Hence lines P Q, CZ, and BX are concurrent,
and let W denote the point of concurrence. Then DXW Z is a
parallelogram, and so DQ = QW and PQ = AW . It suffices to
show that AW
2
= 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
AW C, implying that
AB
AW
=
AW
AC
,
or AW
2
= AB · AC, as desired.

Second Solution:
A
D
X
Y
U
Z
B
C
K
L
M
Figure 2.2.
As shown in Figure 2.2, let K and L be the circumcenters of
triangles ABD and ACD, respectively. Because segment XZ is
symmetric to segment Y U across line KL, XY UZ is an isosceles
trapezoid. Since P, Q are the midpoints of segments Y U and XZ,
respectively, PQ  XY  UZ and 2P Q = XY + UZ. It suffices
to show that (UZ + XY )
2
= 4PQ
2
= 4AB · AC.
Let r
1
and r
2
be the radius of ω
1
and ω

2
, respectively. It is clear
33
that LZ  KX and LU  KY , implying that isosceles triangles
ULZ and Y KX are similar with ratio
r
1
r
2
. Hence XY =
r
1
r
2
·UZ.
It suffices to show that
UZ
2

1 +
r
1
r
2

2
= 4AB · AC,
or
UZ
2r

2
=
√
AB ·AC
r
1
+ r
2
. (∗)
In isosceles triangle U LZ, UZ = 2UL cos ∠LUZ = 2r
2
cos ∠LUZ.
Let M be the foot of the perpendicular from K to line UL.
Because MK ⊥ LU and LK ⊥ UZ (by symmetry), ∠LUZ =
∠MKL. Note that ML = r
2
− r
1
= AL − AK, which implies
that, in right triangle KLM , sin ∠MKL =
ML
KL
and
cos ∠MKL =

1 −sin
2
∠MKL =

1 −


AL −AK
KL

2
By noting that cos ∠MKL = cos ∠LUZ =
UZ
2r
2
, we can rewrite
the equation (∗) as

1 −

AL −AK
KL

2
=
√
AB ·AC
r
1
+ r
2
. (∗
′
)
A
D

B
C
K
L
2g
g
2g
a
a
Figure 2.3.
Note that ∠CLA + 2∠CDA = 360
◦
and ∠BKA = 2∠BDA.
Hence ∠CLA = 2(180
◦
− ∠CDA) = 2∠BDA = ∠BKA. Let
γ = ∠BDA, as indicated in Figure 2.3. Then 2γ = ∠CLA =
34
∠BKA. Because ∠CLA = ∠XKY and both CAL and BAK
are isosceles triangles, they are similar; that is, there is a spiral
similarity centered at A sending triangle ALC to triangle AKB.
Consequently, there is a spiral similarity centered at A sending
triangle ALK to triangle ACB, implying that triangle KAL is
similar to triangle BAC. The equation (∗
′
) becomes

1 −

AC − AB

BC

2
=
√
AB ·AC
r
1
+ r
2
,
or

BC
2
− AC
2
− AB
2
+ 2AC ·AB =
BC
√
AB ·AC
r
1
+ r
2
,
Let α = ∠BAD = ∠DAC. Applying the law of cosines to
triangle ABC yields BC

2
− AC
2
− AB
2
= 2AC · AB cos 2α. By
the double-angle formulas,

BC
2
− AC
2
− AB
2
+ 2AC ·AB
=

2AC · AB(1 −cos 2α) = 2 sin α
√
AC · AB.
It remains to show that
2 sin α
√
AC · AB =
BC
√
AB ·AC
r
1
+ r

2
or
2(r
1
+ r
2
) =
BC
sin α
.
Applying the extended law of sines to triangles ABD and
ACD, we can rewrite the above equation as
AB
sin γ
+
AC
sin γ
=
BD
sin α
+
CD
sin α
,
which is evident, because the first (second) summand on the
left-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 many
prime numbers q such that for every integer n, the number n

p
−p
is not divisible by q.
First Solution: We will need the following fact.
35
Lemma 2. If a, m, n ∈ Z, a > 1, m > 0, n > 0, then
gcd(a
m
− 1, a
n
− 1) = a
gcd(m,n)
− 1
Proof: Induct on m + n ≥ 2. For m = n the statement is
trivial. Suppose that the statement holds for any m
′
, n
′
∈ N with
sum m
′
+ n
′
< m + n. Without loss of generality, we assume that
m > n.
gcd(a
m
− 1, a
n
− 1) = gcd(a

m
− a
n
, a
n
− 1)
= gcd(a
m−n
− 1, a
n
− 1)
= a
gcd(m−n,n)
= a
gcd(m,n)
as m − n + n < m + n and gcd(m − n, n) = gcd(m, n). This
completes the inductive step.
For each prime r = p let F (r) =
(p
rp
−1)(p−1)
(p
p
−1)(p
r
−1)
. By lemma 2
gcd(p
r
− 1, p

p
− 1) = p − 1. Since both p
r
− 1 and p
p
− 1 divide
p
rp
−1, their product divides (p
rp
−1)(p −1). Therefore, F (r) is
an integer.
Lemma 3. For any distinct prime numbers r, s = p
gcd(F (r), F (s)) = 1
Proof: By lemma 2 gcd(p
rp
− 1, p
sp
− 1) = p
p
− 1. Therefore,
gcd(
p
rp
−1
p
p
−1
,
p

sp
−1
p
p
−1
) = 1. Since F (r) |
p
rp
−1
p
p
−1
and F (s) |
p
rp
−1
p
p
−1
,
gcd(F (r), F (s)) = 1.
Lemma 4. For any prime r = p there is a prime divisor q of
F (r) such that p | q −1 but p
2
 |q − 1.
Proof: Clearly, F (r) > 1. First we will show that any prime
divisor q of F (r) is equivalent to 1 modulo p. We know that
q | p
rp
−1. Therefore, q = p and q | p

q− 1
−1 by Fermat’s theorem.
Applying lemma 2, we have q | p
gcd(q−1,rp)
− 1. This means that
either p | q −1 or q | p
r
− 1. However,
q | F (r) |
p
rp
− 1
p
r
− 1
=
p−1

i=0
p
ir
≡ p (mod p
r
− 1)
This means, q cannot divide p
r
− 1 and p | q − 1.
Now we know that all prime divisors of F (r) are equivalent to 1
modulo p. We need to find one that is not equivalent to 1 modulo
36

p
2
. If we assume that all prime divisors of F (r) are 1 modulo
p
2
, then F (r) ≡ 1 (mod p
2
) as their product. On the other hand
F (r) =
(p
rp
−1)(p−1)
(p
p
−1)(p
r
−1)
, where p
rp
− 1 ≡ p
p
− 1 ≡ p
r
− 1 (mod p
2
).
Thus, F (r) ≡ 1 − p (mod p
2
). This contradiction finishes the
proof.

Lemma 5. If r = p is a prime number and q is a prime divisor
of F (r) such that p | q −1 and p
2
 |q − 1, then for any integer n
n
p
− p is not divisible by q.
Proof: Let q = kp+1, where p  |k. Suppose n
p
≡ p (mod q) for
some integer n. Then n
pk
≡ p
k
(mod q). Because clearly q = p,
n
pk
= n
q−1
≡ 1 (mod q) by Fermat’s theorem. This implies
q | p
k
− 1. On the other hand, q | p
rp
− 1. Since p  |k, lemma
2 implies q | p
r
− 1. On the other hand,
q | F (r) |
p

rp
− 1
p
r
− 1
=
p−1

i=0
p
ir
≡ p (mod p
r
− 1)
This means, q cannot divide p
r
−1, which leads to a contradiction.
By lemma 4 for each prime r = p we can pick such prime
number q
r
| F (r) that p | q
r
− 1 and p
2
 |q
r
− 1. Then lemma
5 shows that each q
r
satisfies the desired condition and lemma 3

shows that all q
r
are different since they divide pairwise relatively
prime numbers F(r).
Second Solution: We start with two lemmas. The first is well
known, and the second is problem 6 of IMO 2003.
Lemma 6. Let q be a prime, and let n be a positive integer
relatively prime to q. Denote by d
n
the order of n modulo q, that
is, d
n
is the smallest positive integer such that n
d
n
≡ 1 (mod q).
Then for any positive integer m such that n
m
≡ 1 (mod q), d
n
divides m.
Proof: By the minimality of d
n
, we can write m = d
n
k + r
where k and r are integers with 1 ≤ k and 0 ≤ r < d
n
. Then
1 ≡ n

m
≡ n
d
n
k+r
≡ n
d
n
k
· n
r
= n
r
(mod q).
By the minimality of d
n
, r = 0, that is, d
n
divides m.
37
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 does
not exist. Then for any fixed prime q, there is a positive integer
n such that n
p
− p is divisible by q, that is
n
p

≡ p (mod q). (∗)
If q divides n, then q divides p, and so q = p. We further assume
that q = p. Hence q does not divide n.
By Fermat’s little theorem, n
q−1
≡ 1 (mod q). Thus, by 6, d
n
divides q−1. For the positive integer n, because n
p
≡ p (mod q),
we have n
pd
p
≡ p
d
p
≡ 1 (mod q). Thus, by lemma 6, d
n
divides
both q − 1 and pd
p
, implying that d
n
divides gcd(q − 1, pd
p
).
Now we pick a prime q such that (a) q divides
p
p
−1

p−1
= 1 + p +
··· + p
p−1
, and (b) p
2
does not divide q −1. First we show that
such a q does exist. Note that 1 + p + ··· + p
p−1
≡ 1 + p ≡ 1
(mod p
2
). Hence there is a prime divisor of 1+p+···+p
p−1
that
is not congruent to 1 modulo p
2
, and we can choose that prime
to be our q.
By (a), p
p
≡ 1 (mod q) (and p = q). By lemma 6, d
p
divides
p, and so d
p
= p or d
p
= 1.
If d

p
= 1, then p ≡ 1 (mod q).
If d
p
= p. Then d
n
divides gcd(p
2
, q − 1). By (b), the possible
values of d
n
are 1 and p, implying that n
p
≡ 1 (mod q). By
relation (∗), we conclude p ≡ 1 (mod q).
Thus, in any case, we have p ≡ 1 (mod q). But then by (a),
0 ≡ 1 + p + ··· + p
p−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 such
prime q. We approach indirectly by assuming that there are only
finitely many good primes q
1
, q
2
, . . . , and q
k
. We set

a = pq
p
1
q
p
2
···q
p
k
.
Then
A =
a
p
− 1
a −1
= a
p−1
+ a
p−2
+ ···+ a
2
+ a + 1,
38
implying that A ≡ a + 1 ≡ 1 (mod p
2
). Thus, there exists prime
q divides A and p
2
does not divide q −1. Because q divides a

p
−1,
q is different from p, q
1
, q
2
, . . . , and q
k
.
We claim that q is good. Assume not, there there exist positive
integer n such that n
p
≡ p (mod q). Set s = nq
1
q
2
···q
k
. Thus
s
p
2
≡ (n
p
q
p
1
q
p
2

···q
p
k
)
p
≡ (pq
p
1
q
p
2
···q
p
k
)
p
≡ a
p
(mod q).
Because q divides A, which divides a
p
− 1, it follows that a
p
≡ 1
(mod q). Hence s
p
2
≡ 1 (mod q). By Fermat’s little theorem,
s
q− 1

≡ 1 (mod q). By 6, d
s
divides both q − 1 and p
2
. Because
p
2
does not divide q − 1, d
s
divides p, implying that s
p
≡ 1
(mod q). Consequently,
a ≡ pq
p
1
q
p
2
···q
p
k
≡ n
p
q
p
1
q
p
2

···q
p
k
≡ s
p
≡ 1 (mod q).
We obtain that p ≡ a
p−1
+ a
p−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
q
1
, q
2
, . . . , q
k
are all the good primes. Hence we assumption was
wrong, and there are infinitely many good primes.
Note: The proof can be shortened by starting directly with the
definition of q as in the second half of the above proof. But we
think the argument in the first part provides motivation for the
choice of this particular q.
Many students were able to apply Fermat’s little theorem to
realize that n
pd
p

≡ p
d
p
≡ 1 (mod q). It is also not difficult to
see that there are integers n such that n
d
p
= 1 (mod q), because
of the existence of primitive roots modulo q. By the minimality
of d
p
, we conclude that d
p
= pk, where k is some divisor of d
p
.
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 to
this problem. However, a solution with advanced mathematics
background is available. It involves a powerful prime density
theorem. The prime q satisfies the required condition if and
only if q remains a prime in the field k = Q(
p
√
p). By applying
Chebotarev’s density theorem to the Galois closure of k, we can
show that the set of such q has density
1
p

, implying that there are
39
infinitely many q satisfying the required condition. Of course, this
approach is far beyond the knowledge of most IMO participants.
40
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 equilateral
triangles.)
Solution: Let u be a vertex on the icosahedron, and let u
′
denote the vertex of the icosahedron opposite u
i
. Take a path
P along the edges from u to u
′
. Assume that P travels through
vertices u
1
, u
2
, . . . , u

k
with u
1
= u and u
k
= u
′
. For each i with
1 ≤ i ≤ k, define g(i) = f(u
i
) − 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 negative
value of the value at i = 1, and with each step the value changes
by less than 2 because of the inequalities |f(u
i
)−f(u
i+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 desired
result.

F2.2. Suppose the sequence of nonnegative integers a
1
, a
2
, . . . , a
1997
satisfies
a
i
+ a
j
≤ a
i+j
≤ a
i
+ a
j
+ 1
for all i, j ≥ 1 with i + j ≤ 1997. Show that there exists a real
number x such that a
n
= ⌊nx⌋ (the greatest integer ≤ nx) for all
1 ≤ n ≤ 1997.
Solution: Any x that lies in all of the half-open intervals
I
n
=

a
n

n
,
a
n
+ 1
n

, n = 1, 2, . . . , 1997
will have the desired property. Let
L = max
1≤n≤1997
a
n
n
=
a
p
p
and U = min
1≤n≤1997
a
n
+ 1
n
=
a
q
+ 1
q
.

We shall prove that
a
n
n
<
a
m
+ 1
m
,
41
or, equivalently,
ma
n
< n(a
m
+ 1) (∗)
for all m and n ranging from 1 to 1997. Then L < U, since L ≥ U
implies 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 strong
induction. The base case m = n = 1 is trivial. The induction step
splits into three cases. If m = n, then (∗) certainly holds. If m >
n, then the induction hypothesis gives (m −n)a
n
< n(a
m−n
+ 1),
and adding n(a
m−n

+ a
n
) ≤ na
m
yields (∗). If m < n, then
the induction hypothesis yields ma
n−m
< (n − m)(a
m
+ 1), and
adding ma
n
≤ m(a
m
+ a
n−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 that
inequality and prove your result.
First Solution: The first inequality is true. Observe that
the logarithm function is concave down. We apply Jensen’s

inequality to the points sin x < cos x < sin x+cos x with weights
λ
1
= tan x and λ
2
= 1 − tan x (because 0 < x < π/4, λ
1
and λ
2
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 the
specified 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 in-
equality.
Second Solution: We recall Bernoulli’s inequality: for real
numbers x and a with x > −1 and a > 1 ,
(1 + x)
a

≥ 1 + ax.