Adding equation (4) to equation (3), we find:
(a + b)
2
+ 4(a + b) = 0 ⇒ a + b = 0, −4 (5)
Subtracting equation (4) from equation (3), we find:
(a − b)
2
− 4(a + b) = 1. (6)
But now we see that if a + b = −4, then equation (6) will be false;
thus, a + b = 0. Substituting this into equation (6), we obtain:
(a − b)
2
= 1 ⇒ a − b = ±1 (7)
Since we know that a + b = 0 from equation (5), we now can find
all ordered pairs (a, b) with the help of equation (7). They are
(−1/2, 1/2) and (1/2, −1/2). Therefore, our only solutions (x, y)
are (2, 2), (3, 3), (2, 3), and (3, 2).
2. Consider the sequence of positive integers which satisfies a
n
= a
2
n−1
+
a
2
n−2
+ a
2
n−3
for all n ≥ 3. Prove that if a
k
= 1997 then k ≤ 3.
Solution: We proceed indirectly; assume that for some k > 3,
a
k
= 1997. Then, each of the four numbers a
k−1
, a
k−2
, a
k−3
,
and a
k−4
must exist. Let w = a
k−1
, x = a
k−2
, y = a
k−3
, and
z = a
k−4
. Now, by the given condition, 1997 = w
2
+ x
2
+ y
2
. Thus,
w ≤
√
1997 < 45, and since w is a positive integer, w ≤ 44. But
then x
2
+ y
2
≥ 1997 − 44
2
= 61.
Now, w = x
2
+ y
2
+ z
2
. Since x
2
+ y
2
≥ 61 and z
2
≥ 0, x
2
+ y
2
+
z
2
≥ 61. But w ≤ 44. Therefore, we have a contradiction and our
assumption was incorrect.
If a
k
= 1997, then k ≤ 3.
3. Let k be a positive integer. The sequence a
n
is defined by a
1
= 1, and
a
n
is the n-th positive integer greater than a
n−1
which is congruent
to n modulo k. Find a
n
in closed form.
Solution: We have a
n
= n(2+(n−1)k)/2. If k = 2, then a
n
= n
2
.
First, observe that a
1
≡ 1 (mod k). Thus, for all n, a
n
≡ n
(mod k), and the first positive integer greater than a
n−1
which is
congruent to n modulo k must be a
n−1
+ 1.
4
The n-th positive integer greater than a
n−1
that is congruent to n
modulo k is simply (n − 1)k more than the first positive integer
greater than a
n−1
which satisfies that condition. Therefore, a
n
=
a
n−1
+ 1 + (n − 1)k. Solving this recursion gives the above answer.
4. Given a parallelogram ABCD, inscribe in the angle ∠BAD a circle
that lies entirely inside the parallelogram. Similarly, inscribe a circle
in the angle ∠BCD that lies entirely inside the parallelogram and
such that the two circles are tangent. Find the locus of the tangency
point of the circles, as the two circles vary.
Solution: Let K
1
be the largest circle inscribable in ∠BAD such
that it is completely inside the parallelogram. It intersects the line
AC in two points; let the point farther from A be P
1
. Similarly, let
K
2
be the largest circle inscribable in ∠BCD such that it is com-
pletely inside the parallelogram. It intersects the line AC in two
points; let the point farther from C be P
2
. then the locus is the
intersection of the segments AP
1
and CP
2
.
We begin by proving that the tangency point must lie on line AC.
Let I
1
be the center of the circle inscribed in ∠BAD. Let I
2
be
the center of the circle inscribed in ∠BCD. Let X represent the
tangency point of the circles.
Since circles I
1
and I
2
are inscribed in angles, these centers must
lie on the respective angle bisectors. Then, since AI
1
and CI
2
are
bisectors of opposite angles in a parallelogram, they are parallel;
therefore, since I
1
I
2
is a transversal, ∠AI
1
X = ∠CI
2
X.
Let T
1
be the foot of the perpendicular from I
1
to AB. Similarly,
let T
2
be the foot of the perpendicular from I
2
to CD. Observe that
I
1
T
1
/AI
1
= sin ∠I
1
AB = sin ∠I
2
CD = I
2
T
2
/CI
2
. But I
1
X = I
1
T
1
and I
2
X = I
2
T
2
. Thus, I
1
X/AI
1
= I
2
X/CI
2
.
Therefore, triangles CI
2
X and AI
1
X are similar, and vertical angles
∠I
1
XA and ∠I
2
XC are equal. Since these vertical angles are equal,
the points A, X, and C must be collinear.
The tangency point, X, thus lies on diagonal AC, which was what
we wanted.
Now that we know that X will always lie on AC, we will prove that
any point on our locus can be a tangency point. For any X on our
5
locus, we can let circle I
1
be the smaller circle through X, tangent
to the sides of ∠BAD.
It will definitely fall inside the parallelogram because X is between
A and P
1
. Similarly, we can draw a circle tangent to circle I
1
and
to the sides of ∠BCD; from our proof above, we know that it must
be tangent to circle I
1
at X. Again, it will definitely fall in the
parallelogram because X is between C and P
2
.
Thus, any point on our locus will work for X. To prove that any
other point will not work, observe that any other point would either
not be on line AC or would not allow one of the circles I
1
or I
2
to
be contained inside the parallelogram.
Therefore, our locus is indeed the intersection of segments AP
1
and
CP
2
.
6
1.2 Bulgaria
1. Find all real numbers m such that the equation
(x
2
− 2mx − 4(m
2
+ 1))(x
2
− 4x − 2m(m
2
+ 1)) = 0
has exactly three different roots.
Solution: Answer: m = 3. Proof: By setting the two factors
on the left side equal to 0 we obtain two polynomial equations, at
least one of which must be true for some x in order for x to be a
root of our original equation. These equations can be rewritten as
(x − m)
2
= 5m
2
+ 4 and (x −2)
2
= 2(m
3
+ m + 2). We have three
ways that the original equation can have just three distinct roots:
either the first equation has a double root, the second equation has
a double root, or there is one common root of the two equations.The
first case is out, however, because this would imply 5m
2
+ 4 = 0
which is not possible for real m.
In the second case, we must have 2(m
3
+ m + 2) = 0; m
3
+ m + 2
factors as (m+1)(m
2
−m+2) and the second factor is always positive
for real m. So we would have to have m = −1 for this to occur. Then
the only root of our second equation is x = 2, and our first equation
becomes (x + 1)
2
= 9, i.e. x = 2, −4. But this means our original
equation had only 2 and -4 as roots, contrary to intention.
In our third case let r be the common root, so x − r is a factor of
both x
2
−2mx −4(m
2
+ 1) and x
2
−4x −2m(m
2
+ 1). Subtracting,
we get that x −r is a factor of (2m −4)x−(2m
3
−4m
2
+2m−4), i.e.
(2m−4)r = (2m−4)(m
2
+1). So m = 2 or r = m
2
+1. In the former
case, however, both our second-degree equations become (x −2)
2
=
24 and so again we have only two distinct roots. So we must have
r = m
2
+1 and then substitution into (r −2)
2
= 2(m
3
+m + 2) gives
(m
2
− 1)
2
= 2(m
3
+ m + 2), which can be rewritten and factored
as (m + 1)(m − 3)(m
2
+ 1) = 0. So m = −1 or 3; the first case
has already been shown to be spurious, so we can only have m = 3.
Indeed, our equations become (x − 3)
3
= 49 and (x − 2)
2
= 64 so
x = −6, −4, 10, and indeed we have 3 roots.
2. Let ABC be an equilateral triangle with area 7 and let M, N be
points on sides AB, AC, respectively, such that AN = BM. Denote
7
by O the intersection of BN and CM. Assume that triangle BOC
has area 2.
(a) Prove that MB/AB equals either 1/3 or 2/3.
(b) Find ∠AOB.
Solution:
(a) Let L be on BC with CL = AN , and let the intersections of
CM and AL, AL and BN be P, Q, respectively. A 120-degree
rotation about the center of ABC takes A to B, B to C, C to
A; this same rotation then also takes M to L, L to N, N to
M, and also O to P , P to Q, Q to O. Thus OP Q and M LN
are equilateral triangles concentric with ABC. It follows that
∠BOC = π −∠NOC = 2π/3, so O lies on the reflection of the
circumcircle of ABC through BC. There are most two points
O on this circle and inside of triangle ABC such that the ratio
of the distances to BC from O and from A — i.e. the ratio of
the areas of triangles OBC and ABC — can be 2/7; so once we
show that MB/AB = 1/3 or 2/3 gives such positions of O it will
follow that there are no other such ratios (no two points M can
give the same O, since it is easily seen that as M moves along
AB, O varies monotonically along its locus). If MB/AB =
1/3 then AN/AC = 1/3, and Menelaus’ theorem in triangle
ABN and line CM gives BO/ON = 3/4 so [BOC]/[BNC] =
BO/BN = 3/7. Then [BOC]/[ABC] = (3/7)(CN/CA) =
2/7 as desired. Similarly if MB/AB = 2/3 the theorem gives
us BO/BN = 6, so [BOC]/[BNC] = BO/BN = 6/7 and
[BOC]/[ABC] = (6/7)(CN/AC) = 2/7.
(b) If MB/AB = 1/3 then M ONA is a cyclic quadrilateral since
∠A = π/3 and ∠O = π − (∠P OQ) = 2π/3. Thus ∠AOB =
∠AOM + ∠MOB = ∠ANM + ∠P OQ = ∠ANM + π/3. But
MB/AB = 1/3 and AN/AC = 1/3 easily give that N is the
projection of M onto AC, so ∠ANM = π/2 and ∠AOB =
5π/6.
If MB/AB = 2/3 then MONA is a cyclic quadrilateral as
before, so that ∠AOB = ∠AOM +∠MOB = ∠ANM +∠POQ.
But AMN is again a right triangle, now with right angle at M,
and ∠MAN = π/3 so ∠ANM = π/6, so ∠AOB = π/2.
8
3. Let f(x) = x
2
− 2ax − a
2
− 3/4. Find all values of a such that
|f(x)| ≤ 1 for all x ∈ [0, 1].
Solution: Answer: −1/2 ≤ a ≤
√
2/4. Proof: The graph of f(x)
is a parabola with an absolute minimum (i.e., the leading coefficient
is positive), and its vertex is (a, f(a)). Since f (0) = −a
2
− 3/4, we
obtain that |a| ≤ 1/2 if we want f(0) ≥ −1. Now suppose a ≤ 0;
then our parabola is strictly increasing between x = 0 and x = 1 so
it suffices to check f(1) ≤ 1. But we have 1/2 ≤ a + 1 ≤ 1, 1/4 ≤
(a + 1)
2
≤ 1, 1/4 ≤ 5/4 − (a + 1)
2
≤ 1. Since 5/4 −(a + 1)
2
= f(1),
we have indeed that f meets the conditions for −1/2 ≤ a ≤ 0. For
a > 0, f decreases for 0 ≤ x ≤ a and increases for a ≤ x ≤ 1. So we
must check that the minimum value f(a) is in our range, and that
f(1) is in our range. This latter we get from 1 < (a + 1)
2
≤ 9/4
(since a ≤ 1/2) and so f(x) = −1 ≤ 5/4 − (a + 1)
2
< 1/4. On
the other hand, f(a) = −2a
2
− 3/4, so we must have a ≤
√
2/4 for
f(a) ≥ −1. Conversely, by bounding f (0), f(a), f(1) we have shown
that f meets the conditions for 0 < a ≤
√
2/4.
4. Let I and G be the incenter and centroid, respectively, of a triangle
ABC with sides AB = c, BC = a, CA = b.
(a) Prove that the area of triangle CIG equals |a − b|r/6, where r
is the inradius of ABC.
(b) If a = c + 1 and b = c −1, prove that the lines IG and AB are
parallel, and find the length of the segment IG.
Solution:
(a) Assume WLOG a > b. Let CM be a median and CF be the
bisector of angle C; let S be the area of triangle ABC. Also let
BE be the bisector of angle B; by Menelaus’ theorem on line
BE and triangle ACF we get (CE/EA)(AB/BF )(F I/IC) =
1. Applying the Angle Bisector Theorem twice in triangle
ABC we can rewrite this as (a/c)((a + b)/a)(F I/IC) = 1, or
IC/F I = (a + b)/c, or IC/CF = (a + b)/(a + b + c). Now
also by the Angle Bisector Theorem we have BF = ac/(a + b);
since BM = c/2 and a > b then MF = (a − b)c/2(a + b). So
comparing triangles CMF and ABC, noting that the altitudes
9
to side MF (respectively AB) are equal, we have [CMF]/S =
(a − b)/2(a + b). Similarly using altitudes from M in triangles
CM I and CMF (and using the ratio IC/CF found earlier),
we have [CMI]/S = (a − b)/2(a + b + c); and using altitudes
from I in triangles CGI and CMI gives (since CG/CM = 2/3)
[CGI]/S = (a−b)/3(a + b + c). Finally S = (a+b + c)r/2 leads
to [CGI] = (a − b)r/6.
(b) As noted earlier, IC/CF = (a + b)/(a + b + c) = 2/3 =
CG/CM in the given case. But C, G, M are collinear, as are
C, I, F , giving the desired parallelism (since line M F = line
AB). We found earlier MF = (a − b)c/2(a + b) = 1/2, so
GI = (2/3)(M F ) = 1/3.
5. Let n ≥ 4 be an even integer and A a subset of {1, 2, . . . , n}. Consider
the sums e
1
x
1
+ e
2
x
2
+ e
3
x
3
such that:
• x
1
, x
2
, x
3
∈ A;
• e
1
, e
2
, e
3
∈ {−1, 0, 1};
• at least one of e
1
, e
2
, e
3
is nonzero;
• if x
i
= x
j
, then e
i
e
j
= −1.
The set A is free if all such sums are not divisible by n.
(a) Find a free set of cardinality n/4.
(b) Prove that any set of cardinality n/4+ 1 is not free.
Solution:
(a) We show that the set A = {1, 3, 5, , 2n/4− 1} is free. Any
combination e
1
x
1
+ e
2
x
2
+ e
3
x
3
with zero or two e
i
’s equal
to 0 has an odd value and so is not divisible by n; otherwise,
we have one e
i
equal to 0, so we have either a difference of
two distinct elements of A, which has absolute value less than
2n/4 and cannot be 0, so it is not divisible by n, or a sum
(or negative sum) of two elements, in which case the absolute
value must range between 2 and 4n/4 −2 < n and so again
is not divisible by n.
10
(b) Suppose A is a free set; we will show |A| ≤ n/4. For any k,
k and n −k cannot both be in A since their sum is n; likewise,
n and n/2 cannot be in A. If we change any element k of A to
n −k then we can verify that the set of all combinations
e
i
x
i
taken mod n is invariant, since we can simply flip the sign of
any e
i
associated with the element k in any combination. Hence
we may assume that A is a subset of B = {1, 2, , n/2 − 1}.
Let d be the smallest element of A. We group all the elements
of B greater than d into “packages” of at most 2d elements,
starting with the largest; i.e. we put the numbers from n/2 −
2d to n/2 − 1 into one package, then put the numbers from
n/2 − 4d to n/2 − 2d − 1 into another, and so forth, until we
hit d + 1 and at that point we terminate the packaging process.
All our packages, except possibly the last, have 2d elements; so
let p + 1 be the number of packages and let r be the number
of elements in the last package (assume p ≥ 0, since otherwise
we have no packages and d = n/2 −1 so our desired conclusion
holds because |A| = 1). The number of elements in B is then
2dp+r +d, so n = 4dp+2d+ 2r + 2. Note that no two elements
of A can differ by d, since otherwise A is not free. Also the
only element of A not in a package is d, since it is the smallest
element and all higher elements of B are in packages.
Now do a case analysis on r. If r < d then each complete
package has at most d elements in common with A, since the
elements of any such package can be partitioned into disjoint
pairs each with difference d. Thus |A| ≤ 1 + dp + r and 4|A| ≤
4dp+4r+4 ≤ n (since r+1 ≤ d) so our conclusion holds. If r = d
then each complete package has at most d elements in common
with A, and also the last package (of d elements) has at most
d − 1 elements in common with A for the following reason: its
highest element is 2d, but 2d is not in A since d+d−2d = 0. So
|A| ≤ d(p+1), 4|A| < n and our conclusion holds. If r > d then
we can form r − d pairs in the last package each of difference
d, so each contains at most 1 element of A, and then there
are 2d − r remaining elements in this package. So this package
contains at most d elements, and the total number of elements
in A is at least d(p + 1) + 1, so 4|A| ≤ n and our conclusion
again holds.
11
6. Find the least natural number a for which the equation
cos
2
π(a − x) − 2 cos π(a − x) + cos
3πx
2a
cos
πx
2a
+
π
3
+ 2 = 0
has a real root.
Solution: The smallest such a is 6. The equation holds if a =
6, x = 8. To prove minimality, write the equation as
(cos π(a − x) − 1)
2
+ (cos(3πx/2a) cos(πx/2a + π/3) + 1) = 0;
since both terms on the left side are nonnegative, equality can only
hold if both are 0. From cos π(a − x) − 1 = 0 we get that x is an
integer congruent to a (mod 2). From the second term we see that
each cosine involved must be −1 or 1 for the whole term to be 0; if
cos(πx/2a + π/3) = 1 then πx/2a + π/3 = 2kπ for some integer k,
and multiplying through by 6a/π gives 3x ≡ −2a (mod 12a), while
if the cosine is −1 then πx/2a + π/3 = (2k + 1)π and multiplying by
6a/π gives 3x ≡ 4a (mod 12a). In both cases we have 3x divisible
by 2, so x is divisible by 2 and hence so is a. Also our two cases give
−2a and 4a, respectively, are divisible by 3, so a is divisible by 3.
We conclude that 6|a and so our solution is minimal.
7. Let ABCD be a trapezoid (AB||CD) and choose F on the segment
AB such that DF = CF . Let E be the intersection of AC and BD,
and let O
1
, O
2
be the circumcenters of ADF, BCF . Prove that the
lines EF and O
1
O
2
are perpendicular.
Solution: Project each of points A, B, F orthogonally onto CD to
obtain A
, B
, F
; then F
is the midpoint of CD. Also let the cir-
cumcircles of AF D, BF C intersect line CD again at M, N respec-
tively; then AFMD, BF NC are isosceles trapezoids and F
M =
DA
, NF
= B
C. Let x = DA
, y = A
F
= AF , z = F
B
= F B,
w = B
C, using signed distances throughout (x < 0 if D is be-
tween A
and F
, etc.), so we have x+y = z + w; call this value S, so
DC = 2S. Also let line F E meet DC at G; since a homothety about
E with (negative) ratio CD/AB takes triangle ABE into CDE it
also takes F into G, so DG/GC = F B/AF = F
B
/A
B
= z/y
and we easily get DG = 2zS/(y + z), GC = 2yS/(y + z). Now
12
NF
= w, DF
= S implies DN = z and so DN/DG = (y + z)/2S.
Similarly F
M = x, F
C = S so MC = y and M C/GC = (y +z)/2S
also. So DN/DG = MC/GC, NG/DG = GM/GC and NG ·GC =
DG · GM. Since NC and DM are the respective chords of the
circumcircles of BF C and ADC that contain point G we conclude
that G has equal powers with respect to these two circles, i.e. it is
on the radical axis. F is also on the axis since it is an intersection
point of the circles, so the line F GE is the radical axis, which is
perpendicular to the line O
1
O
2
connecting the centers of the circles.
8. Find all natural numbers n for which a convex n-gon can be di-
vided into triangles by diagonals with disjoint interiors, such that
each vertex of the n-gon is the endpoint of an even number of the
diagonals.
Solution: We claim that 3|n is a necessary and sufficient condition.
To prove sufficiency, we use induction of step 3. Certainly for n = 3
we have the trivial dissection (no diagonals drawn). If n > 3 and 3|n
then let A
1
, A
2
, . . . , A
n
be the vertices of an n-gon in counterclock-
wise order; then draw the diagonals A
1
A
n−3
, A
n−3
A
n−1
, A
n−1
A
1
;
these three diagonals divide our polygon into three triangles and an
(n − 3)-gon A
1
A
2
. . . A
n−3
. By the inductive hypothesis the latter
can be dissected into triangles with evenly many diagonals at each
vertex, so we obtain the desired dissection of our n-gon, since each
vertex from A
2
through A
n−4
has the same number of diagonals in
the n-gon as in the (n −3)-gon (an even number), A
1
and A
n−3
each
have two diagonals more than in the (n − 3)-gon, while A
n−1
has 2
diagonals and A
n
and A
n−2
have 0 each.
To show necessity, suppose we have such a decomposition of a poly-
gon with vertices A
1
, A
2
, . . . , A
n
in counterclockwise order, and for
convenience assume labels are mod n. Call a diagonal A
i
A
j
in our
dissection a “right diagonal” from A
i
if no point A
i+2
, A
i+3
, . . . , A
j−1
is joined to A
i
(we can omit A
i+1
from our list since it is joined
by an edge). Clearly every point from which at least one diagonal
emanates has a unique right diagonal. Also we have an important
lemma: if A
i
A
j
is a right diagonal from A
i
, then within the polygon
A
i
A
i+1
. . . A
j
, each vertex belongs to an even number of diagonals.
Proof: Each vertex from any of the points A
i+1
, . . . , A
j−1
belongs
to an even number of diagonals of the n-gon, but since the diagonals
13
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