Preface
This book is a continuation of Mathematical Olympiads 1996-1997: Olym-piad Problems from Around the World, published by the American Math-ematics Competitions. It contains solutions to the problems from 34 na-tional and regional contests featured in the earlier book, together with selected problems (without solutions) from national and regional contests given during 1998.
This collection is intended as practice for the serious student who wishes to improve his or her performance on the USAMO. Some of the problems are comparable to the USAMO in that they came from na-tional contests. Others are harder, as some countries first have a nana-tional olympiad, and later one or more exams to select a team for the IMO. And some problems come from regional international contests (“mini-IMOs”). Different nations have different mathematical cultures, so you will find some of these problems extremely hard and some rather easy. We have tried to present a wide variety of problems, especially from those countries that have often done well at the IMO.
Each contest has its own time limit. We have not furnished this infor-mation, because we have not always included complete exams. As a rule of thumb, most contests allow a time limit ranging between one-half to one full hour per problem.
Thanks to the following students of the 1998 and 1999 Mathematical Olympiad Summer Programs for their help in preparing and proofreading solutions: David Arthur, Reid Barton, Gabriel Carroll, Chi-Bong Chan, Lawrence Detlor, Daniel Katz, George Lee, Po-Shen Loh, Yogesh More, Oaz Nir, David Speyer, Paul Valiant, Melanie Wood. Without their ef-forts, this work would not have been possible. Thanks also to Alexander Soifer for correcting an early draft of the manuscript.
The problems in this publication are copyrighted. Requests for repro-duction permissions should be directed to:
Dr. Walter Mientka
Secretary, IMO Advisory Board 1740 Vine Street
Contents
1 1997 National Contests: Solutions 3
1.1 Austria . . . 3
1.2 Bulgaria . . . 7
1.3 Canada . . . 24
1.4 China . . . 27
1.5 Colombia . . . 31
1.6 Czech and Slovak Republics . . . 34
1.7 France . . . 38 1.8 Germany . . . 40 1.9 Greece . . . 44 1.10 Hungary . . . 47 1.11 Iran . . . 52 1.12 Ireland . . . 55 1.13 Italy . . . 59 1.14 Japan . . . 62 1.15 Korea . . . 65 1.16 Poland . . . 73 1.17 Romania . . . 78 1.18 Russia . . . 86 1.19 South Africa . . . 105 1.20 Spain . . . 108 1.21 Taiwan . . . 111 1.22 Turkey . . . 118 1.23 Ukraine . . . 121 1.24 United Kingdom . . . 127
1.25 United States of America . . . 130
1.26 Vietnam . . . 136
2 1997 Regional Contests: Solutions 141 2.1 Asian Pacific Mathematics Olympiad . . . 141
2.2 Austrian-Polish Mathematical Competition . . . 145
2.3 Czech-Slovak Match . . . 149
2.4 Hungary-Israel Mathematics Competition . . . 153
2.5 Iberoamerican Mathematical Olympiad . . . 156
2.6 Nordic Mathematical Contest . . . 161
2.7 Rio Plata Mathematical Olympiad . . . 163
3 1998 National Contests: Problems 180
3.1 Bulgaria . . . 180
3.2 Canada . . . 183
3.3 China . . . 184
3.4 Czech and Slovak Republics . . . 185
3.5 Hungary . . . 186 3.6 India . . . 188 3.7 Iran . . . 190 3.8 Ireland . . . 193 3.9 Japan . . . 195 3.10 Korea . . . 196 3.11 Poland . . . 197 3.12 Romania . . . 198 3.13 Russia . . . 200 3.14 Taiwan . . . 207 3.15 Turkey . . . 208 3.16 United Kingdom . . . 209
3.17 United States of America . . . 211
3.18 Vietnam . . . 212
4 1998 Regional Contests: Problems 213 4.1 Asian Pacific Mathematics Olympiad . . . 213
4.2 Austrian-Polish Mathematics Competition . . . 214
4.3 Balkan Mathematical Olympiad . . . 216
4.4 Czech-Slovak Match . . . 217
4.5 Iberoamerican Olympiad . . . 218
4.6 Nordic Mathematical Contest . . . 219
1
1997 National Contests: Solutions
1.1
Austria
1. Solve the system for x, y real:
(x− 1)(y2+ 6) = y(x2+ 1)
(y− 1)(x2+ 6) = x(y2+ 1).
Solution: We begin by adding the two given equations together. After simplifying the resulting equation and completing the square, we arrive at the following equation:
(x− 5/2)2+ (y
− 5/2)2= 1/2. (1)
We can also subtract the two equations; subtracting the second given equation from the first and grouping, we have:
xy(y− x) + 6(x − y) + (x + y)(x − y) = xy(x − y) + (y − x) (x− y)(−xy + 6 + (x + y) − xy + 1) = 0
(x− y)(x + y − 2xy + 7) = 0
Thus, either x− y = 0 or x + y − 2xy + 7 = 0. The only ways to have x− y = 0 are with x = y = 2 or x = y = 3 (found by solving equation (1) with the substitution x = y).
Now, all solutions to the original system where x6= y will be solutions to x + y− 2xy + 7 = 0. This equation is equivalent to the following equation (derived by rearranging terms and factoring).
(x− 1/2)(y − 1/2) = 15/4. (2)
Let us see if we can solve equations (1) and (2) simultaneously. Let a = x− 5/2 and b = y − 5/2. Then, equation (1) is equivalent to:
a2+ b2= 1/2 (3)
and equation (2) is equivalent to:
(a+2)(b+2) = 15/4⇒ ab+2(a+b) = −1/4 ⇒ 2ab+4(a+b) = −1/2. (4)
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 an= a2n−1+
a2n−2+ a2n−3 for all n≥ 3. Prove that if ak= 1997 then k≤ 3.
Solution: We proceed indirectly; assume that for some k > 3, ak = 1997. Then, each of the four numbers ak−1, ak−2, ak−3,
and ak−4 must exist. Let w = ak−1, x = ak−2, y = ak−3, and
z = ak−4. Now, by the given condition, 1997 = w2+ x2+ y2. Thus,
w ≤ √1997 < 45, and since w is a positive integer, w ≤ 44. But then x2+ y2≥ 1997 − 442= 61.
Now, w = x2+ y2+ z2. Since x2+ y2≥ 61 and z2
≥ 0, x2+ y2+
z2
≥ 61. But w ≤ 44. Therefore, we have a contradiction and our assumption was incorrect.
If ak= 1997, then k≤ 3.
3. Let k be a positive integer. The sequence anis defined by a1= 1, and
an is the n-th positive integer greater than an−1 which is congruent
to n modulo k. Find an in closed form.
Solution: We have an= n(2 + (n− 1)k)/2. If k = 2, then an= n2.
First, observe that a1 ≡ 1 (mod k). Thus, for all n, an ≡ n
(mod k), and the first positive integer greater than an−1 which is
The n-th positive integer greater than an−1 that is congruent to n
modulo k is simply (n− 1)k more than the first positive integer greater than an−1 which satisfies that condition. Therefore, an =
an−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 K1 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 P1. Similarly, let
K2 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 P2. then the locus is the
intersection of the segments AP1 and CP2.
We begin by proving that the tangency point must lie on line AC. Let I1 be the center of the circle inscribed in ∠BAD. Let I2 be
the center of the circle inscribed in ∠BCD. Let X represent the tangency point of the circles.
Since circles I1 and I2 are inscribed in angles, these centers must
lie on the respective angle bisectors. Then, since AI1 and CI2 are
bisectors of opposite angles in a parallelogram, they are parallel; therefore, since I1I2 is a transversal, ∠AI1X = ∠CI2X.
Let T1 be the foot of the perpendicular from I1 to AB. Similarly,
let T2be the foot of the perpendicular from I2 to CD. Observe that
I1T1/AI1 = sin ∠I1AB = sin ∠I2CD = I2T2/CI2. But I1X = I1T1
and I2X = I2T2. Thus, I1X/AI1= I2X/CI2.
Therefore, triangles CI2X and AI1X are similar, and vertical angles
∠I1XA and ∠I2XC 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
locus, we can let circle I1 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 P1. Similarly, we can draw a circle tangent to circle I1 and
to the sides of ∠BCD; from our proof above, we know that it must be tangent to circle I1 at X. Again, it will definitely fall in the
parallelogram because X is between C and P2.
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 I1 or I2 to
be contained inside the parallelogram.
Therefore, our locus is indeed the intersection of segments AP1 and
1.2
Bulgaria
1. Find all real numbers m such that the equation
(x2− 2mx − 4(m2+ 1))(x2− 4x − 2m(m2+ 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= 5m2+ 4 and (x− 2)2= 2(m3+ 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 5m2+ 4 = 0 which is not possible for real m.
In the second case, we must have 2(m3+ m + 2) = 0; m3+ m + 2 factors as (m+1)(m2
−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 x2− 2mx − 4(m2+ 1) and x2− 4x − 2m(m2+ 1). Subtracting,
we get that x− r is a factor of (2m − 4)x − (2m3− 4m2+ 2m− 4), i.e.
(2m−4)r = (2m−4)(m2+1). So m = 2 or r = m2+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 = m2+ 1 and then substitution into (r− 2)2= 2(m3+ m + 2) gives
(m2− 1)2 = 2(m3+ m + 2), which can be rewritten and factored
as (m + 1)(m− 3)(m2+ 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
by O the intersection of BN and CM . Assume that triangle BOC has area 2.
(a) Prove that M B/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 − ∠N OC = 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 M B/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 M B/AB = 1/3 then AN/AC = 1/3, and Menelaus’ theorem in triangle ABN and line CM gives BO/ON = 3/4 so [BOC]/[BN C] = BO/BN = 3/7. Then [BOC]/[ABC] = (3/7)(CN/CA) = 2/7 as desired. Similarly if M B/AB = 2/3 the theorem gives us BO/BN = 6, so [BOC]/[BN C] = BO/BN = 6/7 and [BOC]/[ABC] = (6/7)(CN/AC) = 2/7.
(b) If M B/AB = 1/3 then M ON A is a cyclic quadrilateral since ∠A = π/3 and ∠O = π − (∠P OQ) = 2π/3. Thus ∠AOB = ∠AOM + ∠M OB = ∠AN M + ∠P OQ = ∠AN M + π/3. But M B/AB = 1/3 and AN/AC = 1/3 easily give that N is the projection of M onto AC, so ∠AN M = π/2 and ∠AOB = 5π/6.
If M B/AB = 2/3 then M ON A is a cyclic quadrilateral as before, so that ∠AOB = ∠AOM +∠M OB = ∠AN M +∠P OQ. But AM N is again a right triangle, now with right angle at M , and ∠M AN = π/3 so ∠AN M = π/6, so ∠AOB = π/2.
3. Let f (x) = x2 − 2ax − a2 − 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) =−a2
− 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) =−2a2− 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 M F = (a− b)c/2(a + b). So comparing triangles CM F and ABC, noting that the altitudes
to side M F (respectively AB) are equal, we have [CM F ]/S = (a− b)/2(a + b). Similarly using altitudes from M in triangles CM I and CM F (and using the ratio IC/CF found earlier), we have [CM I]/S = (a− b)/2(a + b + c); and using altitudes from I in triangles CGI and CM I 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 M F = (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 e1x1+ e2x2+ e3x3such that:
• x1, x2, x3∈ A;
• e1, e2, e3∈ {−1, 0, 1};
• at least one of e1, e2, e3 is nonzero;
• if xi= xj, then eiej6= −1.
The set A is free if all such sums are not divisible by n. (a) Find a free set of cardinalitybn/4c.
(b) Prove that any set of cardinality bn/4c + 1 is not free.
Solution:
(a) We show that the set A ={1, 3, 5, ..., 2bn/4c − 1} is free. Any combination e1x1 + e2x2+ e3x3 with zero or two ei’s equal
to 0 has an odd value and so is not divisible by n; otherwise, we have one ei equal to 0, so we have either a difference of
two distinct elements of A, which has absolute value less than 2bn/4c 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 4bn/4c − 2 < n and so again is not divisible by n.
(b) Suppose A is a free set; we will show |A| ≤ bn/4c. 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 combinationsP eixi
taken mod n is invariant, since we can simply flip the sign of any eiassociated 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.
6. Find the least natural number a for which the equation cos2π(a− x) − 2 cos π(a − x) + cos3πx
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 O1, O2 be the circumcenters of ADF, BCF . Prove that the
lines EF and O1O2 are perpendicular.
Solution: Project each of points A, B, F orthogonally onto CD to obtain A0, B0, F0; then F0 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 AF M D, BF N C are isosceles trapezoids and F0M =
DA0, N F0 = B0C. Let x = DA0, y = A0F0 = AF , z = F0B0 = F B, w = B0C, using signed distances throughout (x < 0 if D is be-tween A0and F0, 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 = F0B0/A0B0 = z/y and we easily get DG = 2zS/(y + z), GC = 2yS/(y + z). Now
N F0 = w, DF0= S implies DN = z and so DN/DG = (y + z)/2S. Similarly F0M = x, F0C = S so M C = y and M C/GC = (y + z)/2S also. So DN/DG = M C/GC, N G/DG = GM/GC and N G· 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 O1O2connecting 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 A1, A2, . . . , An be the vertices of an n-gon in
counterclock-wise order; then draw the diagonals A1An−3, An−3An−1, An−1A1;
these three diagonals divide our polygon into three triangles and an (n− 3)-gon A1A2. . . An−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 A2 through An−4 has the same number of diagonals in
the n-gon as in the (n− 3)-gon (an even number), A1and An−3each
have two diagonals more than in the (n− 3)-gon, while An−1 has 2
diagonals and An and An−2 have 0 each.
To show necessity, suppose we have such a decomposition of a poly-gon with vertices A1, A2, . . . , An in counterclockwise order, and for
convenience assume labels are mod n. Call a diagonal AiAj in our
dissection a “right diagonal” from Aiif no point Ai+2, Ai+3, . . . , Aj−1
is joined to Ai (we can omit Ai+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 AiAj is a right diagonal from Ai, then within the polygon
AiAi+1. . . Aj, each vertex belongs to an even number of diagonals.
Proof: Each vertex from any of the points Ai+1, . . . , Aj−1 belongs
of the n-gon are nonintersecting these diagonals must lie within our smaller polygon, so we have an even number of such diagonals for each of these points. By hypothesis, Aiis not connected via a
diag-onal to any other point of this polygon, so we have 0 diagdiag-onals from Ai, an even number. Finally evenly many diagonals inside this
poly-gon stem from Aj, since otherwise we would have an odd number of
total endpoints of all diagonals.
Now we can show 3|n by strong induction on n. If n = 1 or 2, then there is clearly no decomposition, while if n = 3 we have 3|n. For n > 3 choose a vertex Ai1 with some diagonal emanating from it,
and let Ai1Ai2 be the right diagonal from Ai1. By the lemma there
are evenly many diagonals from Ai2 with their other endpoints in
{Ai1+1, Ai1+2, . . . , Ai2−1}, and one diagonal Ai1Ai2, so there must
be at least one other diagonal from Ai2 (since the total number of
di-agonals there is even). This implies Ai1Ai2 is not the right diagonal
from Ai2, so choose the right diagonal Ai2Ai3. Along the same lines
we can choose the right diagonal Ai3Ai4 from Ai3, with Ai2 and Ai4
distinct, then continue with Ai4Ai5 as the right diagonal from Ai4,
etc. Since the diagonals of the n-gon are nonintersecting this pro-cess must terminate with some Aik+1 = Ai1. Now examine each of
the polygons AixAix+1Aix+2. . . Aix+1, x = 1, 2, . . . , k (indices x are
taken mod k). By the lemma each of these polygons is divided into triangles by nonintersecting diagonals with evenly many diagonals at each vertex, so by the inductive hypothesis the number of vertices of each such polygon is divisible by 3. Also consider the polygon Ai1Ai2. . . Aik. We claim that in this polygon, each vertex belongs
to an even number of diagonals. Indeed, from Aix we have an even
number of diagonals to points in Aix−1+1, Aix−1+2, . . . , Aix−1, plus
the two diagonals Aix−1Aix and AixAix+1. This leaves an even
num-ber of diagonals from Aix to other points; since Aix was chosen as
the endpoint of a right diagonal we have no diagonals lead to points in Aix+1, . . . , Aix+1−1, so it follows from the nonintersecting
crite-rion that all remaining diagonals must lead to points Aiy for some y.
Thus we have an even number of diagonals from Aix to points Aiy
for some fixed x; it follows from the induction hypothesis that 3|k. So, if we count each vertex of each polygon AixAix+1Aix+2. . . Aix+1
once and then subtract the vertices of Ai1Ai2. . . Aik, each vertex of
our n-gon is counted exactly once, but from the above we have been adding and subtracting multiples of 3. Thus we have 3|n.
9. For any real number b, let f (b) denote the maximum of the function sin x + 2 3 + sin x+ b
over all x∈ R. Find the minimum of f(b) over all b ∈ R.
Solution: The minimum value is 3/4. Let y = 3 + sin x; note y ∈ [2, 4] and assumes all values therein. Also let g(y) = y + 2/y; this function is increasing on [2, 4], so g(2) ≤ g(y) ≤ g(4). Thus 3 ≤ g(y) ≤ 9/2, and both extreme values are attained. It now fol-lows that the minimum of f (b) = max(|g(y) + b − 3|) is 3/4, which is attained by b = −3/4; for if b > −3/4 then choose x = π/2 so y = 4 and then g(y) + b− 3 > 3/4, while if b < −3/4 then choose x =−π/2 so y = 2 and g(y) + b − 3 = −3/4; on the other hand, our range for g(y) guarantees−3/4 ≤ g(y) + b − 3 ≤ 3/4 for b = −3/4. 10. Let ABCD be a convex quadrilateral such that ∠DAB = ∠ABC =
∠BCD. Let H and O denote the orthocenter and circumcenter of the triangle ABC. Prove that H, O, D are collinear.
Solution: Let M be the midpoint of B and N the midpoint of BC. Let E = AB∩ CD and F = BC ∩ AD. Then EBC and F AB are isosceles triangles, so EN∩ F M = 0. Thus applying Pappus’s theorem to hexagon M CEN AF , we find that G, O, D are collinear, so D lies on the Euler line of ABC and H, O, D are collinear. 11. For any natural number n≥ 3, let m(n) denote the maximum
num-ber of points lying within or on the boundary of a regular n-gon of side length 1 such that the distance between any two of the points is greater than 1. Find all n such that m(n) = n− 1.
Solution: The desired n are 4, 5, 6. We can easily show that m(3) = 1, e.g. dissect an equilateral triangle ABC into 4 congruent triangles and then for two points P, Q there is some corner triangle inside which neither lies; if we assume this corner is at A then the circle with diameter BC contains the other three small triangles and so contains P and Q; BC = 1 so P Q ≤ 1. This method will be useful later; call it a lemma.
On the other hand, m(n)≥ n − 1 for n ≥ 4 as the following process indicates. Let the vertices of our n-gon be A1, A2, . . . , An. Take
P1 = A1. Take P2 on the segment A2A3 at an extremely small
distance d2from A2; then P2P1> 1, as can be shown rigorously, e.g.
using the Law of Cosines in triangle P1A2P2 and the fact that the
cosine of the angle at A2 is nonnegative (since n≥ 4). Moreover P2
is on a side of the n-gon other than A3A4, and it is easy to see that
as long as n≥ 4, the circle of radius 1 centered at A4 intersects no
side of the n-gon not terminating at A4, so P2A4 > 1 while clearly
P2A3< 1. So by continuity there is a point P3on the side A3A4with
P2P3= 1. Now slide P3by a small distance d3on A3A4towards A4;
another trigonometric argument can easily show that then P2P3> 1.
Continuing in this manner, obtain P4 on A4A5 with P3P4 = 1 and
slide P4 by distance d4 so that now P3P4 > 1, etc. Continue doing
this until all points Pi have been defined; distances PiPi+1 are now
greater than by construction, Pn−1P1 > 1 because P1 = A1 while
Pn−1 is in the interior of the side An−1An; and all other PiPj are
greater than 1 because it is easy to see that the distance between any two points of nonadjacent sides of the n-gon is at least 1 with equality possible only when (among other conditions) Pi, Pjare endpoints of
their respective sides, and in our construction this never occurs for distinct i, j. So our construction succeeds. Moreover, as all the distances di tend to 0 each Pi tends toward Ai, so it follows that
the maximum of the distances AiPican be made as small as desired
by choosing di sufficiently small. On the other hand, when n > 6
the center O of the n-gon is at a distance greater than 1 from each vertex, so if the Pi are sufficiently close to the Ai then we will also
have OPi > 1 for each i. Thus we can add the point O to our set,
showing that m(n)≥ n for n > 6.
It now remains to show that we cannot have more than n− 1 points at mutual distances greater than 1 for n = 4, 5, 6. As before let the vertices of the polygon be A1, etc. and the center O; suppose we have
n points P1, . . . , Pn with PiPj > 1 for i not equal to j. Since n≤ 6
it follows that the circumradius of the polygon is not greater than 1, so certainly no Pican be equal to O. Let the ray from O through
Pi intersect the polygon at Qiand assume WLOG our numbering is
such that Q1, Q2, . . . , Qnoccur in that order around the polygon, in
the same orientation as the vertices were numbered. Let Q1be on the
also bring Q1into Q01, so triangles Q1Q01O and AkAk+1O are similar.
We claim P2 cannot lie inside or on the boundary of quadrilateral
OQ1Ak+1Q01. To see this, note that P1Q1Ak+1 and P1Ak+1Q01 are
triangles with an acute angle at P1, so the maximum distance from
P1 to any point on or inside either of these triangles is attained
when that point is some vertex; however P1Q1 ≤ OQ1 ≤ 1, and
P1Ak+1 ≤ O1Ak+1 ≤ 1 (e.g. by a trigonometric argument similar
to that mentioned earlier), and as for P1Q01, it is subsumed in the
following case: we can show that P1P ≤ 1 for any P on or inside
OQ1Q01, because n≤ 6 implies that ∠Q1OQ01= 2π/n≥ π/3, and so
we can erect an equilateral triangle on Q1Q01 which contains O, and
the side of this triangle is less than AkAk+1= 1 (by similar triangles
OAkAk+1and OQ1Q01) so we can apply the lemma now to show that
two points inside this triangle are at a distance at most 1. The result of all this is that P2 is not inside the quadrilateral OQ1Ak+1Q01, so
that ∠P1OP2= ∠Q1OP2> 2π/n. On the other hand, the label P1
is not germane to this argument; we can show in the same way that ∠PiOPi+1 > 2π/n for any i (where Pn+1 = P1). But then adding
these n inequalities gives 2π > 2π, a contradiction, so our points Pi
cannot all exist. Thus m(n)≤ n − 1 for n = 4, 5, 6, completing the proof.
12. Find all natural numbers a, b, c such that the roots of the equations x2− 2ax + b = 0
x2− 2bx + c = 0 x2− 2cx + a = 0 are natural numbers.
Solution: We have that a2− b, b2− c, c2− a are perfect squares.
Since a2− b ≤ (a − 1)2, we have b≥ 2a − 1; likewise c ≥ 2b − 1, a ≥
2c− 1. Putting these together gives a ≥ 8a − 7, or a ≤ 1. Thus (a, b, c) = (1, 1, 1) is the only solution.
13. Given a cyclic convex quadrilateral ABCD, let F be the intersection of AC and BD, and E the intersection of AD and BC. Let M, N be the midpoints of AB, CD. Prove that
M N EF = 1 2 AB CD − CD AB .
Solution: Since ABCD is a cyclic quadrilateral, AB and CD are antiparallel with respect to the point E, so a reflection through the bisector of ∠AEB followed by a homothety about E with ratio AB/CD takes C, D into A, B respectively. Let G be the image of F under this transformation. Similarly, reflection through the bisec-tor of ∠AEB followed by homothety about E with ratio CD/AB takes A, B into C, D; let H be the image of F under this trans-formation. G, H both lie on the reflection of line EF across the bisector of ∠AEB, so GH = |EG − EH| = EF |AB/CD − CD/AB|. On the other hand, the fact that ABCD is cyclic implies (e.g. by power of a point) that triangles ABF and DCF are similar with ratio AB/CD. But by virtue of the way the points A, B, G were shown to be obtainable from C, D, F , we have that BAG is also similar to DCF with ratio AB/CD, so ABF and BAG are con-gruent. Hence AG = BF, AF = BG and AGBF is a parallelo-gram. So the midpoints of the diagonals of AGBF coincide, i.e. M is the midpoint of GF . Analogously (using the parallelogram CHDF ) we can show that N is the midpoint of HF . But then M N is the image of GH under a homothety about F with ratio 1/2, so M N = GH/2 = (EF/2)|AB/CD − CD/AB| which is what we wanted to prove.
14. Prove that the equation
x2+ y2+ z2+ 3(x + y + z) + 5 = 0 has no solutions in rational numbers.
Solution: Let u = 2x + 3, v = 2y + 3, w = 2z + 3. Then the given equation is equivalent to
u2+ v2+ w2= 7. It is equivalent to ask that the equation
x2+ y2+ z2= 7w2
has no nonzero solutions in integers; assume on the contrary that (x, y, z, w) is a nonzero solution with |w| + |x| + |y| + |z| minimal.
Modulo 4, we have x2+ y2+ z2≡ 7w2, but every perfect square is
congruent to 0 or 1 modulo 4. Thus we must have x, y, z, w even, and (x/2, y/2, z/2, w/2) is a smaller solution, contradiction.
15. Find all continuous functions f : R→ R such that for all x ∈ R, f (x) = f x2+1 4 .
Solution: Put g(x) = x2+ 1/4. Note that if−1/2 ≤ x ≤ 1/2, then x≤ g(x) ≤ 1/2. Thus if −1/2 ≤ x0 ≤ 1/2 and xn+1 = g(xn) for
n≥ 0, the sequence xn tends to a limit L > 0 with g(L) = L; the
only such L is L = 1/2. By continuity, the constant sequence f (xn)
tends to f (1/2). In short, f is constant over [−1/2, 1/2].
Similarly, if x ≥ 1/2, then 1/2 ≤ g(x) ≤ x, so analogously f is constant on this range. Moreover, the functional equation implies f (x) = f (−x). We conclude f must be constant.
16. Two unit squares K1, K2with centers M, N are situated in the plane
so that M N = 4. Two sides of K1 are parallel to the line M N , and
one of the diagonals of K2 lies on M N . Find the locus of the
mid-point of XY as X, Y vary over the interior of K1, K2, respectively.
Solution: Introduce complex numbers with M = −2, N = 2. Then the locus is the set of points of the form−(w + xi) + (y + zi), where |w|, |x| < 1/2 and |x + y|, |x − y| < √2/2. The result is an octagon with vertices (1 +√2)/2 + i/2, 1/2 + (1 +√2)i/2, and so on. 17. Find the number of nonempty subsets of{1, 2, . . . , n} which do not
contain two consecutive numbers.
Solution: If Fn is this number, then Fn = Fn−1+ Fn−2: such
a subset either contains n, in which case its remainder is a subset of {1, . . . , n−2}, or it is a subset of {1, . . . , n−1}. From F1= 1, F2= 2,
we see that Fn is the n-th Fibonacci number.
18. For any natural number n≥ 2, consider the polynomial Pn(x) = n 2 +n 5 x +n 8 x2+· · · + n 3k + 2 xk,
where k =bn−2 3 c.
(a) Prove that Pn+3(x) = 3Pn+2(x)− 3Pn+1(x) + (x + 1)Pn(x).
(b) Find all integers a such that 3b(n−1)/2c divides Pn(a3) for all
n≥ 3.
Solution:
(a) This is equivalent to the identity (for 0≤ m ≤ (n + 1)/3) n + 3 3m + 2 = 3 n + 2 3m + 2 −3 n + 1 3m + 2 + n 3m + 2 + n 3m− 1 ,
which follows from repeated use of the identity a+1b = a b + a
b−1.
(b) If a has the required property, then P5(a3) = 10 + a3is divisible
by 9, so a ≡ −1 (mod 3). Conversely, if a ≡ −1 (mod 3), then a3+ 1 ≡ 0 (mod 9). Since P2(a3) = 1, P3(a3) = 3,
P4(a3) = 6, it follows from (a) that 3b(n−1)/2c divides Pn(a3)
for all n≥ 3.
19. Let M be the centroid of triangle ABC.
(a) Prove that if the line AB is tangent to the circumcircle of the triangle AM C, then
sin ∠CAM + sin ∠CBM ≤ √2 3.
(b) Prove the same inequality for an arbitrary triangle ABC.
Solution:
(a) Let G be the midpoint of AB, a, b, c the lengths of sides BC, CA, AB, and ma, mb, mcthe lengths of the medians from A, B, C,
respectively. We have c 2 2 = GA2= GM· GC = 1 3m 2 c = 1 12(2a 2+ 2b2 − c2),
whence a2+ b2= 2c2 and m a= √ 3b/2, mb= √ 3a/2. Thus
sin ∠CAM + sin ∠CBM = Kbm1
a + K 1 amb =(a 2+ b2) sin C √ 3ab ,
where K is the area of the triangle. By the law of cosines, a2+ b2= 4ab cos C, so the right side is 2 sin 2C/√3≤ 2/√3.
(b) There are two circles through C and M touching AB; let A1, B1
be the points of tangency, with A1 closer to A. Since G is the
midpoint of A1B1and CM/M G = 2, M is also the centroid of
triangle A1B1C. Moreover, ∠CAM ≤ ∠CA1M and ∠CBM ≤
∠CB1M . If the angles ∠CA1M and ∠CB1M are acute, we are
thus reduced to (a).
It now suffices to suppose ∠CA1M > 90◦, ∠CB1M ≤ 90◦.
Then CM2> CA21+ A1M2, that is,
1 9(2b 2 1+ 2a 2 1− c 2 1) > b 2 1+ 1 9(2b 2 1+ 2c 2 1− a 2 1),
where a1, b1, c1 are the side lengths of A1B1C. From (a), we
have a2
1+ b21 = c21 and the above inequality is equivalent to
a2 1> 7b21. As in (a), we obtain sin ∠CB1M = b1sin ∠B1CA1 a1 √ 3 = b1 a1 √ 3 s 1− a 2 1+ b21 4a1b1 2 . Setting b2 1/a21= x, we get sin ∠CB1M = 1 4√3 p 14x− x2− 1 < 1 4√3 r 2− 1 49− 1 = 1 7, since x < 1/7. Therefore
sin ∠CAM + sin ∠CBM < 1 + sin ∠CB1M < 1 +
1 7 <
2 √ 3. 20. Let m, n be natural numbers and m + i = aib2i for i = 1, 2, . . . , n,
where ai and bi are natural numbers and ai is squarefree. Find all
Solution: Clearly n ≤ 12. That means at most three of the m + i are perfect squares, and for the others, ai ≥ 2, so actually
n≤ 7.
We claim ai 6= aj for i = j. Otherwise, we’d have m + i = ab2i and
m + j = ab2
j, so 6≥ n − 1 ≥ (m + j) − (m + i) = a(b2j− b2i). This
leaves the possibilities (bi, bj, a) = (1, 2, 2) or (2, 3, 1), but both of
those force a1+· · · + an> 12.
Thus the a’s are a subset of{1, 2, 3, 5, 6, 7, 10, 11}. Thus n ≤ 4, with equality only if{a1, a2, a3, a4} = {1, 2, 3, 6}. But in that case,
(6b1b2b3b4)2= (m + 1)(m + 2)(m + 3)(m + 4) = (m2+ 5m + 5)2− 1,
which is impossible. Hence n = 2 or n = 3. One checks that the only solutions are then
(m, n) = (98, 2), (3, 3).
21. Let a, b, c be positive numbers such that abc = 1. Prove that 1 1 + a + b + 1 1 + b + c+ 1 1 + c + a ≤ 1 2 + a+ 1 2 + b+ 1 2 + c. Solution: Brute force! Put x = a + b + c and y = ab + bc + ca. Then the given inequality can be rewritten
3 + 4x + y + x2 2x + y + x2+ xy ≤ 12 + 4x + y 9 + 4x + 2y, or 3x2y + xy2+ 6xy− 5x2 − y2 − 24x − 3y − 27 ≥ 0, or (3x2y− 5x2 − 12x) + (xy2 − y2 − 3x − 3y) + (6xy − 9x − 27) ≥ 0, which is true because x, y≥ 3.
22. Let ABC be a triangle and M, N the feet of the angle bisectors of B, C, respectively. Let D be the intersection of the ray M N with the circumcircle of ABC. Prove that
1 BD = 1 AD+ 1 CD.
Solution: Let A1, B1, C1 be the orthogonal projections of D onto
BC, CA, AB, respectively. Then
DB1= DA sin ∠DAB1= DA sin ∠DAC =
DA· DC
2R ,
where R is the circumradius of ABC. Likewise DA1= DB· DC/2R
and DC1= DA· DB/2R. Thus it suffices to prove DB1= DA1+
DC1.
Let m be the distance from M to AB or BC, and n the distance from N to AC or BC. Also put x = DM/M N (x > 1). Then
DB1 n = x, DC1 m = x− 1, DA1− m n− m = x. Hence DB1 = nx, DC1 = m(x− 1), DA1 = nx− m(x − 1) = DB1− DC1, as desired.
23. Let X be a set of cardinality n + 1 (n ≥ 2). The ordered n-tuples (a1, a2, . . . , an) and (b1, b2, . . . , bn) of distinct elements of X
are called separated if there exist indices i 6= j such that ai = bj.
Find the maximal number of n-tuples such that any two of them are separated.
Solution: If An+1is the maximum number of pairwise separated
n-tuples, we have An+1≤ (n + 1)Anfor n≥ 4, since among pairwise
separated n-tuples, those tuples with a fixed first element are also pairwise separated. Thus An ≤ n!/2. To see that this is optimal,
take all n-tuples (a1, . . . , an) such that adding the missing member
1.3
Canada
1. How many pairs (x, y) of positive integers with x≤ y satisfy gcd(x, y) = 5! and lcm(x, y) = 50!?
Solution: First, note that there are 15 primes from 1 to 50: (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47).
To make this easier, let’s define f (a, b) to be the greatest power of b dividing a. (Note g(50!, b) > g(5!, b) for all b < 50.)
Therefore, for each prime p, we have either f (x, p) = f (5!, p) and f (y, p) = f (50!, p) OR f (y, p) = f (5!, p) and f (x, p) = f (50!, p). Since we have 15 primes, this gives 215 pairs, and clearly x6= y in
any such pair (since the gcd and lcm are different), so there are 214 pairs with x≤ y.
2. Given a finite number of closed intervals of length 1, whose union is the closed interval [0, 50], prove that there exists a subset of the intervals, any two of whose members are disjoint, whose union has total length at least 25. (Two intervals with a common endpoint are not disjoint.)
Solution: Consider
I1 = [1 + e, 2 + e], I2 = [3 + 2e, 4 + 2e], . . . I24 = [47 + 24e, 48 + 24e] where e is small enough that 48 + 24e < 50. To have the union of the intervals include 2k + ke, we must have an interval whose smallest element is in Ik. However, the difference between an element in Ik and Ik + 1 is always greater than 1, so these do not overlap. Taking these intervals and [0, 1] (which must exist for the union to be [0, 50]) we have 25 disjoint intervals, whose total length is, of course, 25. 3. Prove that 1 1999 < 1 2 · 3 4 · · · · · 1997 1998 < 1 44.
Solution: Let p = 1/2· 3/4 · . . . · 1997/1998 and q = 2/3 · 4/5 · . . . · 1998/1999. Note p < q, so p2 < pq = 1/2· 2/3 · . . . · 1998/1999 = 1/1999. Therefore, p < 1/19991/2< 1/44. Also, p = 1998! (999!· 2999)2 = 2−1998 1998 999 , while 21998=1998 0 +· · · +1998 1998 < 19991998 999 . Thus p > 1/1999.
4. Let O be a point inside a parallelogram ABCD such that ∠AOB + ∠COD = π. Prove that ∠OBC = ∠ODC.
Solution: Translate ABCD along vector AD so A0 and D are the same, and so that B0 and C are the same
Now, ∠COD + ∠CO0D = ∠COD + ∠A0O0D0 = 180, so OCO0D is
cyclic. Therefore, ∠OO0C = ∠ODC
Also, vector BC and vector OO0 both equal vector AD so OBCO0 is a parallelogram. Therefore, ∠OBC = ∠OO0C = ∠ODC.
5. Express the sum
n X k=0 (−1)k k3+ 9k2+ 26k + 24 n k
in the form p(n)/q(n), where p, q are polynomials with integer coef-ficients. Solution: We have n X k=0 (−1)k k3+ 9k2+ 26k + 24 n k = n X k=0 (−1)k (k + 2)(k + 3)(k + 4) n k
= n X k=0 (−1)k k + 1 (n + 1)(n + 2)(n + 3)(n + 4) n + 4 k + 4 = 1 (n + 1)(n + 2)(n + 3)(n + 4) n+4 X k=4 (−1)k(k − 3)n + 4 k and n+4 X k=0 (−1)k(k − 3)n + 4 k = n+4 X k=0 (−1)kkn + 4 k − 3 n+4 X k=0 (−1)kn + 4 k = n+4 X k=1 (−1)kkn + 4 k − 3(1 − 1)n+4 = 1 n + 4 n+4 X k=1 (−1)kn + 3 k− 1 = 1 n + 4(1− 1) n+3= 0. Therefore n+4 X k=4 (−1)k(k − 3)n + 4 k = − 3 X k=0 (−1)k(k − 3)n + 4 k = 3n + 4 0 − 2n + 4 1 +n + 4 2 = (n + 1)(n + 2) 2 and the given sum equals 1
1.4
China
1. Let x1, x2, . . . , x1997be real numbers satisfying the following
condi-tions: (a) −√1 3 ≤ xi≤ √ 3 for i = 1, 2, . . . , 1997; (b) x1+ x2+· · · + x1997=−318 √ 3.
Determine the maximum value of x121 + x122 +· · · + x121997.
Solution: Since x12 is a convex function of x, the sum of the
twelfth powers of the xi is maximized by having all but perhaps one
of the xi at the endpoints of the prescribed interval. Suppose n of
the xi equal −√13, 1996− n equal
√
3 and the last one equals −318√3 +√n
3− (1996 − n) √
3. This number must be in the range as well, so
−1 ≤ −318 × 3 + n − 3(1996 − n) ≤ 3.
Equivalently−1 ≤ 4n−6942 ≤ 3. The only such integer is n = 1736, the last value is 2/√3, and the maximum is 1736× 3−6+ 260× 36+
(4/3)6.
2. Let A1B1C1D1 be a convex quadrilateral and P a point in its
in-terior. Assume that the angles P A1B1 and P A1D1 are acute, and
similarly for the other three vertices. Define Ak, Bk, Ck, Dk as the
reflections of P across the lines Ak−1Bk−1, Bk−1Ck−1, Ck−1Dk−1,
Dk−1Ak−1.
(a) Of the quadrilaterals AkBkCkDk for k = 1, . . . , 12, which ones
are necessarily similar to the 1997th quadrilateral?
(b) Assume that the 1997th quadrilateral is cyclic. Which of the first 12 quadrilaterals must then be cyclic?
Solution: We may equivalently define Akas the foot of the
with diameters P Ak, P Bk, P Ck, P Dk give that
∠P AkBk = ∠P Dk+1Ak+1= ∠P Ck+2Dk+2
= ∠P Bk+3Ck+3= ∠P Ak+4Bk+4.
Likewise, in the other direction we have ∠P BkAk = P Bk+1Ak+1
and so on. Thus quadrilaterals 1, 5, 9 are similar to quadrilateral 1997, but the others need not be. However, if quadrilateral 1997 is cyclic (that is, has supplementary opposite angles), quadrilaterals 3, 7, and 11 are as well.
3. Show that there exist infinitely many positive integers n such that the numbers 1, 2, . . . , 3n can be labeled
a1, . . . , an, b1, . . . , bn, c1, . . . , cn
in some order so that the following conditions hold: (a) a1+ b1+ c1=· · · = an+ bn+ cn is a multiple of 6;
(b) a1+· · · + an = b1+· · · + bn = c1+· · · + cn is also a multiple
of 6.
Solution: The sum of the integers from 1 to 3n is 3n(3n + 1)/2, which we require to be a multiple both of 6n and of 9. Thus n must be a multiple of 3 congruent to 1 modulo 4. We will show that the desired arrangement exists for n = 9m. For n = 9, use the
arrangement
8 1 6 17 10 15 26 19 24
21 23 25 3 5 7 12 14 16
13 18 11 22 27 20 4 9 2
(in which the first row is a1, a2, . . . and so on). It suffices to produce
from arrangements for m (without primes) and n (with primes) an arrangement for mn (with double primes):
a00i+(j−1)m= ai+ (m− 1)a0j (1≤ i ≤ m, 1 ≤ j ≤ n)
4. Let ABCD be a cyclic quadrilateral. The lines AB and CD meet at P , and the lines AD and BC meet at Q. Let E and F be the points where the tangents from Q meet the circumcircle of ABCD. Prove that points P, E, F are collinear.
Solution: Let X0 denote the tangent of the circle at a point X on the circle. Now take the polar map through the circumcircle of ABCD. To show P, E, F are collinear, we show their poles are concurrent. E and F map to E0 and F0 which meet at Q. Since P = AB∩ CD, the pole of P is the line through A0∩ B0 and C0∩ D0,
so we must show these points are collinear with Q.
However, by Pascal’s theorem for the degenerate hexagon AADBBC, the former is collinear with Q and the intersection of AC and BD, and by Pascal’s theorem for the degenerate hexagon ADDBCC, the latter is as well.
5. [Corrected] Let A = {1, 2, . . . , 17} and for a function f : A → A, denote f[1](x) = f (x) and f[k+1](x) = f (f[k](x)) for k ∈ N. Find
the largest natural number M such that there exists a bijection f : A→ A satisfying the following conditions:
(a) If m < M and 1≤ i ≤ 17, then
f[m](i + 1)− f[m](i)6≡ ±1 (mod 17). (b) For 1≤ i ≤ 17,
f[M ](i + 1)− f[M ](i)
≡ ±1 (mod 17). (Here f[k](18) is defined to equal f[k](1).)
Solution: The map f (x) = 3x (mod 17) satisfies the required condition for M = 8, and we will show this is the maximum. Note that by composing with a cyclic shift, we may assume that f (17) = 17. Then M is the first integer such that f[M ](1) equals 1 or 16, and likewise for 16. If 1 and 16 are in the same orbit of the permutation f , this orbit has length at most 16, and so either 1 or 16 must map to the other after 8 steps, so M ≤ 8. If they are in different orbits, one (and thus both) orbits have length at most 8, so again M ≤ 8.
6. [Corrected] Let a1, a2, . . . , be nonnegative numbers satisfying an+m≤ an+ am (m, n∈ N). Prove that an≤ ma1+ n m − 1 am for all n≥ m.
Solution: By induction on k, an ≤ kam+ an−mk for k < m/n. Put
n = mk + r with r ∈ {1, . . . , m}; then an ≤ kam+ ar= n− r m am+ ar≤ n− m m am+ ma1 since am≤ ma1 and ar≤ ra1.
1.5
Colombia
1. We are given an m× n grid and three colors. We wish to color each segment of the grid with one of the three colors so that each unit square has two sides of one color and two sides of a second color. How many such colorings are possible?
Solution: Call the colors A, B, C, and let Now let an be the
number of such colorings of a horizontal 1× n board given the col-ors of the top grid segments. For n = 1, assume WLOG the top grid segment is colored A. Then there are three ways to choose the other A-colored segment, and two ways to choose the colors of the remaining two segments for a total of a1= 6 colorings.
We now find an+1 in terms of an. Given any coloring of a 1× n
board, assume WLOG that its rightmost segment is colored A. Now imagine adding a unit square onto the right side of the board to make a 1× (n + 1) board, where the top color of the new square is known. If the new top segment is colored A, then there are two ways to choose the colors of the remaining two segments; otherwise, there are two ways to choose which of the remaining segments is colored A. So, an+1= 2an, so an= 3· 2n.
As for the original problem, there are 3nways to color the top edges
and 3· 2n ways to color each successive row, for a total of 3m+n2mn
colorings.
2. We play the following game with an equilaterial triangle of n(n+1)/2 pennies (with n pennies on each side). Initially, all of the pennies are turned heads up. On each turn, we may turn over three pennies which are mutually adjacent; the goal is to make all of the pennies show tails. For which values of n can this be achieved?
Solution: This can be achieved for all n ≡ 0, 2 (mod 3); we show the positive assertion first. Clearly this is true for n = 2 and n = 3 (flip each of the four possible triangles once). For larger n, flip each possible set of three pennies once; the corners have been flipped once, and the pennies along the sides of the triangle have each been flipped three times, so all of them become tails. Meanwhile, the interior pennies have each been flipped six times, and they form a triangle of side length n− 3; thus by induction, all such n work.
Now suppose n ≡ 1 (mod 3). Color the pennies yellow, red and blue so that any three adjacent pennies are different colors; also any three pennies in a row will be different colors. If we make the corners all yellow, then there will be one more yellow penny than red or blue. Thus the parity of the number of yellow heads starts out different than the parity of the number of red heads. Since each move changes the parity of the number of heads of each color, we cannot end up with the parity of yellow heads equal to that of red heads, which would be the case if all coins showed tails. Thus the pennies cannot all be inverted.
3. Let ABCD be a fixed square, and consider all squares P QRS such that P and R lie on different sides of ABCD and Q lies on a diagonal of ABCD. Determine all possible positions of the point S.
Solution: The possible positions form another square, rotated 45 degrees and dilated by a factor of 2 through the center of the square. To see this, introduce complex numbers such that A = 0, B = 1, C = 1 + i, D = i.
First suppose P and R lie on adjacent sides of ABCD; without loss of generality, suppose P lies on AB and R on BC, in which case Q must lie on AC. (For any point on BD other than the center of the square, the 90-degree rotation of AB about the point does not meet DA.) If P = x, Q = y + yi, then R = (2y− x)i and S = (x− y) + (y − x)i, which varies along the specified square. Now suppose P and R lie on opposite sides of ABCD; again without loss of generality, we assume P lies on AB, R on CD and Q on AC. Moreover, we may assume Q = y + yi with 1/2≤ y ≤ 1. The 90-degree rotation of AB about Q meets CD at a unique point, and so P = 2y− 1, R = i, and S = y − 1 + (1 − y)i, which again varies along the specified square.
4. Prove that the set of positive integers can be partitioned into an in-finite number of (disjoint) inin-finite sets A1, A2, . . . so that if x, y, z, w
belong to Ak for some k, then x− y and z − w belong to the
same set Ai (where i need not equal k) if and only if x/y = z/w.
Solution: Let Ak consist of the numbers of the form (2k− 1)(2n);
x, y, z, w∈ Ak with x > y and z > w. Write x = (2k−1)(2a+b), y = (2k −1)(2a), z = (2k −1)(2c+d), w = (2k −1)(2c). Then x− y = (2k − 1)(2b − 1)(2a), z − w = (2k − 1)(2d − 1)(2c).
Also x/y = 2b, z/w = 2d. Now x/y = z/w if and only if b = d if and only if x− y and z − w have the same largest odd divisor.
1.6
Czech and Slovak Republics
1. Let ABC be a triangle with sides a, b, c and corresponding angles α, β, γ. Prove that the equality α = 3β implies the inequality (a2− b2)(a− b) = bc2, and determine whether the converse also holds.
Solution: By the extended law of sines, a = 2R sin α, b = 2R sin β, c = 2R sin γ, where R is the circumradius of ABC. Thus,
(a2− b2)(a
− b) = 8R3(sin2α
− sin2β)(sin α
− sin β) = 8R3(sin23β− sin2β)(sin 3β− sin β) = 8R3(sin 3β− sin β)2(sin 3β + sin β)
= 8R3(8 cos22β sin2β sin2β cos β) = 8R3(sin2(180◦− 4β))(sin β) = 8R3(sin2γ)(sin β)
= bc2.
The converse is false in general; we can also have α = 3β− 360◦, e.g.
for α = 15◦, β = 125◦, γ = 40◦.
2. Each side and diagonal of a regular n-gon (n ≥ 3) is colored red or blue. One may choose a vertex and change the color of all of the segments emanating from that vertex, from red to blue and vice versa. Prove that no matter how the edges were colored initially, it is possible to make the number of blue segments at each vertex even. Prove also that the resulting coloring is uniquely determined by the initial coloring.
Solution: All congruences are taken modulo 2.
First, changing the order in which we choose the vertices does not affect the end coloring. Also, choosing a vertex twice has no net effect on the coloring. Then choosing one set of vertices has the same effect as choosing its “complement”: the latter procedure is equivalent to choosing the first set, then choosing all the vertices. (Here, in a procedure’s complement, vertices originally chosen an odd number of times are instead chosen an even number of times, and vice versa.)
Label the vertices 1, . . . , 2n + 1. Let ai be the number of blue
seg-ments at each vertex, bibe the number of times the vertex is chosen,
and B be the sum of all bi. When vertex k is chosen, ak becomes
2n− ak ≡ ak; on the other hand, the segment from vertex k to each
other vertex changes color, so the other ai change parity.
Summing the ai gives twice the total number of blue segments; so,
there are an even number of vertices with odd ai— say, 2x vertices.
Choose these vertices. The parity of these aialternates 2x− 1 times
to become even. The parity of the other ai alternates 2x times to
remain even. Thus, all the vertices end up with an even number of blue segments. We now prove the end coloring is unique.
Consider a procedure with the desired results. At the end, ai
be-comes ai+ B− bi (mod 2). All the ai equal each other at the end,
so bj ≡ bk if and only if aj ≡ ak originally. Thus, either bi ≡ 1 if
and only if ai ≡ 1 — the presented procedure — or bi ≡ 1 if and
only if ai ≡ 0 – resuluting in an equivalent coloring from the first
pargraph’s conclusions. Thus, the resulting coloring is unique. This completes the proof.
Note: For a regular 2n-gon, n ≥ 2, choosing a vertex reverses the parities of all of the ai, so it is impossible to have all even ai unless
the ai have equal parities to start with. And even if it is possible to
have all even ai, the resulting coloring is not unique.
3. The tetrahedron ABCD is divided into five convex polyhedra so that each face of ABCD is a face of one of the polyhedra (no faces are di-vided), and the intersection of any two of the five polyhedra is either a common vertex, a common edge, or a common face. What is the smallest possible sum of the number of faces of the five polyhedra? Solution: The smallest sum is 22. No polyhedron shares two faces with ABCD; otherwise, its convexity would imply that it is ABCD. Then exactly one polyhedron P must not share a face with ABCD, and has its faces in ABCD’s interior. Each of P ’s faces must then be shared with another polyhedron, implying that P shares at least 3 vertices with each of the other polyhedra. Also, any polyhedron face not shared with ABCD must be shared with another polyhedron. This implies that the sum of the number of faces is even. Each poly-hedron must have at least four faces for a sum of at least 20. Assume
this is the sum. Then each polyhedron is a four-vertex tetrahedron, and P shares at most 2 vertices with ABCD. Even if it did share 2 vertices with ABCD, say A and B, it would then share at most 2 vertices with the tetrahedron containing ACD, a contradiction. Therefore, the sum of the faces must be at least 22. This sum can indeed be obtained. Let P and Q be very close to A and B, respec-tively; then the five polyhedra AP CD, P QCD, BQCD, ABDP Q, and ABCP Q satisfy the requirements.
4. Show that there exists an increasing sequence {an}∞n=1 of natural
numbers such that for any k ≥ 0, the sequence {k + an} contains
only finitely many primes.
Solution: Let pk be the k-th prime number, k ≥ 1. Set a1 = 2.
For n≥ 1, let an+1be the least integer greater than an that is
con-gruent to−k modulo pk+1 for all k≤ n. Such an integer exists by
the Chinese Remainder Theorem. Thus, for all k ≥ 0, k + an ≡ 0
(mod pk+1) for n≥ k + 1. Then at most k + 1 values in the sequence
{k +an} can be prime; from the k +2-th term onward, the values are
nontrivial multiples of pk+1and must be composite. This completes
the proof.
5. For each natural number n≥ 2, determine the largest possible value of the expression
Vn= sin x1cos x2+ sin x2cos x3+· · · + sin xncos x1,
where x1, x2, . . . , xn are arbitrary real numbers.
Solution: By the inequality 2ab≤ a2+ b2, we get
Vn≤ sin2x1+ cos2x2 2 +· · · + sin2xn+ cos2x1 2 = n 2, with equality for x1=· · · = xn= π/4.
6. A parallelogram ABCD is given such that triangle ABD is acute and ∠BAD = π/4. In the interior of the sides of the parallelogram, points K on AB, L on BC, M on CD, N on DA can be chosen in various ways so that KLM N is a cyclic quadrilateral whose cir-cumradius equals those of the triangles AN K and CLM . Find the
locus of the intersection of the diagonals of all such quadrilaterals KLM N .
Solution: Since the arcs subtended by the angles ∠KLN , ∠KM N , ∠LKM , ∠LN M on the circumcircle of KLM N and the arcs sub-tended by ∠KAN and ∠LCM on the circumcircles of triangles AKN and CLM , respectively, are all congruent, these angles must all be equal to each other, and hence have measure 45◦. The trian-gles SKL and SM N , where S is the intersection of KM and N L, are thus right isosceles triangles homothetic through S. Under the homothety taking K to M and L to N , AB is sent to CD and BC to DA, so S must lie on the segment BD.
1.7
France
1. Each vertex of a regular 1997-gon is labeled with an integer, such that the sum of the integers is 1. Starting at some vertex, we write down the labels of the vertices reading counterclockwise around the polygon. Can we always choose the starting vertex so that the sum of the first k integers written down is positive for k = 1, . . . , 1997? Solution: Yes. Let bk be the sum of the first k integers; then
b1997 = 1. Let x be the minimum of the bk, and find the largest
k such that bk−1 = x; if we start there, the sums will be positive.
(Compare Spain 6.)
2. Find the maximum volume of a cylinder contained in the intersection of a sphere with center O and radius R and a cone with vertex O meeting the sphere in a circle of radius r, having the same axis as the cone.
Solution: Such a cylinder meets the sphere in a circle of some radius s < r. The distance from that circle to the center of the sphere is√R2− s2. The cylinder also meets the cone in a circle of
radius s, whose distance to the center of the sphere is spR2/r2− 1
(since the distance from the circle of radius r to the center of the sphere is√R2− r2). Thus the volume of the cylinder is
πs2(pR2− s2− spR2/r2− 1.
We maximize this by setting its derivative in s to zero: 0 = 2spR2− s2− s 3 √ R2− s2 − 3s 2p R2/r2− 1)
or rearranging and squaring, s4 − 4R2s2+ 4R4 R2− s2 = 9s2R2 − s2r2 r2 . Solving, s2=3R 2+ r2+p(9R2− r2)(R2− r2) 6
and one can now plug s2 into the volume formula given above to get
3. Find the maximum area of the orthogonal projection of a unit cube onto a plane.
Solution: This projection consists of the projections of three mu-tually orthogonal faces onto the plane. The area of the projection of a face onto the plane equals the absolute value of the dot product of the unit vectors perpendicular to the face and the plane. If x, y, z are these dot products, then the maximum area is the maximum of x + y + z under the condition x2+ y2+ z2= 1. However, by Cauchy-Schwarz,px2+ y2+ z2≥ 3(x + y + z) with equality iff x = y = z.
Thus the maximum is√3.
4. Given a triangle ABC, let a, b, c denote the lengths of its sides and m, n, p the lengths of its medians. For every positive real α, let λ(α) be the real number satisfying
aα+ bα+ cα= λ(α)α(mα+ nα+ pα). (a) Compute λ(2).
(b) Determine the limit of λ(α) as α tends to 0. (c) For which triangles ABC is λ(α) independent of α?
Solution: Say m, n, p are opposite a, b, c, respectively, and as-sume a ≤ b ≤ c. It is easily computed (e.g., using vectors) that m2= (2b2+ 2c2 − a2)/4 and so on, so λ(2) =√2 3. If x≤ y ≤ z, then as α→ 0, then x≤ (xα+ yα+ zα)1/α ≤ 31/αx
and so the term in the middle tends to x. We conclude that the limit of λ(α) as α→ 0 is a/p. For λ(α) to be independent of α, we first need a2/p2= 4/3, which reduces to a2+ c2= 2b2. But under that
condition, we have
m = c√3/2, n = b√3/2, p = a√3/2 and so λ(α) is clearly constant for such triangles.
1.8
Germany
1. Determine all primes p for which the system p + 1 = 2x2 p2+ 1 = 2y2 has a solution in integers x, y.
Solution: The only such prime is p = 7. Assume without loss of generality that x, y≥ 0. Note that p + 1 = 2x2is even, so p
6= 2. Also, 2x2≡ 1 ≡ 2y2 (mod p) which implies x
≡ ±y (mod p) since p is odd. Since x < y < p, we have x + y = p. Then
p2+ 1 = 2(p− x)2= 2p2− 4px + p + 1,
so p = 4x− 1, 2x2= 4x, x is 0 or 2 and p is−1 or 7. Of course −1
is not prime, but for p = 7, (x, y) = (2, 5) is a solution.
2. A square Sa is inscribed in an acute triangle ABC by placing two
vertices on side BC and one on each of AB and AC. Squares Sb and
Sc are inscribed similarly. For which triangles ABC will Sa, Sb, Sc
all be congruent?
Solution: This occurs for ABC equilateral (obvious) and in no other cases. Let R be the circumradius of ABC and let xa, xb, xcbe
the side lengths of Sa, Sb, Sc. Finally, let α, β, γ denote the angles
∠BAC, ∠CBA, ∠ACB.
Suppose Sa has vertices P and Q on BC, with P closer to B. Then
2R sin α = BC = BP + P Q + QC = xacot β + xa+ xacot γ
xa =
2R sin α 1 + cot β + cot γ
= 2R sin α sin β sin γ
sin β sin γ + cos β sin γ + cos γ + sin β = 2R sin α sin β sin γ
and similarly for xb and xc. Now xa= xb implies
sin β sin γ + sin α = sin γ sin α + sin β
0 = (sin β− sin α)(sin γ − 1).
Since ABC is acute, we have sin β = sin α, which implies α = β (the alternative is that α + β = π, which cannot occur in a triangle). Likewise β = γ, so ABC is equilateral.
3. In a park, 10000 trees have been placed in a square lattice. Deter-mine the maximum number of trees that can be cut down so that from any stump, you cannot see any other stump. (Assume the trees have negligible radius compared to the distance between adjacent trees.)
Solution: The maximum is 2500 trees. In any square of four adjacent trees, at most one can be cut down. Since the 100× 100 grid can be divided into 2500 such squares, at most 2500 trees can be cut down.
Identifying the trees with the lattice points (x, y) with 0≤ x, y ≤ 99, we may cut down all trees with even coordinates. To see this, note that if a, b, c, d are all even, and p/q is the expression of (d−b)/(c−a) in lowest terms (where p, q have the same signs as d− b, c − a), then one of a + p and b + q is odd, so the tree (a + p, b + q) blocks the view from (a, b) to (c, d).
4. In the circular segment AM B, the central angle ∠AM B is less than 90◦. FRom an arbitrary point on the arc AB one constructs the perpendiculars P C and P D onto M A and M B (C ∈ MA, D ∈ M B). Prove that the length of the segment CD does not depend on the position of P on the arc AB.
Solution: Since ∠P CM = ∠P DM = π/2, quadrilateral P CM D is cyclic. By the Extended Law of Sines, CD = P M sin CM D, which is constant.
5. In a square ABCD one constructs the four quarter circles having their respective centers at A, B, C and D and containing the two adjacent vertices. Inside ABCD lie the four intersection points E,
