Iberoamerican Mathematical Olympiads

1st Iberoamerican 1985

Problem A1

Find all integer solutions to: a + b + c = 24, a2 + b2 + c2 = 210, abc = 440.

Solution

ab + bc + ca = ( (a + b + c)2 - (a2 + b2 + c2) )/2 = 183, so a, b, c are roots of the cubic x3 - 24x2
+ 183x - 440 = 0. But it easily factorises as (x - 5)(x - 8)(x - 11) = 0, so the only solutions are
permutations of (5, 8, 11).

Problem A2

P is a point inside the equilateral triangle ABC such that PA = 5, PB = 7, PC = 8. Find AB.

Solution

Answer: ¥

Let the side length be x. Using the cosine formula, we have cos APB = (74 - x2)/70, cos APC
= (89 - x2)/80, cos BPC = (113 - x2)/112. But cos BPC = cos APC cos BPC - sin APC sin
BPC, so (113 - x2)/112 = (74 - x2)/79 (89 - x2)/80 - ¥  - (74 - x2)2/702) (1 - (89 - x2)2/802) ).

We isolate the square root term, then square. We multiply through by 25.256.49 and, after
some simplification, we get x6 - 138x4 + 1161x2 = 0. Hence x = 0, ±3, ±¥:HGLVFDUGWKH

solution for a point P inside the triangle is x = ¥

zero and negative solutions. x = 3 corresponds to a point P outside the triangle. So the unique

Alternative solution by Johannes Tang

Rotate the triangle about C through 60o. Let P go to P'. We have AP' = 7, CP' = 8 and angle
PCP' = 60o, so PP'C is equilateral. Hence angle CPP' = 60o. Also PP' = 8. Using the cosine
formula on triangle APP' we find angle APP' = 60o. Hence angle APC = 120o. Now applying
cosine formula to triangle APC, we get result.

Problem A3
 Find the roots r1, r2, r3, r4 of the equation 4x4 - ax3 + bx2 -
cx + 5 = 0, given that they are positive reals satisfying
r1/2 + r2/4 + r3/5 + r4/8 = 1.

Solution

= 1/44. But AM/GM gives that (r1/2) (r2/4) (r3/5) (r4/8) ”

We have r1r2r3r4 = 5/4 and hence (r1/2) (r2/4) (r3/5) (r4/8)

( (r1/2 + r2/4 + r3/5 + r4/8)/4 )4 = 1/44 with equality iff

r1/2 = r2/4 = r3/5 = r4/8. Hence we must have r1 = 1/2, r2
= 1, r3 = 5/4, r4 = 2.

Problem B1

The reals x, y, z satisfy x \[\DQG \]- x2)/(1 - x) = (xz - y2)/(1 - y). Show that
(yx - x2)/(1 - x) = x + y + z.

Solution

We have yz - x2 - y2z + yx2 = xz - y2 - x2z + xy2. Hence z(y - x - y2 + x2) = -y2 + xy2 - x2y +
x2. Hence z = (x + y - xy)/(x + y - 1).

So yz = x + y + z - xy - xz, so yz - x2 = x + y + z - x2 - xy - xz = (x + y + z)(1 - x), so (yz -

x2)/(1 - x) = (x + y + z).

Problem B2

The function f(n) is defined on the positive integers and takes non-negative integer values. It
satisfies (1) f(mn) = f(m) + f(n), (2) f(n) = 0 if the last digit of n is 3, (3) f(10) = 0. Find
f(1985).

Solution

If f(mn) = 0, then f(m) + f(n) = 0 (by (1)). But f(m) and f(n) are non-negative, so f(m) = f(n) =
0. Thus f(10) = 0 implies f(5) = 0. Similarly f(3573) = 0 by (2), so f(397) = 0. Hence f(1985)
= f(5) + f(397) = 0.

Problem B3

O is the circumcenter of the triangle ABC. The lines AO, BO, CO meet the opposite sides at
D, E, F respectively. Show that 1/AD + 1/BE + 1/CF = 2/AO.

Solution

Projecting onto the altitude from A, we have AD cos(C - B) = AC sin C = 2R sin B sin C, so
2R/AD = cos(C - B)/(sin B sin C).

Hence 2R/AD + 2R/BE + 2R/CF =cos(C - B)/(sin B sin C) + cos(A - C)/(sin C sin A) + cos(B
- A)/(sin A sin B). So 2R sin A sin B sin C (1/AD + 1/BE + 1/CF) = sin A cos(B - C) + sin B
cos(C - A) + sin C cos(A - B) = 3 sin A sin B sin C + sin A cos B cos C + sin B cos A cos C +
sin C cos A cos B = 3 sin A sin B sin C + sin(A + B) cos C + sin C cos A cos B = 3 sin A sin
B sin C + sin C (cos C + cos A cos B) = 3 sin A sin B sin C + sin C (-cos(A + B) + cos A cos
B) = 4 sin A sin B sin C. Hence 1/AD + 1/BE + 1/CF = 2/R.

2nd Iberoamerican 1987

Problem A1

Find f(x) such that f(x)2f( (1-x)/(1+x) ) = 64x for x not 0, ±1.

Solution

Put x = (1-y)/(1+y), then (1--x)/(1+x) = y, so f( (1-y)/(1+y) )2 f(y) = 64(1-y)/(1+y). Hence f(

(1-x)/(1+x) )2 f(x) = 64(1-x)/(1+x). But f(x)4 f( (1-x)/(1+x) )2 = 642x2, so f(x)3 = 64 x2(1 +
x)/(1 - x). Hence f(x) = 4 ( x2(1 + x)/(1 - x) )1/3.

Problem A2

In the triangle ABC, the midpoints of AC and AB are M and N respectively. BM and CN
meet at P. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching
every side), then ABC is isosceles.

Solution

If the quadrilateral has an inscribed circle then AM + PN = AN + PM (consider the tangents

to the circle from A, M, P, N). But if AB > AC, then BM > CN (see below). We have AN =
AB/2, PM = BM/3, AM = AC/2, PN = CN/3, so it follows that AM + PN < AN + PM.
Similarly, AB < AC implies AM + PN > AN + PM, so the triangle must be isosceles.

To prove the result about the medians, note that BM2 = BC2 + CM2 - 2 BC.CM cos C = (BC -
CM cos C)2 + (CM sin C)2. Similarly, CN2 = (BC - BN cos B)2 + (BN sin B)2. But MN is
parallel to BC, so CM sin C = BN sin B. But AB > AC, so BN > CM and B < C, so cos B >
cos C, hence BN cos B > CM cos C and BC - CM cos C > BC - BN cos B. So BM > CN.

Problem A3

Show that if (2 + ¥ k = 1 + m + n¥IRUSRVLWLYHLQWHJHUVPQNZLWKNRGGWKHQPLVD

perfect square.
Solution

We have (2 + ¥ 4 = 97 + 56¥  ¥ - 1 = 14 (2 + ¥ 2 - 1. Hence (2 + ¥ k+2

(2 + ¥ k - (2 + ¥ k-2. Thus if (2 + ¥ k = ak + bk¥WKHQDk+2 = 14 ak - ak-2.

= 14

Now suppose the sequence ck satisfies c1 = 1, c2 = 5, ck+1 = 4 ck - ck-1. We claim that ck2 - ck-
2
1ck+1 = 6. Induction on k. We have c3 = 19, so c2 - c1c3 = 25 - 19 = 6. Thus the result is true
for k = 2. Suppose it is true for k. Then ck+1 = 4 ck - ck-1, so ck+12 = 4 ckck+1 - ck-1ck+1 = 4 ckck+1
- ck2 + 6 = ck(4 ck+1 - ck) + 6 = ckck+2 + 6, so the result is true for k+1.

Now put dk = ck2 + 1. We show that dk+2 = 14 dk+1 - dk. Induction on k. We have d1 = 2, d2 =
26, d3 = 362 = 14 d2 - d1, so the result is true for k = 1. Suppose it is true for k. We have ck+3 -
4 ck+2 + ck+1 = 0. Hence 12 + 2 ck+3ck+1 - 8 ck+2ck+1 + 2 ck+12 = 12. Hence 2 ck+22 - 8 ck+2ck+1 + 2
ck+12 = 12. Hence 16 ck+22 - 8 ck+2ck+1 + ck+12 + 1 = 14 ck+22 + 14 - ck+12 - 1, or (4 ck+2 - ck+1)2 +
1 = 14 (ck+22 + 1) - (ck+12 + 1), or ck+32 + 1 = 14 (ck+22 + 1) - (ck+12 + 1), or dk+3 = 14 dk+2 - dk+1.
So the result is true for all k.

But a1 = 2, a3 = 26 and a2k+3 = 14 a2k+1 - a2k-1, and d1 = 2, d2 = 26 and dk+1 = 14 dk - dk-1. Hence
a2k-1 = dk = ck2 + 1.

Problem B1

Define the sequence p1, p2, p3, ... as follows. p1 = 2, and pn is the largest prime divisor of p1p2
... pn-1 + 1. Prove that 5 does not occur in the sequence.

Problem B2

Show that the roots r, s, t of the equation x(x - 2)(3x - 7) = 2 are real and positive. Find tan-1r
+ tan-1s + tan-1t.

Solution

Put f(x) = x(x - 2)(3x - 7) - 2 = 3x3 - 13x2 + 14x - 2. Then f(0) = -2, f(1) = 2, so there is a root
between 0 and 1. f(2) = -2, so there is another root between 1 and 2. f(3) = 4, so the third root
is between 2 and 3. f(x) = 0 has three roots, so they are all real and positive.

We have tan(a + b + c) = (tan a + tan b + tan c - tan a tan b tan c)/(1 - (tan a tan b + tan b tan c
+ tan c tan a)). So putting a = tan-1r, b = tan-1s, c = tan-1t, we have, tan(a + b + c) = ( (r + s + t)
- rst)/(1 - (rs + st + tr) ) = (13/3 - 2/3)/(1 - 14/3) = -1. So a + b + c = -Œ/4 + kŒ. But we know
that each of r, s, t is real and positive, so a + b + c lies in the range 0 to 3Œ/2. Hence a + b + c
= 3Œ/4.

Problem B3

ABCD is a convex quadrilateral. P, Q are points on the sides AD, BC respectively such that
AP/PD = BQ/QC = AB/CD. Show that the angle between the lines PQ and AB equals the
angle between the lines PQ and CD.

Solution
If AB is parallel to CD, then it is obvious that PQ is parallel to both. So assume AB and CD
meet at O. Take O as the origin for vectors. Let e be a unit vector in the direction OA and f a
unit vector in the direction OC. Take the vector OA to be ae, OB to be be, OC to be cf, and
OD to be df. Then OP is ( (d - c)ae + (a - b)df)/(d - c + a - b) and OQ is ( (d - c)be + (a -
b)cf)/(d - c + a - b). Hence PQ is (c - d)(a - b)(e + f)/(d - c + a - b). But e and f are unit vectors,
so e+ f makes the same angle with each of them and hence PQ makes the same angle with AB
and CD.

3rd Iberoamerican 1988

Problem A1

The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic
progression. Show that the triangle must be equilateral.

Solution

Let the sides be a, a+d, a+2d with d >= 0. Then the altitudes are k/a •N DG •N DG 
where k is twice the area. We claim that k/a + k/(a+2d) > 2k/(a + d) unless d = 0. This is
equivalent to (a + d)(a + 2d) + a(a + d) > 2a(a + 2d) or 2d2 > 0, which is obviously true. So the
altitudes can only form an arithmetic progression if d = 0 and hence the triangle is equilateral.

Problem A2

The positive integers a, b, c, d, p, q satisfy ad - bc = 1 and a/b > p/q > c/d. Show that q >= b +
d and that if q = b + d, then p = a + c.

Solution

p/q > c/d implies pd > cq and hence pd >= cq + 1, so p/q •FG TG 6LPLODUO\DE!ST
implies a/b •ST ET 6RDE- c/d • TG  TE   EG  TEG %XWDE- c/d = 1/bd.
Hence q •EG

Now assume q = b + d. We have ad - bc = 1 ”GVRDGFG- d ”EFFGDQGKHQFe (a+c-

1)/(b+d) ”FG6RS•DF6LPLODUO\DG- bc ”EVREFEDE•DGDEVR DF  EG 
•DE6RS”DF+HQFHS DF

Problem A3
P is a fixed point in the plane. Show that amongst triangles ABC such that PA = 3, PB = 5, PC
= 7, those with the largest perimeter have P as incenter.

Solution

Given points P, B, C and a fixed circle center P, we show that the point A on the circle which
maximises AB + AC is such that PA bisects angle BAC. Consider a point A' close to A. Then
the change in AB + AC as we move A to A' is AA'(sin PAC - sin PAB) + O(AA'2 ). So for a
maximal configuration we must have sin PAC = sin PAB, otherwise we could get a larger
sum by taking A' on one side or the other. This applies to each vertex of the triangle, so P
must be the incenter.

Problem B1

Points A1, A2, ... , An are equally spaced on the side BC of the triangle ABC (so that BA1 =
A1A2 = ... = An-1An = AnC). Similarly, points B1, B2, ... , Bn are equally spaced on the side CA,
and points C1, C2, ... , Cn are equally spaced on the side AB. Show that (AA12 + AA22 + ... +
AAn2 + BB12 + BB22 + ... + BBn2 + C12 + ... + CCn2) is a rational multiple of (AB2 + BC2 +
CA2).

Solution

Using the cosine formula, AAk2 = AB2 + k2BC2/(n+1)2 - 2k AB.BC/(n+1) cos B. So ™$$k2 =
n AB2 + BC2/(n+1)2 (12 + 22 + ... + n2) - 2 AB.BC cos B (1 + 2 + ... + n)/(n+1). Similarly for
the other two sides.

Thus the total sum is n (AB2 + BC2 + CA2) + n(2n+1)/(6(n+1) ) (AB2 + BC2 + CA2) - n
(AB.BC cos B + BC.CA cos C + CA.AB cos A). But AB.BC cos B = (AB2 + BC2 - CA2)/2,
so AB.BC cos B + BC.CA cos C + CA.AB cos A = (AB2 + BC2 + CA2)/2. Thus the sum is
rational multiple of (AB2 + BC2 + CA2).

Problem B2

Let k3 = 2 and let x, y, z be any rational numbers such that x + y k + z k2 is non-zero. Show
that there are rational numbers u, v, w such that (x + y k + z k2)(u + v k + w k2) = 1.

Solution
We need xu + 2zv + 2yw = 1, yu + xv + 2zw = 0, zu + yv + xw = 0. This is just a
straightforward set of linear equations. Solving, we get: u = (x2 - 2yz)/d, v = (2z2 - xy)/d, w =
(y2 - xz)/d, were d = x3 + 2y3 + 4z3 - 6xyz.

This would fail if d = 0. But if d = 0, then multiplying through by a suitable integer we have
6mnr = m3 + 2n3 + 4r3 for some integers m, n, r. But we can divide by any common factor of
m, n, r to get them without any common factor. But 6mnr, 2n3, 4r3 are all even, so m must be
even. Put m = 2M. Then 12Mnr = 8M3 + 2n3 + 4r3, so 6Mnr = 4M3 + n3 + 2r3. But 6Mnr, 4M3
and 2r3 are all even, so n must be even. Put n = 2N. Then 12MNr = 4M3 + 8N3 + 2r3, so 6MNr
= 2M3 + 4N3 + r3, so r must be even. So m, n, r had a common factor 2. Contradiction. So d
cannot be zero.

Problem B3

Let S be the collection of all sets of n distinct positive integers, with no three in arithmetic
progression. Show that there is a member of S which has the largest sum of the inverses of its
elements (you do not have to find it or to show that it is unique).

Solution

Induction on n. For n = 1, {1} is obviously maximal. Now suppose a1 < a2 < ... < an is a
maximal set for n. Take an+1 to be the smallest integer > an such that {a1, a2, ... , an+1} has no

AP and bn+1 ”Dn+1. There are only finitely many such sequences. So we can find one which is
three members in AP. Now consider the sequences b1 < b2 < ... < bn which have no three in

inverses. It is clearly maximal with respect to sequences whose largest member is ”Dn+1.
maximal. Suppose it is c1 < c2 < ... < cn+1. Now take whichever of ai, ci has the larger sum of

have 1/xn+1 < 1/an+1 and, by induction, 1/x1 + ... + 1/xn ”D1 + ... + 1/an, so 1/x1 + ... + 1/xn+1
Suppose we have a sequence x1 < x2 < ... < xn+1 with no three in AP and xn+1 > an+1. Then we

< 1/a1 + ... + 1/an+1, so it is worse than the sequence we have chosen.

4th Iberoamerican 1989

Problem A1

Find all real solutions to: x + y - z = -1; x2 - y2 + z2 = 1, -x3 + y3 + z3 = -1.

Solution

Answer: (x, y, z) = (1, -1, 1) or (-1, -1, -1).

From the first equation x = z - y - 1. Substituting in the second equation: 2z2 - 2yz + 2y - 2z =
0, so (z - 1)(z - y) = 0. Hence z = 1 or y = z. If z = 1, then from the first equation x + y = 0,
and hence from the last equation, x = 1, y = -1. If y = z, then x = -1, and hence from the last
equation y = z = -1.

Problem A2

Given positive real numbers x, y, z each less than Œ/2, show that Œ/2 + 2 sin x cos y + 2 sin y
cos z > sin 2x + sin 2y + sin 2z.

Solution
We have sin 2x + sin 2y + sin 2z - 2 sin x cos y - 2 sin y cos z = 2 sin x(cos x - cos y) + 2 sin
y(cos y - cos z) + 2 sin z cos z, so we wish to show that sin x(cos x - cos y) + sin y(cos y - cos
z) + sin z cos z < Œ/2 (*).

We have to consider six cases: (1) x ”\”]  [”]”\  \”[”]  \”]”[  ]”
x ”\  ]”\”[7KHILUVWFDVHLVREYLRXVIURPWKHGLDJUDPEHFDXVHWKHOKVUHSUHVHQWVWKH
shaded area, and the rhs represents the whole quarter circle.

In cases (2) and (5) the second term is negative, and - sin y < - sin x, so the sum of the first
two terms is less than sin x (cos x - cos y) + sin x (cos y - cos z) = sin x (cos x - cos z). But by
the same argument as the first case the two rectangles represented by sin x( cos x - cos z) and
sin z cos z are disjoint and fit inside the quarter circle. So we have proved (2) and (5).

In cases (3) and (4), the first term is negative. The remaining two terms represent disjoint
rectangles lying inside the quarter circle, so again the inequality holds.

In case (6) the first two terms are negative. The last term is ½ sin 2z ”½ < Œ/2, so the
inequality certainly holds.

Problem A3

If a, b, c, are the sides of a triangle, show that (a - b)/(a + b) + (b - c)/(b + c) + (c - a)/(a + c) <
1/16.

Solution

Put f(a, b, c) = (a - b)/(a + b) + (b - c)/(b + c) + (c - a)/(a + c). Let A, B, C be the permutation

of a, b, c, with A <= B <= C. If (A, B, C) = (b, a, c), (a, c, b) or (c, b, a), then f(a, b, c) = X,
where X = (B - A)/(B + A) + (C - B)/(C + B) - (C - A)/(A + C). If (A, B, C) = (a, b, c), (b, c,
a) or (c, a, b), then f(a, b, c) = -X.

Put B = A + h, C = B + k = A + h + k, where h, k •6LQFH$%&DUHWKHVLGHVRIDWULDQJOH

we also have A + B > C or A > k. So X = h/(2A + h) + k/(2A + 2h + k) - (h + k)/(2A + h + k)
= hk(h + k)/( (2A + h)(2A + h + k)(2A + 2h + k) ). This is obviously non-negative. We claim
also that it is < 1/20. That is equivalent to: 20h2k + 20hk2 < (2A + h)(2A + h + k)(2A + 2h +
k). Since k < A it is sufficient to show that 20h2k + 20hk2 ” NK NKN NKN 
= 18k3 + 27hk2 + 13h2k + 2h3 or 18k3 + 7hk2 - 7h2k + 2h3 •%XWN2 - 7hk + 2h2 = 7(k -
h/2)2 + h2/4 •DQGKDQGNDUHQRQ-negative, so 18k3 + h(7k2 - 7hk + 2h2) •

Thus we have established that 0 <= X < 1/20, which shows that f(a, b, c) < 1/20, which is
slightly stronger than the required result.

Problem B1

The incircle of the triangle ABC touches AC at M and BC at N and has center O. AO meets
MN at P and BO meets MN at Q. Show that MP.OA = BC.OQ.

Solution

The key to getting started is to notice that angle AQB = 90o.

Angle BAQ = 90o - B/2, so angle OAQ = 90o - B/2 - A/2 = C/2. So OQ = AO sin C/2. Thus
we have to show that MP = BC sin C/2.

Let the incircle touch AB at L and let Y be the midpoint of ML (also the intersection of ML
with AO). Angle NMC = 90o - C/2. It is also A/2 + angle MPY, so angle MPY = 90 - C/2 -
A/2 = B/2. Hence MP = MY/sin B/2. We have MY = MO sin MOA = r cos A/2 (where r is
the inradius, as usual). So MP = (r cos A/2)/sin B/2. We have BC = BN + NC = r (cot B/2 +
cot C/2), so MP/BC = (cos A/2)/( sin B/2 (cot B/2 + cot C/2) ). Hence MP/(BC sin C/2) = (
cos A/2 )/( cos B/2 sin C/2 + sin B/2 cos C/2) = cos A/2 /sin(B/2 + C/2) = 1.

Problem B2

The function f on the positive integers satisfies f(1) = 1, f(2n + 1) = f(2n) + 1 and f(2n) = 3
f(n). Find the set of all m such that m = f(n) for some n.

Solution

We show that to obtain f(n), one writes n in base 2 and then reads it in base 3. For example 12
= 11002, so f(12) = 11003 = 36. Let g(n) be defined in this way. Then certainly g(1) = 1. Now
2n+1 has the same binary expansion as 2n except for a final 1, so g(2n+1) = g(2n) + 1.
Similarly, 2n has the same binary expansion as n with the addition of a final zero. Hence
g(2n) = 3 g(n). So g is the same as f. Hence the set of all m such that m = f(n) for some n is
the the set of all m which can be written in base 3 without a digit 2.

Problem B3
Show that there are infinitely many solutions in positive integers to 2a2 - 3a + 1= 3b2 + b.

Solution

Put A = a - 1 and the equation becomes A(2A + 1) = b(3b + 1). Let d be the greatest common
divisor of A and b. Put A = dx, b = dy. Then x(2dx + 1) = y(3dy + 1). Since x and y are
coprime, x must divide 3dy + 1. So put 3dy + 1 = nx. Then 2dx + 1 = ny. Solving for x and y
in terms of n and d we get x = (n + 3d)/(n2 - 6d2), y = (n + 2d)/(n2 - 6d2).

So we would certainly be home if we could show that there were infinitely many solutions to
n2 - 6d2 = 1. It is not hard to find the first few: 12 - 6.02 = 1, 52 - 6.22 = 1, 492 - 6.202 = 1. We
notice that 492 = 2.52 - 1, so we wonder whether n = 2.492 - 1 might be another solution and
indeed we find it gives d = 1960 = 2.49.20. This suggests we try (2 n2 - 1)2 - 6(2nd)2 = 4n4 -
4n2 + 1 - 24n2d2 = 4n2(n2 - 6d2 - 1) + 1 = 1. So there are indeed infinitely many solutions to n2
- 6d2 = 1 and we are done.

5th Iberoamerican 1990

Problem A1

The function f is defined on the non-negative integers. f(2n - 1) = 0 for n = 0, 1, 2, ... . If m is

not of the form 2n - 1, then f(m) = f(m+1) + 1. Show that f(n) + n = 2k - 1 for some k, and find
f(21990).

Solution

We claim that if 2m <= n < 2m+1, then f(n) = 2m+1 - n - 1. Put r = 2m+1 - n. Then the claim
follows by induction on r. Hence f(21990) = 21990 - 1.

Problem A2

I is the incenter of the triangle ABC and the incircle touches BC, CA, AB at D, E, F
respectively. AD meets the incircle again at P. M is the midpoint of EF. Show that PMID is
cyclic (or the points are collinear).

Solution

∠AEI = ∠AME = 90o, so AEI and AME are similar. Hence AM/AE = AE/AI or AM· AI =
AE2. AE is tangent to the incircle, so AE2 = AP· AD. Hence AM· AI = AP· AD, so if P,M,I,D
are not collinear, then they are cyclic.

Problem A3
f(x) = (x + b)2 + c, where b and c are integers. If the prime p divides c, but p2 does not divide
c, show that f(n) is not divisible by p2 for any integer n. If an odd prime q does not divide c,
but divides f(n) for some n, show that for any r, we can find N such that qr divides f(N).

Solution

The first part is trivial. If p does not divide (x+b), then it does not divide (x+b)2, so it does not
divide (x+b)2+c. On the other hand, if p does divide x+b, then p2 divides (x+b)2, so p2 does
not divide (x+b)2+c.

For the second part, we use induction on r. For r = 1, we are given that q divides f(n). Now
suppose that qr divides f(N) for some N. If qr+1 divides f(N), then we are done. So suppose qr+1
does not divide f(N), so f(N) = qrh where q does not divide h. We have f(N+kqr) = f(N) +
qr(2N+2b)k = qrh + qr(2N+2b)k. Now q divides (N+b)2+c, and does not divide c, so it does
not divide (N+b)2 and hence does not divide N+b. It is odd, so it does not divide 2N+2b.
Hence we can find k such that k(2N+2b) = -h mod q. Then we have qr+1 divides f(N+kqr),
which completes the induction.

Problem B1

The circle C has diameter AB. The tangent at B is T. For each point M (not equal to A) on C
there is a circle C' which touches T and touches C at M. Find the point at which C' touches T
and find the locus of the center of C' as M varies. Show that there is a circle orthogonal to all
the circles C'.

Answer

C' touches T at the intersection of T and the line AM

the locus of the center is a parabola vertex B
the circle center A radius AB is orthogonal to all circles C'

Solution

Let O be the center of C. Let the line AM meet T at N. Let the perpendicular to T at N meet
the line OM at O'. Then ∠O'NM = ∠MAB (O'N parallel to AB, because both perpendicular to
T) = ∠OMA (OM = OA) = ∠O'MN. So O'M = O'N. Hence O' is the center of C'.

Take B to be the origin and A to be the point (2a,0), so O is (a,0) and C has radius a. If O' is
(x,y), then we require that O'O = x+a or (x-a)2+y2 = (x+a)2, or y2 = 4ax, which is a parabola
with vertex B and axis the x-axis.
Triangles AMB, ABN are similar (∠AMB = ∠ABN = 90o), so AM/AB = AB/AN and hence
AM· AN = AB2. Now consider the circle center A radius AB. It must meet the circle C',
because it contains the point M. Suppose it meets at X. Then AX2 = AB2 = AM· AN, so AX is
tangent to C' and hence the circles are orthogonal.

Problem B2

A and B are opposite corners of an n x n board, divided into n2 squares by lines parallel to the
sides. In each square the diagonal parallel to AB is drawn, so that the board is divided into 2n2
small triangles. The board has (n + 1)2 nodes and large number of line segments, each of
length 1 or ¥$Siece moves from A to B along the line segments. It never moves along the
same segment twice and its path includes exactly two sides of every small triangle on the
board. For which n is this possible?

Answer

n=2 only

Solution

The diagram above shows that n=2 is possible (the path is AHEFGHCDIHB). Now suppose n
> 2.

Note that if X is any vertex except A or B, then an even number of segments with endpoint X
must be in the path.

Let F be the bottom left-hand vertex. Two sides of the triangle EFG are in the path, so at least
one of EF and FG is. But EF and EG are the only segments with endpoint F, so an even
number of them must be in the path, so both are in the path. Hence, again considering EFG,
EG is not in the path. Hence, considering EHG, EH and HG are in the path.

E has an even number of segments on the path, so CE is not on the path. Hence (considering
CEH) CH is on the path. Similarly, GJ is not on the path and HJ is on the path. An even
number of segments at H are on the path, so DH and HI are either both on the path or neither
is on the path. But (considering DHI) at least one must be, so they both are. Hence DI is not,
and CD is not.

Since n > 2, C is not the top left vertex. Considering MCD, MC and MD are both on the path.
Considering DLI, DL is on the path. There must be an even number of segments at D, so DP
is on the path. Hence MP is not. Now M cannot be the top left vertex (with n = 3) because
then it should have an odd number of segments, whereas it would have two (MC and MD). So
there must be a vertex N above M. Considering NMP, MN must be in the path. But now M
has an odd number of segments. Contradiction.

Problem B3

f(x) is a polynomial of degree 3 with rational coefficients. If its graph touches the x-axis,
show that it has three rational roots.

Solution
Without loss of generality, f(x) = x3 - ax2 + bx - c, where a, b, c are rational. Since the graph

k = a, 2hk + k2 = b, h2k = c. Hence a2 - 3b = (h - k)2. Put r = ±¥ D2 - 3b), where the sign is

touches the x-axis, there is a repeated root, so we may take the roots to be h, h, k. Hence 2h +

chosen so that h = a/3 + r/3, k = a/3 - 2r/3. We need to show that r is rational. If r is zero there
is nothing to prove, so assume r is non-zero.

We have 9h2 = 2a2 - 3b + 2ar. Hence 27h2k = -2a3 + 9ab + (6b - 2a2)r. But 27h2k = 27c. So r =
(27c + 2a3 - 9ab)/(2(3b - a2)). Note that 3b - 2a2 is non-zero because r is non-zero. So r is a
rational combination of a, b, c and hence is rational.

6th Iberoamerican 1991

Problem 1

The number 1 or the number -1 is assigned to each vertex of a cube. Then each face is given
the product of its four vertices. What are the possible totals for the resulting 14 numbers?

Solution

Answer: 14, 6, 2, -2, -6, -10.

If every vertex is 1, we get 14 and that is clearly the highest possible total. The lowest
possible total cannot be lower than -14, but we cannot even achieve that because if all the
vertices are -1, then all the faces are 1.

If we change a vertex, then we also change three faces. If the vertex and the three faces are all
initially the same, then we make a change of ±8. If three are of one kind and one the opposite,
then we make a change of ±4. If two are of one kind and two the opposite, then we make no
change. Thus any sequence of changes must take us to 14 + 4n for some integer n. But we
have already shown that the total is greater than -14 and at most 14, so the only possibilities
are -10, -6, -2, 2, 6, 10 and 14.

We show that 10 is not possible. If more than 2 vertices are -1, then the vertex total is at most
2, there are only 6 faces, so the total is less than 10. If all vertices are 1, then the total is 14. If
all but one vertex is 1, then the total is 6. So the only possibility for 10 is just two vertices -1.
But however we choose any two vertices, there is always a face containing only one of them,
so at least one face is -1, so the face total is at most 4 and the vertex total is 4, so the total is
less than 10. The other totals are possible, for example:

14: all vertices 1

6: one vertex -1, rest 1

2: three vertices of one face -1, rest 1

-2: all vertices -1

-6: all vertices but one -1

-10: two opposite corners 1, rest -1

Problem A2
Two perpendicular lines divide a square into four parts, three of which have area 1. Show that
the fourth part also has area 1.

Problem A3

f is a function defined on all reals in the interval [0, 1] and satisfies f(0) = 0, f(x/3) = f(x)/2,
f(1 - x) = 1 - f(x). Find f(18/1991).

Problem B1

Find a number N with five digits, all different and none zero, which equals the sum of all
distinct three digit numbers whose digits are all different and are all digits of N.

Solution

Answer: 35964

There are 4.3 = 12 numbers with a given digit of n in the units place. Similarly, there are 12
with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit
numbers is 12.111 (a + b + c + d + e), where n = abcde. So 8668a = 332b + 1232c + 1322d +
1331e. We can easily see that a = 1 is too small and a = 4 is too big, so a = 2 or 3. Obviously e
must be even. 0 is too small, so e = 2, 4, 6 or 8. Working mod 11, we see that 0 = 2b + 2d, so
b + d = 11. Working mod 7, we see that 2a = 3b + 6d + e. Using the mod 11 result, b = 2, d =
9 or b = 3, d = 8 or b = 4, d = 7 or b = 5, d = 6 or b = 6, d = 5 or b = 7, d = 4 or b = 8, d = 3 or
b = 9, d = 2. Putting each of these into the mod 7 result gives 2a - e = 4, 1, 5, 2, 6, 3, 0, 4 mod
7. So putting a = 2 and remembering that e must be 2, 4, 6, 8 and that all digits must be
different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then
straightforward but tiresome to check that none of these give a solution for c. Similarly
putting a = 4, gives a, b, d, e = 3, 4, 7, 8 or 3, 5, 6, 4 as the only possibilities. Checking, we
find the solution above and no others.

Problem B2

Let p(m, n) be the polynomial 2m2 - 6mn + 5n2. The range of p is the set of all integers k such
that k = p(m, n) for some integers m, n. Find which members of {1, 2, ... , 100} are in the
range of p. Show that if h and k are in the range of p, then so is hk.

Answer

1,2,4,5,8,9,10,13,16,17,18, 20,25,26,29,32,34,36,37, 40,41,45,49,50,52,53,58,

61,64,65,68,72,73,74, 80,81,82,85,89,90,97,98,100

Solution

We have p(m,n) = (m-2n)2 + (m-n)2, so p(2a-b,a-b) = a2 + b2. Hence the range of p is just the
sums of two squares.

(a2 + b2)(c2 + d2) = (ac - bd)2 + (ad + bc)2, which establishes that if h and k are in the range,
then so is hk.

Problem B3
Given three non-collinear points M, N, H show how to construct a triangle which has H as
orthocenter and M and N as the midpoints of two sides.

Solution

Take H' so that M is the midpoint of HH'. The circle diameter NH' meets the line through H
perpendicular to MN in two points (in general), either of which we may take as A. Then B is
the reflection of A in M, and C is the reflection of A in N.

To see that this works, note that M is the midpoint of HH' and AB, so AHBH' is a
parallelogram. Hence AH' is parallel to BH and hence perpendicular to AC. In other words
∠NAH' = 90o, so A lies on the circle diameter NH'. MN is parallel to BC, so A lies on the
perpendicular to MN through H.

7th Iberoamerican 1992

Problem A1

an is the last digit of 1 + 2 + ... + n. Find a1 + a2 + ... + a1992.

Solution

It is easy to compile the following table, from which we see that an is periodic with period 20,
and indeed the sum for each decade (from 0 to 9) is 35. Thus the sum for 1992 is 199· 35 + 5 +
6 + 8 = 6984.

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 18 19 20

an 0 1 3 6 0 5 1 8 6 5 5 6 8 1 5 0 6
3 1 0 0

sum 0 1 4 10 10 15 16 24 30 35 40 46 54 55 60 60 66
69 70 70 70

Problem A2

the sum of the lengths of the intervals in which f(x) •

Let f(x) = a1/(x + a1) + a2/(x + a2) + ... + an/(x + an), where ai are unequal positive reals. Find

Answer ™Di

Solution
wlog a1 > a2 > ... > an. The graph of each ai/(x + ai) is a rectangular hyberbola with asymptotes

decreasing parts. For x < -a1, f(x) is negative. For x  -ai, -ai+1), f(x) decreases from ’WR-’
x = -ai and y = 0. So it is not hard to see that the graph of f(x) is made up of n+1 strictly

Finally, for x > -an, f(x) decreases from ’WR7KXVI [  DWQYDOXHVE1 < b2 < ... < bn, and
f(x) •RQWKHQLQWHUYDOV -a1,b1), (-a2,b2), ... , (-an,bn). So the sum of the lengths of these
intervals is ™ Di + bi). We show that ™Ei = 0.

Multiplying f(x) = 1 by š [ + aj) we get a polynomial of degree n:

š [Dj) - ™i (ai šj (x + aj) ) = 0.

L

The coefficient of xn is 1 and the coefficient of xn-1 is ™Dj - ™Di = 0. Hence the sum of the
roots, which is ™Ei, is zero.

Problem A3

ABC is an equilateral triangle with side 2. Show that any point P on the incircle satisfies PA2
+ PB2 + PC2 = 5. Show also that the triangle with side lengths PA, PB, PC has area (¥ 

Solution

Take vectors centered at the center O of the triangle. Write the vector OA as A etc. Then PA2
+ PB2 + PC2 = (P - A)2 + (P - B)2 + (P - C)2 = 3P2 + (A2 + B2 + C2) - 2P.(A + B + C) = 15P2,
since A2 = B2 = C2 = 4P2 and A + B + C = 0. Finally the side is 2, so an altitude is ¥DQGWKH
inradius is (¥  ¥VR3$2 + PB2 + PC2 = 15/3 = 5.

Take Q outside the triangle so that BQ = BP and CQ = AP. Then BQC and BPA are
congruent, so ∠ABP = ∠CBQ and hence ∠PBQ = 60o, so PBQ is equilateral. Hence PQ is
PB and PQC has sides equal to PA, PB, PC. If we construct two similar points outside the
other two sides then we get a figure with total area equal to 2 area ABC and to 3 area PQC +

ABC (PA2+PB2+PC2)/PA2 = (3/4) area ABC = (3¥ 6RDUHD34&  ¥ 

area of three equilateral triangles sides PA, PB, PC. Hence 3 area PQC = 2 area ABC - area

Problem B1

Let an, bn be two sequences of integers such that: (1) a0 = 0, b0 = 8; (2) an+2 = 2 an+1 - an + 2,
bn+2 = 2 bn+1 - bn, (3) an2 + bn2 is a square for n > 0. Find at least two possible values for (a1992,
b1992).

Answer

(1992· 1996, 4· 1992+8), (1992· 1988, -4· 1992+8)

Solution

an satisfies a standard linear recurrence relation with general solution an = n2 + An + k. But an

= 0, so k = 0. Hence an = n2 + An. If you are not familiar with the general solution, then you
can guess this solution and prove it by induction.
Similarly, bn = Bn + 8. Hence an2+bn2 = n4 + 2An3 + (A2+B2)n2 + 16Bn + 64. If this is a
square, then looking at the constant and n3 terms, it must be (n2 + An + 8). Comparing the
other terms, A = B = ±4.

Problem B2

Construct a cyclic trapezium ABCD with AB parallel to CD, perpendicular distance h

between AB and CD, and AB + CD = m.

Problem B3

Given a triangle ABC, take A' on the ray BA (on the opposite side of A to B) so that AA' =
BC, and take A" on the ray CA (on the opposite side of A to C) so that AA" = BC. Similarly
take B', B" on the rays CB, AB respectively with BB' = BB" = CA, and C', C" on the rays AB,
CB. Show that the area of the hexagon A"A'B"B'C"C' is at least 13 times the area of the
triangle ABC.

8th Iberoamerican 1993

Problem 1

A palindrome is a positive integers which is unchanged if you reverse the order of its digits.
For example, 23432. If all palindromes are written in increasing order, what possible prime
values can the difference between successive palindromes take?

Solution

Answer: 2, 11.

Let x be a palindrome and x' the next highest palindrome. If x < 101, then it is easy to see by
inspection that x' - x = 1, 2 or 11, so the only prime differences are 2 and 11.

So assume x > 100. If x and x' have the same final digit, then their difference is divisible by
10 and hence not prime. So they must have different digits. Thus either x = d9...9d and x' =
d'0...0d', where d < 9 and d' = d+1, or x' has one more digit than x and d = 9, d' = 1. In the first
case x' - x = 11. In the second case x' - x = 2. So again the only prime differences are 2 and
11.

Problem 2

Show that any convex polygon of area 1 is contained in some parallelogram of area 2.

Solution

Let the vertices X, Y of the polygon be the two which are furthest apart. The polygon must lie
between the lines through X and Y perpendicular to XY (for if a vertex Z lay outside the line
through Y, then ZY > XY). Take two sides of a rectangle along these lines and the other two
sides as close together as possible. There must be a vertices U and V on each of the other two
sides. But now the area of the rectangle is twice the area of XUYV, which is at most the area
of the polygon. [In the case of a triangle one side of the rectangle will be XY.]
Problem A3

Find all functions f on the positive integers with positive integer values such that (1) if x < y,
then f(x) < f(y), and (2) f(y f(x)) = x2f(xy).

Solution

Answer: f(x) = x2.

Note that (1) implies f is (1, 1).

Put y = 1. Then f( f(x) ) = x2 f(x).

Put y = f(z), then f( f(z) f(x) ) = x2 f(x f(z) ) = x2z2 f(xz) = f( f(xz) ). But f is (1, 1) so f(xz) =
f(x) f(z).

Now suppose f(m) > m2 for some m. Then by (1), f( f(m) ) > f(m2 = f(m.m) = f(m)2. But f(
f(m) ) = m2 f(m), so m2 > f(m). Contradiction.

Similarly, suppose f(m) < m2. Then m2 f(m) = f( f(m) ) < f(m2) = f(m)2, so m2 < f(m).
Contradiction. So we must have f(m) = m2.

Problem B1

ABC is an equilateral triangle. D is on the side AB and E is on the side AC such that DE
touches the incircle. Show that AD/DB + AE/EC = 1.

Solution

Put BD = x, CE = y, BC = a. Then since the two tangents from B to the incircle are of equal
length, and similarly the two tangents from D and E, we have ED + BC = BD + CE, or ED =
x + y - a. By the cosine law, ED2 = AE2 + AD2 - AE.AD. Substituting and simplifying, we get
a = 3xy/(x + y). Hence AD/DB = (2y - x)/(x + y) and AE/EC = (2x - y)/(x + y) with sum 1.

Problem B2

If P and Q are two points in the plane, let m(PQ) be the perpendicular bisector of PQ. S is a
finite set of n > 1 points such that: (1) if P and Q belong to S, then some point of m(PQ)
belongs to S, (2) if PQ, P'Q', P"Q" are three distinct segments, whose endpoints are all in S,
then if there is a point in all of m(PQ), m(P'Q'), m(P"Q") it does not belong to S. What are the
possible values of n?

Answer

n = 3 (equilateral triangle), 5 (regular pentagon).

Solution

on at most two bisectors, so 2n •Q Q-1)/2. Hence n ”

There are n(n-1)/2 pairs of points. Each has a point of S on its bisector. But each point of S is
The equilateral triangle and regular pentagon show that n = 3, 5 are possible.

Consider n = 4. There are 6 pairs of points, so at least one point of S must be on two bisectors.
wlog A is on the bisectors of BC and BD. But then it is also on the bisector of CD.
Contradiction.

Problem B3

We say that two non-negative integers are related if their sum uses only the digits 0 and 1. For

that: (1) if a $DQGE %WKHQDDQGEDUHUHODWHG  LIFLVUHODWHGWRHYHU\PHPEHURI$

example 22 and 79 are related. Let A and B be two infinite sets of non-negative integers such

then it belongs to B, (3) if c is related to every member of B, then it belongs to A. Show that
in one of the sets A, B we can find an infinite number of pairs of consecutive numbers.

Solution

Suppose there is a member of A with last digit d. Then every member of B must have one of
two possible last digits. Suppose there are members of B with both possibilities. Then every
member of A must have last digit d. So either every member of A has the same last digit or
every member of B has the same last digit (or both). Suppose every member of A has the
same last digit d.

But now if n belongs to B and n + d has last digit 0, then n+1 + d has last digit 1. Moreover, if
m is any member of A, then m+n has last digit 0 and other digits all 0 or 1. Hence m+n+1 last
last digit 1 and other digits all 0 or 1, so n+1 must also belong to B. Similarly, if n is in B and
n+d has last digit 1, then n-1 must also belong to B. So in either case there are infinitely many
pairs of consecutive numbers in B.

9th Iberoamerican 1994

Problem A1

Show that there is a number 1 < b < 1993 such that if 1994 is written in base b then all its
digits are the same. Show that there is no number 1 < b < 1992 such that if 1993 is written in
base b then all its digits are the same.

Solution

Any even number 2n can be written as 22 in base n-1. In particular 1994 = 22996.

We have to show that we cannot write 1993 = aaa ... ab. If the number has n digits, then 1993
= a(1 + b + ... + bn-1) = a(bn - 1)/(b - 1). But 1993 is prime, so a must be 1. Hence bn-1 + ... + b
- 1992 = 0. So b must divide 1992 = 233.83. We cannot have n = 2, for then b = 1992 and we
require b < 1992. So n > 2. But 832 = 6889 > 1993, so b must divide 24. Hence b = 2, 3, 4, 6,
8, 12, or 24. But we can easily check that none of these work:

1 + 2 + 22 + ... + 29 = 1023, 1 + ... + 210 = 2047.

1 + 3 + ... + 36 = 1093, 1 + ... + 3^7 = 3280

1 + 4 + ... + 45 = 1365, 1 + ... + 46 = 5461

1 + 6 + ... 64 = 1555, 1 + ... + 65 = 9331

1 + 8 + 82 + 83 = 585, 1 + ... + 84 = 4681

1 + 12 + 122 + 123 = 1885, 1 + ... + 124 = 22621

1 + 24 + 242 = 601, 1 + ... + 243 = 14425

Problem A2

ABCD is a cyclic quadrilateral. A circle whose center is on the side AB touches the other
three sides. Show that AB = AD + BC. What is the maximum possible area of ABCD in terms
of |AB| and |CD|?

Answer

(h/2 + k/2) ¥ KN- h2/4), where h = |CD|, k = |AB|

Solution

Let the circle have center O on AB and radius r. Let ∠OAD = , ∠OBC = 3. Since ABCD is
cyclic, ∠ADC = 180o-3, so ∠ODA = 90o-3/2. If AD touches the circle at X, then AD = AX +
XD = r cot  + r tan(3/2). Similarly, BC = r cot 3 + r tan(/2). Put t = tan(/2). Then cot  =
(1-t2)/2t, so cot  + tan(/2) = (1+t2)/2t = 1/sin . Similarly for 3, so AD + BC = r/sin  + r/sin
3 = AO + OB = AB.

Suppose AD and BC meet at H (we deal below with the case where they are parallel). Then
HCD and HAB are similar, so area HCD = (CD2/AB2) area HAB and area ABCD = (1 -
CD2/AB2) area HAB. Also AB/CD = HA/HC = HB/HD = (HA+HB)/(HC+HD) =
(HA+HB)/(HB-BC+HA-DA) = (HA+HB)/(HA+HB-AB). Hence HA+HB = AB2/(AB-CD),
which is fixed. Now for fixed HA+HB we maximise the area of HAB by taking HA = HB and
hence AD = BC.

Put h = CD, k = AB. So k cos  + h = k. Hence cos  = (1-h/k). Hence sin  = ¥ KN- h2/k2).
So area ABCD = ½(h+k) ½ k sin  = (h/2 + k/2) ¥ KN- h2/4) (*).

If AD and BC are parallel then A and B must lie on the circle, so that ∠DAB = ∠ABC = 90o.
But ABCD is cyclic, so it must be a rectangle. Hence AB = CD and area ABCD = k2/2. In this
case (*) still gives the correct answer.

Problem A3
There is a bulb in each cell of an n x n board. Initially all the bulbs are off. If a bulb is
touched, that bulb and all the bulbs in the same row and column change state (those that are
on, turn off, and those that are off, turn on). Show that it is possible by touching m bulbs to
turn all the bulbs on. What is the minimum possible value of m?

Answer

n odd, n is minimum
n even, n2 is minimum

Solution

If n is odd, touch each bulb in the first column. Then bulbs in the first column are each
switched n times, which is odd and so end up on. All other bulbs are switched just once, and
so end up on. n is obviously minimal, because if m < n, then there is a bulb which is not
switched at all (there must be a column with no bulb touched and a row with no bulb touched,
so the bulb in that column and row is not switched).

In n is even, touch each bulb. Then each bulb is switched 2n-1 times, so ends up on. We show
that it is not possible to do better.

Note first that there is no benefit in touching a bulb more than once, so each must be touched
zero of one times. Thus we can represent the scheme as an array of 0s and 1s, where 0 means
that the corresponding bulb is not touched, and 1 means that it is touched.

Let A, B, C, D be four values at the corners of a rectangle. We claim that A+B has the same
parity as C+D. Let LAB be the number of 1s in the row AB are touched, similarly LBC (the
number of 1s in the column BC), LCD, LDA. Since bulb A is switched we must have LAB + LDA
+ A odd (note that LAB + LDA double-counts the no. of touches of A). Similarly, LBC + LCD +
C is odd, so A + C + (LAB + LBC + LCD + LDA) is even. Similarly, considering B and D, we
find that B + D + (LAB + LBC + LCD + LDA) is even, so A+C and B+D have the same parity.

= D and B = C, or A 'DQG%'
Adding B+C to both, we get that A+B and C+D have the same parity. It follows that either A

Keeping A and B fixed, we can now vary C (and hence D). It follows that either the row
through B matches that through A, or it has every cell different (to the corresponding cell in
row A). Similarly for the other rows. So we have k rows of one type and n-k rows which are
equal to its "complement". Suppose first that k = n, so that all rows are the same. If we have
all 1s, then we have a solution. If we have all 0s, we obviously do not have a solution. So
suppose there is a 0 and a 1 in each row. Then the total count at a 1 is n-1 higher than at a 0
(because of the extra n-1 1s in the same column). So they cannot both be odd (because n is
even). Contradiction.

Finally suppose there is a row and a complement row. So position A in one is 1, then position
B in the same column in the other has 0. If a row has h 1s, then a complement row has n-h 1s.
The column has z 1s, so A has z+h-1 or z+n-h-1 1s, and B has z+h or z+n-h 1s. But since n is
even, z+h and z+n-h have the same parity, so A and B have opposite parity. Contradiction. So
the only solution for n even is all 1s.

Problem B1
ABC is an acute-angled triangle. P is a point inside its circumcircle. The rays AP, BP, CP
intersect the circle again at D, E, F. Find P so that DEF is equilateral.

Solution

Let the angle bisector of A meet BC at A'. Let the perpendicular bisector of AA' meet the line
BC at X. Take the circle center X through A and A'. Similarly, let the angle bisector of B meet
AC at B' and let the perpendicular bisector of BB' meet the line AC at Y. Take the circle
center Y through B and B'. The two circles meet at a point P inside the triangle, which is the
desired point.

PAB and PED are similar, so DE/AB = PD/PB. Similarly, DF/AC = PD/PC, so DE/DF =
(AB/AC)(PC/PB). Thus we need PB/PC = AB/AC. So P must lie on the circle of Apollonius,
which is the circle we constructed with center X. Similarly, it must lie on the circle of
Apollonius with center Y and hence be one of their points of intersection. It also lies on the
third circle and hence we choose the point of intersection inside the triangle.

Problem B2

... , Ar of {0, 1, 2, ... , n-1} each with k elements such that each integer 0 ”PQFDQEH
n and r are positive integers. Find the smallest k for which we can construct r subsets A1, A2,

written as a sum of one element from each of the r subsets.

Answer

smallest integer such that kr •Q.

Solution

We can form at most kr distinct sums, so kr must be •Q

Now consider A1 = {0, 1, 2, ... , k-1}, A2 = {0, k, 2k, ... , (k-1)k}, A3 = {0, k2, 2k2, ... , (k-
1)k2}, ... , Ar = {0, kr-1, 2kr-1, ... , (k-1)kr-1}. Then for any non-negative integer m < kr, we can

element from each Ai. This subset works for (k-1)kr-1 < n ”Nr. For smaller n above (k-1)r we
write m with r digits in base k (using leading zeros as necessary) and hence as a sum of one

cannot use all the elements given above, but we do not need them, so we just replace the
elements which are too large by arbitrary elements under n.
For example, suppose n = 17, r = 4. We need k = 3. So we form A1 = {0, 1, 2}, A2 = {0, 3, 6},
A3 = {0, 9, 18}, A4 = {0, 27, 54}. Now 18, 27, 54 are unnecessary, so we pad out A3 and A4
with other elements. We could take A3 = {0, 1, 9}, A4 = {0, 1, 2}.

Problem B3

Show that given any integer 0 < n ”1000000 we can find at set S of at most 1100000 positive
integers such that S includes 1 and n and every element of S except 1 is a sum of two
(possibly equal) smaller elements of S.

10th Iberoamerican 1995

Problem A1

Find all possible values for the sum of the digits of a square.

Solution

Answer: any non-negative integer = 0, 1, 4 or 7 mod 9.

02 = 0, (±1)2 = 1, (±2)2 = 4, (±3)2 = 0, (±4)2 = 7 mod 9, so the condition is necessary.

We exhibit squares which give these values.

0 mod 9. Obviously 02 = 0. We have 92 = 81, 992 = 9801 and in general 9...92 = (10n - 1)2 =
102n - 2.10n + 1 = 9...980...01, with digit sum 9n.

1 mod 9. Obviously 12 = 1 with digit sum 1, and 82 = 64 with digit sum 10. We also have 982
= 9604, 9982 = 996004, and in general 9...982 = (10n - 2)2 = 102n - 4.10n + 4 = 9...960...04,
with digit sum 9n+1.

4 mod 9 Obviously 22 = 4 with digit sum 4, and 72 = 49 with digit sum 13. Also 972 = 9409
with digit sum 22, 9972 = 994009 with digit sum 31, and in general 9...972 = (10n - 3)2 = 102n -
6.10n + 9 = 9...940...09, with digit sum 9n+4.

7 mod 9 Obviously 42 = 16, with digit sum 7. Also 952 = 9025, digit sum 16, 9952 = 990025
with digit sum 25, and in general 9...952 = (10n - 5)2 = 102n - 10n+1 + 25 = 9...90...025, with
digit sum 9n-2.

Problem A2

Find all solutions in real numbers x1, x2, ... , xn+1 all at least 1 such that: (1) x11/2 + x21/3 + x31/4
+ ... + xn1/(n+1) = n xn+11/2; and (2) (x1 + x2 + ... + xn)/n = xn+1.

Answer

The only solution is the obvious, all xi = 1.

Solution
By Cauchy-Schwartz, (™[i1/2)2 ” ™ VXP[i), with equality iff all xi equal. In other
words, if we put xn+1 = (x1 + x2 + ... + xn)/n, then ™[i1/2 ”Q[n+11/2. But since all xi •ZH
have x11/2 + x21/3 + x31/4 + ... + xn1/(n+1) ”™[i1/2 with equality iff x2 = x3 = ... = xn = 1. Hence
x11/2 + x21/3 + x31/4 + ... + xn1/(n+1) ”[n+11/2 with equality iff all xi = 1.

Problem A3

L and L' are two perpendicular lines not in the same plane. AA' is perpendicular to both lines,
where A belongs to L and A' belongs to L'. S is the sphere with diameter AA'. For which
points P on S can we find points X on L and X' on L' such that XX' touches S at P?

Problem B1

ABCD is an n x n board. We call a diagonal row of cells a positive diagonal if it is parallel to

AC. How many coins must be placed on an n x n board such that every cell either has a coin
or is in the same row, column or positive diagonal as a coin?

Answer

smallest integer • Q-1)/3

[so 2m-1 for n = 3m-1, 2m for n = 3m, 2m+1 for n = 3m+1]

Solution

There must be at least n-k rows without a coin and at least n-k columns without a coin. Let r1,
r2, ... , rn-k be cells in the top row without a coin which are also in a column without a coin. Let

coin. Each of the 2n-2k-1 ri and cj are on a different positive diagonal, so we must have k •
r1, c2, c3, ... , cn-k be cells in the first column without a coin which are also in a row without a

2n-2k-1 and hence k • 2n-1)/3.

Let (i,j) denote the cell in row i, col j. For n = 3m-1, put coins in (m,1), (m-1,2), (m-2,3), ... ,
(1,m) and in (2m-1,m+1), (2m-2,m+2), ... , (m+1,2m-1). It is easy to check that this works.
For n = 3m, put an additional coin in (2m,2m), it is easy to check that works. For n = 3m+1
we can use the same arrangement as for 3m+2.

Problem B2

The incircle of the triangle ABC touches the sides BC, CA, AB at D, E, F respectively. AD
meets the circle again at X and AX = XD. BX meets the circle again at Y and CX meets the
circle again at Z. Show that EY = FZ.
Problem B3

f is a function defined on the positive integers with positive integer values. Use f m(n) to mean
f(f( ... f(n) ....)) = n where f is taken m times, so that f 2(n) = f(f(n)), for example. Find the
largest possible 0 < k < 1 such that for some function f, we have f m(n) QIRUP 
[kn], but f m(n) = n for some m (which may depend on n).

Answer

we can get k arbitrarily close to 1

Solution

The basic idea is to take a block of integers m+1, m+2, ... , M and to define f(m+1) = m+2,
f(m+2) = m+3, ... , f(M-1) = M, f(M) = m+1. Then for any integer h in the block we have fn(h)
KIRUQ 0-m-1 and fM-m(h) = h. Note that the ratio (M-m)/h is worst (smallest) for
h = M.

For example, take the first block to be 1, 2, ... , N, the second block to be N+1, ... , N2, the
third block, N2+1, ... , N3 and so on. Then for any integer n we have fm(n) QIRUPNQ
where k = 1 - 1/N.

11th Iberoamerican 1996

Problem A1

Find the smallest positive integer n so that a cube with side n can be divided into 1996 cubes
each with side a positive integer.

Solution

Answer: 13.

Divide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit
cube, so the large cube contains at least 1996 unit cubes. But 123 = 1728 < 1996 < 2197 = 133,
so it is certainly not possible for n < 13.

It can be achieved with 13 by 1.53 + 11.23 + 1984.13 = 133 (actually packing the cubes
together to form a 13 x 13 x 13 cube is trivial since there are so many unit cubes).

Problem 2

M is the midpoint of the median AD of the triangle ABC. The ray BM meets AC at N. Show
that AB is tangent to the circumcircle of NBC iff BM/BN = (BC/BN)2.

Solution
Applying Menelaus to the triangle ADC, we have (AM/MD)(BD/DC)(CN/NA) = 1, so
(CN/NA) = 2. Hence AN/AC = 1/3. Applying Menelaus to the triangle BNC, we have
(BM/MN)(AN/AC)(CD/DB) = 1, so BM/MN = 3. That is true irrespective of whether AB is
tangent to the circle NBC.

If AB is tangent, then AB2 = AN.AC = 1/3 AC2. Also angle ABN = angle BCN, so triangles
ANB and ABC are similar. Hence BC/BN = AC/AB. Hence (BC/BN)2 = 3 = BM/BN.

Conversely, if (BC/BN)2 = BM/BN, then (BC/BN)2 = 3.

Now applying the cosine formula to AMN and AMB and using cos AMN + cos AMB = 0, we
have (3AN2 - 3AM2 - 3MN2) + (AB2 - AM2 - BM2) = 0, so AB2 + AC2/3 = AD2 + 3/4 BN2.
Similarly from triangles ADC and ADB we get AB2 + AC2 = 2AD2 + BC2/2. So using BN2 =
BC2/3 we get 2AB2 + 2/3 AC2 = AB2 + AC2 and hence (AC/AB)2 = 3 = (BC/BN)2. So AC/AB
= BC/BN. Note that is not enough to conclude that triangles ABC and BNC are similar,
because the common angle C is not between AC and AB. However, we have AN/AB = (1/3)
AC/AB = AB/AC, so AB2 = AN.AC, so AB is tangent to the circle NBC.

Problem A3

n = k2 - k + 1, where k is a prime plus one. Show that we can color some squares of an n x n
board black so that each row and column has exactly k black squares, but there is no rectangle
with sides parallel to the sides of the board which has its four corner squares black.

Solution

We can regard the rows as lines and the columns as points. Black squares denote incidence.
So line 3 contains point 4 iff square (3, 4) is black. The condition about rectangles then means
that there is at most one line through two distinct points.

Suppose we take the points to be (a, b, c), where a, b, c are residues mod p, not all zero, and
the coordinates are homogeneous, so that we regard (a, b, c), (2a, 2b, 2c), ... , ( (p-1)a, (p-1)b,
(p-1)c ) as the same point. That gives (p3 - 1)/(p-1) = p2 + p + 1 points, which is the correct
number.
We can take lines to be lx + my + nz = 0, where the point is (x, y, z). In other words, the lines
are also triples (l, m, n), with l, m, n residues mod p, not all zero and (l, m, n), (2l, 2m, 2n), ... ,
( (p-1)l, (p-1)m, (p-1)n ) representing the same line.

One way of writing the points is p2 of the form (a, b, 1), p of the form (a, 1, 0) and lastly (1, 0,
0). Similarly for the lines. We must show that (1) each point is on p+1 lines (so each column
has p+1 black squares), (2) each line has p+1 points (so each row has p+1 black squares, (3)
two lines meet in just one point (so no rectangles).

(1): Consider the point P (a, b, 1) with a non-zero. Then for any m, there is a unique l such
that la + mb + 1.1 = 0, so there are p lines of the form (l, m, 1) which contain P. Similarly,
there is a unique l such that la + 1b + 0.1 = 0, so one line of the form (l, 1, 0) contains P. The
line (1, 0, 0) does not contain P. So P lies on just p+1 lines. Similarly for (a, b, 1) with b non-
zero. The point (0, 0, 1) does not lie on any lines (l, m,, 1), but lies on (l, 1, 0) and (1, 0, 0), so
again it lies on p+1 lines.

Consider the point Q (a, 1, 0) with a non-zero. For any m, there is a unique l such that Q lies
on (l, m, 0). There is also a unique l such that Q lies on (l, 1, 0). Q does not lie on (1, 0, 0), so
it lies on just p+1 lines. Similarly, the point (0, 1, 0) lies on the p lines (l, 0, 0) and on (1, 0,
0), but no others.

Finally, the point (1, 0, 0) lies on the p lines (0, m, 1), the line (0, 1, 0) and no others. Thus in
all cases a point lies on just p+1 lines. The proof of (2) is identical.

(3). Suppose the lines are (l, m, n) and (L, M, N). If l and L are non-zero, then we can take the
lines as (1, m', n') and (1, M', N'). So any point (x, y, z) on both satisfies x + m'y + n'z = 0 (*)
and x + M'y + N'z = 0. Subtracting, (m' - M')y + (n' - N')z = 0. The coefficients cannot both be
zero, since the lines are distinct. So the ratio y : z is fixed. Then (*) gives the ratio x : y. So the
point is uniquely determined. If just one of l, L is non-zero, then we can take the lines as (0,
m', n'), (1, M', N'). We cannot have both m' and n' zero, so the ratio y : z is determined, then
the other line determines the ratio x : y. So again the point is uniquely determined. Finally,
suppose l and L are both zero. Then since the lines are distinct y and z must both be zero. So
the unique point on both lines is (1, 0, 0).

Problem B1

b ”QDQGDE!Q6KRZWKDWWKHVXPRIDEWDNHQRYHUDOOVXFKSDLUVLV
n > 2 is an integer. Consider the pairs (a, b) of relatively prime positive integers, such that a <

Solution

Induction on n. It is obvious for n = 3, because the only pairs are (1, 3) and (2, 3), and 1/3 +
1/6 = 1/2. Now suppose it is true for n. As we move to n+1, we introduce the new pairs (a,
n+1) with a relatively prime to n+1 and we lose the pairs (a, n+1-a) with a relatively prime to
n+1-a and hence to n+1. So for each a relatively prime to n+1 and < (n+1)/2 we gain (a, n+1)
and (n+1-a, n+1) and lose (a, n+1-a). But 1/a(n+1) + 1/( (n+1-a)(n+1) ) = ( n+1-a + a)/( a(n+1-
a)(n+1) ) = 1/( a(n+1-a) ).

Problem B2
An equilateral triangle of side n is divided into n2 equilateral triangles of side 1 by lines
parallel to the sides. Initially, all the sides of all the small triangles are painted blue. Three
coins A, B, C are placed at vertices of the small triangles. Each coin in turn is moved a
distance 1 along a blue side to an adjacent vertex. The side it moves along is painted red, so
once a coin has moved along a side, the side cannot be used again. More than one coin is
allowed to occupy the same vertex. The coins are moved repeatedly in the order A, B, C, A,
B, C, ... . Show that it is possible to paint all the sides red in this way.

Solution

We use induction. It is obvious for n = 1 and 2 - see diagram

above. Note that A, B, C start and end at vertices of the large
triangle.

Now assume that for n we can find a solution with A, B, C starting and ending at the vertices
of the large triangle. Take n+1. We start with the paths shown which bring A, B, C to A', B',
C' at the vertices of a triangle side n-1. Now by induction we can continue the paths so that we
bring A, B, C, back to the vertices of that triangle after tracing out all its edges. Finally, note
that for each of the points A', B', C' there is a path length 2 over untraced segments to a vertex
of the large triangle. So we get a solution for n+1 and hence for all n.

Problem B3

that the square of the distance between Ai and Aj (for i M LVNi + kj. Show that n is at most 4
A1, A2, ... , An are points in the plane. A non-zero real number ki is assigned to each point, so

and that if n = 4, then 1/k1 + 1/k2 + 1/k3 + 1/k4 = 0.

Solution

Suppose we have four points A, B, C, D with associated numbers a, b, c, d. Then AB2 = a + b,

AC2 = a + c, so AB2 - AC2 = b - c. Similarly, DB2 - DC2 = b - c, so AB2 - AC2 = DB2 - DC2.
Let X be the foot of the perpendicular from A to BC, and Y the foot of the perpendicular from
D to BC. Then AB2 - AC2 = (AX2 + XB2) - (AX2 + XC2) = XB2 - XC2. Similarly for D, so
XB2 - XC2 = YB2 - YC2. Hence X = Y, so AD is perpendicular to BC. Similarly, BD is
perpendicular to AC, and CD is perpendicular to AB. Hence D is the (unique) orthocenter of
ABC. So n <= 4.
Suppose n = 4, so we have four points A, B, C, D with associated numbers a, b, c, d. We have
AB2 + AC2 - BC2 = (a + b) + (a + c) - (b + c) = 2a. But by the cosine formula it is also 2 AB
AC cos BAC. Hence a = AB AC cos BAC. Similarly for A, B, D etc. Hence ab/cd = (AB AC
cos BAC)(BA BD cos ABD)/( (CA.CD cos ACD)(DB DC cos BDC) ) = (AB2/CD2) (cos
BAC/cos BDC) (cos ABD/cos ACD).

Take ABC to be acute with D inside. Then angle ABD = angle ACD ( = 90o - angle BAC),
and angle BDC = 90o + angle ACD = 180o - angle BAC. So cos BAC/cos BDC = -1. Thus
ab/cd = - AB2/CD2 = - (a + b)/(c + d). Hence ab(c + d) + cd(a + b) = 0, so 1/a + 1/b + 1/c +
1/d = 0.

12th Iberoamerican 1997

Problem A1

k >= 1 is a real number such that if m is a multiple of n, then [mk] is a multiple of [nk]. Show
that k is an integer.

Solution

Suppose k is not an integer. Take an integer n such that nk > 1, but nk is not an integer. Now
take a positive integer c such that 1/(c+1) <= nk - [nk] < 1/c. Then 1 <= (c+1)nk - (c+1) [nk] <
1 + 1/c. Hence [(c+1)nk] = (c+1) [nk] + 1. Put m = (c+1)n. Then m is a multiple of n. But if
[mk] is a multiple of [nk], then [mk] - (c+1) [nk] = 1 is a multiple of [nk], which is impossible
since nk > 1. So we have a contradiction. So k must be an integer.

Problem 2

I is the incenter of the triangle ABC. A circle with center I meets the side BC at D and P, with
D nearer to B. Similarly, it meets the side CA at E and Q, with E nearer to C, and it meets AB
at F and R, with F nearer to A. The lines EF and QR meet at S, the lines FD and RP meet at T,
and the lines DE and PQ meet at U. Show that the circumcircles of DUP, ESQ and FTR have
a single point in common.

Solution

D and P are the reflections of Q and E respectively in the line CI. Hence PQ and DE meet at a
point on CI. So U lies on CI. So ∠PIU = 1/2 ∠PIE = ∠PDE (I is center of circle through D, P,
E) = ∠PDU (same angle). Hence PDIU is cyclic. In other words, I lies on the circumcircle of
DUP. Similarly, it lies on the circumcircles of ESQ and FTR.

But the same argument shows that ∠DPT = ∠DIT, so DPIT is cyclic. So T lies on the circle
through D, P and I and hence on the circumcircle of DUP. Similarly, for the other circles. So
the circumcircles of CUP and FTR meet at T and I. Similarly, the circumcircles of FTR and
ESQ meet at S and I, and the circumcircles of ESQ and DUP meet at U and I. So the three
circumcircles have just one point in common, namely I.

Problem A3

n > 1 is an integer. Dn is the set of lattice points (x, y) with |x|, |y| <= n. If the points of Dn are
colored with three colors (one for each point), show that there are always two points with the
same color such that the line containing them does not contain any other points of Dn. Show
that it is possible to color the points of Dn with four colors (one for each point) so that if any
line contains just two points of Dn then those two points have different colors.

Solution

Consider the 4 points shown in the diagram. In each case the segment joining them is the
diagonal of an m x 1 parallelogram or rectangle, so it cannot contain any other lattice points.
The next points along each line are obviously outside set Dn. That proves the first part.

The second part is the standard parity argument. Color (x, y) with color 1 if x and y are both
even, 2 if x is even and y is odd, 3 if x is odd and y is even, and 4 if x and y are both odd.
Then if two points are the same color, that means the first coordinates are the same parity and
their second coordinates are the same parity. Hence the midpoint of the segment joining them
is also a lattice point and they are not the only two points of Dn on the line.

Problem B1

Let o(n) be the number of 2n-tuples (a1, a2, ... , an, b1, b2, ... , bn) such that each ai, bj = 0 or 1
and a1b1 + a2b2 + ... + anbn is odd. Similarly, let e(n) be the number for which the sum is even.
Show that o(n)/e(n) = (2n - 1)/(2n + 1).

Solution

We prove by induction that o(n) = 22n-1 - 2n-1. For n = 1, this reads o(1) = 21 - 20 = 1, which is
obviously true - the only such 2-tuple is (1, 1). Suppose it is true for n.
If (a1, a2, ... , an, b1, b2, ... , bn) gives an odd sum, then we can take (an+1, bn+1) to be any of (0,
0), (0, 1), (1, 0) and still get an odd sum for (a1, a2, ... , an+1, b1, b2, ... , bn+1). On the other hand
if (a1, a2, ... , an, b1, b2, ... , bn) is even, then we must have an+1 = bn+1 = 1 to get an odd sum.
Thus o(n+1) = 3 o(n) + e(n). But o(n) = 22n-1 - 2n-1 and e(n) = (o(n) + e(n) ) - o(n) = 22n - 22n-1
+ 2n-1 = 22n-1 + 2n-1. So o(n+1) = 4.22n-1 - 2.2n-1 = 22(n+1)-1 - 2(n+1)-1, which establishes the result
for n+1 and hence for all n.

Hence e(n) = 22n - o(n) = 22n-1 + 2n-1 and o(n)/e(n) = (2n - 1)/(2n + 1).

Problem B2

ABC is an acute-angled triangle with orthocenter H. AE and BF are altitudes. AE is reflected

in the angle bisector of angle A and BF is reflected in the angle bisector of angle B. The two
reflections intersect at O. The rays AE and AO meet the circumcircle of ABC at M and N
respectively. P is the intersection of BC and HN, R is the intersection of BC and OM, and S is
the intersection of HR and OP. Show that AHSO is a parallelogram.

Solution

We show first that O is the circumcenter of ABC. ∠ABF = 90o - A. The line BC is the
reflection in BD of the line BA and the line BF' is the refection of BF, so angle CBF' = 90o -
A. But if O' is the circumcenter, then ∠BO'C = 2 ∠BAC = 2 A, so ∠O'BC = 90o - A. Hence
O' lies on BF'. Similarly, it lies on AE' (the reflection of AE in the angle bisector of A). Hence
O = O'.

∠MBC = ∠MAC = 90o - C (since AH is an altitude) = ∠FBC (since BF is an altitude) =

∠HBC (same angle). So triangles HBE and MBE are congruent and HE = EM. [Note: this
should be a familiar result.].
AN is a diameter, so angle AMN = 90o = angle AEC, so BC and MN are parallel. Hence P is
the midpoint of HN and of BC. So OP is perpendicular to BC. So AH and OS are parallel.

Since R lies on BC, triangles HER and MER are congruent, so ∠EHR = ∠EMR = ∠AMO
(same angle) = ∠MAO. Hence HS and AO are parallel. So AHSO is a parallelogram.

Problem B3

Given 1997 points inside a circle of radius 1, one of them the center of the circle. For each
point take the distance to the closest (distinct) point. Show that the sum of the squares of the
resulting distances is at most 9.

Solution

Let the points be Pi for i = 1, 2, ... , 1997. Take P1 to be the center of the given unit circle. Let

radius xi/2. Then Ci and Cj cannot overlap by more than one point because xi and xj ”3iPj.
xi be the distance from Pi to the closest of the other 1996 points. Let Ci be the circle center Pi

Also xi ”VLQFH31Pi ”7KXV&i is entirely contained in the circle center P1 radius 3/2.

3/2. Hence (x12 + x22 + ... + x19972)/4 ”

Since the circles Ci do not overlap, their total area cannot exceed the area of the circle radius

13th Iberoamerican 1998

Problem A1

There are 98 points on a circle. Two players play alternately as follows. Each player joins two
points which are not already joined. The game ends when every point has been joined to at
least one other. The winner is the last player to play. Does the first or second player have a
winning strategy?

Solution

Answer: the first player has a winning strategy.

Assume there are n points. The first to play so that n-2 points each have at least one segment
loses, because the other player simply joins the last two points and the game ends. But there
are N = (n-3)(n-4)/2 possible plays amongst the first n-3 points to get a segment. For n = 1 or
2 mod 4, N is odd and for n = 0 or 3 it is even. So the first player wins for n = 1 or 2 mod 4
(and in particular for n = 98) and the second player for n = 0 or 3 mod 4.

Problem A2

The incircle of the triangle ABC touches BC, CA, AB at D, E, F respectively. AD meets the
circle again at Q. Show that the line EQ passes through the midpoint of AF iff AC = BC.

Solution
∠AQM = ∠EQD (opposite angle) = ∠EDC (CD tangent to circle EQD) = (180o - ∠C)/2 =
∠A/2 + ∠B/2 (*).

MF2 = MQ.ME (MF tangent to circle FQE). So AM = AF is equivalent to AM2 = MQ.ME or

AM/MQ = ME/AM. But since triangles AMQ and EMA have a common angle M, AM/MQ =
ME/AM iff they are similar, and hence iff ∠AQM = ∠A. Using (*) AM = AF iff ∠A = ∠B.

Problem A3

Find the smallest number n such that given any n distinct numbers from {1, 2, 3, ... , 999},
one can choose four different numbers a, b, c, d such that a + 2b + 3c = d.

Solution

Answer: n = 835.

Consider the set S = {166, 167, ... , 999}. The smallest possible value for a + 2b + 3c, for
distinct a, b, c in S is 168 + 2.167 + 3.166 = 1000. So we cannot find distinct a, b, c, d in S
with a + 2b + 3c = d. So the smallest n > 834.

to be m = a1 < a2 < ... < a835 = M. Obviously M •P•VR-3m •- 3M and

Now suppose S is any subset of 835 elements which satisfies the condition. Take it elements

hence M - 3m •- 2M •- 2.999 = 504. Put k = M - 3m.

There are at least 167 disjoint pairs (a, b) of numbers taken from {1, 2, ... , 999} with a + 2b =
k, namely

(k - 2, 1)

(k - 4, 2)

(k - 6, 3)

...

(k-334, 167) - note that in the extreme case k = 504 this is (170, 167)

At least one number from each pair must either (1) be M or m or (2) not belong to S - or
otherwise we would have a + 2b + 3m = M for distinct elements a, b, m and M in S. None of
the numbers can be M and at most one of them can be m, so we have at least 166 numbers
which are not in S. That means S contains at most 999 - 166 = 833 numbers. Contradiction.
So S cannot have 835 elements. Nor can it have more than 835 elements (or we just take a
subset of 835 elements, which must also satisfy the condition, and get a contradiction).

Problem B1

Representatives from n > 1 different countries sit around a table. If two people are from the
same country then their respective right hand neighbors are from different countries. Find the
maximum number of people who can sit at the table for each n.

Solution

Answer: n2.

Obviously there cannot be more than n2 people. For if there were, then at least one country
would have more than n representatives. But there are only n different countries to choose
their right-hand neighbours from. Contradiction.

Represent someone from country i by i. Then for n = 2, the arrangement 1122 works. [It
wraps round, so that the second 2 is adjacent to the first 1.] Suppose we have an arrangement
for n. Then each of 11, 22, ... , nn must occur just once in the arrangement. Replace 11 by
1(n+1)11, 22 by 2(n+1)22, ... , and (n-1)(n-1) by (n-1)(n+1)(n-1)(n-1). Finally replace nn by
n(n+1)(n+1)nn. It is easy to check that we now have an arrangement for n+1. We have added
one additional representative for each of the countries 1 to n and n+1 representatives for
country n+1, so we have indeed got (n+1)2 people in all. We have also given a representative
of each country 1 to n a neighbour from country n+1 on his right and we have given the (n+1)
representatives from country n+1 neighbours (on their right) from each of the other countries.
Otherwise we have left the seating unchanged.

Problem B2

P1, P2, ... , Pn are points in the plane and r1, r2, ... , rn are real numbers such that the distance
between Pi and Pj is ri + rj (for i not equal to j). Find the largest n for which this is possible.

Solution

Answer: n = 4.

Draw a circle radius ri at Pi. Then each pair of circles must touch. But that is possible iff n ”

Problem B3
k is the positive root of the equation x2 - 1998x - 1 = 0. Define the sequence x0, x1, x2, ... by x0
= 1, xn+1 = [k xn]. Find the remainder when x1998 is divided by 1998.

Solution

Put p(x) = x2 - 1998x - 1. Then p(1998) = -1, p(1999) = 1998, so 1998 < k < 1999. Also k is
irrational (using the formula for the root of a quadratic). We have xn = [k xn-1], so xn < k xn-1
and > k xn-1 - 1. Hence xn/k < xn-1 < xn/k + 1/k, so [xn/k] = xn-1 - 1.

k = (1998k + 1)/k = 1998 + 1/k. Hence kxn = 1998xn + x/k. Hence xn+1 = [kxn] = 1998xn +
[xn/k] = 1998xn + xn-1 - 1. Hence xn+1 = xn-1 - 1 mod 1998. So x1998 = 1 - 999 = 1000 mod
1998.

14th Iberoamerican 1999

Problem A1

Find all positive integers n < 1000 such that the cube of the sum of the digits of n equals n2.

Solution

n < 1000, so the sum of the digits is at most 27, so n2 is a cube not exceeding 273. So we are
looking for m3 which is also a square. That implies m is a square. So the only possibilities are
m = 1, 4, 9, 16, 25. Giving n = 1, 8, 27, 64, 125. The corresponding cubes of the digit sums
are 1, 512, 729, 1000, 512, whereas the corresponding squares are 1, 64, 729, 4096, 15625.
Thus the only solutions are n = 1, 27.

Problem A2

Given two circles C and C' we say that C bisects C' if their common chord is a diameter of C'.
Show that for any two circles which are not concentric, there are infinitely many circles which
bisect them both. Find the locus of the centers of the bisecting circles.

Solution

Let C, C' have center O, O' respectively and radius r, r' respectively. Let a circle center P

C. Hence PA2 = OP2 + r2. Conversely, the circle center P, radius ¥ 232 + r2) bisects C. So P
bisect C. Suppose it meets C at A and B. Then AB is perpendicular to OP and is a diameter of

will bisect C and C' iff OP2 + r2 = OP'2 + r'2.

It is well-known that the locus of points P' with equal tangents to C and C' is the radical axis.
Call the radical axis R. For a point P' on the radical axis we have P'O2 - r2 = P'O'2 - r'2. If we
reflect P' in the perpendicular bisector of OO' to get P, then PO = P'O' and PO' = P'O, so PO'2
- r2 = PO2 - r'2 and hence PO2 + r2. Call the reflection of the R in the perpendicular bisector of
OO' the line R'. We have established that points on R' form part of the locus. Conversely, if P'
is such that there is a circle center P' bisecting both circles, then OP'2 + r2 = O'P'2 + r'2, so if P
is the reflection of P' then OP2 - r2 = OP'2 - r'2 and hence P lies on the radical axis R. Hence P'
must lie on R'.

Radical axis
We have PT2 = PO2 - r2 = PX2 + OX2 - r2 , and similarly PT'2 = PX2 + O'X2 - r'2. So PT = PT'
iff OX2 - r2 = O'X2 - r'2. There is evidently a unique point X for which that is true, so the locus
of such P is the line through X perpendicular to OO'

If the circles intersect, then the point X evidently lies on the line joining the two common
points, because OX2 - r2 = -XY2 = O'X2 - r'2. In any case the midpoint of each common
tangent evidently lies on the line, so that provides a way of constructing it.

Problem A3

Given points P1, P2, ... , Pn on a line we construct a circle on diameter PiPj for each pair i, j and
we color the circle with one of k colors. For each k, find all n for which we can always find
two circles of the same color with a common external tangent.

Solution

Answer: n > k+1.

There are n-1 circles with diameter PiPi+1. Obviously, each pair has a common tangent. If n-1
> k, then two of them must have the same color.

If n-1 ”NWKHQFRORUDOOFLUcles with diameter PiPj and i < j with color i. Then if two circles
have the same color, then both have a tangent at one of the points. Hence one lies inside the
other and they do not have a common external tangent.

Problem B1
factor •
Show that any integer greater than 10 whose digits are all members of {1, 3, 7, 9} has a prime

Solution

Such a number cannot be divisible by 2 (or its last digit would be even) or by 5 (or its last
digit would be 0 or 5). So if the result is false then the number must be of the form 3m7n for
non-negative integers m, n. But we claim that a number of this form must have even 10s digit.

It is easy to prove the claim by induction. It is true for 3 and 7 (the digit is 0 in both cases).
But if we multiply such a number by 3 or 7, then the new 10s digit has the same parity as the
carry from the units digit. But multiplying 1, 3, 7, 9 by 3 gives a carry of 0, 0, 2, 6
respectively, which is always even, and multiplying by 7 gives a carry of 0, 2, 4, 6, which is
also always even. So the new number also has an even 10s digit.

Problem B2

O is the circumcenter of the acute-angled triangle ABC. The altitudes are AD, BE and CF.
The line EF cuts the circumcircle at P and Q. Show that OA is perpendicular to PQ. If M is
the midpoint of BC, show that AP2 = 2 AD· OM.

Solution

Let OA and PQ meet at T. ∠AEH = ∠AFH = 90o, so AEHF is cyclic, so ∠AFT = ∠AFE
(same angle) = ∠AHE = 90o- ∠HAE = 90o - ∠DAC (same angle) = ∠C. But ∠TAF = ∠OAF
(same angle) = 90o - (1/2) ∠AOB = 90o - ∠C. Hence ∠AFT = 90o, which establishes that OA
and PQ are perpendicular.

Let the circumradius be R and let AA' be a diameter. We have AF = AC cos A = 2R sin B cos
A. Hence AT = AF cos OAB = AF sin C = 2R cos A sin B sin C. Now PT2 = PT· TQ =
AT.A'T = AT(2R - AT). Hence AP2 = 2R· AT = 4R2 cos A sin B sin C.

We have AD = AC sin C = 2R sin B sin C, and OM = OC cos COM = R cos A. Hence 2

AD· OM = AP2.

Problem B3
Given two points A and B, take C on the perpendicular bisector of AB. Define the sequence
C1, C2, C3, ... as follows. C1 = C. If Cn is not on AB, then Cn+1 is the circumcenter of the
triangle ABCn. If Cn lies on AB, then Cn+1 is not defined and the sequence terminates. Find all
points C such that the sequence is periodic from some point on.

Solution

Answer: any C such that ∠ACB = 180o r/s, with r and s relatively prime integers and s not a
power of 2.

Let ∠ACnB = xn, where the angle is measured clockwise, so that xn is positive on one side of
AB and negative on the other side. Then xn uniquely identifies Cn on the perpendicular
bisector.

We have xn+1 = 2xn. To make this work in all cases we have to take it mod 180o (so that if
ACnB is obtuse, then Cn+1 lies on the other side of AB). If xn is eventually periodic then xm+1
= xn+1, for some n > m, so (2n - 2m)x1 = 0 mod 180. Hence x1 = 180 r/s for some relatively
prime integers r, s. Also s cannot be a power of 2 for then we would have xk = 180r for some
k, in which case the sequence would terminate rather than be periodic.

Conversely, suppose x1 = 180 r/s, with r and s relatively prime and s not a power of 2. Then
xn+1 = 180 2nr/s cannot be 0 mod 180, so the sequence does not terminate. Put s = 2bc, with c
odd. Let d = 3(c), where 3(m) is Euler's phi function, so that 2d = 1 mod c. Then xb+1 = 180 r/c
mod 180 and 2b+d = 2b mod c, so xb+d+1 = 180 r/c mod 180. Hence the sequence is periodic.

15th Iberoamerican 2000

Problem A1

Label the vertices of a regular n-gon from 1 to n > 3. Draw all the diagonals. Show that if n is
odd then we can label each side and diagonal with a number from 1 to n different from the
labels of its endpoints so that at each vertex the sides and diagonals all have different labels.

Solution

Labeling the diagonal/side between i and j as i+j (reduced if necessary mod n) almost works.
The labels for all the lines at a given vertex will be different. But the line between i and n will
have label i, the same as one endpoint. However, we are not using the label 2i for the lines
from vertex i. So for the line between i and n we use 2i instead of i+n. The only points that
need checking are (1) whether a line from i to n has a label different from n, and (2) whether
all the lines at n have different labels. Both points are ok because n is odd.

Problem A2

Two circles C and C' have centers O and O' and meet at M and N. The common tangent closer
to M touches C at A and C' at B. The line through B perpendicular to AM meets the line OO'
at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear.

Solution
A neat coordinate solution by Massaki Yamamoto (a competitor) is as follows.

Take AB as the x-axis and the perpendicular line through M as the y-axis. Choose the unit of
length so that M has coordinates (0, 1). Let A be (-m, 0) and B be (n, 0). Then considering the
right-angled triangle O'MK, where K is (n, 1) we find that O' is (n, (n2+1)/2 ). Similarly, O is
(-m, (m2+1)/2 ) ).

The gradient of the lie AM is 1/m, so the gradient of the line BD is -m and hence its equation
is mx + y = mn. The gradient of the line OO' is (n-m)/2, so its equation is 2y - x(n-m) = mn+1.
These intersect at ( (mn-1)/(m+n), (mn2+m)/(m+n) ). B' is (n, n2+1). It is now easy to check
that the lines MB' and MD both have gradient n, so M, D, B' are collinear.

Problem A3

Find all solutions to (m + 1)a = mb + 1 in integers greater than 1.

Answer

(m, a, b) = (2, 2, 3).

Solution

Taking equation mod m+1 we get (-1)b = -1, so b is odd. Hence we can divide the rhs by m+1
to get mb-1 - mb-2 + ... - m + 1. This has an odd number of terms. If m is odd, then each term is
odd and so the total is odd, but (m+1)a-1 is even (note that a > 1). Contradicton, so m is even.

We have mb = (m+1)a - 1. Expanding the rhs by the binomial theorem, and using b > 1, we see
that m must divide a. So a is even also. Put a = 2A, m = 2M. We can factorise (m+1)a - 1 as (
(m+1)A + 1) ( (m+1)A - 1). The two factors have difference 2, so their gcd divides 2, but both
factors are even, so their gcd is exactly 2.

If M = 1 or a power of 2, then the smaller factor 3A - 1 must be 2, so A = 1 and we have 3A +

1 = 4, so (2M)b = 8. Hence M = 1 and b = 3 and we have the solution (m, a, b) = (2, 2, 3).

If M is not a power of 2, then Mb > 2b, so we must have the larger factor 2· Mb and the smaller
factor 2b-1. But the larger factor is now > 2b+1, so the difference between the factors is at least
3· 2b-1 > 2. Contradiction.

Problem B1
Some terms are deleted from an infinite arithmetic progression 1, x, y, ... of real numbers to
leave an infinite geometric progression 1, a, b, ... . Find all possible values of a.

Solution

Answer: the positive integers.

If a is negative, then the terms in the GP are alternately positive and negative, whereas either
all terms in the AP from a certain point on are positive or all terms from a certain point on are
negative. So a cannot be negative. If a is zero, then all terms in the GP except the first are

the AP must have infinitely many positive terms and hence x •
zero, but at most one term of the AP is zero, so a cannot be zero. Thus a must be positive, so

Let d = x - 1, so all terms of the AP have the form 1+nd for some positive integer n. Suppose
a = 1 + md, a2 = 1 + nd, then (1 + md)2 = 1 + nd, so d = (n - 2m)/m2, which is rational. Hence
a is rational. Suppose a = b/c, where b and c are relatively prime positive integers and c > 1.
Then the denominator of the nth term of the GP is cn, which becomes arbitrarily large as n
increases. But if d = h/k, then all terms of the AP have denominator at most k. So we cannot
have c > 1. So a must be a positive integer.

On the other hand, it is easy to see that any positive integer works. Take x = 2, then the AP
includes all positive integers and hence includes any GP with positive integer terms.

Problem B2

Given a pile of 2000 stones, two players take turns in taking stones from the pile. Each player
must remove 1, 2, 3, 4, or 5 stones from the pile at each turn, but may not take the same
number as his opponent took on his last move. The player who takes the last stone wins. Does
the first or second player have a winning strategy?

Solution

The first player has a winning strategy. He takes 4 on his first move leaving 7 mod 13 (2000 =
153.13 + 7 + 4). Now we claim that the first player can always leave: (1) 0 mod 13, (2) 3 mod
13 by taking away 3, (3) 5 mod 13 by taking away 5, or (4) 7 mod 13, and that the second
player can never leave 0 mod 13.

Let us look at each of these in turn. If the first player leaves 0 mod 13, then the second player
can take 3 and leave 10. In that case the first player takes 5 (a type (3) move). If the second
player takes 1, 2, 4 or 5, leaving 12, 11, 9 or 8 mod 13, then the first player takes 5, 4, 2, 1
(respectively) and leaves 7 mod 13 (a type (4) move).

If the first player leaves 3 mod 13 by taking away 3, then the second player cannot leave 0
mod 13, because he cannot take 3 stones. If he takes 1, 2 leaving 2, 1 mod 13 respectively,
then the first player takes 2, 1 leaving 0 mod 13 (a type (1) move). If the second player takes
4, 5 leaving 12, 11 mod 13, then the first player takes 5, 4 leaving 7 mod 13 (a type (4) move).

If the first player leaves 5 mod 13 by taking 5, then the second player cannot leave 0 mod 13,
because he cannot take 5 stones. If he takes 1, 2, 3, 4 stones, leaving 4, 3, 2, 1 mod 13, then
the first player takes 4, 3, 2, 1 stones leaving 0 mod 13 (a type (1) move).
Finally, if the first player leaves 7 mod 13, and the second player takes 1 stone, then the first
player takes 3 stones leaving 3 mod 13 (a type (2) move). If the second player takes 2, 3, 4, or
5 stones leaving 5, 4, 3, 2 mod 13, then the first player takes 5, 4, 3, 2 stones leaving 0 mod 13
(a type (1) move).

So the second player can never leave 0 mod 13 and hence, in particular, can never take the
last stone. But we have shown that the first player can always make a move of one of the four
types, so can always move and hence must win (since after less than 2000 moves there will be
no stones left).

Problem B3

all the vertices of the hexagon. Show that there is a unit of area k for any 0 < k ”:KDWLV
A convex hexagon is called a unit if it has four diagonals of length 1, whose endpoints include

the largest possible area for a unit?

Solution

Answer: We can get arbitrarily close to (but not achieve) (3¥  DSSUR[ 1.3) by:

To prove the first part, consider the diagram below. Take AB = AC = 1 and angle BAC = 2.
Take DE = DF = 1 and take the points of intersection X and Y such that AX = DX = AY =
DY = 2/3. It is easy to check that the area of the hexagon is sin 2. So by taking  in the
interval (0, Œ/4] we can get any area 0 < k ”

It is easy to check that there are six possible configurations for the unit diagonals, as shown in
the diagram below.
Consider case 1.

The area of the hexagon is area AEDC + area AFE + area BAC. The part of the segment BF
that lies inside AEDC is wasted. The rest goes to provide height for the triangles on bases AE
and AC. So area AFE + area BAC can be maximised by taking F close to A and ∠BAC as
close to a right angle as possible, so that the height of the triangle BAC (on the base AC) is as
large as possible. We can then get arbitrarily close to the area of:

We obviously make AEB a straight line. Now area ADE + area ADC = area ACE + area
CDE. So if we regard every point except D as fixed, then we maximise the area by taking
∠EAD = ∠CAD, so that D is the maximum distance from CE. Thus a maximal configuration
must have ∠AED = ∠CAD. Similarly, it must have ∠CAD = ∠CAB, so all three angles must
be equal. That disposes of case 1.

In cases 2 and 6 we find by a similar (but more tedious argument) the same maximum,
although in one case we have to use the argument at the end for the final optimisation. In the
other cases the maximum is smaller.

However, all these details would take an already long solution way over length. Does anyone
have a better approach?

No. 6 (second case) can be made arbitrarily close to the figure below (with AB = AC = BD =
1). To optimise it, suppose ∠ACB = . Area ABDC = area ABC + area BCD. If we fix , then
BC is fixed, so to maximise area BCD we must take ∠CBD = 90o. But  cannot be optimal
unless also ∠CAD = 90o. We have BA = BD and hence ∠BAD = ∠BDA = 45o - /2. Hence
90o = ∠CAD = ∠BAC - ∠BAD = (180o - 2) - (45o - /2). Hence  = 30o. So ∠ACD =
∠BDC = 60o and ∠CAB = ∠ABD = 120o. It is easy to check that this has area (3¥ 

16th Iberoamerican 2001

Problem A1

Show that there are arbitrarily large numbers n such that: (1) all its digits are 2 or more; and
(2) the product of any four of its digits divides n.

Solution

3232 = 16 x 202 and 10000 = 16 x 625. So any number with 3232 as its last 4 digits is
divisible by 16. So consider N = 22223232. Its sum of digits is 18, so it is divisible by 9.
Hence it is divisible by 9.16 = 144. But any four digits have at most four 2s and at most two
3s, so the product of any four digits divides 144 and hence N. But now we can extend N by
inserting an additional 9m 2s at the front. Its digit sum is increased by 18m, so it remains
divisible by 144 and it is still divisible by the product of any four digits.

Alternative solution

The number 111111111 with nine 1s is divisible by 9. Hence the number with twenty-seven
1s which equals 111111111 x 1000000001000000001 is divisible by 27. So N, the number
with twenty-seven 3s, is divisible by 34. Now the number with 27n 3s is divisible by N and
hence by 34.

Problem A2

ABC is a triangle. The incircle has center I and touches the sides BC, CA, AB at D, E, F
respectively. The rays BI and CI meet the line EF at P and Q respectively. Show that if DPQ
is isosceles, then ABC is isosceles.

Solution
AF = AE, so ∠AFE = 90o - A/2. Hence ∠BFP = 90o + A/2. But ∠FBP = B/2, so ∠FPB = C/2.
But BFP and BDP are congruent (BF = BD, BP common, ∠FBP = ∠FDP), so ∠DPB = C/2
and ∠DPQ = C. Similarly, ∠DQP = B. Hence ∠PDQ = A. So DQP and ABC are similar. So
if one is isosceles, so is the other.

Problem A3

Let X be a set with n elements. Given k > 2 subsets of X, each with at least r elements, show
that we can always find two of them whose intersection has at least r - nk/(4k - 4) elements.

Problem B1

Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest
number of trios that can be subsets of a set of n distinct real numbers?

Answer

(m-1)m for n = 2m
m2 for n = 2m+1

Solution

Let X be one of the elements. What is the largest number of trios that can have X as middle
element? Obviously, at most max(b,a), where b is the number of elements smaller than X and
a is the number larger. Thus if n = 2m, the no. of trios is at most 0 + 1 + 2 + ... + m-1 + m-1 +
m-2 + ... + 1 + 0 = (m-1)m. If n = 2m+1, then the no. is at most 0 + 1 + 2 + ... + m-1 + m + m-
1 + ... + 1 + 0 = m2.

These maxima can be achieved by taking the numbers 1, 2, 3, ... , n.

Problem B2

Two players play a game on a 2000 x 2001 board. Each has one piece and the players move
their pieces alternately. A short move is one square in any direction (including diagonally) or
no move at all. On his first turn each player makes a short move. On subsequent turns a player
must make the same move as on his previous turn followed by a short move. This is treated as
a single move. The board is assumed to wrap in both directions so a player on the edge of the
board can move to the opposite edge. The first player wins if he can move his piece onto the
same square as his opponent's piece. For example, suppose we label the squares from (0, 0) to
(1999, 2000), and the first player's piece is initially at (0, 0) and the second player's at (1996,
3). The first player could move to (1999, 2000), then the second player to (1996, 2). Then the
first player could move to (1998, 1998), then the second player to (1995, 1). Can the first
player always win irrespective of the initial positions of the two pieces?

Problem B3

Show that a square with side 1 cannot be covered by five squares with side less than 1/2.
17th Iberoamerican 2002

Problem A1

The numbers 1, 2, ... , 2002 are written in order on a blackboard. Then the 1st, 4th, 7th, ... ,
3k+1th, ... numbers in the list are erased. Then the 1st, 4th, 7th, ... 3k+1th numbers in the
remaining list are erased (leaving 3, 5, 8, 9, 12, ... ). This process is carried out repeatedly
until there are no numbers left. What is the last number to be erased?

Solution

Answer: 1598.

Let an be the first number remaining after n iterations, so a0 = 1, a1 = 2, a3 = 3, a4 = 5 etc. We

claim that:

an+1 = 3/2 an if an is even, and

3/2 (an + 1) - 1 if an is odd.

We use induction on n. Suppose an = 2N. Consider the number 3N. There are initially N
smaller numbers = 1 mod 3. So after the first iteration, it will lie in 2Nth place. Hence, it will
lie in first place after n+1 iterations. Similarly, suppose an = 2N+1. Consider 3N+2. There are
initially N+1 smaller numbers = 1 mod 3. So after the first iteration, it will lie in 2N+1st
place. Hence, it will lie in first place after n+1 iterations. That completes the induction.

We may now calculate successively the members of the sequence: 1, 2, 3, 5, 8, 12, 18, 27, 41,
62, 93, 140, 210, 315, 473, 710, 1065, 1598, 2397. Hence 1598 is the last surviving number
from 1, 2, ... , 2002.

Problem A2

Given a set of 9 points in the plane, no three collinear, show that for each point P in the set,
the number of triangles containing P formed from the other 8 points in the set must be even.

Solution

Join each pair of points, thus dividing the plane into polygonal regions. If a point P moves
around within one of the regions then the number of triangles it belongs to does not change.
But if it crosses one of the lines then it leaves some triangles and enters others. Suppose the
line is part of the segment joining the points Q and R of the set. Then it can only enter or
leave a triangle QRX for some X in the set. Suppose x points in the set lie on the same side of
the line QR as P. Then there are 6 - x points on the other side of the line QR. So P leaves x
triangles and enters 6-x. Thus the net change is even. Thus if we move P until it is in the outer
infinite region (outside the convex hull of the other 8 points), then we change the number of
triangles by an even number. But in the outside region it belongs to no triangles.

Problem A3

ABC is an equilateral triangle. P is a variable interior point such that $3& o. The ray
CP meets AB at M, and the ray AP meets BC at N. What is the locus of the circumcenter of
the triangle MBN as P varies?
Solution

Answer: the segment of the perpendicular bisector of BG (where G is the center of the
triangle) which forms a rectangle with AC.

∠MPN = ∠APC = 120o and ∠MBN = 60o, so MBNP is cyclic, in other words, P lies on the
circumcircle of BMN.

P also lies on the circle AGC, so ∠CPG = ∠CAG (if P is on the same side of AG as A) = 30o
= ∠MBG. So PMBG is cyclic. In other words, G also lies on the circumcircle of BMN. If P
lies on the other side, the same conclusion follows from considering ∠APG.

Since B and G lie on the circumcircle, the center O must lie on the perpendicular bisector of
BG. But it is clear that the extreme positions of O occur when P is at A and B and that these
are the feet of the perpendiculars from A and B to the perpendicular bisector.

Problem B1

ABC is a triangle. BD is the an angle bisector. E, F are the feet of the perpendiculars from A,
C respectively to the line BD. M is the foot of the perpendicular from D to the line BC. Show
that ∠DME = ∠DMF.

Solution
Let H be the foot of the perpendicular from D to AB. ∠AHD = ∠AED = 90o, so AHED is
cyclic. Hence ∠DAE = ∠DHE. But M is the reflection of H is the line BD, so ∠DME =
∠DAE.

AE is parallel to CD, so ∠DAE = ∠DCF. ∠DFC = ∠DMC, so DMCF is cyclic. Hence ∠DCF
= ∠DMF. Hence ∠DME = ∠DMF.

Problem B2

The sequence an is defined as follows: a1 = 56, an+1 = an - 1/an. Show that an < 0 for some n
such that 0 < n < 2002.

Solution

Note that whilst an remains positive we have a1 > a2 > a3 > ... > an. Hence if am and am+n are in
this part of the sequence, then am+1 = am - 1/am, am+2 = am+1 - 1/am+1 < am+1 - 1/am = am - 2/am.
By a trivial induction am+n < am - n/am.

If we use one step then we need 562 = 3136 terms to get a1+3136 < 56 - 562/56 = 0, which is not
good enough. So we try several steps.

Thus suppose that an > 0 for all n<= 2002. Then we get successively:
a337 < 56 - 336/56 = 50
a837 < 50 - 500/50 = 40
a1237 < 40 - 400/40 = 30
a1537 < 30 - 300/30 = 20
a1737 < 20 - 200/20 = 10
a1837 < 10 - 100/10 = 0.

Contradiction. So we must have an < 0 for some n < 2002.

Problem B3

A game is played on a 2001 x 2001 board as follows. The first player's piece is the policeman,
the second player's piece is the robber. Each piece can move one square south, one square east
or one square northwest. In addition, the policeman (but not the robber) can move from the
bottom right to the top left square in a single move. The policeman starts in the central square,
and the robber starts one square diagonally northeast of the policeman. If the policeman
moves onto the same square as the robber, then the robber is captured and the first player
wins. However, the robber may move onto the same square as the policeman without being
captured (and play continues). Show that the robber can avoid capture for at least 10000
moves, but that the policeman can ultimately capture the robber.

Solution

Color the squares with three colors as follows:

0 1 2 0 1 2 0 ... 2

1 2 0 1 2 0 1 ... 0

2 0 1 2 0 1 2 ... 1
0 1 2 0 1 2 0 ... 2

1 2 0 1 2 0 1 ... 0

...

2 0 1 2 0 1 2 ... 1

The middle square is color 2 (moving 999+1 squares E from the top left increases the color by
1, then moving 999+1 S increases it by another 1) and the square immediately NE of it is also
2. So both P and R start on color 2. Note that any move increases the color by 1 mod 3, except
for P's special move which changes the color from 1 to 0.

Until P has made this move, after each move of P, P's color is always 1 more than R's color
(mod 3), so P cannot win (irrespective of the moves made by either player). Immediately after
he makes the special move for the first time, P is on color 0 and R is on color 1, so
immediately after his move P's color is now 1 less than R's color mod 3. Again P cannot win.
But after P has made the special move for the second time, P's color is the same as R's (mod
3) immediately after P's move.

Note that it takes P at least 2001 moves to complete his special move for the first time and at
least 6002 moves (in total) to complete his special move for the second time. This solves the
first part of the question. Suppose R just moves down to the bottom right and then moves in
small circles (one move NW, one move S, one move E) waiting for P. It takes P at least 6002
+ 3999 (moving from top left to the capture square, one square short of the bottom right) =
10001 to capture him, so R makes at least 10000 moves before being captured.

We claim that P wins if he can get into any of the positions shown below relative to R, with R
to move (*):

x P x x x

P x x P x

x x R x x

x P x x P

x x x P x

If follows that P can also win from the four positions below (**):
x x x P x x x

x x x x x x x

x x x x x x x

P x x R x x P

x x x x x x x

x x x x x x x

x x x P x x x
For in each case at least one of R's possible moves allow P to move immediately into one of
the winning positions at (*). But R can only make the other moves a limited number of times
before running into the border. [That is obvious if the other two moves are E and S. If they are
NW and E, then every NW move takes R closer to the top border, but his total number of E
moves can never exceed his total number of NW moves by more than 2000 because of the
right border. Similarly, for NW and S.]

Now let d be the number of rows plus the number of columns that R and P are apart. It is easy
to check that the positions in (*) and (**) represent the only possibilities for d = 2 and 3. We
show that P can always get to d = 2 or 3. For P can always copy R's move, so he can certainly
move so that d never increases. But one of R's moves will always allow P to decrease d by 1
or 2. There are three cases to consider:
Case 1. If P is east of R and R moves E, then P moving NW will decrease d by 1 or 2. That is
not possible if P is in the top row, but then moving S will decrease d by 2 unless R is also in
the top row. If both are in the top row, then P moves S. Now after R's next move, P moves
NW which reduces d by 2.
Case 2. If P is south of R and R moves S, then a similar argument, shows that P can always
decrease d by 1 or 2 in one or two moves.
Case 3. If P is not south or east or R, and R moves NW, then P can always decrease d by 1 or
2 by moving S or E.

But repeated decreases by 1 or 2 must bring d ultimately to 2 or 3 and hence to one of (*) or
(**). So P can always win.

It remains to prove the claim that (*) are winning positions. The reason is that in each case R
has one move blocked off, so must make one of the other two. P then copies R's move, so next
turn R has the same move blocked off. Repeated use of the other two moves will bring him
ultimately to one of the sides.

We start with the easiest case: in the two following positions. R cannot move to z, so he must
move east or south on each move. Hence he will (after at most 4000 moves) reach the bottom
right corner. He then loses moving out of it.

x P x

P z x

x x R

The other cases of (*) are slightly more complicated. Starting from either of the two positions
below, we show that R must eventually reach the extreme left column.
w x P x

x R z x

x y x P

R cannot move to z, so he can only make NW and S moves. But his total number of S moves
can never exceed his total number of NW moves by more than 2000 because he cannot move
off the bottom of the board, so he must eventually reach the extreme left column. [If he
reaches the bottom row at y, then P can always move to z to preserve the configuration. If R
reaches the top row by moving to w, then P can always move to z to preserve the
configuration.]
Having reached the extreme left column he is forced to move south. Eventually moving to y
will take him to the corner. P then moves to z and R is captured on his next move.

The final case to consider is the two positions below. R cannot move to z, so must move E or
NW. A similar argument to the previous case shows that he must eventually reach the top
row. Having reached it at w, P moves to z. So R is forced to move right along the top row.
When he reaches the corner at y, P moves to z and R is captured when he moves out of the
corner.

w x x

x R y

P z x

x x P

18th Iberoamerican 2003

Problem A1

Let A, B be two sets of N consecutive integers. If N = 2003, can we form N pairs (a, b) with a
∠ A, b ∠ B such that the sums of the pairs are N consecutive integers? What about N = 2004?

Answer

Yes, no.

Solution

wlog A = B = {1, 2, ... , N} - if we have a solution for A = {a+1, a+2, ... , a+N} and B = {b+1,
b+2, ... , b+N}, then subtracting a from every element of A and b from every element of b
gives a solution for A = B = {1, 2, ... , N}. Suppose the sum set is (m+1), (m+2), ... , (m+N).
It has sum N(2m+N+1)/2 and A and B each have sum N(N+1)/2, so we must have 2m = N+1,
hence N must be odd. So we cannot do it for N = 2004.

Suppose N = 2M+1, take the pairs (1,M+1), (3,M), (5,M-1), ... , (2M+1,1), (2,2M+1), (4,
2M), ... , (2M, M+2).

Problem A2

C is a point on the semicircle with diameter AB. D is a point on the arc BC. M, P, N are the
midpoints of AC, CD and BD. The circumcenters of ACP and BDP are O, O'. Show that MN
and OO' are parallel.

Solution
Let the center of the circle be X and the radius r. Let ∠AXM = , ∠BXN = 3. Note that O is
the intersection of XM and the perpendicular to CD at Q, the midpoint of CP. We have XM =
r cos . Let CD and XM meet at Y. Then ∠PYX = 90o - ∠PXY = 90o - ∠PXC - ∠CXM =  +
3 - 3 = . Hence OX = PQ sec 3, so OX/XM = PQ/(r cos  cos 3). Similarly, O'X/ON = PQ/(r
cos  cos 3), so OO' and MN are parallel.

Problem A3

satisfies x0 = 1, 0 ”[i ”[i-1 for 1 ”L”DQGZKLFKPD[LPLVHV S. Unfortunately he could

Pablo was trying to solve the following problem: find the sequence x0, x1, x2, ... , x2003 which

not remember the expression for S, but he knew that it had the form S = ± x1 ± x2 ± ... ± x2002
+ x2003. Show that he can still solve the problem.

Solution

For any combination of signs the maximum is obtained by taking all xi as large as possible.
Suppose we have a different set of xi. Then for some k we must have xk < 2xk-1 and xi = 2xi-1
for all i > k. Suppose 2xk-1 - xk = h > 0. Then we can increase xk by h, xk+1 by 2h, xk+2 by 4h,
... . So the sum will be increased by h(± 1 ± 2 ± ... ± 2m-1 + 2m) for some m •%XW“““
... ± 2m-1 •-(1 + 2 + ... + 2m-1) = -2m + 1, so the overall sum will be increased by at least 1. So
the set of xi was not maximal.

Problem B1

A ^`GRHVQRWFRQWDLQVL[FRQVHFXWLYHLQWHJHUV)LQGWKHODUgest possible value

of |A|. How many such subsets are there (of the maximum size)?

Answer

max = 41; no. ways 495

Solution

18}, ... , {43, ... , 48}. So |A| ”%XWDYDOXHRILVFHUWDLQO\SRVVLEOHIRUH[DPSOH

We must exclude at least one element of each of the 8 sets {1, 2, ... , 6}, {7, ... , 12}, {13, ... ,

exclude 2, 8, 14, ... , 44.

The largest excluded element must be at least 44 (or we have the 6 consecutive elements 44,
45, 46, 47, 48, 49). The smallest excluded element must be at most 6. If we exclude 2 and 44,
then the difference between them is 7· 6 and so the other 6 excluded elements are fixed. But if
we exclude 3 and 44, for example, then there are several possible choices for the other
elements.
There are 5 ways of choosing the smallest and largest excluded element to get a difference of
7· 6 between them (2 and 44, 3 and 45, 4 and 46, 5 and 47, 6 and 48). There are 4 ways to get a
difference of 7· 6 - 1 (3 and 44, 4 and 45, 5 and 46, 6 and 47). There are 3 ways to get a
difference of 7· 6 - 2 (4 and 44, 5 and 45, 6 and 46), 2 ways to get a difference of 7· 6 - 3 (5 and
44, 6 and 45), and 1 way to get a difference of 7· 6 - 4 (6 and 44).

If the difference is 7· 6 - 1, then we can shorten any of the 7 gaps, so there are 7 possibilities.
For example, with 3 and 44, we could shorten the first gap, so excluding 3, 8, 14, 20, 26, 32,
38 and 44, or the second gap, so excluding 3, 9, 14, 20, 26, 32, 38 and 44, and so on.

If the difference is 7· 6 - 2, then we can shorten one gap by two (7 possibilities) or two gaps by
one (21 possibilities), total 28. If the difference is 7· 6 - 3, then we can shorten on gap by three
(7), one by two and one by one (42) or three by one (35), total 84. Finally, if the difference is
7· 6 - 4, we can shorten one by four (7), one by three and one by 1 (42), two by two (21), one
by two and two by one (105), or four by one (35), total 210.

So the total number of possibilities is 5· 1 + 4· 7 + 3· 28 + 2· 84 + 1· 210 = 495.

Problem B2

ABCD is a square. P, Q are points on the sides BC, CD respectively, distinct from the
endpoints such that BP = CQ. X, Y are points on AP, AQ respectively. Show that there is a
triangle with side lengths BX, XY, YD.

Solution

We have DY < BY ”%;;< WKLVLVDOPRVWREYLRXVEXWWRSURYHIRUPDOO\XVHWKHFRVLQH

formula for BAY and DAY and notice that ∠BAY > ∠DAY). Similarly, BX < DX ”'<
YX. So it remains to show that XY < BX + DY.

Take Q' on the extension of BC so that BQ' = DQ, as shown in the diagram. Take Y' on AQ'
so that AY' = AY. Then XY' ”%;%<  %;'<1RZZHFODLPWKDW∠PAQ' > ∠ PAQ,
so it follows by the same observation as above that XY' > XY. But the claim is almost
obvious. Note that PQ' = AB
So take P' on AD with ∠P'PQ' = 90o. Then A lies inside the circle P'PQ', so extend PA to meet
it again at A'. Then ∠PA'Q' = ∠PP'Q' = 45o, so ∠PAQ' = ∠PA'Q' + ∠AQ'Q' > 45o. But
∠PAQ' + ∠PAQ = 90o, so ∠PAQ' > ∠PAQ as claimed.

Problem B3

The sequences a0, a1, a2, ... and b0, b1, b2, ... are defined by a0 = 1, b0 = 4, an+1 = an2001 + bn,
bn+1 = bn2001 + an. Show that no member of either sequence is divisible by 2003.

Solution

2003 is prime, so a2002 = 1 mod 2003 for any a not divisible by 2003. Thus an+1 = an-1 + bn
mod 2003, bn+1 = bn-1 + an mod 2003. Put cn = anbn. Then cn+1 = cn + 1/cn + 2 = (cn + 1)2/cn
mod 2003. So if cn PRGWKHQFn+1 PRGXQOHVVFn = -1 mod 2003. Then if
cn+1 = -1 mod 2003, we must have (cn2 + 3cn + 1)/cn = 0 mod 2003, so cn2 + 3cn + 1 = 0 mod
2003. Note that c0 = 4. So it is sufficient to show that there are no solutions to x2 + 3x + 1 = 0
mod 2003, or equivalently to (x - 1000)2 = 10002 - 1 = 502 mod 2003. In other words, we
have to show that 502 is a quadratic non-residue mod 2003.

The easiest way to do that is to use the law of quadratic reciprocity, but that is almost
certainly outside the syllabus. We note that 4· 502 = 5 mod 2003, so 502 is a square iff 5 is a
square. It is sufficient to show that 51001 = -1 mod 2003, for then if we had x2 = 5, we would
have x2002 = -1 mod 2003, whereas we know that x2002 = 1 mod 2003. We note that 1001 =
7· 11· 13. We start by showing that 57 = 8 mod 2003. We have 55 = 3125 = 1122 mod 2003, so
56 = 5610 = 1604 mod 2003, so 57 = 8020 = 8 mod 2003.

We calculate successively 211 = 2048 = 45 mod 2003, so 222 = 2025 = 22 mod 2003.
Multiplying by 22 is relatively easy, so 244 = 484, 266 = 10648 = 633, 288 = 13926 = -95, 2110
= -2090 = -87, 2132 = -1914 = 89, 2143 = 4005 = -1 all mod 2003. Hence 811· 13 = -1 mod 2003,
so 51001 = -1 mod 2003, as required, and we are done.