3rd Chinese Mathematical Olympiad 1988 Problems & Solutions

A1.  a₁, ... , a_n are reals, not all 0, such that there exist b_i so that ∑₁ⁿ b_i(x_i - a_i) ≤ √(∑₁ⁿ x_i²) - √(∑₁ⁿ a_i²) for all real x_i. Find the b_i (in terms of the a_i).

A2.  ABCD is a cyclic quadrilateral. Its circumcircle has center O and radius R. The rays AB, BC, CD, DA meet the circle center O radius 2R at A', B', C', D' respectively. Show that A'B' + B'C' + C'D' + D'A' ≥ 2(AB + BC +CD + DA). When does equality hold?

A3.  x₁, x₂, ... , x_n are real numbers, at least one of which is greater than 1988. Select all terms x_i such that for some j >= i the average of the terms from x_i to x_j (inclusive) exceeds 1988. Show that the average of the selected terms exceeds 1988.

B1.  (1) The positive reals a, b, c satisfy (a² + b² + c²)² > 2(a⁴ + b⁴ + c⁴). Show that there is a triangle with sides a, b, c. (2) The n > 3 positive reals a₁, a₂, ... , a_n satisfy (a₁² + ... + a_n²)² > (n - 1)(a₁⁴ + ... + a_n⁴). Show that every three of the a_i can form the sides of a triangle.

B2.  XABC, X'A'B'C' and X"A"B"C" are tetrahedra. X, X' and X'' are collinear. The planes through A normal to XA, through B normal to XB and through C normal to XC meet at O. Points O' and O" are defined similarly. If O, O' and O" coincide what is the intersection of the circumspheres of the three tetrahedra?

B3.  Define f on the positive integers as follows: f(1) = f(2) = f(3) = 2. For n > 3, f(n) is the smallest positive integer which does not divide n. Define f₁ to be f and f_k+1(n) = f(f_k(n) ). Let g(n) be the smallest k such that f_k(n) = 2. Determine g(n) as explicitly as possible. 

Solutions

Problem A1

a₁, ... , a_n are reals, not all 0. Find b_i such that ∑₁ⁿ b_i(x_i - a_i) ≤ √(∑₁ⁿ x_i²) - √(∑₁ⁿ a_i²) for all real x_i.

Solution

Put A = √ ∑₁ⁿ a_i². Take x_j = a_j for all j except i, so that the inequality becomes b_i(x_i - a_i) ≤ √(A² + x_i² - a_i²) - A = (x_i² - a_i²)/(√(A² + x_i² - a_i²) + A).

Take x_i > a_i, so that x_i - a_i > 0, then we deduce b_i ≤ (x_i + a_i)(√(A² + x_i² - a_i²) + A). Now let x_i tend to a_i (from above), and we deduce that b_i ≤ a_i/A. Similarly, taking x_i tending to a_i from below gives b_i ≥ a_i/A. Hence b_i = a_i/A.

It remains to prove that the b_i work. Note first that ∑ b_ia_i = 1/A ∑ a_i² = A, so we need to show that ∑ b_ix_i ≤ (∑x_i²)^1/2. But this follows immediately from Cauchy-Schwartz since ∑ b_i² = 1. 

Problem A2

ABCD is a cyclic quadrilateral. Its circumcircle has center O and radius R. The rays AB, BC, CD, DA meet the circle center O radius 2R at A', B', C', D' respectively. Show that A'B' + B'C' + C'D' + D'A' ≥ 2(AB + BC +CD + DA). When does equality hold?

Solution

Answer: Equality iff ABCD is a square.

We use Ptolemy's inequality. Applying it to the quadrilateral OAD'A' we have OD'·AA' ≤ OA·A'D' + OA'·AD' and hence 2AA' ≤ A'D' + 2AD' or 2AB + 2BA' ≤ A'D' + 2AD'. Adding to the three similar inequalities gives: 2(AB + BC + CD + DA) + 2(BA' + CB' + DC' + AD') ≤ (A'B' + B'C' + C'D' + D'A') + 2(BA' + CB' + DC' + AD') and hence the required inequality. We can only have equality if OAD'A', OBA'B', OCB'C', ODC'D' are all cyclic.

If OAD'A' is cyclic, then ∠OAA' = ∠OD'A' = ∠OA'D' (since OA' = OD') = 180 deg - OAD' = OAD. So OA bisects angle DAB. Similarly O lies on the other angle bisectors of ABCD. Hence ∠OAB = ∠OAD (angle bisector) = ∠ODA (OA = OD) = ∠ODC = ... = ∠OBA. Hence also ∠AOB = ∠BOC = ∠COD = ∠DOA. So A, B, C, D are equally spaced on the circle and hence form a square. It is obvious that if A, B, C, D form a square, then we have equality. 

Problem A3

x₁, x₂, ... , x_n are real numbers, at least one of which is greater than 1988. Select all terms x_i such that for some j >= i the average of the terms from x_i to x_j (inclusive) exceeds 1988. Show that the average of the selected terms exceeds 1988.

Solution

Let x_i be the first selected term. Take the smallest j ≥ i such that the average of the terms from x_i to x_j exceeds 1988. Then we claim that the average of the terms from x_k to x_j exceeds 1988 for any k satisfying i ≤ k ≤ j. It is clearly true for k = i. Suppose it was false for some k > i. Then the average of the terms x_i to x_k-1 would have to exceed 1988 (otherwise the average of the terms x_i to x_j would not exceed 1988). But that contradicts the minimality of j. So all the terms from i to j are selected and they have average exceeding 1988. If there are any further selected terms, then take x_i' to be the first, and so on. Thus we can group the selected terms into batches, with the average of each batch exceeding 1988.

Problem B1

(1) The positive reals a, b, c satisfy (a² + b² + c²)² > 2(a⁴ + b⁴ + c⁴). Show that there is a triangle with sides a, b, c.

(2) The n > 3 positive reals a₁, a₂, ... , a_n satisfy (a₁² + ... + a_n²)² > (n - 1)(a₁⁴ + ... + a_n⁴). Show that every three of the a_i can form the sides of a triangle.

Solution

Multiplying out, we find that (a + b + c)(a + b - c)(a - b + c)(-a + b + c) = 2(a²b² + b²c² + c²a²) - (a⁴ + b⁴ + c⁴). So if the inequality (1) holds, then (a + b + c)(a + b - c)(a - b + c)(-a + b + c) > 0. We are given that a, b, c are positive, so a + b + c > 0. We may assume that a ≥ b ≥ c, so that a - b + c > 0 and a + b - c ≥ 0. Hence also -a + b + c > 0, so a, b, c can form a triangle.

In the general case (2) we use induction on n. Suppose the result is true for < n. The Cauchy-Schwartz inequality gives ∑ 1.a ≤ (∑ 1)^1/2 (∑a²)^1/2, so (∑a)² ≤ n ∑a². Put 2b² = a₁² + a₂² + a₃². Then we have (∑ a_i²)² = (b² + b² + ∑₄ⁿ a_i²)² ≤ (n - 1)( 2b⁴ + ∑₄ⁿ a_i⁴). So if (n - 1)(a₁⁴ + ... + a_n⁴) < (∑₁ⁿ a_i²)², then (n - 1)(a₁⁴ + a₂⁴ + a₃⁴) < (n - 1) 2b⁴. Hence 2(a₁⁴ + a₂⁴ + a₃⁴) < (a₁² + a₂² + a₃²)². So there is a triangle with sides a₁, a₂, a₃. Similarly, for any other a_i, a_j, a_k. 

Problem B2

XABC, X'A'B'C' and X"A"B"C" are tetrahedra. X, X' and X'' are collinear. The planes through A normal to XA, through B normal to XB and through C normal to XC meet at O. Points O' and O" are defined similarly. If O, O' and O" coincide what is the intersection of the circumspheres of the three tetrahedra.

Solution

OA is perpendicular to XA, so A must lie on the sphere diameter OX. So do B and C, so it must be the circumsphere of XABC. Similarly, the circumsphere of X'A'B'C' is the sphere diameter OX', and the circumsphere of X"A"B"C" is the sphere diameter OX". We now distinguish three cases. (1) if X = X' = X'', then the three spheres coincide and there intersection is the sphere diameter OX. (2) if O lies on the line XX'X'', then the three spheres are nested and intersect only at O. (3) O does not lie on the line XX'X". The centers of the three spheres are the midpoints of OX, OX', OX" and so they must be collinear. So the three spheres must intersect in a circle with center on the line of their centers. Since O lies on this circle, it must be the circle obtained by rotating O about the line of centers. 

Problem B3

Define f on the positive integers as follows: f(1) = f(2) = f(3) = 2. For n > 3, f(n) is the smallest positive integer which does not divide n. Define f₁ to be f and f_k+1(n) = f(f_k(n) ). Let g(n) be the smallest k such that f_k(n) = 2. Determine g(n) explicitly.

Solution

Suppose a prime p divides f(n). Then f(n) must be a power of p, for if f(n) = p^rm with m > 1 not a multiple of p, then by the minimality of f(n), p^r and m both divide n and hence f(n) does also. Contradiction. If f(n) is not a power of 2, then f(f(n)) = 2, so g(n) = 2. If f(n) is a power of 2, then either f(n) = 2, in which case g(n) = 1, or f( f(n) ) = 3, in which case g(n) = 3.

Clearly f(n) = 2 iff n is odd. There is no convenient explicit expression for n such that f(n) is a power of 2 and at least 2². For example, f(n) = 4 iff n is a multiple of 6 but not 4, in other words n = 6 mod 12. Similarly, f(n) = 8 iff n is a multiple of 3·4·5·7, but not 8, f(n) = 16 iff n is a multiple of 3·5·7·11·13·8 but not 16, and so on.

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2nd Chinese Mathematical Olympiad 1987 Problems & Solutions

A1.  n is a positive integer. Show that zⁿ⁺¹ - zⁿ - 1 = 0 has a root on the unit circle |z| = 1 iff n is congruent to 4 mod 6.

A2.  An equilateral triangle side n is divided into n² equilateral triangles of side 1 by lines parallel to its sides. The n(n+1)/2 vertices of the triangles are each labeled with a real number, so that if ABC and BCD are small triangles then the sum of the labels on A and D equals the sum of the labels on B and C. The labels on the vertices of the original triangle are a, b and c. What is the shortest distance between a vertex with the largest label and a vertex with the smallest label? What is the sum of all the labels?

A3.  In a tournament each player plays every other player. Every game results in win or loss. A prize is awarded to every player X who fulfils the following condition: for each player Z who beats X there must be a player Y such that X beats Y and Y beats Z. Only one prize is awarded. Show that the recipient must have won every game.

B1.  Given five points inside an equilateral triangle of area 1, show that one can find three equilateral triangles with total area at most 0.64 and sides parallel to the original triangle, so that each of the five points is inside one or more of the new triangles.

B2.  A tetrahedron has the following properties. There is a sphere, center X, which touches each edge of the tetrahedron; there are four spheres with centers the vertices of the tetrahedron which each touch each other externally; and there is another sphere center X which touches all four spheres. Prove that the tetrahedron is regular.

B3.  A set of distinct positive integers has sum 1987. What is the maximum possible value for three times the total number of integers plus the number of odd integers? 

Solutions

Problem A1

n is a positive integer. Show that zⁿ⁺¹ - zⁿ - 1 = 0 has a root on the unit circle |z| = 1 iff n is congruent to 4 mod 6.

Solution

If z is a root, we have zⁿ(z - 1) = 1. Hence if also |z| = 1, then |z - 1| = 1. So 0, 1, z must form the vertices of an equilateral triangle. Hence z = exp(± i π/3), so that (z - 1) = exp(± 2iπ/3). But zⁿ(z - 1) = 1, so n+2 must be a multiple of 6 and it is easy to see that this is also a sufficient condition. 

Problem A2

An equilateral triangle side n is divided into n² equilateral triangles of side 1 by lines parallel to its sides. The n(n+1)/2 vertices of the triangles are each labeled with a real number, so that if ABC and BCD are small triangles then the sum of the labels on A and D equals the sum of the labels on B and C. The labels on the vertices of the original triangle are a, b and c. What is the shortest distance between a vertex with the largest label and a vertex with the smallest label? What is the sum of all the labels?

Solution

Take any three adjacent small triangles ABC, BCD, BDE. wlog the labels on A, B, C are d, d+x, d+y. Then the label on D must be d+x+y and the label on E must be d+2x. It is now an easy induction to show that the vertices along the line DBE are an arithmetic progression with difference x. Another induction shows that the labels along any line of vertices parallel to DBE is also an arithmetic progression with the same difference. If the two vertices of the large triangle along such a line have labels a and b, then we see that x must be (b - a)/n. Similarly, the vertices along a line parallel to the side a c must have labels in arithmetic progression with difference (a - c)/n, and vertices along a line in the third direction are in arithmetic progression with difference (c - b)/n. If all a, b, c are distinct, then it follows that the smallest and largest values must occur at the three vertices of the large triangle and hence their distance apart must be n. If just two are distinct, then one extreme value occurs just at a vertex of the large triangle and the other occurs at all vertices along the opposite side. So the shortest distance is n (√3)/2 for n even and √(1/4 + 3n²/4) for n odd. If a = b = c, then all the vertices have the same label and so the shortest distance is 1 (or 0 according to taste).

The sum of the vertices along the side from a to b is (n + 1)(a + b)/2. Similarly, for the other two sides, so the sum around the perimeter is n(a + b + c) = no. of points x (a + b + c)/2. If a = b + nx, b = c + ny, then the vertices of the large triangle formed by the remaining points have labels a - 2x - y, b + x - y, c + x + 2y, so the sum around the perimeter is again (no. of points) x (a + b + c)/2. Thus the sum of all the labels = no. of points x (a + b + c)/3 = (1 + 2 + ... + n+1)(a + b + c)/3 = (n + 1)(n + 2)(a + b + c)/6. 

Problem A3

In a tournament each player plays every other player. Every game results in win or loss. A prize is awarded to every player X who fulfils the following condition: for each player Z who beats X there must be a player Y such that X beats Y and Y beats Z. Only one prize is awarded. Show that the recipient must have won every game.

Solution

We show first that if no player has a higher score than X, then X fulfils the condition. For suppose no player has a higher score than X. If Y beats X, and beats everyone whom X beats, then X has a lower score than Y contradiction. So if Y beats X, then there must be someone whom X beats who beats Y, so X fulfils the condition.

Suppose A receives the only prize. Let S be the set of players who beat A. Let B have the highest score from the games between players in S. Then we have just shown that given any other player C in S, B either beats C or beat someone who beat C. But B also beat A who beat everyone outside S. So B is eligible for a prize. Contradiction. So S must be empty. 

Problem B1

Given five points inside an equilateral triangle of area 1, show that one can find three equilateral triangles with total area at most 0.64 and sides parallel to the original triangle, so that each of the five points is inside one or more of the new triangles.

Solution

Divide the triangle into 25 triangles area 1/25 by lines parallel to its sides. If 3 points lie inside one of the three triangles made up of 16 small triangles (and area 0.64), then we can cover the other two with triangles of zero area. So we can assume that at least 3 points lie inside the 9 triangles along each side. That means there are two possibilities: (1) two corner parallelograms (each made up of two small triangles) each contain 2 points and the 5th point is in the other corner parallelogram; or (2) one corner parallelogram contains 2 points, two contain 1 each and the 5th point lies in the five small triangles lying between them. In case (1) we can take a triangle area 4/25 containing each corner parallelogram (total area 0.48). In case (2) we can take a triangle area 9/25 containing the point on the side and the nearest corner parallelogram, a small triangle area 1/25 containing the other single corner point and a triangle area 4/25 containing the corner parallelogram with two points (total area 0.56). 

Problem B2

A tetrahedron has the following properties. There is a sphere, center X, which touches each edge of the tetrahedron; there are four spheres with centers the vertices of the tetrahedron which each touch each other externally; and there is another sphere center X which touches all four spheres. Prove that the tetrahedron is regular.

Solution

Let AB be a side of the tetrahedron. Let the spheres center A and B touch at R on AB, and let the sphere center X touch AB at W. The key insight is that R = W.

Let C be another vertex of the tetrahedron. Suppose the vertex spheres meet at U on BC and V on CA. Then AV = AW, BU = BW, CU = CV, so AB = AV + CU, BC = BU + CU, CA = CU + AV, so AV + BU + CU = (AB + BC + CA)/2, so AV = (AB - BC + CA)/2, BU = (AB + BC - CA)/2, CU = (-AB + BC + CA)/2.

Suppose the sphere center X touches BC at P and CA at Q. Then AR and AQ are tangents to the sphere center X, so they are equal. Similarly, BP = BR, CP = CQ. So exactly the same argument as above shows that AQ = (AB - BC + CA)/2 etc. Hence the points coincide: P = U, Q = V, R = W.

Now consider the right-angled triangle XAR. We have AX² = AR² + XR². Suppose the sphere center X touching the vertex spheres has radius R and the sphere center X touching the sides has radius r. If the sphere radius R encloses one of the vertex spheres then it must enclose them all (since they touch at a different point). So either XA = R + AR (and similarly for XB, XC, XD) or XA = R - AR (and similarly for XB etc). Hence R² ± 2R AR + AR² = AR² + r², so AR = ± (R² - r²)/R. Similarly for BR, CP etc with the same sign in all cases. Hence all the sides have the same length and the tetrahedron is regular. 

Problem B3

A set of distinct positive integers has sum 1987. What is the maximum possible value for three times the total number of integers plus the number of odd integers?

Solution

1 + 2 + ... + 62 = 62·63/2 = 1953, 1 + 2 + ... + 63 = 63·64/2 = 2016. So we can find at most 62 distinct integers with sum 1987. We can take 35 of these to be odd and 27 even without penalty: 1 + 2 + ... + 54 (so far 27 of each) + 55 + 57 + 59 + 61 + 63 + 65 + 67 + 69 = 1981. But if we increase the number of odd any further we must decrease the total number (replacing 54 by 71 is an increase of 17). If we replace 4 even numbers we cannot replace them by 3 odd numbers (71 + 73 + 75 - 54 - 52 - 50 - 48 = 15 > 6). So this is the best we can do. To get a total actually equal to 1987 just increase one of the odd numbers by 6, eg 75 instead of 69. That gives 4·35 + 3·27 = 221.

The argument that 221 is optimal is perhaps not entirely convincing. So suppose we have m odd numbers and n even numbers. The total of the odd numbers cannot be less than the sum of the first m odd numbers which is m² and the sum of the even numbers cannot be less than the sum of the first n even numbers which is n(n+1). So we must show that if 4m + 3n >= 222, then m² + n(n+1) > 1987. But 4m + 3n ≥ 222 implies 4m + 3(n + 1/2) ≥ 223 1/2 and hence, by Cauchy-Schwartz, 5√(m² + (n+ 1/2)²) > 223 1/2, so m² + n² + n > (447/10)² - 1/4 > 1987.

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1st Chinese Mathematical Olympiad 1986 Problems & Solutions

A1.  a₁, a₂, ... , a_n are reals. Show that if the sum of any two is non-negative, then for any non-negative real x₁, x₂, ... , x_n with sum 1, we have a₁x₁ + a₂x₂ + ... + a_nx_n ≥ a₁x₁² + a₂x₂² + ... + a_nx_n². Show that the converse is also true.

A2.  ABC is a triangle. The altitude from A has length 12, the angle bisector from A has length 13. What is are the possible lengths for the median from A if the angle A is (1) acute, (2) obtuse, (3) a right-angle?

A3.  The n complex numbers z_i satisfy |z₁| + |z₂| + ... + |z_n| = 1. Show that we can find a subset whose sum has modulus 1/6 or more.

B1.  PQRS is a convex quadrilateral which lies inside the triangle ABC of area 1. Show that three of its vertices form a triangle of area 1/4 or less.

B2.  The sequence a₁, a₂, .... , a₃₉₇₂ includes each of the numbers from 1 to 1986 twice. Can the terms be rearranged so that there are just n numbers between the two n's?

B3.  Each point in the plane is colored black or red. Show that we can find three points of the same color with each pair a distance 1 apart or three points of the same color with each pair a distance √3 apart. 

Solutions

Problem A1

a₁, a₂, ... , a_n are reals. Show that if the sum of any two is non-negative, then for any non-negative real x₁, x₂, ... , x_n with sum 1, we have a₁x₁ + a₂x₂ + ... + a_nx_n ≥ a₁x₁² + a₂x₂² + ... + a_nx_n². Show that the converse is also true.

Solution

We have ∑ a_ix_i = ∑ a_ix_i ∑ x_i = ∑ a_ix_i² + ∑ (a_i + a_j)x_ix_j ≥ ∑ a_ix_i².

The converse is obvious: take x_i = x_j = 1/2, others 0. 

Problem A2

ABC is a triangle. The altitude from A has length 12, the angle bisector from A has length 13. What is are the possible lengths for the median from A if the angle A is (1) acute, (2) obtuse, (3) a right-angle?

Solution

Let M be the midpoint of BC. Let AD be the altitude from A and AE the angle bisector. Let the median AM have length k. Take MN to be the perpendicular bisector of BC with N on the opposite side of BC to A and lying on the circumcircle of ABC. If ∠A is 90^o, then the circumcenter is at M. Then 12/5 = AD/DE = EM/MN = EM/AM = √(k² - 144) - 5)/k. Solving the quadratic, k = 2028/119 = 17.04.

If A is obtuse, then the circumcenter lies on MN, so the same argument shows that k > 2028/119. Evidently any such value is achievable, because we can make DM (and hence AM) arbitrarily large. Then construct N, then pick O on MN such that AO = ON. Then construct B and C on the line DM so that OB = OC = OA.

Similarly, if A is acute, the circumcenter lies on the opposite side of BC to N, so k < 2028/119. On the other hand E must lie between D and M, so k > 13 (or equal if we allow the degenerate triangle with B = C = E = M). 

Problem A3

The n complex numbers z_i satisfy |z₁| + |z₂| + ... + |z_n| = 1. Show that we can find a subset whose sum has modulus 1/6 or more.

Solution

Divide the complex plane into 120^o sectors with center the origin. At least one sector must contain a subset with modulus sum at least 1/3. The sum of the projections of this subset onto the center line bisecting the sector must must be at least 1/6 (because cos k ≥ 1/2 for k ≤ 60^o). So the sum has a projection length ≥ 1/6 and hence must have modulus at least 1/6. 

Problem B1

PQRS is a convex quadrilateral which lies inside the triangle ABC of area 1. Show that three of its vertices form a triangle of area 1/4 or less.

Solution

We prove first that a parallelogram PQRS inside ABC has area at most 1/2. Assume the parallelogram has maximal area. If the parallelogram has less than 2 vertices on the sides of the triangle then we can expand it (about the point in contact, if any, or about its center if none) until it has two points in contact. That would increase its area. So it must have at least 2 vertices on the sides. If it had only 2, then we could translate it a small distance away from the vertex where those two sides meet. That would give it no vertices on the sides. Contradiction. So it must have at least 3 vertices on the sides. Suppose it has only 3. Let vertex P not be on a side. Translate P and Q along the line PQ preserving their distance apart until P is on a side (whilst Q remains in the triangle). That does not change the area. Now translate PR slightly along the side. Then neither Q nor S will be on a side and we could increase the distance between PR and QS. Contradiction. So we may assume that P and Q lie on BC, R on CA and S on AB. Now translate PQ along BC until P is at B. This does not change the area.

Now triangles ASR and CQR are similar to ABC and their areas are AR²/AC² and CR²/AC². Since PQRS has maximal area, the position or R on AC must be such as to minimise AR² + CR² = AR² + (AC - AR)² = 2(AR - AC/2)² + AC²/2. Hence R must be the midpoint of AC and the area of PQRS must be 1/2.

Now if PQRS is any convex quadrilateral it contains a point X such that three of the vertices and X form a parallelogram. For take the vertex P such that the ray PQ does not meet the line RS and the ray PS does not meet the line QR. Then X to be the intersection of lines parallel to PQ through S and parallel to PS through Q. Now the triangle PQS has half the area of the parallelogram PQXS, which is at most half the area of the triangle. 

Problem B2

The sequence a₁, a₂, .... , a₃₉₇₂ includes each of the numbers from 1 to 1986 twice. Can the terms be rearranged so that there are just n numbers between the two n's?

Solution

Generalising from 1986 to n, no such arrangement is possible for n = 1 or 2 mod 4. Hence, in particular, it is not possible for 1986.

Color the positions alternately black and white. Odd numbers must occupy positions of the same color, even numbers occupy positions of the same color. But the total number of positions is even, so the number of each color is equal and hence there must be an even number of odd numbers.

Comment. This is known as Langford's problem. The fact that sequences exist for n = 0 or 3 mod 4 can be demonstrated explicitly. For n = 4m, we have 4m-4 ... even ... 2m 4m-2 2m-3 ... odd ... 1 4m-1 1 ... odd ... 2m-3 2m ... even ... 4m-4 4m 4m-3 ... odd ... 2m+1 4m-2 2m-2 ... even ... 2 2m-1 4m-1 2 ... even 2m-2 2m+1 ... odd ... 4m-3 2m-1 4m. For n = 4m-1 we have 4m-4 ... even ... 2m 4m-2 2m-3 ... odd 1 4m-1 1 ... odd 2m-3 2m ... even ... 4m-4 2m-1 4m-3 ... odd 2m+1 4m-2 2m-2 ... even ... 2 2m-1 4m-1 2 ... even ... 2m-2 2m+1 ... odd 4m-3. 

Problem B3

Each point in the plane is colored black or red. Show that we can find three points of the same color with each pair a distance 1 apart or three points of the same color with each pair a distance √3 apart.

Solution

Assume we cannot find an equilateral triangle side 1 with all vertices the same color. Take an equilateral triangle side 1. Two of the vertices must have different colors. Take a point a distance 2 from each. Then it must have a different color from one of them, so we have points AB a distance 2 apart, with A black and B white. wlog the midpoint M is white. The two points a distance 1 from B and M must both be black. But then they form an equilateral triangle side √3 with A.

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