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.