7th Iberoamerican Mathematical Olympiad Problems 1992

A1.  a_n is the last digit of 1 + 2 + ... + n. Find a₁ + a₂ + ... + a₁₉₉₂.

A2.  Let f(x) = a₁/(x + a₁) + a₂/(x + a₂) + ... + a_n/(x + a_n), where a_i are unequal positive reals. Find the sum of the lengths of the intervals in which f(x) ≥ 1.

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 (√3)/4.

B1.  Let a_n, b_n be two sequences of integers such that: (1) a₀ = 0, b₀ = 8; (2) a_n+2 = 2 a_n+1 - a_n + 2, b_n+2 = 2 b_n+1 - b_n, (3) a_n² + b_n² is a square for n > 0. Find at least two possible values for (a₁₉₉₂, b₁₉₉₂).

B2.  Construct a cyclic trapezium ABCD with AB parallel to CD, perpendicular distance h between AB and CD, and AB + CD = m.

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. 

Solutions

Problem A1

a_n 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 a_n 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 

Let f(x) = a₁/(x + a₁) + a₂/(x + a₂) + ... + a_n/(x + a_n), where a_i are unequal positive reals. Find the sum of the lengths of the intervals in which f(x) ≥ 1.

Answer

∑ a_i

Solution

wlog a₁ > a₂ > ... > a_n. The graph of each a_i/(x + a_i) is a rectangular hyberbola with asymptotes x = -a_i and y = 0. So it is not hard to see that the graph of f(x) is made up of n+1 strictly decreasing parts. For x < -a₁, f(x) is negative. For x ∈ (-a_i, -a_i+1), f(x) decreases from ∞ to -∞. Finally, for x > -a_n, f(x) decreases from ∞ to 0. Thus f(x) = 1 at n values b₁ < b₂ < ... < b_n, and f(x) ≥ 1 on the n intervals (-a₁,b₁), (-a₂,b₂), ... , (-a_n,b_n). So the sum of the lengths of these intervals is ∑ (a_i + b_i). We show that ∑ b_i = 0.

Multiplying f(x) = 1 by ∏(x + a_j) we get a polynomial of degree n:

∏(x + a_j) - ∑_i (a_i ∏_j≠i(x + a_j) ) = 0.

The coefficient of xⁿ is 1 and the coefficient of x^n-1 is ∑ a_j - ∑ a_i = 0. Hence the sum of the roots, which is ∑ b_i, is zero.

Thanks to Juan Ignacio Restrepo

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 (√3)/4.

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 √3 and the inradius is (√3)/3 = 1/√3, so PA² + 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 = 60^o, 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 + area of three equilateral triangles sides PA, PB, PC. Hence 3 area PQC = 2 area ABC - area ABC (PA²+PB²+PC²)/PA² = (3/4) area ABC = (3√3)/4. So area PQC = (√3)/4.

Thanks to Johann Peter Gustav Lejeune Dirichlet

Problem B1

Let a_n, b_n be two sequences of integers such that: (1) a₀ = 0, b₀ = 8; (2) a_n+2 = 2 a_n+1 - a_n + 2, b_n+2 = 2 b_n+1 - b_n, (3) a_n² + b_n² 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

a_n satisfies a standard linear recurrence relation with general solution a_n = n² + An + k. But a_n = 0, so k = 0. Hence a_n = n² + An. If you are not familiar with the general solution, then you can guess this solution and prove it by induction.

Similarly, b_n = Bn + 8. Hence a_n²+b_n² = 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.

Thanks to Juan Ignacio Restrepo