14th Asian Pacific Mathematics Olympiad 2002 Problems

14th Asian Pacific Mathematics Olympiad 2002 Problems

A1.  x_i are non-negative integers. Prove that x₁! x₂! ... x_n! ≥ ( [(x₁ + ... + x_n)/n] ! )ⁿ (where [y] denotes the largest integer not exceeding y). When do you have equality?

A2.  Find all pairs m, n of positive integers such that m² - n divides m + n² and n² - m divides m² + n.

A3.  ABC is an equilateral triangle. M is the midpoint of AC and N is the midpoint of AB. P lies on the segment MC, and Q lies on the segment NB. R is the orthocenter of ABP and S is the orthocenter of ACQ. The lines BP and CQ meet at T. Find all possible values for angle BCQ such that RST is equilateral.

A4.  The positive reals a, b, c satisfy 1/a + 1/b + 1/c = 1. Prove that √(a + bc) + √(b + ca) + √(c + ab) ≥ √(abc) + √a + √b + √c.

A5.  Find all real-valued functions f on the reals which have at most finitely many zeros and satisfy f(x⁴ + y) = x³f(x) + f(f(y)) for all x, y. 

14th APMO 2002 Problem 1

x_i are non-negative integers. Prove that x₁! x₂! ... x_n! ≥ ( [(x₁ + ... + x_n)/n] ! )ⁿ (where [y] denotes the largest integer not exceeding y). When do you have equality?

Solution

Answer: Equality iff all x_i equal.

For given 2m the largest binomial coefficient is (2m)!/(m! m!) and for 2m+1 the largest binomial coefficient is (2m+1)!/( m! (m+1)!). Hence for fixed x_i + x_j the smallest value of x_i! x_j! is for x_i and x_j as nearly equal as possible.

If x₁ + x₂ + ... + x_n = qn + r, where 0 < r < n, then we can reduce one or more x_i to reduce the sum to qn. This will not affect the rhs of the inequality in the question, but will reduce the lhs. Equalising the x_i will not increase the lhs (by the result just proved). So it is sufficient to prove the inequality for all x_i equal. But in this case it is trivial since k! = k! . 

14th APMO 2002 Problem 2

Find all pairs m, n of positive integers such that m² - n divides m + n² and n² - m divides m² + n.

Solution

Assume n ≥ m.

(m+1)² - m = (m+1) + m². Clearly n² increases faster than n, so n² - m > n + m² for n > m+1 and hence there are no solutions with n > m+1. It remains to consider the two cases n = m and n = m+1.

Suppose n = m. Then we require that n² - n divides n² + n. If n > 3, then n² > 3n, so 2(n² - n) > n² + n. Obviously n² - n < n² + n, so if n > 3, then n² - n cannot divide n² + n. It is easy to check that the only solutions (with n = m) less than 3 are n = 2 and n = 3.

Finally suppose n = m+1. We require m² - m - 1 divides m² + 3m + 1. If m >= 6, then m(m - 5) > 3, so 2(m² - m - 1) > m² + 3m + 1. Obviously m² - m - 1 < m² + 3m + 1, so m² - m - 1 cannot divide m² + 3m + 1 for m >= 6. Checking the smaller values, we find the solutions less than 6 are m = 1 and m = 2.

Summarising, the only solutions are: (n, m) = (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2). 

14th APMO 2002 Problem 3

ABC is an equilateral triangle. M is the midpoint of AC and N is the midpoint of AB. P lies on the segment MC, and Q lies on the segment NB. R is the orthocenter of ABP and S is the orthocenter of ACQ. The lines BP and CQ meet at T. Find all possible values for angle BCQ such that RST is equilateral.

Answer 15^o.

Solution

 

Suppose CP < BQ. Since R is the intersection of BM and the perpendicular from P to AB, and S is the intersection of CN and the perpendicular from Q to AC, we have MR > NS. Hence (treating BC as horizontal), R is below S. But T must be to the right of the midpoint of BC. Hence T is to the right of the perpendicular bisector of RS, so RST cannot be equilateral. Contradiction. Similarly if CP > BQ. So CP = BQ. 

Let L be the midpoint of BC. Put ∠CBP = x and ∠RAM = y. So RM = AM tan y, TL = BL tan x = AM tan x. But ∠APB = 60^o + x, so y = 30^o - x. So if x ≠ 15^o, then TL ≠ RM. However, RST equilateral implies TL = RM, so x and hence also ∠BCQ = 15^o.

14th APMO 2002 Problem 4

The positive reals a, b, c satisfy 1/a + 1/b + 1/c = 1. Prove that √(a + bc) + √(b + ca) + √(c + ab) ≥ √(abc) + √a + √b + √c.

Solution

Thanks to Suat Namli for this.

Multiplying by √(abc), we have √(abc) = √(ab/c) + √(bc/a) + √(ca/b). So it is sufficient to prove that √(c + ab) ≥ √c + √(ab/c).

Squaring, this is equivalent to c + ab ≥ c + ab/c + 2√(ab) or c + ab ≥ c + ab(1 - 1/a - 1/b) + 2√(ab) or a + b >= 2√(ab) or (√a - √b)² ≥ 0.