14th Asian Pacific Mathematics Olympiad 2002 Problems



14th Asian Pacific Mathematics Olympiad 2002 Problems

A1.  xi are non-negative integers. Prove that x1! x2! ... xn! ≥ ( [(x1 + ... + xn)/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

xi are non-negative integers. Prove that x1! x2! ... xn! ≥ ( [(x1 + ... + xn)/n] ! )n (where [y] denotes the largest integer not exceeding y). When do you have equality?
Solution
Answer: Equality iff all xi 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 xi + xj the smallest value of xi! xj! is for xi and xj as nearly equal as possible.
If x1 + x2 + ... + xn = qn + r, where 0 < r < n, then we can reduce one or more xi 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 xi will not increase the lhs (by the result just proved). So it is sufficient to prove the inequality for all xi 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 15o.
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 = 60o + x, so y = 30o - x. So if x ≠ 15o, then TL ≠ RM. However, RST equilateral implies TL = RM, so x and hence also ∠BCQ = 15o.

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.


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