8th Asian Pacific Mathematics Olympiad 1996 Problems

8th Asian Pacific Mathematics Olympiad 1996 Problems

A1.  ABCD is a fixed rhombus. P lies on AB and Q on BC, so that PQ is perpendicular to BD. Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD. The distance between PQ and P'Q' is more than BD/2. Show that the perimeter of the hexagon APQCQ'P' depends only on the distance between PQ and P'Q'.

A2.  Prove that (n+1)^mn^m ≥ (n+m)!/(n-m)! ≥ 2^mm! for all positive integers n, m with n ≥ m.

A3.  Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.

A4.  For which n in the range 1 to 1996 is it possible to divide n married couples into exactly 17 single sex groups, so that the size of any two groups differs by at most one.

A5.  A triangle has side lengths a, b, c. Prove that √(a + b - c) + √(b + c - a) + √(c + a - b) ≤ √a + √b + √c. When do you have equality?

ABCD is a fixed rhombus. P lies on AB and Q on BC, so that PQ is perpendicular to BD. Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD. The distance between PQ and P'Q' is more than BD/2. Show that the perimeter of the hexagon APQCQ'P' depends only on the distance between PQ and P'Q'.

Solution

BPQ and DQ'P' are similar. Let PQ meet BD at X and P'Q' meet BD at Y. XY is fixed, so BX + DY is fixed. Hence also, BP + DQ' and BQ + DP' and PQ + P'Q' are fixed. So PQ + P'Q' - BP - BQ - DP' - DQ' is fixed, so PQ + P'Q' + (AB - BP) + (BC - BQ) + (CD - DP') + (DA - DQ') is fixed, and that is the perimeter of the hexagon.

Prove that (n+1)^mn^m ≥ (n+m)!/(n-m)! ≥ 2^mm! for all positive integers n, m with n ≥ m.

Solution

For any integer k ≥ 1, we have (n + k)(n - k + 1) = n² + n - k² + k ≤ n(n + 1). Taking the product from k = 1 to m we get (n + m)!/(n - m)! ≤ (n + 1)^mn^m.

For k = 1, 2, ... , m, we have n ≥ k and hence n + k ≥ 2k. Taking the product from k = 1 to m, we get (n + m)!/(n - m)! ≥ 2^mm! .

Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.

Solution

Take the points as A, B, C, D in that order. Let I be the incenter of ABC. The ray CI bisects the angle ACB, so it passes through M, the midpoint of the arc AB. Now ∠MBI = ∠MBA + ∠IBA = ∠MCA + ∠IBA = (∠ACB + ∠ABC)/2 = 90^o - (∠CAB) /2 = 90^o - ∠CMB/2 = 90^o - ∠IMB/2. So the bisector of ∠IMB is perpendicular to IB. Hence MB = MI. Let J be the incenter of ABD. Then similarly MA = MJ. But MA = MB, so the four points A, B, I, J are concyclic (they lie on the circle center M). Hence ∠BIJ = 180^o - ∠BAJ = 180^o - ∠BAD/2.

Similarly, if K is the incenter of ADC, then ∠BJK = 180^o - ∠BDC/2. Hence ∠IJK = 360^o - ∠BIJ - ∠BJK = (180^o - ∠BIJ) + (180^o - ∠BJK) = (∠BAD + ∠BDC)/2 = 90^o. Similarly, the other angles of the incenter quadrilateral are 90^o, so it is a rectangle.

For which n in the range 1 to 1996 is it possible to divide n married couples into exactly 17 single sex groups, so that the size of any two groups differs by at most one.

Solution

Answer: 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 27, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 45, 46, 47, 48, 54, 55, 56, 63, 64, 72.

If n = 17k, then the group size must be 2k. Hence no arrangement is possible, because one sex has at most 8 groups and 8.2k < n.

If 2n = 17k+h with 0 < h < 17, then the group size must be k or k+1. One sex has at most 8 groups, so 8(k+1) ≥ n. Hence 16k + 16 ≥ 17k + h, so 16 - h ≥ k (*). We also require that 9k ≤ n. Hence 18k < 2n = 17k + h, so k ≤ h (**). With (*) this implies that k ≤ 8. So n ≤ 75.

Each group has at least one person, so we certainly require n ≥ 9 and hence k ≥ 1. It is now easiest to enumerate. For k = 1, we can have h = 1, 3, ... 15, giving n = 9-16. For k = 2, we can have h = 2, 4, ... 14, giving n = 18-24. For k = 3, we can have h = 3, 5, ... 13, giving n = 27-32. For k = 4, we can have h = 4, 6, ... 12, giving n = 36-40. For k = 5 we can have h = 5, 7, 9, 11, giving n = 45-48. For k = 6, we can have h = 6, 8, 10, giving n = 54, 55, 56. For k = 7, we can have h = 7, 9, giving n = 63, 64. For k = 8, we can have h = 8, giving n = 72.

A triangle has side lengths a, b, c. Prove that √(a + b - c) + √(b + c - a) + √(c + a - b) ≤ √a + √b + √c. When do you have equality?

Solution

Let A² = b + c - a, B² = c + a - b, C² = a + b - c. Then A² + B² = 2c. Also A = B iff a = b. We have (A - B)² ≥ 0, with equality iff A = B. Hence A² + B² ≥ 2AB and so 2(A² + B²) ≥ (A + B)² or 4c ≥ (A + B)² or 2√c ≥ A + B, with equality iff A = B. Adding the two similar relations we get the desired inequality, with equality iff the triangle is equilateral.