7th Asian Pacific Mathematics Olympiad 1995 Problems

7th Asian Pacific Mathematics Olympiad 1995 Problems

A1.  Find all real sequences x₁, x₂, ... , x₁₉₉₅ which satisfy 2√(x_n - n + 1) ≥ x_n+1 - n + 1 for n = 1, 2, ... , 1994, and 2√(x₁₉₉₅ - 1994) ≥ x₁ + 1.

A2.  Find the smallest n such that any sequence a₁, a₂, ... , a_n whose values are relatively prime square-free integers between 2 and 1995 must contain a prime. [An integer is square-free if it is not divisible by any square except 1.]

A3.  ABCD is a fixed cyclic quadrilateral with AB not parallel to CD. Find the locus of points P for which we can find circles through AB and CD touching at P.

A4.  Take a fixed point P inside a fixed circle. Take a pair of perpendicular chords AC, BD through P. Take Q to be one of the four points such that AQBP, BQCP, CQDP or DQAP is a rectangle. Find the locus of all possible Q for all possible such chords.

A5.  f is a function from the integers to {1, 2, 3, ... , n} such that f(A) and f(B) are unequal whenever A and B differ by 5, 7 or 12. What is the smallest possible n? 

Solution

A1. Find all real sequences x₁, x₂, ... , x₁₉₉₅ which satisfy 2√(x_n - n + 1) ≥ x_n+1 - n + 1 for n = 1, 2, ... , 1994, and 2√(x₁₉₉₅ - 1994) ≥ x₁ + 1.

Solution

Answer: the only such sequence is 1, 2, 3, ... , 1995.

Put x₁₉₉₅ = 1995 + k. We show by induction (moving downwards from 1995) that x_n ≥ n + k. For suppose x_n+1 ≥ n + k + 1, then 4(x_n - n + 1) ≥ (x_n+1- n + 1)² ≥ (k+2)² ≥ 4k + 4, so x_n ≥ n + k. So the result is true for all n ≥ 1. In particular, x₁ ≥ 1 + k. Hence 4(x₁₉₉₅ - 1994) = 4(1 + k) ≥ (2 + k)² = 4 + 4k + k², so k² ≤ 0, so k = 0.

Hence also x_n ≥ n for n = 1, 2, ... , 1994. But now if x_n = n + k, with k > 0, for some n < 1995, then the same argument shows that x₁ ≥ 1 + k and hence 4 = 4(x₁₉₉₅ - 1994) ≥ (x₁ + 1)² ≥ (2 + k)² = 4 + 4k + k² > 4. Contradiction. Hence x_n = n for all n ≤ 1995.

Find the smallest n such that any sequence a₁, a₂, ... , a_n whose values are relatively prime square-free integers between 2 and 1995 must contain a prime. [An integer is square-free if it is not divisible by any square except 1.]

Solution

Answer: n = 14.

We can exhibit a sequence with 13 terms which does not contain a prime: 2·101 = 202, 3·97 = 291, 5·89 = 445, 7·83 = 581, 11·79 = 869, 13·73 = 949, 17·71 = 1207, 19·67 = 1273, 23·61 = 1403, 29·59 = 1711, 31·53 = 1643, 37·47 = 1739, 41·43 = 1763. So certainly n ≥ 14.

If there is a sequence with n ≥ 14 not containing any primes, then since there are only 13 primes not exceeding 41, at least one member of the sequence must have at least two prime factors exceeding 41. Hence it must be at least 43·47 = 2021 which exceeds 1995. So n =14 is impossible.

ABCD is a fixed cyclic quadrilateral with AB not parallel to CD. Find the locus of points P for which we can find circles through AB and CD touching at P.

Solution

Answer: Let the lines AB and CD meet at X. Let R be the length of a tangent from X to the circle ABCD. The locus is the circle center X radius R. [Strictly you must exclude four points unless you allow the degenerate straight line circles.]

Let X be the intersection of the lines AB and CD. Let R be the length of a tangent from X to the circle ABCD. Let C₀ be the circle center X radius R. Take any point P on C₀. Then considering the original circle ABCD, we have that R² = XA·XB = XC·XD, and hence XP² = XA·XB = XC·XD.

If C₁ is the circle through C, D and P, then XC.XD = XP², so XP is tangent to the circle C₁. Similarly, the circle C₂ through A, B and P is tangent to XP. Hence C₁ and C₂ are tangent to each other at P. Note that if P is one of the 4 points on AB or CD and C₀, then this construction does not work unless we allow the degenerate straight line circles AB and CD.

So we have established that all (or all but 4) points of C₀ lie on the locus. But for any given circle through C, D, there are only two circles through A, B which touch it (this is clear if you consider how the circle through A, B changes as its center moves along the perpendicular bisector of AB), so there are at most 2 points on the locus lying on a given circle through C, D. But these are just the two points of intersection of the circle with C₀. So there are no points on the locus not on C₀. 

Take a fixed point P inside a fixed circle. Take a pair of perpendicular chords AC, BD through P. Take Q to be one of the four points such that ASCQ, ASDQ, BSCQ or BSDQ is a rectangle. Find the locus of all possible Q for all possible such chords.

Solution

Let O be the center of the fixed circle and let X be the center of the rectangle ASCQ. By the cosine rule we have OQ² = OX² + XQ² - 2·OX·XQ cos θ and OP² = OX² + XP² - 2·OX·XP cos(θ+π), where θ is the angle OXQ. But cos(θ+π) = -cos θ, so OQ² + OP²= 2OX² + 2XQ². But since X is the center of the rectangle XQ = XC and since X is the midpoint of AC, OX is perpendicular to AC and hence XO² + XC² = OC². So OQ² = 2OC² - OP². But this quantity is fixed, so Q must lie on the circle center O radius √(2R² - OP²), where R is the radius of the circle.

Conversely, it is easy to see that all points on this circle can be reached. For given a point Q on the circle radius √(2R² - OP²) let X be the midpoint of PQ. Then take the chord AC to have X as its midpoint.

f is a function from the integers to {1, 2, 3, ... , n} such that f(A) and f(B) are unequal whenever A and B differ by 5, 7 or 12. What is the smallest possible n?

Solution

Answer: n = 4.

Each pair of 0, 5, 12 differ by 5, 7 or 12, so f(0), f(5), f(12) must all be different, so n ≥ 3.

We can exhibit an f with n = 4. Define f(m) = 1 for m = 1, 3, 5, 7, 9, 11 (mod 24), f(m) = 2 for m = 2, 4, 6, 8, 10, 12 (mod 24), f(m) = 3 for m = 13, 15, 17, 19, 21, 23 (mod 24), f(m) = 4 for m = 14, 16, 18, 20, 22, 0 (mod 24).