2nd Australian Mathematical Olympiad Problems 1981

A1.  Show that in any set of 27 distinct odd positive numbers less than 100 we can always find two with sum 102. How many sets of 26 odd positive numbers less 100 than can we find with no two having sum 102?

A2.  Given a real number 0 < k < 1, define p(x) = (x - k)(x - k²) ... (x - kⁿ)/( (x + k)(x + k²) ... (x + kⁿ) ). Show that if kⁿ⁺¹ ≤ x < 1, then p(x) < p(1).

A3.  ABC is isosceles and M is the midpoint of the base BC. A circle center M touches AB and AC. X lies on the segment AB and Y lies on the segment AC. Show that XY touches the circle iff 4 XB·CY = BC².

B1.  Let N = 915! + 916!/1! + 917!/2! + ... + 1980!/1065! Show that if 1065 < n < 1982, then n divides N.

B2.  A series of parallel lines in three directions divides the plane into equilateral triangles of side 1. Show that if there are vertices a distance h apart and vertices a distance k apart, then there are vertices a distance hk apart. Are there vertices a distance √1981 apart?

B3.  In an archery contest, the two contestants A and B start at x = 0 and walk towards the target at x = 1 together. Each chooses the distance from which he shoots. A has probability x² of hitting the target from x. B has probability x of hitting the target from x. If both hit the target, then the first to shoot wins. What is A's best strategy?

Solutions

 

Problem A1

Show that in any set of 27 distinct odd positive numbers less than 100 we can always find two with sum 102. How many sets of 26 odd positive numbers less 100 than can we find with no two having sum 102?

Answer

2²⁴

Solution

The odd numbers under 100 can be divided into 24 pairs with sum 102 (3+99, 5+97, ... , 49+53) and two other numbers 1, 51. If we take 27, then we must take at least 25 from the pairs and hence both members of at least one pair.

Such a set of 26 odd numbers must include 1 and 51 and one member of each pair. So there are 2²⁴ possibilities. 

Problem A2

Given a real number 0 < k < 1, define p(x) = (x - k)(x - k²) ... (x - kⁿ)/( (x + k)(x + k²) ... (x + kⁿ) ). Show that if kⁿ⁺¹ ≤ x < 1, then p(x) < p(1).

Solution

Note that 0 < kⁿ⁺¹ < kⁿ < k^n-1 < ... < k < 1, so there are various possible cases according to where x lies. If x = kⁱ, then lhs = 0, so the inequality is certainly true. If k < x < 1, then for any 0 < y < x we have (x-y)/(x+y) = 1 - 2y/(x+y) < 1 - 2y/(1+y) = (1-y)/(1+y). We may apply this with y = each kⁱ and multiply to get the result. So we may assume that for some i, we have kⁱ⁺¹ < x < kⁱ. Note that each of (x - k), (x - k²), ... , (x - kⁱ) is negative, whereas each of (x - kⁱ⁺¹), ... , (x - kⁿ⁺¹) is positive, so if i is odd, then lhs is negative and the result certainly holds. So assume i is even.

Now for j > i we have y = k^j < x, so (x-y)/(x+y) < (1-y)/(1+y) - as above. Hence ∏_i<j<n (x-k^j)/(x+k^j) < ∏_i<j<n (1-k^j)/(1+k^j). For the other terms, since there are an even number we may switch the order of x and k^j, so that ∏_1≤j≤i (x-k^j)/(x+k^j) = ∏_1≤j≤i (k^j-x)/(k^j+x). But each term (k^j-x)/(k^j+x) is now positive and > (k^j-kⁱ⁺¹)/(k^j+kⁱ⁺¹) and so ∏_1≤j≤i (x-k^j)/(x+k^j) < ∏_1≤j≤i (k^j-kⁱ⁺¹)/(k^j+kⁱ⁺¹) = ∏_1≤j≤i (1-k^j)/(1+k^j) - we can cancel a power of k from each term and then invert their order.

Finally, we multiply the two inequalities together to get the required result. 

Problem A3

ABC is isosceles and M is the midpoint of the base BC. A circle center M touches AB and AC. X lies on the segment AB and Y lies on the segment AC. Show that XY touches the circle iff 4 XB·CY = BC².

Solution

Suppose XY touches the circle at E. Let AB, AC touch the circle at D, F respectively. Note that by symmetry about AM, ∠BMD = ∠CMF. Also ∠DMX = ∠XME and ∠EMY = ∠YMF. Hence ∠BMD + ∠DMX + ∠YMF = 90^o. Hence ∠CYM = 90^o - ∠YMF = ∠BMX. But ∠B = ∠C, so triangles BXM and CMY are similar. Hence BX/BM = CM/CY, which is the required result.

Now suppose that the equality holds. Take XY' tangent to the circle, then the equality also holds with Y' in place of Y. Hence CY = CY' and so Y must coincide with Y'. Hence XY is tangent.

 Problem B1

Let N = 915! + 916!/1! + 917!/2! + ... + 1980!/1065! Show that if 1065 < n < 1982, then n divides N.

Solution

We have N/915! = 915C915 + 916C915 + ... + 1980C915. But 915C915 = 916C916 and 916C916 + 916C915 = 917C916. Then 917C916 + 917C915 = 918C916 and so on. So N/915! = 1981C916. Hence 916N = 1981!/1065! = 1981·1980 ... 1066.

Note that 2·916 = 1832 appears on the rhs, so for any n ≠ 1832 (and 1065 < n < 1982) we certainly have n a factor of N. But the rhs includes the factors 1603 = 7·229, 1600 = 4·400 and 1832. Hence we can divide through by 916 = 4·229 and conclude that N is also divisible by 1832. 

Problem B2

A series of parallel lines in three directions divides the plane into equilateral triangles of side 1. Show that if there are vertices a distance h apart and vertices a distance k apart, then there are vertices a distance hk apart. Are there vertices a distance √1981 apart?

Answer

Yes.

Solution

We may take the vertices to be the points m + nω in the complex plane, where ω is a complex cube root of 1, so 1 + ω + ω² = 0. Now suppose m + nω and M + Nω are two vertices, then (m + nω)(M + Nω) = mM + mNω + nMω + nNω² = (mN - nN) + (mN + nM - nN)ω, which is also a vertex. So if there are vertices a distance h apart, then by translation, we can take one of them to be the origin and the other to be z. Similarly, if there are vertices a distance k apart, then we can take one to be the origin and the other z'. But then the origin and the vertex zz' are a distance hk apart.

We can take ω = (-1 + i√3)/2, so |m + nω| = √((m - n/2)² + 3n²/4) = √(m² + n² - mn). Thus |3 + ω| = √7, |19 + 6ω| = √283. Their product gives the distance √1981.

Problem B3

In an archery contest, the two contestants A and B start at x = 0 and walk towards the target at x = 1 together. Each chooses the distance from which he shoots. A has probability x² of hitting the target from x. B has probability x of hitting the target from x. If both hit the target, then the first to shoot wins. What is A's best strategy?

Solution

If the first player to shoot hits the target then he wins. If he misses, then the other player waits until he is up against the target and cannot miss, so if the first player to shoot misses, then he loses. So if A shoots first at x, then he has a prob x² of winning, whereas if B shoots first at x, then A has a prob 1-x of winning. Put k = (√5-1)/2 = 0.618... . Then for x < k, we have x² < 1-x, so it is better for A if B shoots first. Thus whilst x < k, A does nothing and hopes that B shoots. Conversely, if x > k, then x² > 1-x, so it is worse for A if B shoots first. Moreover, the longer he waits the worse his position if B manages to shoot first.

Thus A's strategy is as follows. If B shoots at x < k, then he waits until he reaches the target before shooting. If B has not shot at x = k, then A shoots.