24th Canadian Mathematical Olympiad Problems 1992
1. Show that n! is divisible by (1 + 2 + ... + n) iff n+1 is not an odd prime.
2. Show that x(x - z)2 + y(y - z)2 ≥ (x - z)(y - z)(x + y - z) for all non-negative reals x, y, z. When does equality hold?
3. ABCD is a square. X is a point on the side AB, and Y is a point on the side CD. AY meets DX at R, and BY meets CX at S. How should X and Y be chosen to maximise the area of the quadrilateral XRYS?
4. Find all real solutions to x2(x + 1)2 + x2 = 3(x + 1)2.
5. There are 2n+1 cards. There are two cards with each integer from 1 to n and a joker. The cards are arranged in a line with the joker in the center position (with n cards each side of it). For which n < 11 can we arrange the cards so that the two cards with the number k have just k-1 cards between them (for k = 1, 2, ... , n)?
Solutions
Problem 1
Show that n! is divisible by (1 + 2 + ... + n) iff n+1 is not an odd prime.
Solution
For n = 1, we have to show that 1 divides 1, which is obvious. So assume n > 1. We have 1 + 2 + ... + n = n(n+1)/2. So we have to show that n+1 divides (n-1)! for n+1 not a prime. But that is obvious except possibly in the case n+1 = p2. However, in that case p and 2p < n, so p2 still divides (n-1)! . [The converse, that if n+1 is an odd prime then it does not divide n!, is obvious.]
Problem 2
Show that x(x - z)2 + y(y - z)2 >= (x - z)(y - z)(x + y - z) for all non-negative reals x, y, z. When does equality hold?
Solution
Answer: equality iff x = y = z or x = y and z = 0.
Put u = x - z, v = y - z. Then x(x - z)2 + y(y - z)2 - (x - z)(y - z)(x + y - z) = (u + z)u2 + (v + z)v2 - uv(u + v + z) = (u + v)(u - v)2 + z(u2 - uv + v2). If u and v are both non-negative, then this is obviously non-negative (since u2 - uv + v2 = (u - v)2 + uv) and zero only if u = v = 0 or u = v and z = 0 (in other words, x = y = z or x = y and z = 0).
If just one of u, v is negative, then the rhs of the original expression is negative or zero and the rhs is positive, so strict inequality holds.
If both u and v are negative, then we may assume x + y > z (otherwise the lhs of the original expression is positive and the rhs negative or zero). But in that case z > -(u + v) and u2 - uv + v2 > (u - v)2, so z(u2 - uv + v2) > -(u + v)(u - v)2.
Problem 3
ABCD is a square. X is a point on the side AB, and Y is a point on the side CD. AY meets DX at R, and BY meets CX at S. How should X and Y be chosen to maximise the area of the quadrilateral XRYS?
Solution
Answer: the maximum area of (area ABCD)/4 is achieved whenever XY is parallel to AD.
Area XRYS = area XRS + area YRS. Area XRS = (XR/XD) area XSD = (XR/XD) (XS/XC) area XDC = (XR/XD) (XS/XC) (area ABCD)/2. Similarly, area YRS = (YS/BY) (RY/AY) (area ABCD)/2. But XR/XD = 1 - YR/AY, XS/XC = 1 - YS/BY. So if we put XR/XD = (1/2 + x), XS/XC = (1/2 + y), then area XRYS = (1/4 + xy) area ABCD.
But triangles ARX and YRD are similar, so XR > DR iff AX > YD. Similarly, XS > SC iff BX > CY. Hence XR/XD > 1/2 iff XS/XC < 1/2 and XR/XD = 1/2 iff XS/XC = 1/2. Thus either x = y = 0 or xy is negative. If x = y = 0, then XY is parallel to AD.
Problem 4
Find all real solutions to x2(x + 1)2 + x2 = 3(x + 1)2.
Solution
Expanding, x4 + 2x3 - x2 - 6x - 3 = 0. Factorising, (x2 - x - 1)(x2 + 3x + 3) = 0. The first quadratic has real roots x = 1/2 ± (√5)/2. The second quadratic is (x + 3/2)2 + 3/4, so has no real roots.
Problem 5
There are 2n+1 cards. There are two cards with each integer from 1 to n and a joker. The cards are arranged in a line with the joker in the center position (with n cards each side of it). For which n < 11 can we arrange the cards so that the two cards with the number k have just k-1 cards between them (for k = 1, 2, ... , n)?
Solution
Consider the number of pairs (i, j) where i lies between the two j's or j lies between the two i's (if both i's or both j's do then we count both). The number must be 0 or 2, because we have either i ... i ... j ... j or i ... j ... i ... j or j ... i ... i ... j or i ... j ... j ... i or j ... i ... j ... i or j ... j ... i ... i. So the sum over all such pairs is even. But the sum is also 1 + 2 + ... + (n-1) = (n-1)n/2 less the number of pairs which straddle the joker. If n is even, then an even number of pairs must straddle the joker and if n is odd an odd number of pairs. So if n = 4m+1, then the sum is even less odd = odd, and so 4m+1 is impossible. Similarly, if n = 4m+2, then the sum is odd less even = odd, so 4m+2 is impossible. Thus we must have n = 0 or 3 mod 4. In particular, n = 1, 2, 5, 6, 9, 10 are impossible. We show below that n = 3, 4, 7, 8 are possible.
1 1 3 J 2 3 2
1 1 3 4 J 3 2 4 2
1 1 2 5 2 6 7 J 5 3 4 6 3 7 4
1 1 2 6 2 5 7 8 J 6 5 3 4 7 3 8 4
Comment. See also Chinese 86/B2. This is a variant on Langford's problem (which is the same except that there is no joker and we require k numbers between the two k's (not k-1).
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