11th Canadian Mathematical Olympiad Problems 1979
1. If a > b > c > d is an arithmetic progression of positive reals and a > h > k > d is a geometric progression of positive reals, show that bc ≥ hk.
2. Show that two tetrahedra do not necessarily have the same sum for their dihedral angles.
3. Given five distinct integers greater than one, show that the sum of the inverses of the four lowest common multiples of the adjacent pairs is at most 15/16. [Two of the numbers are adjacent if none of the others lies between them.]
4. A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.
5. The lattice is the set of points (x, y) in the plane with at least one coordinate integral. Let f(n) be the number of walks of n steps along the lattice starting at the origin. Each step is of unit length from one intersection point to another. We only count walks which do not visit any point more than once. Find f(n) for n =1, 2, 3, 4 and show that 2n < f(n) ≤ 4·3n-1.
Solutions
Problem 1
If a > b > c > d is an arithmetic progression of positive reals and a > h > k > d is a geometric progression of positive reals, show that bc > hk.
Solution
Let a - d = 3t. Then bc = (d + 2t)(d + t) = d2 + 3td + 2t2. Also a/d = 1 + 3t/d, so hk = d2(1 + 3t/d) = d2 + 3td < bc.
Problem 2
Show that two tetrahedra do not necessarily have the same sum for their dihedral angles.
Solution
Take AB = BC = BD and AB normal to the plane BCD. Let M be the midpoint of CD. Then the sum of the dihedral angles is 90o + 90o + ∠AMB + ∠CBD. So as CBD tends to a straight line the sum tends to 180o + 90o + 180o = 450o. As CBD tends to zero the sum tends to 180o + 45o + 0o = 225o.
Problem 3
Given five distinct integers greater than one, show that the sum of the inverses of the four lowest common multiples of the adjacent pairs is at most 15/16. [Two of the numbers are adjacent if none of the others lies between.]
Solution
Let the numbers be a > b > c > d > e > 1. We need to show that 1/lcm(a, b) + 1/lcm(b, c) + 1/lcm(c, d) + 1/lcm(d, e) ≤ 15/16.
If the numbers are 2, 3, 4, 5, 6 then the sum is 1/6 + 1/12 + 1/20 + 1/30 = (10 + 5 + 3 + 2)/60 = 1/3 < 15/16. Otherwise we must have lcm(a, b) ≥ b ≥ 3, lcm(b, c) ≥ c ≥ 4, lcm(c, d) ≥ d ≥ 5, lcm(d, e) ≥ e ≥ 7, so sum ≤ 1/3 + 1/4 + 1/5 + 1/7 = (140 + 105 + 84 + 60)/420 = 389/420 < 15/16.
Problem 4
A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.
Solution
Let the edge be the circle C. Let the center be O. Let AB be a diameter of C. Let S be the midpoint of the arc AB. Let P be any point on the minor arc SB. Let C' be the circle on diameter OB and let its center (the midpoint of OB) be O'. The rabbit runs from S towards B around C. We show that the dog runs from O to B around the arc OB closer to the arc SB. Let OP intersect C' at Q. Then angle QO'O = 2 x angle POS, so the distance along the circle C from S to P equals the distance along the circle C' from O to Q.
Problem 5
The lattice is the set of points (x, y) in the plane with at least one coordinate integral. Let f(n) be the number of walks of n steps along the lattice starting at the origin. Each step is of unit length from one intersection point to another. We only count walks which do not visit any point more than once. Find f(n) for n =1, 2, 3, 4 and show that 2n < f(n) ≤ 4·3n-1.
Solution
f(1) = 4 (any of 4 directions), f(2) = 12 (any of 3 directions for step 2), f(3) = 36 (any of 3 directions for step 3), f(4) = 100 (any of 3 for step 4, except for the 8 paths after step 3 which go around 3 sides of a square - only 2 possible directions for each of those).
You cannot go back the way you came, so for steps 2 to n there is a choice of at most 3 directions. Hence f(n) <= 4.3n-1. If you always go N or E, then you can never visit a point more than once, so f(n) > 2n.
Solutions
Problem 1
If a > b > c > d is an arithmetic progression of positive reals and a > h > k > d is a geometric progression of positive reals, show that bc > hk.
Solution
Let a - d = 3t. Then bc = (d + 2t)(d + t) = d2 + 3td + 2t2. Also a/d = 1 + 3t/d, so hk = d2(1 + 3t/d) = d2 + 3td < bc.
Problem 2
Show that two tetrahedra do not necessarily have the same sum for their dihedral angles.
Solution
Take AB = BC = BD and AB normal to the plane BCD. Let M be the midpoint of CD. Then the sum of the dihedral angles is 90o + 90o + ∠AMB + ∠CBD. So as CBD tends to a straight line the sum tends to 180o + 90o + 180o = 450o. As CBD tends to zero the sum tends to 180o + 45o + 0o = 225o.
Problem 3
Given five distinct integers greater than one, show that the sum of the inverses of the four lowest common multiples of the adjacent pairs is at most 15/16. [Two of the numbers are adjacent if none of the others lies between.]
Solution
Let the numbers be a > b > c > d > e > 1. We need to show that 1/lcm(a, b) + 1/lcm(b, c) + 1/lcm(c, d) + 1/lcm(d, e) ≤ 15/16.
If the numbers are 2, 3, 4, 5, 6 then the sum is 1/6 + 1/12 + 1/20 + 1/30 = (10 + 5 + 3 + 2)/60 = 1/3 < 15/16. Otherwise we must have lcm(a, b) ≥ b ≥ 3, lcm(b, c) ≥ c ≥ 4, lcm(c, d) ≥ d ≥ 5, lcm(d, e) ≥ e ≥ 7, so sum ≤ 1/3 + 1/4 + 1/5 + 1/7 = (140 + 105 + 84 + 60)/420 = 389/420 < 15/16.
Problem 4
A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.
Solution
Let the edge be the circle C. Let the center be O. Let AB be a diameter of C. Let S be the midpoint of the arc AB. Let P be any point on the minor arc SB. Let C' be the circle on diameter OB and let its center (the midpoint of OB) be O'. The rabbit runs from S towards B around C. We show that the dog runs from O to B around the arc OB closer to the arc SB. Let OP intersect C' at Q. Then angle QO'O = 2 x angle POS, so the distance along the circle C from S to P equals the distance along the circle C' from O to Q.
Problem 5
The lattice is the set of points (x, y) in the plane with at least one coordinate integral. Let f(n) be the number of walks of n steps along the lattice starting at the origin. Each step is of unit length from one intersection point to another. We only count walks which do not visit any point more than once. Find f(n) for n =1, 2, 3, 4 and show that 2n < f(n) ≤ 4·3n-1.
Solution
f(1) = 4 (any of 4 directions), f(2) = 12 (any of 3 directions for step 2), f(3) = 36 (any of 3 directions for step 3), f(4) = 100 (any of 3 for step 4, except for the 8 paths after step 3 which go around 3 sides of a square - only 2 possible directions for each of those).
You cannot go back the way you came, so for steps 2 to n there is a choice of at most 3 directions. Hence f(n) <= 4.3n-1. If you always go N or E, then you can never visit a point more than once, so f(n) > 2n.
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