2nd Canadian Mathematical Olympiad Problems 1970

2nd Canadian Mathematical Olympiad Problems 1970

1.  Find all triples of real numbers such that the product of any two of the numbers plus the third is 2.

2.  The triangle ABC has angle A > 90^o. The altitude from A is AD and the altitude from B is BE. Show that BC + AD ≥ AC + BE. When do we have equality?

3.  Every ball in a collection is one of two colors and one of two weights. There is at least one of each color and at least one of each weight. Show that there are two balls with different color and different weight.

4.  Find all positive integers whose first digit is 6 and such that the effect of deleting the first digit is to divide the number by 25. Show that there is no positive integer such that the deletion of its first digit divides it by 35.

5.  A quadrilateral has one vertex on each side of a square side 1. Show that the sum of the squares of its sides is at least 2 and at most 4.

6.  Given three non-collinear points O, A, B show how to construct a circle center O such that the tangents from A and B are parallel.

7.  Given any sequence of five integers, show that three terms have sum divisible by 3.

8.  P lies on the line y = x and Q lies on the line y = 2x. Find the locus for the midpoint of PQ, if |PQ| = 4.

9.  Let a₁ = 0, a_2n+1 = a_2n = n. Let s(n) = a₁ + a₂ + ... + a_n. Find a formula for s(n) and show that s(m + n) = mn + s(m - n) for m > n.

10.  A monic polynomial p(x) with integer coefficients takes the value 5 at four distinct integer values of x. Show that it does not take the value 8 at any integer value of x. 

Solutions

Problem 1

Find all triples of real numbers such that the product of any two of the numbers plus the third is 2.

Solution

Answer: (1, 1, 1) and (-2, -2, -2).

Let the numbers by x, y, z, so x + yz = 2 (1), y + zx = 2 (2), z + xy = 2 (3). Subtracting (2) from (1) gives (x - y)(1 - z) = 0, so x = y or z = 1. If x = y, then x + xz = 2 and z + x² = 2, so x³ - 3x + 2 = 0. Factorising, (x - 1)²(x + 2) = 0, so x = 1, y = 1, z = 1 or x = -2, y = -2, z = -2. If z = 1, then x + y = 2 and xy = 1. Hence x² - 2x + 1 = 0. Factorising (x - 1)² = 0, so x = 1. 

Problem 2

The triangle ABC has angle A > 90^o. The altitude from A is AD and the altitude from B is BE. Show that BC + AD ≥ AC + BE. When do we have equality?

Solution

This inequality is awkward because it is weak. In fact BC + AD ≥ AC + AB.

We have BC² = AB² + AC² - 2AB.AC cos A, 2AD·BC = 2 area ABC = AB·AC sin A, so (BC + AD)² ≥ BC² + 2AD·BC ≥ AB² + AC² + 2AB·AC (sin A - cos A). So it is sufficient to show that sin A - cos A ≥ 1. Put A = x + 3π/4, where -π/4 ≤ x <= π/4. Then sin A - cos A = sin x cos 3π/4 + cos x sin 3π/4 - cos x cos 3π/4 + sin x sin 3π/4 = √2 cos x ≥ 1 with equality iff x = ±π/4 or A = π/2 or π. Thus BC + AD ≥ AB + AC with equality iff AD = 0 and ∠A = π, in other words iff B lies on the ray CA. Then AB ≥ BE with equality iff AE = 0, so BC + AD ≥ AC + BE with equality iff B = A. 

Problem 3

Every ball in a collection is one of two colors and one of two weights. There is at least one of each color and at least one of each weight. Show that there are two balls with different color and different weight.

Solution

Let the colors be red and blue and the weights 1 and 2. A ball can be R1, R2, B1 or B2 (with the obvious notation). If the result is false, then we cannot have R1 and B2 and we cannot have R2 and B1. That means all balls must be R1 and R2, or R1 and B1, or R2 and B2, or B1 and B2. But in the first case there is no blue ball, in the second case no weight 2, in the third case no weight 1, and in the last case no red ball. 

Problem 4

Find all positive integers whose first digit is 6 and such that the effect of deleting the first digit is to divide the number by 25. Show that there is no positive integer such that the deletion of its first digit divides it by 35.

Solution

Let the smaller number be N and have n digits. Then the larger number is N + 6.10ⁿ. So 25N = N + 6.10ⁿ, so 4N = 10ⁿ and N = 250...0. It is easily checked that 6250...0 (625 or 625 followed by one or more zeros) work.

In the second case we would have 35N = N + k.10ⁿ, where k is one of 1, 2, ... , 9. But then 17N = k·10ⁿ, which is not possible because the rhs is not divisible by 17. 

Problem 5

A quadrilateral has one vertex on each side of a square side 1. Show that the sum of the squares of its sides is at least 2 and at most 4.

Solution

Let the quadrilateral be ABCD. Suppose A is a distance w, 1-w from the two nearest vertices of the square. Define x, y, z similarly. Then the sum of the squares of the sides of the quadrilateral is w² + (1-w)² + x² + (1-x)² + y² + (1-y)² + z² + (1-z)². But w² + (1-w)² = 2(w - 1/2)² + 1/2 which is at least 1/2 and at most 1. Similarly for the other pairs of terms. 

Problem 6

Given three non-collinear points O, A, B show how to construct a circle center O such that the tangents from A and B are parallel.

Solution

Take D so that AOBD is a parallelogram. Let the line perpendicular to DO through O meet the circles on diameters AO and BO (again) at P and Q respectively. Then the required circle has radius OP.

Obviously PQ is perpendicular to AP and BQ, so it is sufficient to show that OP = OQ. Now ∠OPA = 90^o (AO diameter of circle OPA), and ∠POD = 90^o (by construction), so PA is parallel to OD, so OP = perpendicular distance of A from line OD. Similarly OQ = perpendicular distance of B from line OD. But the two distances are the same since AOBD is a parallelogram. Hence OP = OQ. 

Problem 7

Given any sequence of five integers, show that three terms have sum divisible by 3.

Solution

We can replace each integer by its remainder on division by 3. If three remainders are the same, then we can take those three integers. Similarly, if three remainders are different, then we can take those three integers. But one or the other must be true. 

Problem 8

P lies on the line y = x and Q lies on the line y = 2x. Find the equation for the locus for the midpoint of PQ, if |PQ| = 4.

Solution

Let P be (a, a), Q be (b, 2b). Then the midpoint is ( (a+b)/2, (a+2b)/2) = (x, y). We have (a - b)² + (a - 2b)² = 16, so 2a² - 6ab + 5b² = 16. We want to recast this as an equation connecting x and y. We have a = 2(2x - y), b = 2(y - x), so 8(2x - y)² - 24(2x - y)(y - x) + 20(y - x)² = 15, or 2(4x² - 4xy + y²) + 6(2x² - 3xy + y²) + 5(x² - 2xy + y²) = 4. So 25x² -36xy + 13y² = 4. The locus is a long, thin ellipse centered on the origin. 

Problem 9

Let a₁ = 0, a_2n+1 = a_2n = n. Let s(n) = a₁ + a₂ + ... + a_n. Find a formula for s(n) and show that s(m + n) = mn + s(m - n) for m > n.

Solution

We have 1 + 2 + ... + n = n(n+1)/2, so s(2n+1) = n(n+1), s(2n) = s(2n+1) - n = n². We may combine these into a single formula: s(n) = [n²/4].

m and n have the same parity. If they are both even, then s(m+n) - s(m-n) = (m+n)²/4 - (m-n)²/4 = mn. If they are both odd, then s(m+n) = (m+n)²/4 - 1/4, s(m-n) = (m-n)²/4 - 1/4 and s(m+n) - s(m-n) = mn. 

Problem 10

A monic polynomial p(x) with integer coefficients takes the value 5 at four distinct integer values of x. Show that it does not take the value 8 at any integer value of x.

Solution

We have p(x) - 5 = (x-a)(x-b)(x-c)(x-d) q(x) for some polynomial q(x) with integer coefficients and some integers a, b, c, d. Now suppose x is an integer. If x equals a, b, c, or d, then p(x) = 5, not 8. If x is not equal to a, b, c or d, then each of x-a, x-b, x-c, x-d is a distinct integer. So the smallest possible absolute value for the product (x-a)(x-b)(x-c)(x-d) is |1|.|-1|.|2|.|-2| = 4. The smallest possible value for |q(x)| is 1, so |p(x) - 5| is at least 4. So p(x) cannot take the value 8 (if x is an integer).