16th Canadian Mathematical Olympiad Problems 1984

16th Canadian Mathematical Olympiad Problems 1984

1.  Show that the sum of 1984 consecutive positive integers cannot be a square.

2.  You have keyring with n identical keys. You wish to color code the keys so that you can distinguish them. What is the smallest number of colors you need? [For example, you could use three colors for eight keys: R R R R G B R R. Starting with the blue key and moving away from the green key uniquely distinguishes each of the red keys.]

3.  Show that there are infinitely many integers which have no zeros and which are divisible by the sum of their digits.

4.  An acute-angled triangle has unit area. Show that there is a point inside the triangle which is at least 2/(3^3/4) from any vertex.

5.  Given any seven real numbers show we can select two, x and y, such that 0 ≤ (x - y)/(1 + xy) ≤ 1/√3. 

Solutions

Problem 1

Show that the sum of 1984 consecutive positive integers cannot be a square.

Solution

(n+1) + (n+2) + ... + (n+1984) = (n+1984)(n+1985)/2 - n(n+1)/2 = 1984n + 1984.1985/2 = 992(2n + 1985). But 992 = 31.32 and (2n+1985) is odd, so the highest power of 2 dividing 992(2n+1985) is 32 = 2⁵. Hence it is not a square. 

Problem 2

You have keyring with n identical keys. You wish to color code the keys so that you can distinguish them. What is the smallest number of colors you need? [For example, you could use three colors for eight keys: R R R R G B R R. Starting with the blue key and moving away from the green key uniquely distinguishes each of the red keys.]

Solution

Answer: n = 2, n > 5 two colors. n = 3, 4 or 5, three colors.

For n even: take m red keys, then two blue keys, then n-m-3 red keys, then one blue key. Provided m > n-m-3 this uniquely identifies each key. So this works for n > 5 (eg take m = 1). For n = 1, only one color is needed. For n = 2, two colors are needed. For n = 3, we need three colors, because if we only use two we cannot distinguish the keys with same color. Similarly for n = 4, we need three colors (and three are sufficient eg RRGB - one red is next to blue and the other is not). We also need three for n = 5, although that takes a little more effort to demonstrate. 

Problem 3

Show that there are infinitely many integers which have no zeros and which are divisible by the sum of their digits.

Solution

We claim that the number a_n with 3ⁿ digits, all 1, is divisible by 3ⁿ. This is an easy induction. It is true for n = 1: 111 = 3·37. But a_n+1 = a_n x 10...010...01 and 10...010...01 has sum of digits 3 and hence is divisible by 3, so a_n+1 is divisible by one more power of 3 than a_n. 

Problem 4

An acute-angled triangle has unit area. Show that there is a point inside the triangle which is at least 2/(3^3/4) from any vertex.

Solution

Let the triangle be ABC. Let R be the circumradius and O the circumcenter. Since the triangle is acute-angled, O lies inside the triangle. We show that it is the desired point. Angle AOB = 2 angle C, so AB = 2R sin C. Similarly, BC = 2R sin A. Hence area ABC = (1/2) AB·BC sin B = 2R² sin A sin B sin C = 1. So it is sufficient to show that sin A sin B sin C ≤ 3^3/2/8. By the AM/GM we have sin A sin B sin C ≤ (sin A + sin B + sin C)³/27. But sin x is convex for 0 <= A, B, C ≤ 180^o, so sin A + sin B + sin C ≤ 3 sin(A/3 + B/3 + C/3). One can also prove this directly by noting that sin A + sin B = sin( (A+B)/2 + (A-B)/2 ) + sin( (A+B)/2 - (A-B)/2) = 2 sin(A+B)/2 cos(A-B)/2 ≤ 2 sin(A+B)/2 with equality iff A = B. Thus sin A sin B sin C ≤ sin³ 60^o = 3^3/2/8 as required. 

Problem 5

Given any seven real numbers show we can select two, x and y, such that 0 ≤ (x - y)/(1 + xy) ≤ 1/√3.

Solution

Take the arctan of each number. The resulting 7 numbers lie in the open interval (-π/2, π/2). Hence there must be two with difference < π/6. Suppose they are a >= b. Then 0 ≤ tan(a - b) < 1/√3. So if the corresponding original numbers are x = tan a, y = tan b, then x and y satisfy the required inequalities.