1st Canadian Mathematical Olympiad Problems 1969

1st Canadian Mathematical Olympiad Problems 1969

1.  a, b, c, d, e, f are reals such that a/b = c/d = e/f; p, q, r are reals, not all zero; and n is a positive integer. Show that (a/b)ⁿ = (p aⁿ + q cⁿ + r eⁿ)/(p bⁿ + q dⁿ + r fⁿ ).

2.  If x is a real number not less than 1, which is larger: √(x+1) - √x or √x - √(x-1)?

3.  A right-angled triangle has longest side c and other side lengths a and b. Show that a + b ≤ c√2. When do we have equality?

4.  The sum of the distances from a point inside an equilateral triangle of perimeter length p to the sides of the triangle is s. Show that s √12 = p.

5.  ABC is a triangle with |BC| = a, |CA| = b. Show that the length of the angle bisector of C is (2ab cos C/2)/(a + b).

6.  Find 1.1! + 2.2! + ... + n.n! .

7.  Show that there are no integer solutions to a² + b² = 8c + 6.

8.  f is a function defined on the positive integers with integer values. Given that (1) f(2) = 2, (2) f(mn) = f(m) f(n) for all m,n, and (3) f(m) > f(n) for all m, n such that m > n, show that f(n) = n for all n.

9.  Show that the shortest side of a cyclic quadrilateral with circumradius 1 is at most √2.

10.  P is a point on the hypoteneuse of an isosceles, right-angled triangle. Lines are drawn through P parallel to the other two sides, dividing the triangle into two smaller triangles and a rectangle. Show that the area of one of these component figures is at least 4/9 of the area of the original triangle.

Solutions

Problem 1

a, b, c, d, e, f are reals such that a/b = c/d = e/f; p, q, r are reals, not all zero; and n is a positive integer. Show that (a/b)ⁿ = (p aⁿ + q cⁿ + r eⁿ)/(p bⁿ + q dⁿ + r fⁿ ).

Solution

If a/b = c/d, then aⁿ/bⁿ = cⁿ/dⁿ. So it is sufficient to show that if A/B = C/D then (pA + qC)/(pB + qD) = A/B, which is trivially true. [Apply it once with A = aⁿ, B = bⁿ, C = cⁿ, D = dⁿ, then again with p = 1.] 

Problem 2

If x is a real number not less than 1, which is larger: √(x+1) - √x or √x - √(x-1)?

Solution

Answer: √x - √(x-1) is larger.

4x² - 4x + 1 > 4x² - 4x. Taking the positive square root of each side gives 2x - 1> 2√(x²-x). Hence x² + x > x² - x + 1 + 2√(x²-x). Taking the positive square root, we get √(x²+x) > √(x²-x) + 1. Hence x + (x - 1) - 2√(x²-x) > (x + 1) + x - 2√(x²+x). Taking the positive square root, √x - √(x-1) > √(x+1) - √x. 

Problem 3

A right-angled triangle has longest side c and other side lengths a and b. Show that a + b ≤ c√2. When do we have equality?

Solution

Answer: equality iff a = b.

We have 2a² + 2b² = 2c² (Pythagoras). Also a² - 2ab + b² = (a - b)² >= 0, with equality iff a = b. Hence a² + 2ab + b² <= 2c² with equality iff a = b. 

Problem 4

The sum of the distances from a point inside an equilateral triangle of perimeter length p to the sides of the triangle is s. Show that s √12 = p.

Solution

Let the triangle be ABC and the point inside be X. Let the perpendiculars from X to BC, CA, AB be XD, XE, XF respectively. The area of triangle ABX = base x height/2 = AB·XD/2. Similarly for BCX and CAX. But area ABX + area BCX + area CAX = area ABC. Hence AB·XD/2 + BC·XE/2 + CA·XF/2 = AB²(√3)/4. So s = AB (√3)/2 = p (√3)/6 = p/√12. 

Problem 5

ABC is a triangle with |BC| = a, |CA| = b. Show that the length of the angle bisector of C is (2ab cos C/2)/(a + b).

Solution

Let the bisector be CD. Then the length of the perpendicular from C to BC is CD sin C/2. The perpendicular from C to AC has the same length. Hence area DBC = (a CD sin C/2)/2 and area DAC = (b CD sin C/2)/2. But area DBC + area DAC = area ABC = (ab sin C)/2 = ab sin C/2 cos C/2. Hence CD = (2ab cos C/2)/(a + b). 

Problem 6

Find 1.1! + 2.2! + ... + n.n! .

Solution

We show by induction that 1·1! + 2·2! + ... + n·n! = (n+1)! - 1. It is obviously true for n = 1. Suppose it is true for n, then 1.1! + ... + (n+1) (n+1)! = (n+1)! - 1 + (n+1) (n+1)! = (n+2)! - 1, so it is true for n+1. 

Problem 7

Show that there are no integer solutions to a² + b² = 8c + 6.

Solution

Squares are always congruent to 0, 1 or 4 mod 8. So a² + b² must be congruent to 0, 1, 2, 4, or 5 mod 8. But 8c + 6 is congruent to 6 mod 8. 

Problem 8

f is a function defined on the positive integers with integer values. Given that (1) f(2) = 2, (2) f(mn) = f(m) f(n) for all m,n, and (3) f(m) > f(n) for all m, n such that m > n, show that f(n) = n for all n.

Solution

We show by induction that f(m) = m for all m such that 2^n-1n <= m <= 2ⁿ.

We have f(2.1) = f(2) f(1), so f(1) = 1, so the result is true for n = 1. Suppose it is true for n. Then f(2ⁿ) = 2ⁿ. Hence f(2ⁿ⁺¹) = f(2ⁿ2ⁿ) = f(2ⁿ) f(2ⁿ) = 2ⁿ2ⁿ = 2ⁿ⁺¹. But there are 2ⁿ - 1 values strictly between 2ⁿ and 2ⁿ⁺¹. So, by (3), the result is true for n+1. 

Problem 9

Show that the shortest side of a cyclic quadrilateral with circumradius 1 is at most √2.

Solution

Let the quadrilateral be ABCD and let the circumcenter be O. We have AB = 2 sin AOB/2. Now ∠AOB + ∠BOC + ∠COD + ∠DOA = 360^o. So the smallest of the angles AOB/2, BOC/2, COD/2, DOA/2 is not greater than 45^o. But sin x is increasing over the range 0 to 45^o, so the shortest side has length at most 2 sin 45^o = √2. 

Problem 10

P is a point on the hypoteneuse of an isosceles, right-angled triangle. Lines are drawn through P parallel to the other two sides, dividing the triangle into two smaller triangles and a rectangle. Show that the area of one of these component figures is at least 4/9 of the area of the original triangle.

Solution

Let the triangle be ABC with hypoteneuse BC. Let the feet of the perpendiculars from P to AB and AC by X, Y respectively. If BP >= 2/3 BC, then area XBP is at least 4/9 area ABC. Similarly, if BP < 1/3 BC, then area YPC is at least 4/9 area ABC. The sides of the rectangle are equal to BP and BC, so its area is BP(BC - BP) = BC²/4 - (BP - BC/2)². Hence if BP lies between BC/3 and 2BC/3, then the area of the rectangle is at least BC²(1/4 - 1/36) = 2/9 BC² = 4/9 area ABC.