27th British Mathematical Olympiad 1991 Problems

27th British Mathematical Olympiad 1991 Problems

1.  ABC is a triangle with ∠B = 90^o and M the midpoint of AB. Show that sin ACM ≤ 1/3.

2.  Twelve dwarfs live in a forest. Some pairs of dwarfs are friends. Each has a black hat and a white hat. Each dwarf consistently wears one of his hats. However, they agree that on the nth day of the New Year, the nth dwarf modulo 12 will visit each of his friends. (For example, the 2nd dwarf visits on days 2, 14, 26 and so on.) If he finds that a majority of his friends are wearing a different color of hat, then he will immediately change color. No other hat changes are made. Show that after a while no one changes hat.

3.  A triangle has sides a, b, c with sum 2. Show that a² + b² + c² + 2abc < 2.

4.  Let N be the smallest positive integer such that at least one of the numbers x, 2x, 3x, ... , Nx has a digit 2 for every real number x. Find N. Failing that, find upper and lower bounds and show that the upper bound does not exceed 20. 

Solutions

Problem 1

ABC is a triangle with ∠B = 90^o and M the midpoint of AB. Show that sin ACM ≤ 1/3.

Solution

Take AM = MB = 1 and let BC = x. Then CM² = x² + 1, CA² = x² + 4. Applying the cosine rule to triangle ACM, we get cos ACM = (2 + x²)/(x⁴ + 5x² + 4)^1/2. We claim that this is at least 2√2)/3 with equality iff x = √2. Certainly angle ACM < 90^o, so cos ACM is positive so the claim is equivalent to 9(x⁴ + 4x² + 4) ≥ 8(x⁴ + 5x² + 4), or x⁴ - 4x² + 4 ≥ 0, or (x² - 2)² ≥ 0, which establishes the result. Thus the maximum value of sin ACM is (1 - 8/9)^1/2 = 1/3.

Alternative solution

Let N be the midpoint of AC. Let G be the centroid (intersection of BN and CM). Let P be the foot of the perpendicular from N to CM. Then sin ACM = NP/NC = NP/NB (angle B is 90^o, so N is the center of the circle ABC) = NP/3NG. But NP/NG ≤ 1 with equality iff BN and CM are perpendicular. 

Problem 2

Twelve dwarfs live in a forest. Some pairs of dwarfs are friends. Each has a black hat and a white hat. Each dwarf consistently wears one of his hats. However, they agree that on the nth day of the New Year, the nth dwarf modulo 12 will visit each of his friends. (For example, the 2nd dwarf visits on days 2, 14, 26 and so on.) If he finds that a majority of his friends are wearing a different color of hat, then he will immediately change color. No other hat changes are made. Show that after a while no one changes hat.

Solution

Consider the number of friendly pairs who wear a different color. It must increase if a dwarf changes his hat. But it cannot increase indefinitely. [The details of the visiting rota are irrelevant.] 

Problem 3

A triangle has sides a, b, c with sum 2. Show that a² + b² + c² + 2abc < 2.

Solution

Each side must be shorter than the sum of the other two. So a < 1, b < 1, c < 1. Hence 2(1 - a)(1 - b)(1 - c) > 0, or 2 - 2(a + b + c) + 2(ab + bc + ca) - 2abc > 0, or -2 + 2(ab + bc + ca) - 2abc > 0. Subtracting (a + b + c)² = 4 gives the required result. 

Problem 4

Let N be the smallest positive integer such that at least one of the numbers x, 2x, 3x, ... , Nx has a digit 2 for every real number x. Find N. Failing that, find upper and lower bounds and show that the upper bound does not exceed 20.

Solution

Answer: 12.

A real number may have infinitely many digits. The strategy therefore has to be to exhibit some rational number which requires N to get a 2, and then to show that for any real number some special digit, such as the first, is a 2 for some multiple <= N.

A little experimentation reveals that for most integers fairly small multiples suffice. We notice that 15 requires 8, but most seem to need less.

If x has first digit 7, 8 or 9, then 3x has first digit 2. If x has first digit 5 or 6, then 4x has first digit 2. If x has first digit 4, then 5x has first digit 2. If x has first digit 3, then 7x has first digit 2. Obviously, if x has first digit 2, then 1x has first digit 2. So we need only consider numbers with first digit 1.

If x has second digit 2, we are home. If x has second digit 0 (including of course the case x has no second digit), 1, 3 or 4, then 2x has first digit 2. If x has second digit 5, then 8x has second digit 2.

It is convenient to notice that x and 10x behave the same way, so we can restrict x to be between 1 and 10. The analysis above shows that we need only consider 1.6 ≤ x < 1.9. If 5/3 ≤ x < 2, then 20 ≤ 12x < 30.

We might also look at 5/3. Its multiples are 1.66... , 3.33... , 5, 6.66... , 8.33... , 10, 11.66... , 13.33... , 15, 16.66... , 18.33... , 20. So it requires N = 12.

So we are left with 1.6 ≤ x < 5/3. The first digit is not 2 for N up to 12. But we notice that 2x has a 2 for 1.6 ≤ x < 1.65, and for 1.66 ≤ x < 1.665 and for 1.666 ≤ x ≤ 1.6665 and so on. Also 5x has a 2 for 1.64 ≤ x < 1.66 and for 1.664 ≤ x < 1.666 and so on. Thus 2x or 5x has a 2 for any x in the range 1.6 ≤ x < 5/3.

Note that there are other x which require N = 12, such as 1.6795 and 1.6835.