26th Canadian Mathematical Olympiad Problems 1994

26th Canadian Mathematical Olympiad Problems 1994

1.  Find -3/1! + 7/2! - 13/3! + 21/4! - 31/5! + ... + (1994² + 1994 + 1)/1994!

2.  Show that every power of (√2 - 1) can be written in the form √(k+1) - √k.

3.  25 people sit in circle. They vote for or against an issue every hour. Each person changes his vote iff his vote was different from both his neighbours on the previous vote. Show that after a while no one's vote changes.

4.  AB is the diameter of a circle. C is a point not on the line AB. The line AC cuts the circle again at X and the line BC cuts the circle again at Y. Find cos ACB in terms of CX/CA and CY/CB.

5.  ABC is an acute-angled triangle. K is a point inside the triangle on the altitude AD. The line BK meets AC at Y, and the line CK meets AB at Z. Show that ∠ADY = ∠ADZ.

Solutions

Problem 1

Find -3/1! + 7/2! - 13/3! + 21/4! - 31/5! + ... + (1994² + 1994 + 1)/1994!

Solution

∑ (-1)ⁿ (n/n! + 1/n!) = -1 + 1/1994! (the series telescopes). Similarly, ∑ (-1)ⁿ n²/n! = ∑ (-1)ⁿ n/(n-1)! = ∑ (-1)ⁿ (1/(n-2)! + 1/(n-1)! ) = 1/1993! . So expression given is -1 + 1/1994! + 1/1993! = -1 + 1995/1994! .

Problem 2

Show that every power of (√2 - 1) can be written in the form √(k+1) - √k.

Solution

Put a = (√2 - 1)ⁿ, b = (√2 + 1)ⁿ. Put c = (b + a)/2, d = (b - a)/2, so that a = c - d. Expanding by the binomial theorem, we see that for n even, c and d/√2 are both integers, and for n odd, c/√2 and d are integers. So for any n, c² and d² are integers. But c² - d² = ab = 1. Thus putting k = d², we have a = √(k+1) - √k.

Comment. There are many different approaches to this problem, all of which work!

Problem 3

25 people sit in circle. They vote for or against an issue every hour. Each person changes his vote iff his vote was different from both his neighbours on the previous vote. Show that after a while no one's vote changes.

Solution

Let S_n be the set of people who do not change their votes at round n (for n > 1). Obviously S_n is a subset of S_n+1. Suppose S_n is non-empty but does not include everyone. Then we can find two adjacent people x in S_n and y not in S_n. Since y voted differently in round n, y must have voted the opposite way to x in round n-1. But x votes the same way in round n, so y must vote the same way as x in round n. Hence y belongs to S_n+1. So if S₂ is non-empty then S_n keeps on growing until it includes everyone.

That is true for any n. But if n is even, then it is possible to have S₂ empty. As you move around the circle people vote alternately for and against. If n is odd, that is impossible.

Problem 4

AB is the diameter of a circle. C is a point not on the line AB. The line AC cuts the circle again at X and the line BC cuts the circle again at Y. Find cos ACB in terms of CX/CA and CY/CB.

Solution

Answer: cos²ACB = (CX/CA).(CY/CB). It is the positive root for C outside the circle and the negative root for C inside the circle.

For C outside the circle, we have cos ACB = CX/CB = CY/CA (one should check that this is true in all configurations). Hence result.

For C on the circle it is still true because cos ACB = 0 and CX = CY = 0. For C inside the circle, cos ACB = - cos ACY = - CY/CA and also = - cos BCX = - CX/CB.

Problem 5

ABC is an acute-angled triangle. K is a point inside the triangle on the altitude AD. The line BK meets AC at Y, and the line CK meets AB at Z. Show that ∠ADY = ∠ADZ.

Solution

Let the rays DZ, DY meet the line through A parallel to BC at U, V respectively. Then UZA is similar to DZB, so UA/ZA = DB/ZB. Similarly VA/YA = DC/YC. Hence UA/VA = ZA (DB/ZB) (YC/DC) 1/YA = (DB/DC) (YC/YA) (ZA/ZB). But that is 1 by Ceva's theorem. Hence UA = VA, so ∠ADZ = ∠ADY.