olymmath

CMOQR 2014

Canadian Mathematical Olympiad Qualifying Repêchage 2014.

Other problems

Problem 1

Let f:Z→Z+f : \mathbb{Z} \rightarrow \mathbb{Z}^+ be a function, and define h:Z×Z→Z+h : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}^+ by h(x,y)=gcd⁡(f(x),f(y))h(x, y) = \gcd (f(x), f(y)) . If h(x,y)h(x, y) is a two-variable polynomial in xx and yy , prove that it must be constant.

Problem 2

Alphonse and Beryl play a game involving nn safes. Each safe can be opened by a unique key and each key opens a unique safe. Beryl randomly shuffles the nn keys, and after placing one key inside each safe, she locks all of the safes with her master key. Alphonse then selects mm of the safes (where m<nm < n ), and Beryl uses her master key to open just the safes that Alphonse selected. Alphonse collects all of the keys inside these mm safes and tries to use these keys to open up the other n−mn - m safes. If he can open a safe with one of the mm keys, he can then use the key in that safe to try to open any of the remaining safes, repeating the process until Alphonse successfully opens all of the safes, or cannot open any more. Let Pm(n)P_m(n) be the probability that Alphonse can eventually open all nn safes starting from his initial selection of mm keys.

(a) Show that P2(3)=23P_2(3) = \frac23 .

(b) Prove that P1(n)=1nP_1(n) = \frac1n .

(c) For all integers n≥2n \geq 2 , prove that P2(n)=2n⋅P1(n−1)+n−2n⋅P2(n−1).P_2(n) = \frac2n \cdot P_1(n-1) + \frac{n-2}{n} \cdot P_2(n-1). (d) Determine a formula for P2(n)P_2 (n) .

Problem 3

Let 1000≤n=ABCD10≤99991000 \leq n = \text{ABCD}_{10} \leq 9999 be a positive integer whose digits ABCD\text{ABCD} satisfy the divisibility condition: 1111∣(ABCD+AB×CD).1111 | (\text{ABCD} + \text{AB} \times \text{CD}). Determine the smallest possible value of nn .

Problem 4

In △ABC\triangle ABC , the interior sides of which are mirrors, a laser is placed at point A1A_1 on side BCBC . A laser beam exits the point A1A_1 , hits side ACAC at point B1B_1 , and then reflects off the side. (Because this is a laser beam, every time it hits a side, the angle of incidence is equal to the angle of reflection). It then hits side ABAB at point C1C_1 , then side BCBC at point A2A_2 , then side ACAC again at point B2B_2 , then side ABAB again at point C2C_2 , then side BCBC again at point A3A_3 , and finally, side ACAC again at point B3B_3 .

(a) Prove that ∠B3A3C=∠B1A1C\angle B_3A_3C = \angle B_1A_1C .

(b) Prove that such a laser exists if and only if all the angles in △ABC\triangle ABC are less than 90∘90^{\circ} .