2001COMC加拿大数学公开赛真题答案免费下载

历年 Canadian Open Mathematics Challenge加拿大数学公开赛

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2001 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由两部分题目组成

Part A共8题，每题5分

Part B共4题，每题10分

共计12题，满分80分

不可使用任何计算器

完整版下载链接见文末

部分真题预览：

Part A Solutions:

A3) Two solutions:

1. Solution 1:Let ABCDEF be a regular hexagon with a side length of 1. Each angle is 120° . Thus, if we join FC, EB, DA, each of the interior angles is bisected, and so each part is 60° . Thus the hexagon is decomposed into 6 equilateral triangles, as shown. The maximum distance between any two points on the hexagon is the distance between two opposite vertices. Since each of the triangles is equilateral with a side length of 1, the diagonal distance is 2, ie. the maximum possible length of PQ is 2.
   • Brief version of Solution 1: A regular hexagon with side length 1 can be decomposed into 6 equilateral triangles with a side length of 1, as shown. The maximum distance between any two points is between opposite vertices, and this distance is 2.

Part B Solutions:

B2)

1. Two Solutions:
   1. Solution 1:We define a “losing position” to be a number of cards, such that if a player receives this number of cards at the beginning of his or her turn, he or she is guaranteed to lose assuming best play by both players. A “winning position” is defined similarly.
      Clearly, by the rules of the game, 1 is a losing position.
      For a player to receive 1 card at the beginning of a turn, the previous player must start with 2 cards. (This is true since a player can never remove more than half of the deck, so the number of cards at the beginning of the previous turn can never be more than double that of the current turn.) Therefore, 2 is a winning position, since a player starting with 2 cards can only remove 1, and so passes 1 card to the other player, who loses.
      Is 3 a winning position or a losing position?
      Given a pack of 3 cards, the rules of the game say that a player can only remove 1 card, and so pass a pack of 2 cards (a winning position) to the other player. Therefore, 3 is a losing position.
      We can then see that 4, 5 and 6 are all winning positions, as given 4, 5 or 6 cards, a player can remove 1, 2 or 3 cards respectively to pass the other player 3 cards, a losing position.Therefore, 7 is a losing position, since a player removing 7 cards must remove 1, 2 or 3 cards, thus giving the other player 6, 5 or 4 cards respectively, any of which is a winning position. So if Alphonse starts with 7 cards, Beryl can always win.
   2. Summary of Beryl’s Strategy
      • She will receive 4, 5 or 6 cards from Alphonse.
      • Remove 1, 2 or 3 cards in order to pass 3 cards back to Alphonse.
      • Alphonse is forced to remove 1 only, and pass back 2 to Beryl.
      • Beryl removes 1 and passes 1 back, so Alphonse loses.
2. Two Solutions:
   1. Solution 1: We must determine if 52 is a winning position or a losing position. By a similar argument to above, since 7 is a losing position, 8 through 14 are all winning positions, since they can all be reduced to 7 in one turn. Therefore, 15 is a losing position, since given 15 cards, a player is forced to reduce to some number between 8 and 14, since no more than 7 cards can be removed.
      Similarly, 16 through 30 are winning positions, 31 is a losing position, and 32 through 62 are winning positions.
      Therefore, 52 is a winning position, so Alphonse has a winning strategy.
   2. Summary of Alphonse’s strategy
      • Alphonse removes 21 cards from original 52, and pass 31 cards to Beryl.
      • If Beryl removes b₁ cards with 1 ≤ b₁ ≤ 15, Alphonse removes 16 1 − b cards to reduce the pack to 15 cards. [Notice that this is always a legal move, since 2 (16 - b₁) = 32 -2b₁≤31 − b₁ , so 16 1 − b is never more than half of the pack.]
      • If Beryl removes b2 cards with 1 ≤ b₂ ≤ 7, Alphonse removes 8 − b₂ to reduce the pack to 7 cards. [This move is always legal by a similar argument.]
      • Beryl now has 7 cards, so Alphonse should adopt Beryl’s strategy from (a).

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