2023 AMC 12A 第 13 题

先试着解答 2023 AMC 12A 第 13 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2023 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

13.

在一场乒乓球锦标赛中,每位参赛者都与其他每位参赛者恰好比赛一次。虽然右手选手的人数是左手选手的两倍,但左手选手赢的场数比右手选手赢的场数多 40%40\%。(没有平局,也没有双利手选手。)总共进行了多少场比赛?
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was 40%40\% more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
1515
3636
4545
4848
6666
答案:B
知识点:组合整除性
小提示:
设有 LL 名左手选手和 2L2L 名右手选手,所以总人数为 3L3L。
Let there be LL left-handed and 2L2L right-handed players, so 3L3L players total
大提示:
若右手选手赢 RR 场,则总场数为 2.4R2.4R,所以总场数必须是 1212 的倍数
If right-handers win RR games, the total is 2.4R,2.4R, so the total must be a multiple of 1212
解答:
设有 LL 名左手选手和 2L2L 名右手选手,则共有 3L3L 人,总场数为 (3L2)\binom{3L}{2}。
若右手选手赢 RR 场,则左手选手赢 1.4R1.4R 场,所以总场数为 2.4R=125R2.4R=\tfrac{12}{5}R。要使它是整数场数,总场数必须是 1212 的倍数。
左手选手至多能赢下所有至少有一名左手选手参加的比赛,即 (L2)+2L2=5L2−L2\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2} 场。另一方面,他们必须赢 712(3L2)=7L(3L−1)8\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8 场。比较这两个量可得 21L−7≤20L−421L-7\le 20L-4,所以 L≤3L\le 3。
依次试 L=1,2,3L=1,2,3,总场数为 3,15,363,15,36;其中只有 3636 是 1212 的倍数,且可实现:33 名左手选手可以赢下全部 1818 场混合对局以及左手内部的 33 场,共 21=1.4⋅1521=1.4\cdot 15 场。
所以正确答案是 B。
Let there be LL left-handed and 2L2L right-handed players, for 3L3L players and (3L2)\binom{3L}{2} games total.
If right-handers win RR games, left-handers win 1.4R,1.4R, so the total is 2.4R=125R.2.4R=\tfrac{12}{5}R. For this to be an integer count, the total number of games must be a multiple of 12.12.
The left-handers can win at most every game involving at least one left-hander, namely (L2)+2L2=5L2−L2.\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}. On the other hand, they must win 712(3L2)=7L(3L−1)8\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8 games. Comparing these quantities gives 21L−7≤20L−4,21L-7\le 20L-4, so L≤3.L\le 3.
Testing L=1,2,3L=1,2,3 gives totals 3,15,36;3,15,36; only 3636 is a multiple of 12.12. It is achievable: the 33 left-handers can win all 1818 mixed games and their 33 internal games, giving 21=1.4⋅1521=1.4\cdot 15 wins.
Thus, the correct answer is B.
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