2026 AIME I 第 1 题

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1.

Patrick 从学校沿一条笔直的路以恒定速度步行去公园。Patrick 出发一小时后,Tanya 从学校沿同一条 笔直的路以恒定速度跑向公园,速度比 Patrick 步行快每小时 22 英里。Tanya 出发一小时后, José 从学校沿同一条笔直的路以恒定速度骑车去公园,速度比 Tanya 跑步快每小时 77 英里。 三人同时到达公园。学校到公园的距离是 mn\frac{m}{n} 英里,其中 mmnn 为互质正整数。求 m+nm + n

Patrick started walking at a constant speed along a straight road from his school to the park. One hour after Patrick left, Tanya started running at a constant speed of 22 miles per hour faster than Patrick walked, following the same straight road from the school to the park. One hour after Tanya left, José started bicycling at a constant speed of 77 miles per hour faster than Tanya ran, following the same straight road from the school to the park. All three people arrived at the park at the same time. The distance from the school to the park is mn\frac{m}{n} miles, where mm and nn are relatively prime positive integers. Find m+n.m + n.

答案:277
知识点:路程、速度与时间方程组
难度评级:1840
解答:

设 Patrick 的速度为每小时 vv 英里,行程时间为 TT 小时。于是 Tanya 用 T1T - 1 小时以速度 v+2v + 2, 行进,José 用 T2T - 2 小时以速度 v+9v + 9 行进 (这比 Tanya 的速度多 77)。因为三人走过同一段距离, vT=(v+2)(T1)=(v+9)(T2). \begin{aligned} vT &= (v+2)(T-1) \\ &= (v+9)(T-2). \end{aligned}

展开第一个等式得 0=2Tv20 = 2T - v - 2,所以 v=2T2v = 2T - 2。展开第二个等式得 0=9T2v180 = 9T - 2v - 18,所以 2v=9T182v = 9T - 18。代入可得 4T4=9T184T - 4 = 9T - 18,因此 T=145T = \frac{14}{5},且 v=185v = \frac{18}{5}

距离为 vT=185145=25225vT = \frac{18}{5} \cdot \frac{14}{5} = \frac{252}{25},已经是最简分数, 所以 m+n=252+25=277m + n = 252 + 25 = 277

Let vv be Patrick's speed in miles per hour and TT his travel time in hours. Then Tanya travels for T1T - 1 hours at speed v+2,v + 2, and José travels for T2T - 2 hours at speed v+9v + 9 (which is 77 more than Tanya's speed). Since all three cover the same distance, vT=(v+2)(T1)=(v+9)(T2). \begin{aligned} vT &= (v+2)(T-1) \\ &= (v+9)(T-2). \end{aligned}

Expanding the first equality gives 0=2Tv2,0 = 2T - v - 2, so v=2T2.v = 2T - 2. Expanding the second gives 0=9T2v18,0 = 9T - 2v - 18, so 2v=9T18.2v = 9T - 18. Substituting, 4T4=9T18,4T - 4 = 9T - 18, hence T=145T = \frac{14}{5} and v=185.v = \frac{18}{5}.

The distance is vT=185145=25225,vT = \frac{18}{5} \cdot \frac{14}{5} = \frac{252}{25}, which is in lowest terms, so m+n=252+25=277.m + n = 252 + 25 = 277.

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