返回题库

AIME 2026 I · 第 1 题

AIME 2026 I — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

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, Jose 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+nm+n.

解析

Solution 1

We set up an equation in units of time from the info given, with Patrick's speed as ss and distance to the park as dd:

ds=ds+2+1=ds+9+2\frac{d}{s} = \frac{d}{s+2}+1 = \frac{d}{s+9}+2

From the first two expressions, you get ds+2d=sd−s2−2sds+2d=sd-s^2-2s From the first and third expressions, you get ds+9d=sd−s2−9sds+9d=sd-s^2-9s

After solving this system of equations, we get d=25225d = \frac{252}{25}, or 277\boxed{277}.

~ Logibyte and Bowen824

Solution 2

Let Patrick's speed be R and his time spent walking to the part be T. Let the distance of his school to the park be D. So, we can write the equation D=RT.D=RT. After going through the problem, we can write that D=(R+2)(T−1)D=(R+2)(T-1) and D=(R+9)(T−2).D=(R+9)(T-2). We can substitute the last two equations to get

(R+2)(T−1)=(R+9)(T−2)(R+2)(T-1)=(R+9)(T-2) RT+2T−R−2=RT+9T−2R−18.RT+2T-R-2=RT+9T-2R-18. Simplifying this, we can find R=7T−16.R=7T-16. We can plug the two equations D=RTD=RT and D=(R+2)(T−1)D=(R+2)(T-1) to get

RT=RT+2T−R−2.RT=RT+2T-R-2. We can substitute R=7T−16R=7T-16 to get

(7T−16)T=(7T−16)T+2T−(7T−16)−2.(7T-16)T=(7T-16)T+2T-(7T-16)-2. Simplifying, we have

5T=145T=14 or T=145.T=\frac{14}{5}. We can now find that R=7(145)−16=185.R=7\left(\frac{14}{5} \right)-16=\frac{18}{5}. Since D=RT,D=RT, D=185⋅145=25225.D=\frac{18}{5} \cdot \frac{14}{5}=\frac{252}{25}. That means that m+n=277.m+n=\boxed{277}.

~gogogo2022

Solution 3

Let Patrick’s speed be pp. Tanya runs 22 mph faster, and Jose bikes 99 mph faster.

Patrick walks for 11 hour before Tanya starts, creating a gap of pp miles. Since Tanya gains on him at 22 mph, the time it takes her to catch Patrick is

p2\frac{p}{2} .

Tanya then runs for 11 hour before Jose starts, creating a gap of p+2p+2 miles. Jose gains on her at 77 mph, so the time it takes him to catch Tanya is

p+27\frac{p+2}{7} .

Since they arrive at the park at the same time and Jose started 11 hour later,

p2−p+27=1\frac{p}{2} - \frac{p+2}{7} = 1 .

Solving,

7p−2(p+2)=145p=18p=185\begin{aligned} 7p - 2(p+2) &= 14 \\ 5p &= 18 \\ p &= \frac{18}{5} \end{aligned} Patrick’s total travel time is p2+1=145\frac{p}{2}+1=\frac{14}{5} hours, so the distance is

D=185⋅145=25225D = \frac{18}{5}\cdot\frac{14}{5}=\frac{252}{25} Thus,

m+n=277.m+n=\boxed{277}. ~matchas (Made by a ninth grader if any issues feel free to edit!)

Interesting note

This problem is identical to the 2020 Purple Comet Middle School #8: https://purplecomet.org/views/data/2020MSSolutions.pdf

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #1

piacademyus.org

Video Solution (Similar to Solution 2)

https://www.youtube.com/watch?v=sNd5zkIMgHM

Video Solution (Intuitive)

https://www.youtube.com/watch?v=twFpo62THNA - Continuum Math