Two Cyclists and a Bird: cracking a tough problem in two simple steps

Problem: Two cyclists are approaching each other with speeds of 2 mph and 3 mph. When they are 10 miles apart, a bird starts flying back and forth between them at a speed of 12 mph, maintaining this pattern until the cyclists meet. What distance will the bird travel?

Answer: 24 miles

Fig. 1. Two cyclists approaching each other with a bird flying between them..

Solution. If you think this problem requires complex math, you're in for a pleasant surprise. This is more of a logic problem than a mathematical one, meaning it can be solved with basic arithmetic in two simple steps. Its elegant solution highlights a relationship between speed, distance, and time, three fundamental concepts used in physics. If you understand how this relationship works, you will find this problem a doddle.

A bird joins the cyclists on their journey when it aligns with the first one, as shown in Fig. 1. This moment marks the distance between the cyclists becoming 10 miles and initiates the countdown to the end of their shared journey. Since all three start and finish at the same time, the time they spend on the trip is the same. This is the crucial factor that will lead to a quick solution. We can reverse the bird's direction and align it first with cyclist B. This won't alter the outcome because it will change neither the cyclists' speed nor the distance between them, so that the travel time will remain the same.

Now that we have established that the travel time is the same for everyone, we can calculate it using the closing speed formula. The closing speed is the rate at which two cyclists close the gap between them. Assuming D is the distance between cyclists, V1 and V2 are their speeds, we estimate the travel time as:

T = D ÷ (V1+V2) = 10 miles ÷ 5 mph = 2 hours

Knowing the bird's speed, V3, and the travel time, T, we can find the distance, d, it travels:

d = V3 x T = 12 mph x 2 hours = 24 miles

The bird will cover a distance of 24 miles.  

The puzzle's solution, which otherwise would require calculating the infinite series of back-and-forth trips, is commonly credited to Hungarian mathematician Paul Erdos. A condition is applied to the puzzle: the bird's speed can't be less than the cyclists'. If it were slower, the faster cyclist would disrupt the bird's free flight and drag it along until reaching the other cyclist. This would prevent the bird from moving back and forth at its speed, compromising the puzzle's integrity and reducing it to a regular problem with a predictable solution.