# Flight distance

1. Apr 1, 2005

### TSN79

Two people live 20 km from eachother. They start cycling towards eachother with a speed of 10 km/h. A fly starts from one of them and flies to the other person and then back to the first person. It has an average speed of 15 km/h. It continues to do this until the two people meet. How far will the fly have flown when the two people finally meet? Apparently this has one very complex solution as well as being a simple grade school/high school problem, but I'm having trouble seeing the simplicity I think...abyone?

Could it be that the answer is indeed 15 km...? It just hit me...

Last edited: Apr 1, 2005
2. Apr 1, 2005

### moose

Well considering that the fly will fly for an hour exactly, i think that it is 15km

3. Apr 2, 2005

### vikasj007

it should be 15 km, but can somebody give the other method for solving it, i have done it before but just cant recollect it right now.

4. Apr 2, 2005

### Berislav

Sum the infinite series.

5. Apr 2, 2005

### ceptimus

It's interesting to run this one in reverse. The two cyclists start out touching each other, with the fly trapped between them. Then they start cycling away from each other, and the fly goes back and forth between them. At the end of an hour, how far from each of the cyclists will the fly be? Obviously the two figures will add together to equal 20km.

6. Apr 2, 2005

### BicycleTree

The fly could be any distance from each of the cyclists (adding up to 20 km) because if you run it forwards, any starting location for the fly between the two cyclists ends up in the same situation.

7. Apr 3, 2005

### kleinwolf

I know this is wrong, but we could also say : in this problem, the fly is always flying at 25km/h towards the other cyclist. So the time needed is : 20/25=4/5h...Hence the distance the fly has flown : 4/5*15=12km.

8. Apr 3, 2005

### T@P

the grade school solution is as follows:

the cyclists meet in the middle, so they each travel 10 km.
they go at 10 km/h, so it takes them 1 hour to meet.

fly can only 'fly' (not intended) while the two havent met, or it has to stop when they do meet. they meet in one hour, and the fly is constantly moving until they meet. hence, total distance is speed * time, or 15 km/h * 1 hour = 15 km.

the complex solution is the infinite series. there is even a joke about this, some famous math guy was asked this question and he comes up with the answer in like 5 seconds. so the people that ask him are like "wow you figured out the 'trick' so fast" and hes like "no i did the infinite series really fast..."

haha ok it would be more funny if i remembered who did that... :rofl:

9. Apr 3, 2005

### BicycleTree

10. Apr 3, 2005

### Jimmy Snyder

I first heard this one when I was in High School, about 40 years ago. I worked very hard to sum the series, but I did not have the tools to solve it. Eventually, the poser told me the answer.

As a graduate student in math, I heard the following story. Someone asked this riddle of John Von Neuman (a mathematician) who gave the correct answer after a few moments reflection. The poser said "Oh, you've heard this one before". Professor Von Neuman said "No, I just summed the series".

When I heard that Professor Von Neuman could sum the series in his head, I decided that surely I could do it with paper and pencil. Sure enough, it is a straightforward exercise. I recommend it to you.

11. Apr 4, 2005

### T@P

YES haha it was Neuman who did the series in his head (thanks jimmysnyder)

12. Apr 4, 2005

### Jimmy Snyder

As the fly is traveling at 15 km/hr toward a bicyclist who is traveling at 10 km/hr, they are closing at 25 km/hr. As they are 20 km apart at the outset, it will take (20 km) / (25 km/hr) = .8 hr to meet. At that time the fly will have travelled (.8 hr) x (15 km/hr) = 12 km. The two cyclists will have travelled (.8 hr) x (10 km/hr) = 8 km each. At that point the fly will turn around. The situation now is the same as at the outset except that the separation between the cyclists is now 4 km instead of 20 km. Therefore it will take 1/5 as long for the fly to close with the first cyclist as before and it will travel 1/5 the distance. That is .16 hr and 2.4 km. Also the cylists will travel 1/5 as far, or 1.6 km each. At that point the fly will turn around again and the situation will be the same as at the outset except that the distance between the cyclists will now be .8 km, that is 1/25 or (1/5) squared.

Therefore, the total travel time will be $.8 hr \times (1 + \frac{1}{5} + \frac{1}{5}^2 + \frac{1}{5}^3 + \dots) = .8 hr \times \frac{5}{4} = 1 hr.$

And the distance travelled by the fly will be $12 km \times (1 + \frac{1}{5} + \frac{1}{5}^2 + \frac{1}{5}^3 + \dots) = 12 km \times \frac{5}{4} = 15 km$