Treebeard's Stumper Answer 5 November 1999

Piece of Cake

The Dunn Middle School Piece of Cake bike ride this weekend is a slow ride up the big hill along Santa Rosa Creek from Cambria, followed by a fast downhill to Morro Bay. I don't know the actual distances up and down, but they're close, so we can assume they're equal for this stumper. It's steep, so you pedal (and push) up the hill at an average speed of just 5 miles per hour. Then you zoom down to the end as fast as you can. How fast do you have to ride down so that your average speed for the whole trip is 10 miles per hour? Careful, this is a stumper!

DMS riders on top of Piece of Cake.

I ride up the hill at just 5 miles per hour. But even if I ride the same distance down the hill at the speed of light, it's not fast enough for me to average 10 mph for the whole ride! The distance doesn't matter for this problem, so pick an easy one, say 20 miles each way. At 5 mph, it takes 4 hours to ride the 20 miles up. But to average 10 mph, I must ride the whole 40 miles in (40 miles / 10 mph) = 4 hours. That's impossible since I've already used up all my time. This stumper shows how even familiar concepts can contain unexpected loopholes!

Notes:

Without exception, the DMS kids guessed I have to ride down the hill at 15 miles per hour. That answer seems so easy since (5 + 15) / 2 = 10. But it's wrong:

d 20 miles Time to ride up = -- = -------- = 4 hours vup 5 mph d 20 miles Time to ride down = -- = -------- = 1.3 hours vdown 15 mph d 40 miles Average speed = - = --------- = 7.5 miles per hour, not 10! t 5.3 hours

This problem is tricky because we're really working with an average of averages, and that changes everything.

Graybear sent this answer:

The speed o' light! What can I say? If you averaged 6 mph up and wanted to average 10 mph for the whole trip, I would have had to think about it a minute. (30 mph) How about creating a graph of speed up the hill vs. speed down to help explain the math?

Ok Graybear, here's the graph:

The dotted red lines show the asymptotes up and down. As my uphill speed gets closer and closer to 5 mph (from above), my downhill speed must get larger without limit. And similarly, as my uphill speed gets larger and larger, my downhill speed gets closer and closer to 5 miles per hour. Yes, there is a mirror image of this graph in the quadrant below (5,5), but that involves negative speeds and has no physical meaning. What other math function has this shape?

Use the familiar formulas d=vt, v=d/t, and t=d/v for this problem. Say my speed up the Piece of Cake hill is 6 miles per hour, and the distance is 20 miles up and 20 miles down (though it really doesn't matter). I want to average 10 miles per hour for the whole trip, so I must ride the whole way in (2 * 20 miles / 10 mph) = 4 hours. I figure that I spend (20 miles / 6 mph) = 3.3 hours pedaling up. That leaves me (4 - 3.3) = 0.7 hours to ride the 20 miles downhill. So my downhill speed is d / t = (20 miles / 0.7 hours) = 30 miles per hour as shown on the graph.

We can generalize this with some simple algebra. Showing algebra in HTML is much harder than doing it! I did my best with this, but if it doesn't work on your browser, I also have this in plain text.

d	=	distance up (and down) the hill, so the total distance is 2d
t	=	time, with separate tup, tdown, and ttotal times defined below
vavg	=	my average speed for the whole ride
vup	=	my average speed up the hill
vdown	=	my average speed down the hill

My time to go up the hill is:

          d
    tup = ----
          vup

To achieve my desired average speed, I must do the whole ride in a total time of:

            2d
    ttotal = ----
            vavg

So the time remaining to ride the downhill is:

                         2d      d
    tdown = ttotal - tup = ----- - ----
                         vavg     vup

And so my downhill speed must be:

             d           d
    vdown =  ---- =  ------------
             tdown    2d     d   
                     ---- - ---
                     vavg    vup

We can simplify this by applying the distributive property in the denominator to pull out the common factor d:

                          d
                 = ----------------
                        2     1
                    d ( --- - --- )
                        vavg   vup

The distance d now cancels out. This is why it doesn't matter what the distance is for this problem!

                        1
                 = -----------
                    2     1
                    --- - ---
                    vavg   vup


                        1
                 = -------------
                    2vup - vavg
                    -----------
                      vavg vup

So we finally arrive at a general solution:

             vavg vup
    vdown = ------------
            2vup - vavg

Look at the term under the division line. If my average velocity for the whole trip (v_avg) is greater than or equal to twice my average uphill speed (v_up), then my downhill speed (v_down) is negative or undefined. You can't average twice your average speed for the first half of any distance. Q.E.D.

Here are some more links about this stumper:

• DMS parent Lisa Stewart did the 1999 Piece of Cake bike ride and wrote about it for the DMS Friday Newsnote. You can read her account of our ride. More Piece of Cake pictures will soon be available on the DMS Photo page.

• This puzzle has come up before on the Web, e.g. on the Guinness Mind Benders page.

• I used this puzzle once before in my Two DVT Quickies (24 Jan 96) stumper. Why not? It's a perfect middle school math stumper that is still surprising to me! My Up This Hill and Down (22 January 1999) stumper is similar.

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