We should be flying back home from Iceland about this time. Hopefully everyone has had fun with the puzzles while we have been gone. If you haven't had a chance to solve the puzzle about the Iceland Ring Road yet, take a look at last Friday's post and don't read any further. But if you want the answer, read on...
A: Let A be the speed of the first couple and B be the speed of the second couple. After an equivalent amount of time T, one couple has traveled AT miles and the other travels BT miles. For the return, the first couple now travels BT miles in 9 hours, while the other couple travels AT miles in 16 hours.
A = BT/9
B = AT/16
9A = BT
16B = AT
T = 9A/B
T = 16B/A
9A^2 = 16B^2
Take the square root of both sides (which is okay because both are positive)
3A = 4B
This tells us the ratio of their speeds is 4 to 3. In other words, over the same time, the faster couple will travel 4/7 of the ring road, the slower couple will travel 3/7. The difference is 120 miles. And if 1/7 is 120 miles, the whole road is 840 miles.
wow, that is a LOT cleaner than what I did (and I suspect ralph loizzo also did also), which was figure out the distance of the two paths first. It involved a huge trinomial which I had to plug into the quadratic equation. Of course it resulted in the same answer, the shorter path = 360 miles, longer path = 480 miles, thus the whole road is 840 miles.
ReplyDeleteQuite the elegant solution, Blaine. I went down the Happy Steve road as well and then, seeing my nasty equations (and, literally out of back of the envelope space), put it all off for later (which became actually never). I guess the real ah ha moment is when you figured that each couple's second day travel is equivalent to the other's first day. I knew that as I drew my figure and set up my equations but I never used it to my advantage. Big time saver that and much simpler equations to boot.
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