Soccer or Math?
Bailey, who is almost late for soccer practice, ties Duke’s collar directly to the corner of a square platform next to the practice field. Unfortunately, Bailey doesn’t realize that the platform is actually on wheels, and when the action gets exciting, Duke tries to run between the soccer goal posts, which are the regulation 7.32 meters apart. (The practice field has no nets on the goals.) Even worse, the platform is eight meters on a side, so Duke gets stuck, unable to get between the posts, and only able to slide back and forth in such a way that two adjacent sides of the square always touch the goal posts. (Duke is too excited at the prospect of joining Bailey’s game to think of backing up.)
What is the length (rounded down to a whole number of meters) of the path formed by the points that Duke can reach once he has become stuck in this way?
Belt, Suspenders, and Braces
For the next practice, Bailey again has Duke along, but this time he is prepared. There happen to be three trees near the field planted exactly at the vertices of an equilateral triangle six meters on a side. Bailey attaches Duke’s collar to each of the three trees with a separate six-meter-long chain, thinking “That should hold even if one of the chains breaks.”
How much more area will Duke have in which to roam if one of the chains does break than if all three hold?
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Solutions to week 87
Queen Bee. We can solve this one just by tracing backward through the generations. The queen has a mother and a father (one generation back; we need to reach six generations back). The father contributes just one grandmother and the mother contributes a grandfather and a grandmother for a total of one grandfather and two grandmothers (two generations back). The grandfather contributes one great-grandmother and the the two grandmothers contribute two great-grandfathers and two great-grandmothers, for a total of two great-grandfathers and three great-grandmothers (three generations back). Continuing in this way, the two great-grandfathers contribute two female ancestors and the three great-grandmothers contribute three male and three female ancestors in the previous generation, for a total of three male and five female four generations back. These in turn contribute three female and five male/five female for a total of five male and eight female ancestors five generations back. Finally, all thirteen of these ancestors in generation five must have a mother six generations back, so there are 13 great-great-great-great-grandmother bees.
OK, that was easy enough. But do you detect a famous sequence in there? How do you think it arises? More on this next week.
Coco-not. If the six choices prior to Coach Taylor’s are all the same kind of candy, she has a one in six chance of getting a coconut chocolate. For that to happen, on the second choice one of the five candies of the same type out of the eleven remaining must be chosen, a 5/11 chance, followed by a 4/10 chance on the next candy, and so on down to a 1/7 chance on the sixth candy, for a probability of 5×4×3×2/11×10×9×8×7 = 1/11×7×3×2 = 1/462.
Otherwise, if any coconut candy has been chosen, Coach Taylor will definitely not choose a coconut candy. The overall chance of a coconut not being chosen prior to Coach Taylor is (10/12)×(9/11)×(8/10)×(7/9)×(6/8)×(5/7) = 5/22. So there is a 17/22 – 1/462 = 356/462 = 178/231 chance of being in this case.
The 5/22 of the time when no coconut has been chosen breaks down into situations in which the type of chocolate with the most left has five remaining, four remaining, or three remaining. In this case, the first six choosers are randomly picking a subset of six chocolates out of the ten non-coconut chocolates, five dark and five milk. It’s equivalent to randomly choose four of these ten chocolates that remain after the first six; there are ten choose four or 210 ways that can happen. There are only ten such sets that contain four chocolates of the same type (five sets of four dark and five sets of four milk). There are 50 such sets that consist of three dark and one milk (ten ways of picking three milk and five ways of picking one dark), for 100 sets that contain three of one type and one of the other once we add in the three milk and one dark possibility. And finally, there are 100 such sets that contain two chocolates of each type (ten ways to pick two out of five milks times ten ways to pick two out of five darks.
Thus in 10/210 = 1/21 of the 5/22-chance case when Coach Taylor is in jeopardy of getting a coconut, there are five of one type left and she has a 1/5 chance of getting a coconut. In 100/210 = 10/21 of that 5/22 chance, there are four of one type left, so she has a 1/4 chance of getting a coconut. And in the remaining 10/21 portion of that 5/22 chance, there are three of each type left, so she has a 1/3 chance of getting a coconut.
Now we have analyzed all of the possibilities, so we can get the grand total of Coach Taylor’s chances of getting a coconut. That calculation goes as follows: She has a
(1/462)×(1/6) + (178/231)×0 + (5/22)((1/21)×(1/5) + (10/21)×(1/4) + (10/21)×(1/3))
chance of getting a coconut chocolate. This simplifies to
1/2772 + (5/22)(12/1260 + 150/1260 + 200/1260) = 1/2772 + (5/22)(362/1260) = 1/2772 + 181/(22×126) = 1/2772 + 181/2772 = 182/2772 = 91/1386 = 13/198.
So Coach Taylor has a 13/198 chance of being stuck with a coconut candy. Not a bad strategy, is it? That’s less than a seven percent chance of ending up with coconut.
Links to all of the puzzles and solutions are on the Complete Varsity Math page.
Come back next week for answers and more puzzles.