Varsity Math, Week 43

Varsity Math 43

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In the third week of vacation, most of the team stops by Finley’s family reunion to join in the festivities.

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Relation Riddle

As the team minibus is heading to the reunion, Finley says, “I need to tell you about the people who will be at the reunion today. Not counting me, there will be one grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, three sisters, one son, three daughters, one father-in-law, one mother-in-law, one brother-in-law, and one son-in-law.” “Ugh,” groan the teammates, “how will we keep track of 25 people?” “Oh, no,” reassures Finley, “there are many fewer people than that today. In fact, there will be the minimal number of people possible there, consistent with my description. And all but one of them are blood relatives of mine.” (Note that although everybody is someone’s son or daughter, in this problem an attendee only counts as a “daughter” if a parent of hers is also at the reunion, and similarly for the other relationships. Also, there are only opposite-sex marriages in this particular scenario.)

How are each of the attendees at the reunion that day related to Finley?

The Magic Castle

Crusty Conundrum

On the second day of the reunion, Uncle Sylvester brings his signature spherical sourdough. (The loaf is in fact perfectly spherical, and uniformly coated with crust. How does he do that?) The number of attendees that day is equal to the solution to last week’s Triangulating the Difference puzzle, and Sylvester cuts the sphere into exactly that number of equal-volume pieces using perfectly parallel planar cuts. The youngest relative shouts out, “I love the crust! I want the piece with the most crust!”

Which piece would the youngest relative like best, according to the crust criterion?

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Solution to week 42

The Old Coin Game. Just three weighings always suffice to determine whether there is a fake coin and if so, which one it is and whether it is lighter or heavier. How is that possible? One must take care to use all of the possible outcomes of the weighings; each time, either the left-hand pan can be lighter (L), heavier (H), or they can balance (B). So a possible outcome, for example, would be LBH, meaning that on the first weighing, the left-hand pan would be lighter, on the second they would balance, and on the third weighing they would be heavier. We want to arrange the coins on each weighing so that each outcome means something different. There are 3×3×3 = 27 possible outcomes, and we have to distinguish 27 possible states: no fake, or each of thirteen coins being lighter or heavier (1 + 13×2 = 27 also; note the 14th coin we already know is genuine). The clever idea is to label each coin with the outcome that will identify it as counterfeit, and then also use the labels to figure out which pan to put it on each time. So for example, if we put the coin labeled LBH on the left hand side for the first weighing, off the scales altogether for the second weighing, and on the right-hand side in the last weighing (to try to make the left-hand pan heavier than the right), and if that’s the only coin in each of those conditions, then if the weighings actually come out as LBH, we know that coin is fake and lighter. What if it is heavier? That flips everything, so an outcome of HBL would tell us that this coin is the fake and it is heavier than a real coin. So we have to be careful, we can’t use any label and its opposite label, otherwise they would clash on the outcomes. To find a set of labels with no interference between the label of one and the opposite of another, we make a table, attempting to be systematic:

Coin Label Opposite
1 LLL HHH
2 HHB LLB
3 LLH HHL
4 HBH LBL
5 LBB HBB
6 HBL LBH
7 LHL HLH
8 HLB LHB
9 LHH HLL
10 BLL BHH
11 BHB BLB
12 BBL BBH
13 BLH BHL

OK, so we have used up all of the possible outcomes, except for “BBB” which will correspond to the case that there is no fake, so everything always balances because all of the coins weigh the same amount. There is just one problem left; the above labels for the coins ends up with five coins in the left pan and four coins in the right pan on every weighing, which will clearly always make the left pan come out heavier. In order for the balance to be able to respond to one coin being slightly off, we must have the same number of coins on each side. Fortunately, we have the 14th known, good coin: we simply leave that on the right-hand pan as a counterweight through all of the weighings. Now when we perform the weighings, if it comes out as one of the labels, that coin is lighter; as the opposite of one of the labels, that coin is heavier, and as “BBB”, the coins are all fair.

Wow, that just barely worked, adds Coach Newton. To show you how close it is, it can be proven that if you don’t have that 14th known good coin, there’s no way to find the counterfeit in three weighings. Not because there aren’t enough possible outcomes, just because there’s no way to arrange all the weighings to take advantage of them.

Recent Weeks

Week 42: Triangulating the Difference & The Old Coin Game, solution to A Day Late and A Fence Short

Week 41: Fractured Fruit & A Day Late and A Fence Short, solutions to Relay 11: Duplicate Division, Stamp Stumper, Awkward Positions, & Sine of Success

Week 40: Awkward Positions & Sine of Success, solution to Spiral Notebook

Week 39: Stamp Stumper & Spiral Notebook, solution to Phenomenal Packing

Week 38: Phenomenal Packing & Duplicate Division, solutions to Hat Trick, Possible Parabolas, Equal Laterals, & Isocircules

Week 37: Equal Laterals & Isocircules, solution to Circle Crossing

Week 36: Possible Parabolas & Circle Crossing, solution to Heart Trick

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

Come back next week for answers and more puzzles.

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