It’s just an average week for Varsity Math, so Coach Newton challenges the team with two problems about averages.
Oakley is exploring different ways to separate the digits from one to nine into two groups, with each digit in one group or the other and neither group empty. Oakley notices that whenever two digits, one from each group, are swapped, the numerical average of the digits in one group goes up and the other goes down. However, moving a single digit from one group to the other can cause the average of both groups to increase.
What is the largest average of one of the starting two groups of digits that allows both averages to be increased in this way?
Square of Harmony
Harmony likes taking averages of pairs of numbers. She likes to take the usual average, by adding the numbers and dividing by two, and she likes to take what she calls the “upside-down” average, by dividing 2 by the sum of the reciprocals of the numbers. She says to Oakley, “I drew a lovely rectangle yesterday. It had the same area as a square with an integer side length.” Oakley asks, “What were the length and width of your rectangle?” Harmony replies, “That’s the problem. I don’t remember. But I remember that their average was 9 and their upside-down average was 4.”
What was the side length of the square that had the same area as Harmony’s rectangle?
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Solutions to week 58
Nearly Invincible. Let’s call the person who is in every winning coalition but one the “power voter.” If there’s only one winning coalition the power voter is not in, it must be the coalition of all of the other shareholders, since that is the coalition the power voter is not in that has the most votes. Also, a winning coalition is one with at least 49 shares (since that is the smallest majority of the 97 shares). Therefore, the other four shareholders have the property that no three of them own 49 shares. The way for them to own the maximal number of shares without any three of them having 49 is for each of them to have 1/3 of 48, or 16, shares. (Yes, one person could have more shares than 16, say 17; but then no two of the others could add up to 32 shares, so the most they could have would be 16, 15, and 15, respectively, and 17+16+15+15 < 16+16+16+16. Things get even worse if you try to make one of the other shareholders' number of shares even larger.) So the other four shareholders have a total of 64 shares at most, and so the power voter must own at least 33 shares.
Don’t think it works the other way, though. For example, if one shareholder owned 33 shares, it is still possible that a different shareholder owns 49 (and the other three add up to 15 shares). Then there are lots of coalitions that don’t include the 33-share owner — any group that includes the majority owner. So, you need 33 shares to be a power voter in this scenario, but just because someone owns 33 shares does not necessarily create a power voter. It only happens when all of the other shareholders are perfectly split as described above.
Extremely Equal. Let’s say there are v students voting, and the “winning” topic A gets a votes as best. Say the other topics get b and c votes as best, with a > b, a > c, and a+b+c = v. So b + c ≤ 2a-2, so v ≤ 3a-2. On the other hand, topic A gets at most v–a votes as worst, but that is more than half of v, so v > 2a. Therefore it is impossible for a to be 1, since then v ≤ 1 but also v > 2. Similarly if a were 2, then v ≤ 4 but also v > 4, again impossible. So a must be at least three, in which case v ≤ 7 and v > 6, i.e. v = 7, and the votes would be a=3, b=2, and c=2, with all four of the B and C voters listing A as worst. That scenario could work; let’s check what happens with eight voters. In order for A to have the most best votes, a would have to be at least four, but then there are at most four voters left, so topic A cannot get a majority of worst votes. On the other hand, for any larger odd number v of voters, we could have a = (v-1)/2, b = (v-3)/2, and c=2 with all of the B and C voters listing A as worst. And for any larger even number of voters, we could have a = v/2 – 1, b = v/2-2, and c=3, and again all of the B and C voters listing A as worst. So the only number of voters where this strange best/worst combination can occur, but one more voter prevents it from occurring, is seven voters, in which case topic A gets four votes as worst.
Links to all of the puzzles and solutions are on the Complete Varsity Math page.
Come back next week for answers and more puzzles.