# Math Problem

Please complete the Will of the People problem according to the new instructons by tonight at 8 pm EST.
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Let’s consider only the rightmost circle, as shown in below figure:
Applying the Pythagorean’s theorem with x  d , y  r 
x2  y 2   R  y 
2
2
d2 
d
R 
4 
2
2
d
d2
 d2 
 R 2  Rd 
4
4
2
2
 R  Rd  d  0
 d2 
Dividing d2 to the left side, we have
2
R R
    1  0
d d
which gives

R 1
 1 5
d 2

d
, we have
2
Solution:
The only condition that ensures two triangles are congruent is SASAS. One example eliminates almost all
of the other conditions
In the above figure we have two triangles satisfying ASASA that are not congruent. The triangle formed
by the points (0,0), (2,0), and (2,2) has sides of lengths 2, 2, and 4 and angles of measure 1, 1, and 2 tradians. The triangle formed by the points (0,0), (2,0), and (1,-1) has sides of length 2 and angles of
measure 1, 1, and 2. These two triangles satisfy the ASASA condition but are not congruent. This also
eliminates the ASA, SAS, and AAS conditions as well the possibility for a SSA or AAA condition.
Hence, the SSS and SSSA conditions fail.
Similarly, the last remaining condition, SASAS, actually does hold.
A decision method that would result in the selection of site e can be described as follow:
We eliminate the site with the greatest 1st place votes- that’s a with 3. Since no remaining site has three
1st site votes, we remove c with two 1st placed votes. This leaves b and e, and now using the Borda
count we have e wins.
Since X is favored 9 to 8 over both Y and Z, the numbers a > b > c such that X will get more points than Y
is (a,b,c) = (8,7,1).
If two had voted b > c > a shifted a to the top of their ballots and voted a > b > c, we have the following
ballot tally:
6 abc
2 acb
3 bca
3 cab
1 cba
Hence, the ballot would be used to give a the victory with an 11(=6 + 2 + 3) to 4(=3 + 1) majority over b.
If their Yamaha Pogosticks malfunctioned and they hadn’t reached the polls in time to submit their 2anti-a b > c > a ballots, the outcome would still be the victory for a with smaller ratio of 10(=6 + 2 + 2) to
7(=5 + 2) majority over b.
(a) If each pair of people Di , D j have the temperature t ij that satisfied both of them, we have
tij   Li , Ri    L j , R j  . Hence,  Li , Ri    L j , R j    , for all 1  i, j  N .
Therefore,
N
k 1
 Lk , Rk  
 L , R    L , R   is a non-empty set.
i
i, j
i
That means there is some t 
j
N
k 1
j
 Lk , Rk  , which is the temperature that satisfies everyone.
(b) Let t1 be the temperature most overlapped by residents’ comfort zone. Now, remove all residents
whose comfort zones contain t1. Let t2 be the temperature most overlapped by the remaining comfort
zones. By our construction, we have that every comfort zone contains t1 or t2 (or both), as desired.
(c) Suppose that there is no temperature agreeable to at least 50% of the residents. Since each
residence must contain either t1 or t 2 , we have t1 is agreeable to less than 50% of the residents and so
is t 2 . Hence, t1 , t2  is agreeable to less than 50% + 50% = 100% of the residents, which contradicts to
part (b).
Therefore, there is a temperature (either t1 or t 2 ) agreeable to at least 50% of the residents.
SWC(3,4) means for any 4 residents there is a temperature agreeable to (at least) 3 of them.
For SWC(3,5), it may imply that there are 3 temperatures, called t1 , t2 and t3 , such that every
resident’s comfort zone contains 2 of them (or all). Moreover, there is a temperature agreeable to at
least
2
of the residents.
3
For SWC(11,17), it may imply that there are 11 temperatures, called t1 , t2 ,
, t11 , such that every
resident’s comfort zone contains 6 of them (or all). Moreover, there is a temperature agreeable to at
least
6
of the residents.
11
There must be a restaurant at which at least 2457 Pilleywogians eat
The rext draft algorithm comes to an end with each player having an offer from a different team
because if there are two teams offer to a player, then he will turn down   the offer from his team and
hence get the offer from different team. The process ends when all players have offer from their
different team.
The rext draft is not daft because since each player having an offer from a different team, there is no
team Y that prefers a player y not assigned to it than its assigned player, and since y had turned
down   the team Y if there was another team offering him, so y does not prefer Y over the
assigned team.
If any team has a player it prefers to that produced by a rext draft, let’s call the player y . Since there is
a team Y that prefers a player y not assigned to it than its assigned player, and y prefers Y over his
assigned team. Therefore, by definition, that draft is daft. Hence, the rext draft is team-optimal, as
desired.
Yes, there is a player-optimal algorithm.
Examples of tables for which the team-optimal and player-optimal assignment are different.
Examples of tables for which the team-optimal and player-optimal assignment are the same.
About the N! possible assignments in the latter case, there also exists both team-optimal and playeroptimal assignment among those possible ones.
The rext draft algorithm might last 3 days longer to reach a non-draft assignment. An upper bound for
the duration of the rext draft algorithm on an NxN preference table is 2N. That is because for each pair
of team and player, it takes 1 days for the team to offer a job and may need 1 day for the player to turn
down   the offer.
Yes, these questions suggest generalization: The rext draft algorithm for NxN preference table will come
to an end with each player having an offer from a different team, and it lasts at most 2N days.
Other of possible applications of variants of this model can be performing the rext draft algorithm to get
each player having an offer from a different team so that it prevents the baseball teams from getting
smaller in the near future.
It’s not possible to find a non-daft partition of a, b, c, d into 2 teams, because there is no information

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