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Math Circle Contest # 2 Solutions.
1.
By using simple arithmetic, we know that there must be at least 12 of each color.
We need to get 3 of each color. Thus, let’s suppose that two colors occur with the least
amount possible- 12 colors. If we pick 87 balls, then we might happen to pick the 87 that
are not of the two colors, thus getting only two colors. Thus, one needs to pick 88 balls
to get at least three colors.
2. The triangles ABF and ADF are equal, because angles BAF and DAF are equal
(bisector) and angles AFB and AFD are equal (right angles) and they share a side AF.
Thus, AB=AD, and so we know that AB has to be half the length of AC (because of the
median drawn). So we know that the two sides are 2 and 4. We need to find the third
side. The only way you can have integer coefficients is if the third side is length 3,
because we know that it has to be smaller than 4, but bigger than 2 (you can use the
cosine rule or the triangle sum inequality to see that 1, 2, 4, or any other number bigger
than 4 is not possible for that side because of the angle requirements).
3.
The number of chips in a particular row must be a multiple of 3, since there are
twice as many white pieces as black pieces, and thus any row contains at least 6 pieces.
Therefore, the maximum number of black and white chips which may be placed on the
chessboard is 48: 32 white and 16 black.
The pattern is shown here:
W
W
B
W
B
W
W
B
W
W
B
W
This can be repeated for the other 3 quarters of the chessboard.
4. the solution of the coin problem is to pick any
distinct two pairs of coins and weigh them (so that would
be 2 weightings already). Then there are 2 possibilities:
1. If the two pairs weigh the same amount, then you know
that the counterfeit coin in the third pair. Thus, you
weigh one of the coins in the third pair- if that coin
weighs half as much as one of the first two pairs, then the
counterfeit coin is the one that you didn't weigh. If that
coin does not weigh half as much as the first two pairs,
then the coin itself is counterfeit.
2. If the two pairs weigh a different amount, then you
know that the counterfeit coin is in one of these two
pairs. Then you weigh any two coins from two different pairs
plus the one coin that's in the third pair. Let a be the
weight of the correct coin and b be the weight of the
counterfeit coin. Then that weight is either either =
to 3a+b or 2a+b. But we already know 3a+b (one counterfeit
coin and 3 correct one in the first two pairs). Thus, if we
subtract 3a+b plus that weight that either = to 3a+b or
to 2a+b, we either get a or b. In order to find out which
one it is, we could look at the first two pairs and see if
that weight is equal to a weight that is 1/2 as much as any
pair. If it is, then that weight is the weight of the "correct"
coin. If it is not, it is the weight is counterfeit.
Now you need to find the counterfeit coin:
1. If you got the weight of the counterfeit coin, then
you weighted 3a on the scale the third time, and so the
counterfeit coin would be the coin not chosen to be weighed
in the pair with one "correct coin" and one counterfeit coin
(the one that does not weigh 2*b).
2. If you got the weight of the "correct" coin, then the
counterfeit coin would be the one picked for the third
weighting that does not belong to the pair which weighs
twice as much as the "correct" coin.
5. First of all, let’s rank numbers a, b, c, d, e according to the pair sum. So according to the
pair sums, we know that a+b=0 and d+e=17, since these are the smallest and biggest pair
sums respectively. Now since we know that each number was used 4 times, then we
know that the sum of the 5 numbers numbers must be 4 times less than the sum of all the
pair sums. Thus, we get 20 as the total sum of the five numbers. Because we already
know a+b and d+e, we know that c=3. After plugging in different values for a and b, we
get that a= -1, b=1, c=3, d=6, e=11 is the only possible solution.