Problem #9

[This problem is worth 3 points.]

In convex pentagon ABCDE, angle A and angle B are both 120 degrees. Sides EA, AB, and BC are all length 2, and sides CD and DE are length 4. What is the area of the pentagon?

Solution #7

Let the first term of the geometric sequence be a, and the ratio be r. Then the first five terms of the sequence are a, ar, ar², a³, and a⁴.

So we have:

ar - a = 9
ar⁴-ar³=576

From the first, we have a(r-1)=9.

Factoring the second, we have:

r³a(r-1)=576

Substitution 9 for a(r-1), we have:

r³=576
r=4

So 4a - a = 9, and a = 3.

Thus the first five members of the sequence are 3, 12, 48, 192, and 768, and their sum is 1023.

Problem #8

[This problem is worth 4 points.]

Find the sum of all real numbers x such that:

(2^x - 4)³ + (4^x - 2)³ = (2^x + 4^x - 6)³

Solution #6

First we specify 12 arrangements that work, and then we prove that these are the only arrangements that work.

Suppose all of the numbers that are 0 mod 3 are put in one box, all of the numbers that are 1 mod 3 are put in another box, and all of the numbers that are 2 mod 3 are put in the third box. Then when two numbers are drawn from two boxes, their sum mod 3 determines their origin.

If their sum is 0 mod 3, they came from the 1 mod 3 and the 2 mod 3 box. If their sum is 1 mod 3, they came from the 0 mod 3 and the 1 mod 3 box. If their sum is 2 mod 3, they came from the 0 mod 3 and the 2 mod 3 box.

This is then sufficient for the magician to determine the unused box. But there are six ways for the magician to distribute the numbers in this way, since there are three ways to choose the box into which to put the 0 mod 3 numbers, and then 2 ways to choose the box into which to put the 1 mod 3 numbers.

Suppose instead that 1 is put into one box, 2 - 99 are put in a second box, and 100 is put in the third box.

Then if the sum is 3 - 100, the first two boxes were chosen. If it is 101, the first and third box were chosen. If it is 102-199, the second and third boxes were chosen.

Again, this is sufficient for the magician to determine the unused box. And again there are six ways of achieving this arrangement, for a total of 12.

We must now prove that these 12 are the only workable arrangements. We will in fact prove that this is true for any number n of numbers, assuming n is at least 4. We do this by induction on n.

We begin with the case n = 4. There are (ignoring the selection of box color) only six ways to arrange {1,2,3,4} into the three boxes:

{1},{2},{3,4}
{1},{2,3},{4}
{1},{2,4}{3}
{1,2},{3},{4}
{1,3},{2},{4}
{1,4},{2},{3}

Checking each case, we find that only the second and the sixth work, which fits our claim.

We now assume that only our specified arrangements work for n, and consider n+1.

We separate two cases: the number n+1 is in a box by itself, or it is not.

Suppose n+1 is in a box by itself. Now suppose that n and k are in different boxes, for some k > 1. Then n+k can be formed by selecting them. But n+k can also be formed by selecting n+1 and k-1, and these will come from different boxes than n and k (since n+1 is by itself). Thus 2 through n must be in the same box, which means 1 is in a box by itself. This is the arrangement that has 1 in a box, the largest number in a box, and all the other numbers together in a single box.

Now suppose n+1 is not in a box by itself. Then if n+1 is removed, we have a permissible arrangement of 1 through n. This arrangement must either be by remainder mod 3, or the smallest-largest-other arrangement.

If it is by mod 3, n+1 must go in the box matching its mod 3. If it does not, then it is in a different box from n-2, and the sum 2n-1 can be made by adding the two of these. But n and n-1 are in different boxes, and also sum to 2n-1, which is impermissible.

If it is by smallest-largest-other, then n+1 cannot be put anywhere. If it is put with 1, it can be added to 2 to form n+3. But n is in a box by itself, and can be added to 3 to make n+3. If it is put with n or with {2,...,n-1}, it can be added to 1 to form n+2, and n can be added to 2 to form n+2.

Thus in both cases, we have to preserve one of our permissible arrangements, showing that these are the only permissible arrangements. This shows that 12 is the maximal number (and, in fact, is the maximal number for any n greater than or equal to 4).

Problem #7

[This problem is worth 3 points.]

In a geometric sequence of positive numbers, the fifth term is 576 larger than the fourth term, and the second term is 9 larger than the first term. What is the sum of the first five terms?

Problem #6

[This problem is worth 10 points]

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes: a red box, a white box, and a blue box. Each box then contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each of those two boxes, and announces their sum to the magician. Given this sum, the magician identifies the box from which no card has been chosen.

How many different ways are there to put the cards into the boxes so that this trick always works?

Week 1 Standings

Here are the point totals after the first week. If your score doesn't look right to you, send me an email and let me know.

MK: 14
SS: 13
AG: 12
JP: 6
TM: 4
PG: 4
BW: 3
ML: 3
SD: 3
MD: 1