Tuesday, August 7, 2012

Problem #41

[This problem is worth 5 points.]

A fair coin is flipped 10 times. Find the probability that there are no consecutive tosses that come up heads.

Monday, August 6, 2012

Problem #40

[This problem is worth 4 points.]

Let k be an integer such that 36 + k, 300 + k, and 596 + k are the squares of three consecutive terms of an arithmetic sequence. Find k.

Thursday, August 2, 2012

Problem #39

[This problem is worth 3 points.]

The polynomial x2 + bx + c is a factor of both x4 + 6x2 + 25 and 3x4 + 4x2 + 28x + 5. What are b and c?

Wednesday, August 1, 2012

Solution #36

We factor into (π - 3)(π - 4). The first factor is positive and the second is negative, so the product is negative.

Problem #38

[This problem is worth 6 points.]

What is the area of the largest square in the Cartesian coordinate plane such that the interior of the square contains at most 3 points of the form (a,b), where a and b are both integers?

Tuesday, July 31, 2012

Problem #37

[This problem is worth 4 points.]

Find the area enclosed by the graph of |x-60| + |y| = |x/4|.

Monday, July 30, 2012

Solution #35

The point (2,2) can be reached in either 4 or 6 steps. If it is reached in 4 steps, those steps must be R, R, U, U, in some order. There are 4C2 = 6 ways of placing the two "R" moves, and hence 6 ways of producing the required four-step sequence. There are 44 possible four-step sequences, so there is a 6/(44) probability that the bug reaches (2,2) after four steps.

To reach (2,2) in six steps, the steps must include R, R, U, U, and then must also include a "cancelling pair" of the form R, L or U, D. We consider the first case; the second will be equivalent.
There are (6C3)(3C2) ways of arranging R, R, R, L, U, U, for a total of 60 ways. However, some of these ways will have the bug reach (2,2) in four steps, and must be eliminated. As we saw above, there are 6 ways to arrange R, R, U, U, and each such arrangement can then be completed either R, L, or L, R. Thus 12 arrangements must be eliminated, leaving 48.

Arrangements of R, R, U, U, U, D will produce another 48 ways, for a total of 96. There are 46 arrangements of six steps, so there is a 96/(46) probability that the bug reaches (2,2) in exactly six steps.

The final probability is thus 6/(44) + 96/(46) = 3/64.