Problem #37

[This problem is worth 4 points.]

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

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.

Solution #34

We have log₆a + log₆b + log₆c = 6, so log₆abc = 6.

Thus abc = 6⁶. Because a, b, c is an geometric sequence, we have b = ra, and c = r²a for some r > 1.

Thus abc = r³a³, so r³a³ = 6⁶, and ra = 36.

ra = b, so b is 36. The difference between b and a is a perfect square, so the possible values for a are 35, 32, 27, 20, and 11.

Each value for a generates a value for r, and we can use that r value to see if it produces an integer value for c. The only value that works is 27, which makes r = 4/3, and makes c = 48. Thus a + b + c = 27 + 36 + 48 = 111.

Problem #36

[This problem is worth 1 point.]

Without the use of a calculator, determine whether the following expression is positive, negative, or zero.


π² - 7π + 12

Solution #33

The sine function is bounded between -1 and 1, so we first consider when the monotonically increasing log function is in this range. We have:

-1 <= (log₂x)/5 <= 1
-5 <= log₂x <= 5
1/32 <= x <= 32

We then note that sin(5πx) has a cycle length of 2/5. So between 0 and 32, it completes 80 cycles. Because the log function is roughly horizontal, it will intersect a given cycle twice.

We now have to check the endpoints carefully. Because sin(5πx) is 0 at x =0, and increasing, we do not miss any intersections between 0 and 1/32. However, (log₂x)/5 is 0 at x = 2, where sin(5πx) is also 0. This means that we miss one intersection point, leaving a total of 159.

Problem #35

[This problem is worth 4 points.]

A bug is on the Cartesian coordinate plane at the point (0,0). Each second, the bug crawls either up, down, left, or right for 1 unit. What is the probability that the bug will be on the point (2,2) in 6 or fewer seconds?

Problem #34

[This problem is worth 5 points.]

Let a, b, and c be positive integers such that:

(i) a, b, c, is an increasing geometric sequence.
(ii) b-a is a perfect square.
(iii) log₆ a + log₆b + log₆c = 6

Find a + b + c