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

Solution #32

Consider the following diagram:

PM is the median, and PN is the perpendicular to QR. MR = x, so QM also is x. Letting MN be y, QN is then x-y.

We use the Pythagorean theorem on triangle PNQ to obtain PN = √ 16 - (x-y)²  . We also use the Pythagorean theorem ontriangle PNM to obtain PN = √ (7/2)² - y²  .

Setting these two equal and squaring both sides, we have:

16 - (x-y)² = 49/4 - y²
16 - x² + 2xy - y² = 49/4 - y²
15/4 = x² -2xy

We can also use the Pythagorean theorem on triangle PNR to obtain PN = √ 49 - (x+y)²  . Equating this with the first expression for PN and squaring both sides, we have:

16 - x² + 2xy - y² = 49 - x² - 2xy - y²

33 = 4xy

Substituting in the first expression, we have:

15/4 = x² -33/2
x² = 81/4
x = 9/2

Thus QR = 9.

Problem #33

[This problem is worth 4 points.]

For how many real numbers x does (log₂ x)/5 = sin(5πx)?