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

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)?

Solution #31

AC is the diagonal of the square, and has length 3 √ 2  . The triangle must have area 9, so letting the required altitude be x, we have:

(1/2)3x √ 2  = 9

x = 3 √ 2 

Problem #32

[This problem is worth 5 points.]

PQR is a triangle with PQ = 4 and PR = 7. PM is the median to side QR, and PM = 7/2. What is QR?

Problem #31

[This problem is worth 2 points.]

Given a square ABCD of side length 3, a scalene triangle is constructed with AC as one of the sides. If the area of the triangle is the same as the area of the square, what is the length of the altitude of the triangle to the base AC?

Week 6 Standings

Point totals through problem #30:

MK: 93
SS: 87
AG: 33
JP: 7
BW: 7
ML: 6
TM: 5
PG: 4
SD: 3
SG: 3
MD: 1

Solution #30

Factor m³ + 6m² + 5m into m(m² + 6m + 5) = m(m+5)(m+1). We can then see that this product is always a multiple of 3. If m = 0 mod 3, then m is divisible by 3. If m = 1 mod 3, then m + 5 is a multiple of 3. If m = 2 mod 3, then m + 1 is a multiple of 3. So in any case, some factor is a multiple of 3.

On the other hand, 27n³ + 9n² + 9n + 1 is never a multiple of 3. Each of the first three terms is always a multiple of 3, so the entire expression is always 1 mod 3.

Thus there are no integer values of m and n giving the two expressions the same value.

Solution #29

Recall that i² = -1, i³ = -i, and i⁴ = 1. Thus i^-1 = i³ = -i.

So we have (i - i^-1)^-1 = (i - -i)^-1 = (2i)^-1 = (1/2)(i^-1) = (1/2)(-i) = -i/2.

Problem #30

[This problem is worth 5 points.]

Find all pairs (m,n) of integers such that m³ + 6m² + 5m = 27n³ + 9n² + 9n + 1.

Solution #28

To find the shortest path for the gecko, we unfold the room into a net, and then look for a short line path from gecko to fly. However, there is more than one way to unfold into a net, and different unfoldings will produce different straight line distances, so we must make sure we have found the most efficient route.

The optimum unfolding is as follows:

The gecko is one foot horizontally and vertically from the lower left corner of the upper 10 x 8 rectangle, and the fly is one foot horizontally and vertically from the upper right corner of the lower 8 x 10 rectangle.

The horizontal distance from G to F is thus 9 + 7 = 16, and the vertical distance is 1 + 12 + 1 = 14. The total distance is then √ 14² + 16²  = √ 452  = 2 √ 113  .

Problem #29

[This problem is worth 2 points.]

Simplify as much as possible (i - i^-1)^-1.

Solution #27

We want to find all x1 such that x1, x2, ..., x20 is an f-generated sequence in which x20 is the first 1.

If x20 is 1, then x19 must be 10. From here, we can work backward either by subtracting 1 or by multiplying by 10.

However, if we subtract 1 nine times in a row, we will be back to 1. So in the backtracking from x19 to x10, there must be at least one multiplying by 10. With two options at each step, we get 2⁹ ways of backtracking from x19 to x10, and after subtracting the option of subtracting 1 nine times in a row, we are down to 2⁹-1.

Now we must backtrack nine more steps to x1. Given that there must have been a multiplying by 10 somewhere in the backtracking from x19 to x10, x10 is greater than 10. Thus we can subtract 1 each time without danger of dropping to 1, and both options are available as we move from x10 to x1, giving 2⁹ ways of doing this backtracking.

However, there cannot be a string of ten consecutive backtracking moves in which we subtract 1 -- if there were, some element in that string would be a multiple of 10, which would require the "divide by 10" rule going forward.

So we have a preliminary total of 2⁹(2⁹ - 1) possibilities, from which we need to subtract the possibilities in which there are ten consecutive -1 moves. But we have already blocked the option of having ten consecutive -1 moves at the start of the backtracking, so we only need to count the "ten consecutive -1" options that are preceded by a "multiply by 10" backtracking move.

There are a total of 18 backtracking moves, not counting the forced initial move of multiplying by 10. We require a sequence "*10, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1", but the other seven moves are free. Those seven moves can be filled in 2⁷ ways, and the sequence can be placed in 8 ways. So 8*2⁷ arrangements must be eliminated.

Thus gives a total of 2⁹(2⁹ - 1) - 8*2⁷ = 260,608 sequences.

Problem #28

[This problem is worth 3 points.]

A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is currently on a 10 x  8 side wall, one foot from the ceiling and one foot from the back 12 x 8 wall. The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly, assuming it can only walk across the ceiling and walls?

Solution #26

We will show that if n >= 2^k, then player B can guarantee a victory.

Suppose B, over a sequence of k+1 guesses, guesses the sets S₁,...,S_k+1. For the j-th guess, let T_j be S_j if A's answer is "yes", and {1,...,N} - S_j if A's answer is "no".

If A's j-th answer is truthful, then x is in T_j. Since at least one answer in a string of k+1 answers must be truthful, x must be in the union of T₁ through T_k+1.

Suppose now that B has more than 2^k uneliminated possibilities. We will show how B can eliminate another possibility. Given what was said above, it suffices to show that B can make a sequence of k+1 guesses such that the union of the T_j resulting from those guesses does not exhaust the uneliminated possibilities -- since x must be in that union, anything outside that union can be eliminated, and if there is at least one thing outside, B will have succeeded.

We now describe B's guessing procedure. Let S₁ be a subset of the uneliminated possibilities of size at least 2^k-1, and which also omits at least 2^k-1 of the uneliminated possibilities.

If A's answer is "yes", then B proceeds to pick S₂ so that S₂'s overlap with C(S₁) (the complement of S₁ -- the uneliminated possibilities not in S₁) is of size at least 2^k-2, and so that C(S₂)'s overlap with C(S₁) is also of size at least 2^k-2.

If A's answer is "no", B proceeds in the same way, but with C(S₁) replaced with S₁.

S₃ is determined in the same way, with S₁ in the previous method replaced by S₂.

B continues in this way through S_k. Given the construction, it follows that the size of the complement of the union of the S_j is at least 1. For his k+1-st guess, B then takes the singleton set of some member of that complement.

If A says "no" to that guess, then that single element can be eliminated, because the complement of that singleton is added to the union where x must be.

If A says "yes" to that guess, B repeats the above procedure starting with the uneliminated possibilities minus the single element on which the previous procedure ended. After k steps of that procedure, we are guaranteed that the complement of the determined sets, along with C(S_k+1), will be non-empty. We then eliminate anything in that complement.

So B can reduce any collection of more than 2^k uneliminated possibilities. Thus B can reduce the uneliminated possibilities to size 2^k. If B's allotted number of guesses n is at least 2^k, then, B is guaranteed victory with proper play.

Problem #27

[This problem is worth 7 points.]

Define the function f(x) as follows:

(A) If x = 1, f(x) = 1
(B) If x is divisible by 10, f(x) = x/10
(C) Otherwise, f(x) = x+1

We can then use repeated application of f to generate a sequence from any starting number. So, for example, f(5) = 6, f(6) = 7, f(7) = 8, f(8) = 9, f(9) = 10, f(10) = 1, and f(1) = 1. This produces the sequence 5, 6, 7, 8, 9, 10, 1, 1, 1, ...

We then use this sequence to define a new function d(x). For any x, d(x) is the number of terms in the f-generated sequence until the first occurrence of 1 in the sequence. So from the previous example, we see that d(5) = 7, since the 7th term of the sequence 5, 6, 7, 8, 9, 10, 1, 1, ... is the first 1.

For how many values of x does d(x) = 20?

Problem #26

[This problem is worth 10 points.]

A and B play the following game:

Step 1: Two integers k and n are chosen, and revealed to both players.

Step 2: Player A selects two further integers x and N, with 1 <= x <= N. Player A tells N to B, but not x.

Step 3: Player B now asks A questions of the following form: "Does x belong to the set ... of positive integers", for any choice of set.
B may ask as many questions of this form as he wants.
A must give either a "yes" or a "no" answer to each question. The answers, however, do not have to be truthful.
The only constraint on A is that in any sequence of k+1 consecutive answers, at least one must be truthful.

Step 4: After B has asked as many questions as he wishes, he specifies a set X of at most n positive integers. If x is in X, B wins; otherwise, A wins.

Give a non-trivial condition on n, k, and N such that satisfaction of that condition guarantees that B can win the game, and prove that the condition is sufficient.

Solution #25

First, note that 104729 is prime. For brevity, denote it as p.

Then we want x and y such that 2/p = 1/x + 1/y, or 2xy = p(y + x).

From this, it follows that p is a factor of 2, of x, or of y. It is not a factor of 2. Assume first that it is a factor of x. Then x = pa, for some positive integer a.

Thus 2apy = p(y + ap), or 2ay = y + ap, or (2a-1)y = ap. But the greatest common factor of a and 2a-1 is 1, so a must be a factor of y. Thus y = ab, for some positive integer b.

Thus (2a-1)ab = ap, or (2a-1)b = p. Thus either 2a-1 = p, or b=p. Suppose first that b=p. Then 2a-1 = 1, and a = 1. Then x = y = p.

Suppose instead that 2a-1 = p. Then b = 1. Then a = (p+1)/2, so x = p(p+1)/2, and y = (p+1)/2. In our particular case, x = 5484134085, and y = 52365.

If, instead, we assume p is a factor of y, we get the same options reversed. Thus the possible ordered pairs are (104729,104729), (5484134085,52365), and (52365,5484134085).

Solution #24

First we determine the probability that all the red chips are drawn, then all the blue, then all the green.

The probability that the first chip is red is 2/9, and that the second is red is 1/8. Then the probability that the third chip is blue is 3/7, that the fourth is blue is 2/6, and that the fifth is blue is 1/5. At this point, only green chips will remain.

So the probability of getting this order is 2/9 * 1/8 * 3/7 * 2/6 * 1/5 = 1/1260.

There are then six orders in which the three blocks of colors can be placed. Each order has equal probability, so the final answer is 6/1260 = 1/210.

Problem #25

[This problem is worth 5 points.]

Find all ordered pairs (x,y) of positive integers such that 2/104729 can be written as 1/x + 1/y.

Solution #23

2¹⁰²⁴-1 is a difference of squares, so it can be factored as (2⁵¹²+10(2⁵¹²-1).

2⁵¹²-1 is again a difference of squares, so it can be factored as (2²⁵⁶+1)(2²⁵⁶-1).

Again we have a difference of squares. Continuing in this manner, we eventually get to 2⁴-1, which factors as (2²+1)(2²-1) = 5*3. So 5 and 3 are both prime factors. Since 2¹⁰²⁴-1 is odd, 2 is not a prime factor of it, and 5 and 3 must be the smallest prime factors. Thus the answer is 5*3 = 15.

Problem #24

[This problem is worth 1 point.]

A bag contains 2 red, 3 blue, and 4 green chips. Chips are drawn from the bag with equal probability and without replacement. What is the probability that the chips are drawn in continuous color blocks (i.e., all the red, then all the blue, then all the green; or, all the blue, then all the green, then all the red; and so on)?

Solution #22

Situate the circle in the cartesian plane, using one chord as the x-axis and the other as the y-axis, and the point of intersection of the two chords as the origin. This gives four points on the circle: (-3,0), (4,0), (0,6), and (0,-2).

The center of the circle is at the same x-coordinate as the midpoint of the horizontal chord, and at the same y-coordinate as the midpoint of the vertical chord. Thus the center is at (1/2, 2).

We then calculate the distance from (1/2,2) to any point on the circle, such as (4,0). The distance is then √ ((4-1/2)2+ (2-0)2)  = √ (49/4+ 4) =√ (65/4)  = √ 65 /2. The diameter is then twice this distance, or √ 65 .

Problem #23

[This problem is worth 2 points.]

What is the product of the two smallest prime factors of 2¹⁰²⁴-1?

Solution #21

Think of each club member as a point, and each committee as a line (the committee members will then be the points of intersection between lines). Then we want a four-line diagram in which each pair of lines intersect, and in which we never have more than two lines intersecting at the same point.

So each pair of lines must determine a unique intersection point. Since there are 4C2 = 6 pairs of lines, there must be 6 intersection points, and hence 6 members of the club.

Problem #22

[This problem is worth 4 points.]

Two chords of a circle intersect perpendicularly. One chord is divided by the intersection into sections of length 3 and 4; the other chord is divided by the intersection into section of length 6 and 2. What is the diameter of the circle?

Week 4 Standings

Standings through the first 20 questions:

MK: 69
SS: 67
AG: 31
JP: 7
BW: 7
ML: 6
TM: 5
PG: 4
SD: 3
SG: 3
MD: 1

Problem #21

[This problem is worth 5 points.]

A club of n members is organized into four committees following two rules:

(1) Each member belongs to exactly two committees.
(2) Each pair of committees has exactly one member in common.

Find all possible values of n.

Solution #20

The Star Tours ride consists of one section of 2 options, and three sections of 3 options.

After n rides, the probability of getting a particular option on the 2-option section each time is 1/2ⁿ. There are two options, so the probability of getting the same option every time is 2(1/2ⁿ) = 1/2^n-1. So the probability of having seen both of the two options after n rides is 1-1/2^n-1.

Next consider the sections with 3 options. Picking two particular options, the probability of getting one of them on a particular ride is 2/3, so the probability of getting one of them each ride for n rides is (2/3)ⁿ. There are three selections of two options, so it looks like we can triple this number to get the probability of getting some set of two for all n rides, yielding 2ⁿ/3^n-1.

However, this is not quite right. Let the three options be A, B, and C. When we pick A and B, and consider the possibility of getting a sequence of n rides that draws only from A and B, we include the possibility of getting only A, and also the possibility of getting only B. When we pick B and C, we include the possibility of only B, and the possibility of only C. This creates a double-counting of the possibility of only B. Each single-option run is thus double-counted, and its probability needs to be subtracted off.

After n rides, the probability of getting a particular option on a three-ride section each time is 1/3ⁿ. There are three options to be subtracted off, so the probability to be removed is  is 3(1/3ⁿ)=1/3^n-1.

Thus the probability of getting at most two options of the three in n rides is  2ⁿ/3^n-1 - 1/3^n-1. The probability of getting all three is then 1 minus this.

The four ride sections are independent, so to find the probability of getting all the options in each section, we just multiply the probabilities. This yields:

(1-1/2^n-1)(1 - 2ⁿ/3^n-1 + 1/3^n-1)³

Running the numbers, this yields a probability of 89.86% after 11 rides, and goes above 90% to 93.17% after 12 rides.

(In fact, Sophia managed to see all eleven segments after a mere 7 rides.)

Solution #19

If n is the difference of two squares, then n can be written in the form a²-b² = (a+b)(a-b).

Suppose n is odd, and hence of the form 2m+1. Then let a = m+1 and b = m. Then (a+b)(a-b) = (m+1+m)(m+1-m) = 2m + 1 = n. So all odd numbers between 1 and 1000 are differences of two squares.

If n is even, at least one of a+b and a-b must be even. But if one is, they both are (because the difference between them is 2b, which is even). So n is then a multiple of 4. So no number than is 2 mod 4 can be written as the difference of two squares.

Suppose n is 0 mod 4. Then n = 4m for some m. Let a = m+1 and b = m-1, and we have (m+1+m-1)(m+1-(m-1)) = (2m)(2) = 4m.

Thus all odd numbers are all numbers 0 mod 4 between 1 and 1000 can be written as the difference of two squares, for a total of 750 numbers.

Solution #18

We can rewrite Σ_n=1⁴⁴cos n as Σ_n=1⁴⁴cos (45-x). (This just reverses the direction of the addends.) We then use the cosine angle sum identity to rewrite cos (45-x) as (cos 45)(cos x) + (sin 45)(sin x). Since cos 45 = sin 45 = 1/ √ 2  , we can rewrite this as 1/ √ 2 (cos x + sin x).

This gives us:

Σ_n=1⁴⁴cos n = 1/ √ 2 Σ_n=1⁴⁴(cos n + sin n) = 1/ √ 2(Σ_n=1⁴⁴cos n) + 1/ √ 2(Σ_n=1⁴⁴sin n).

Thus:

(1 - 1/ √ 2)Σ_n=1⁴⁴cos n = 1/ √ 2(Σ_n=1⁴⁴sin n)

(Σ_n=1⁴⁴cos n)/(Σ_n=1⁴⁴sin n) = 1/ √ 2 /(1 - 1/ √ 2) = 1 + √ 2

Problem #20

[This problem is worth 4 points.]

The Star Tours ride at Disneyland is composed of four parts, and the content of each part is randomly determined. In the first part, the rider encounters either (a) Darth Vader or (b) Hans Solo and the Millennium Falcon. In the second part, the rider crash lands on either (a) Hoth, (b) Tatooine, or (c) Kashyyk. In the third part, the rider receives a video message from either (a) Admiral Ackbar, (b) Princess Leia, or (c) Yoda. In the fourth and final part, the rider is chased through either (a) Coruscant, (b) Naboo, or (c) an under-construction Death Star orbiting Geonosis.

Sophia wants to experience all of the eleven component pieces of the Star Tours ride. She doesn't care whether she experiences all 54 possible ride combinations, but wants all of the component pieces. Assuming that in each part, the encounter is randomly determined with equal probabilities, and that the selection of component piece is independent between parts, how many times does Sophia need to ride Star Tours to have at least a 90% chance of experiencing all eleven component pieces?

Solution #17

2 miles per minute is a speed of 120 miles per hour, and 2 minutes per mile is a speed of 30 miles per hour. Let d be the distance from A to B. Then the time in one direction is d/120, and in the other direction is d/30. This is a total travel time of 5d/120, and a total distance of 2d. The speed is thus 2d/(5d/120) = 240/5 = 48.

Problem #19

[This problem is worth 3 points.]

How many of the integers between 1 and 1000 (inclusive) can be written as the difference of two perfect squares?

Solution #16

(5x - 11)/(2x²+x-6) = (5x-11)/((2x - 3)(x+2)).

We then write (5x-11)/((2x-3)(x+2)) = A/(2x-3) + B/(x+2).

Multiplying both sides by (2x-3)(x+2), we have:

5x - 11 = A(x+2) + B(2x-3)

5x - 11 = Ax + 2A + 2Bx - 3B

5x - 11 = (A+2B)x + (2A-3B)

Equating coefficients of like terms, we have:

A + 2B = 5
2A - 3B = -11

Doubling the first equation gives 2A + 4B = 10, and subtracting the second from the first gives 7B = 21, and B = 3. Then A = -1, so our sum is:

-1/(2x-3) + 3/(x+2)

Problem #18

[This problem is worth 8 points.]

Find (Σ_n=1⁴⁴cos n^°)/(Σ_n=1⁴⁴sin n^°)

Problem #17

[This problem is worth 1 point.]

A car travels from point A to point B at a speed of 2 miles per minute. It then returns from point B to point A at a speed of 2 minutes per mile. What is its average speed for the entire journey, in miles per hour?

Problem #16

[This problem is worth 2 points.]

Express the fraction (5x - 11)/(2x²+x-6) as the sum of fractions each of which has as its denominator a polynomial of degree 1.

Solution #15

First note that triangles BCM and BPM are both right, because CM is chosen perpendicular to BK. Because BK is the angle bisector of angle B, BCM and BPM are similar. Because they share side BM, they are congruent.

Thus M is the midpoint of CP, and BC = MP = 120.

Similarly, triangles AQN and ACN are congruent. Thus N is the midpoint of CQ, and AQ = AC = 117.

Because M and N are respectively the midpoints of PC and QC, by considering triangle PCQ we see that MN is half of PQ.

AQ + BP = AB + PQ, so PQ = AQ + BP - AB = 120 + 117 - 125 = 112. Thus MN = 56.

Week 3 Standings

Through problem #15:

SS: 54
MK: 46
AG: 29
JP: 7
BW: 7
ML: 6
TM: 5
PG: 4
SD: 3
SG: 3
MD: 1