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.

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.

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?

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?

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

Solution #14

Recall that, using the Binomial Theorem, the expansion of (x+y)ⁿ is the sum of all terms of the form nCk*x^n-ky^k.

We want a term in the expansion of  (a-1/ √ a  )⁷ in which the a term has an exponent of -1/2. Using the above general form, the k^th term will have an exponent of 7-k (from the a term) plus (-1/2)k (from the 1/ √ a  term). So we need 7 - k - k/2 = -1/2, or 15/2 = 3k/2, or k = 5. The coefficient is then (-1)⁵*7C5 = -21.

Problem #15

[This problem is worth 7 points.]

Let ABC be a triangle with AB = 125, AC = 117, and BC = 120. Let the angle bisector of A intersect BC at L, and the angle bisector of B intersect AC at K. Let M be the point of intersection of the perpendicular from C to BK, and let N be the point of intersection of the perpendicular from C to AL. Find MN.

Solution #13

Factoring 12! + 14! we have 12!(1 + 14*13) = 12!(183). The largest prime factor of 12! is clearly 11. 183 = 3*61. Thus 61 is the largest prime factor of 12! + 14!.

Solution #12

Note that triangle ABD and triangle AA'B have bases of the same length (since AD = AA' = 6). Their altitudes from B are also the same length, so they have the same area.

Because AB = BB', the altitude from B' to AD is twice the length of the altitude from B to AD. So triangle AA'B' as twice the area of triangle ABD.

By similar reasoning:

The area of triangle BB'C is twice the area of triangle BAC.
The area of triangle CC'D is twice the area of triangle CBD.
The area of triangle DD'A is twice the area of triangle DCA.

The area of A'B'C'D' is: [AA'B] + [BB'C] + [CC'D] + [DD'A] + [ABCD]

From the above, this is:

2[ABD] + 2[BAC] + 2[CBD] + 2[DCA] + 10

But [ABD] + [CBD] = [ABCD] = 10, and [BAC] + [DCA] = [ABCD] = 10. So:

[A'B'C'D'] = 2*10 + 2*10 + 10 = 50.

Problem #14

[This problem is worth 3 points.]

In the expansion of (a-1/ √ a  )⁷, what is the coefficient of a^-1/2?

Solution #11

Let x be the radius of the circle. Then the diameter is 2x, and the circumference is 2xπ. Thus 4/2xπ = 2x, so 4 = 4x²π, and x²π = 1. But x²π is the area of the circle, so the area is 1.

Problem #13

[This problem is worth 1 point.]

What is the largest prime factor of 12! + 14!?

Problem #12

[This problem is worth 7 points.]

Convex quadrilateral ABCD has an area of 10. Side AB has a length of 6, BC of 7, CD of 8, and DA of 9. Each side of the quadrilateral is then extended beyond one of its endpoints, with AB extended beyond B to a new point B', BC extended beyond C to a new point C', and so on. Each of the extended sides is twice the length of the original side (so AB' is of length 12). Find the area of the quadrilateral A'B'C'D'.

Week 2 Standings

Standings through the end of the second week:

SS: 41
MK: 31
AG: 19
JP: 6
BW: 6
ML: 6
TM: 5
PG: 4
SG: 3
SD: 3
MD: 1

Problem #11

[This problem is worth 1 point.]

Four times the reciprocal of the circumference of a certain circle equals the diameter of that circle. What is the area of the circle?

Solution #10

From xy + x + y = 71, we have xy + x + y + 1 = 72, or (x+1)(y+1) = 72.

From x²y+xy² = 880 we have xy(x+y) = 880.

880 = 2⁴*5*11, so each of x, y, and x+y must have its factors among these.

If (x+1)(y+1) = 72, then the pair of x+1 and y+1 must be (in some order), one of (1,72), (2,36), (3,24), (4,18), (6,12), or (8,9).

These pairs then give possible values for x and y of (0,71), (1,35), (2,23), (3,17), (5,11), and (7,8). (0,71) can be eliminated because x and y are both positive. (1,35), (2,23), (3,17), and (7,8) all include prime factors for x or y not in our permitted list. Thus we must have x and y be 5 and 11. x² + y² is then 121 + 25 = 146.

(Note that our answer does not tell us which of x and y is 5, and which is 11. This is to be expected, given the symmetry of the starting information in x and y.)

Solution #9

 

Add point F as shown above, so that AF = BF = 2. Triangle ABF is then equilateral with side length 2.

Draw EC, and note that EC is parallel to AB, because angles EAB and ABC are equal. Thus angles AEC and ECB are both 60 degree angles, and triangles CDE and CEF are both equilateral with side length 4.

The area of the pentagon is then the area of the two equilateral triangles of side length 4, minus the area of the equilateral triangle of side length 2.

The area of an equilateral triangle is  s² √ 3  /4, where s is the side length. So our total area is 2*(4² √ 3  /4) - 2² √ 3  /4 = 7 √ 3  .

Solution #8

Let a = 2^x - 4, and let b = 4^x - 2. Then our equation can be rewritten as:

a³ + b³ = (a+b)³

Expanding the right-hand side, we have:

a³ + b³ = a³ + 3a²b + 3ab² + b³

And hence:

3a²b + 3ab² = 0
ab(a+b) = 0

We thus have three cases: a = 0, b = 0, a = -b.

If a = 0, then 2^x = 4, and x = 2.

If b = 0, then 4^x = 2, and x = 1/2. If a = -b, then 2^x - 4 = 2 - 4^x, or:

4^x + 2^x - 6 = 0

4^x = (2^x)², so this factors to:

(2^x + 3)(2^x - 2)= 0

So 2^x = -3, or 2^x = 2. But 2^x is never negative, so we ignore that solution. 2^x yields x = 1.

The three solutions are thus 2, 1, and 1/2, and their sum is 7/2.

Problem #10

[This problem is worth 4 points.]

x and y are positive integers such that xy + x + y = 71, and xy²+x²y = 880. Find x² + y².

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

Solution #5

The maximal distance will occur when we draw the line segment connecting the two centers, and then extend it so that it intersects the two spheres on their opposite sides.

The distance will then be the sum of (i) the distance between the two centers, (ii) the radius of the larger sphere, and (iii) the radius of the smaller sphere. The two radii are 19 and 87, and their sum is 106.

To find the distance between the two centers, we use the three-dimensional distance formula. The distance is

	_______
√	(12 - -2)2 + (8 - -10)2 + (-16 - 5)2

= √ 196+324+441  = √ 961  =31.

Thus the maximal distance is 106+31 = 137.

Solution #4

We calculate the slopes between pairs of points:

(-3,-7) to (15,4): slope = 18/11
(-3,-7) to (3,-3): slope = 6/4 = 3/2
(-3,-7) to (12,3): slope = 15/10 = 3/2

So (-3,-7), (3,-3), and (12,3) are all on a line of slope 3/2. Thus (15,4) is not on a line with the other three.

Problem #5

[This problem is worth 3 points.]

What is the greatest possible distance between two points, one on a sphere of radius 19 centered at (-2,-10,5) and the other on a sphere of radius 87 centered at (12,8,-16)?

Solution #3

Because AEB and DFC are both 5-12-13 triangles, they are both right. Now add corresponding triangles on sides AD and BC:

The resulting quadrilateral is a rectangle, because all of its angles are right. And it is a square, because each of its sides is composed of two right triangle legs, one of length 5 and one of length 12.

Thus it is a square of side length 17. EF is a diagonal of that square, and hence has a length of 17 √ 2 

Solution #2

We want to find x³+y³. Factoring, we have:

x³+y³=(x+y)(x²-xy+y²)

Since x + y = 1, this is equal to x²-xy+y².

We then have x²-xy+y²=(x²+2xy+y²)-3xy

=(x+y)²-3xy

=1²-3*1

=-2

Alternatively, (x+y)³ = 1³=1. Expanding, (x+y)³ = x³ + 3x² + 3xy² + y³ = (x³+y³) + 3xy(x+y) = (x³+y³) + 3.
So (x³+y³) +3 = 1, and x³+y³=-2.

Problem #4

[This problem is worth 1 point.]

Which of the following four points is not on the same line as the other three: (-3,-7), (15,4), (3,-3), or (12,3)?

Solution #1

24 = 2^3 * 3, and 70 = 2 * 5 * 7.

We have 24x divisible by 70, so 2, 5, and 7 all need to be factors of 24x. 2 is already a factor of 24, but 5 and 7 are not. So 5 and 7 need to be factors of x.

We have 70x divisible by 24, so 2^3 and 3 need to be factors of 70x. 2 is already a factor of 70, so we need an additional 2^2 and a 3 in x.

So x needs to contain factors of 2^2, 3, 5, and 7. To minimize x, we use just these factors. Thus x = 2^2 * 3 * 5 * 7 = 420.

Problem #3

[This problem is worth 5 points.]

ABCD is a square with side length 13. E and F are points exterior to the square, with AE = CF = 5 and BE = DF = 12. Find EF.

Problem #2

[This problem is worth 4 points]

Given that x+y = 1, and xy = 1, what is the sum of the cubes of x and y?

Problem #1

[This problem is worth 2 points.]

The three positive integers 24, 70, and x have the property that the product of any two is divisible by the third. What is the minimum possible value of x?

2012 Summer Math Problem Solving Ultramarathon

I'll be running a ten-week, fifty-problem math competition for math team members (and anyone else interested in participating) this summer, starting Monday June 11 and ending Friday August 17. The structure:

1. A new problem will be posted on this blog each weekday during the competition. Answers to questions should be e-mailed to me. I'll only accept one answer per person (to avoid scatter-shot problem solving methods), so be sure you've got your final answer before sending it.

2. You will have 48 hours from the time of posting of the question to send me an answer. After 48 hours, a solution to the problem will be posted, and I won't accept further answers.

3. Each problem will be worth between 1 and 10 points, based on my assessment of its difficulty. Anyone who sends a correct answer will get that number of points added to their running score. The first person to send a correct answer will get a bonus point. [Note: last year I gave double points for the first correct answer. I've changed that this year to a single bonus point, because the bonus points were overwhelming other factors, making the competition into a speed event.]

4. Comments will not be open on the question posts, but they will be on the solution posts. You're encouraged to post alternative solutions in the comments, or to ask questions about the solution if it doesn't make sense to you.

5. At the end of each week, I'll post a list of the standings (using just first initials, to preserve anonymity). The top three finishers from math team (either middle school or high school) will receive some nominal prize when math team starts back in the fall.

6. Calculators are permitted on questions unless I specify otherwise. Many of the questions posted will be drawn from old math competitions, and you could probably find most of them with a bit of googling. I don't have any good mechanism to stop people from doing that, so you're just on the honor system -- your answers should be entirely your own.

The first question will appear here Monday. Good luck to everyone!