Summer Problem Solving Marathon Question #47

[Value = 3 points]

A certain polynomial has a remainder of 3 when divided by x-1, and a remainder of 5 when divided by x-3. What is its remainder when divided by x² - 4x + 3?

Wrapping Up the Summer Problem Solving Marathon

We're going to end the Summer Problem Solving Marathon at the end of this week, which will take us through Problem #50. Math team meetings will resume on Tuesday September 6.

Summer Problem Solving Marathon Question #46

[Value = 3 points]

What is the surface area of a cube with a space diagonal of length 1? (The space diagonal of a cube is the line segment from a vertex of the cube to the most distant vertex from that vertex.)

Summer Problem Solving Marathon Question #45

(This question should have been posted Thursday. Sorry for the delay.)

[Value = 6 points]

A fair coin is tossed 12 times. What is the probability that there are no two consecutive heads in the string of tosses?

Summer Problem Solving Marathon Solution #43

Let log₈(log₂ x) = y. Then log₂ x = 8^y = 2^3y, and x = 2^23y.

We then also have log₂(log₈ x) = y, so log₈ = 2^y, and x = 8^2y = 2^3(2y).

Since log₈(log₂ x) = log₂(log₈ x), we have 2^23y = 2^3(2y). Thus 3^3y = 3(2^y), and 2^2y = 3, and 2y = log₂ 3, and y = (log₂3)/2.

Since x = 2^3(2y), log₂ x = 3(2^y), and (log₂ x)² = 9(2^2y).

We now substitute (log₂3)/2 for y in this expression. This gives 9(2^log₂3) = 9 x 3 = 27.

Summer Problem Solving Marathon Solution #42

Let x be the probability of getting heads (H) on the biased coin. Then 1-x is the probability of getting tails (T).

We first calculate the probability of getting exactly one head. There are five outcomes with exactly one head (matching the five ways of picking a location for the H in the sequence). Each of those five ways has probability x(1-x)⁴ of occurring. So p(H=1) = 5x(1-x)⁴.

Next we calculate the probability of getting exactly two heads. There are ten outcomes with exactly two heads (matching the ten ways of picking two locations for the two Hs in the sequence). Each of these ten ways has probability x²(1-x)³ of occurring. So p(H=2)=10x²(1-x)³.

Since p(H=1) = p(H=2), we have:

5x(1-x)⁴ = 10x²(1-x)³

Simplifying (given that 0 < x < 1), we have:

1 - x = 2x
x = 1/3

We now calculate p(H=3). There are ten ways of picking locations for the three heads. Each of those ten ways has probability x³(1-x)² of occurring. So our probability is:

10(1/3)³(2/3)² = 40/243

Summer Problem Solving Marathon Question #44

[Value = 5 points]

Simplify (56 + 6√43)^3/2 - (56 - 6√43)^3/2 to an integer value. No calculators allowed for this question!

Summer Problem Solving Marathon Question #43

[Value = 6 points]

Log₂(log₈ x) = log₈(log₂ x). Find (log₂ x)².

Summer Problem Solving Marathon Question #42

[Value = 4 points]

A certain biased coin has a fixed probability of coming up heads, where that probability is neither 0 nor 1. When the coin is flipped five times, the probability of getting exactly one head is the same as the probability of getting exactly two heads. What is the probability of getting exactly three heads?

Summer Problem Solving Marathon Solution #41

Adding all three equations, we obtain:

(k+2)x + (k+2)y + (k+2)z = 3k

or:

(k+2)(x+y+z) = 3k

When k = -2, this becomes 0 = -6, which is not true for any values of x, y, and z. So there is no solution to the system of equations when k = -2.

For any other value of k, we have x + y + z = 3k/(k+2). So by setting x = y = z = k/(k+2), we obtain a solution.'

Thus k = -2 is the only value for which there is no solution in x, y, and z to the system of equations.

Summer Problem Solving Marathon Solution #40

Because 3 and 5 are both odd, 3¹¹ and 5¹³ are also both odd. Thus 3¹¹ + 5¹³ is even, and is divisible by 2. So 2 is the smallest prime factor of the sum.

Summer Problem Solving Marathon Solution #39

Consider first the horizontal movement of the bug. The bug moves horizontally on each odd-numbered move, with the horizontal movements alternating positive and negative directions. On the first move, the bug moves 1 unit horizontally, on the third -1/4 horizontally, on the fifth 1/16 horizontally, on the seventh -1/64 horizontally, and so on.

Combining successive positive and negative movements, we have movement of 3/4, and then 3/64, and so on. This is an infinite geometric sequence with initial term 3/4 and common ration 1/16. Its sum is thus (3/4)/(1 - 1/16) = (3/4)/(15/16) = 4/5.

Next consider the vertical movement of the bug. The bug moves vertically on each even-numbered move, with the vertical movement alternating positive and negative directions. On the second move, the bug moves 1/2 unit vertically, on the fourth -1/8, on the sixth 1/32, on the eighth -1/128, and so on.

Combining successive positive and negative movements, we have movements of 3/8, and then 3/128, and so on. This is an infinite geometric sequence with initial terms 3/8 and common ratio 1/16. Its sum is thus (3/8)/(1 - 1/16) = (3/8)/(15/16) = 2/5.

In the limit, then, the bug approaches the point (4/5, 2/5).

Summer Problem Solving Marathon Question #41

[Value = 5 points]

Find the set of all values of k such that the following system of equations has no solutions in x, y, and z:

kx + y + z = k
x + ky + z = k
x + y + kz = k

Summer Problem Solving Marathon Question #40

[Value = 1 point]

What is the smallest prime factor of 3¹¹ + 5¹³?

Summer Problem Solving Marathon Solution #38

Let x + iy be an arbitrary root of the polynomial. Then the reciprocal of that root is 1/(x + iy). We then convert the denominator to a real number by multiplying by the complex conjugate x - iy, to obtain:

(x - iy)/((x + iy)(x - iy)) = (x - iy)/(x² + y²)

But because the root is on the circle centered at 0 + 0i with radius 1, we have, by the Pythagorean theorem, x² + y² = 1. So the reciprocal of the root is x - iy.

But this is the complex conjugate of the original root. Complex roots of polynomials with real-valued coefficients always comes in pairs by complex conjugates, so the reciprocal of the original root is another root of the polynomial.

Thus the reciprocals of the roots of the polynomial are just the roots of the polynomial. We thus want the sum of the roots of the polynomial. But the sum of the roots of any polynomial with a leading coefficient of 1 is the negative of the coefficient of the next-to-largest power of the variable. Hence the sum of the roots, which is also the sum of the reciprocals of the roots, is -a.

Summer Problem Solving Marathon Solution #37

FLOOR(log₂N) will always be equal to the exponent of the largest power of 2 less than or equal to N. So:

FLOOR(log₂N) = 0 for N = 1
FLOOR(log₂N) = 1 for 2 ≤ N ≤ 3
FLOOR(log₂N) = 2 for 4 ≤ N ≤ 7
FLOOR(log₂N) = 3 for 8 ≤ N ≤ 15
FLOOR(log₂N) = 4 for 16 ≤ N ≤ 31
FLOOR(log₂N) = 5 for 32 ≤ N ≤ 63
FLOOR(log₂N) = 6 for 64 ≤ N ≤ 127
FLOOR(log₂N) = 7 for 128 ≤ N ≤ 255
FLOOR(log₂N) = 8 for 256 ≤ N ≤ 511
FLOOR(log₂N) = 9 for 512 ≤ N ≤ 1023
FLOOR(log₂N) = 10 for N = 1024

So the total we want is 2*1 + 4*2 + 8*3 + 16*4 + 32*5 + 64*6 + 128*7 + 256*8 + 512*9 + 10 = 8204.

Summer Problem Solving Marathon Question #39

[Value = 4 points]

A bug is at the origin of the Cartesian coordinate plane. It travels one unit right, to the point (1,0). It then makes a 90 degree turn counterclockwise, and travels .5 units to (1,.5). It continues in this manner -- making 90 degree turns counterclockwise, and each time travelling half as far as on the previous move. In the limit, what point does the bug approach?

Summer Problem Solving Marathon Question #38

[Value = 5 points]

Suppose that z⁴ + az³ + bz² + cz + d is a fourth-degree polynomial with real-valued coefficients. Suppose also that each root of this polynomial is a complex number lying on the circle in the complex plane centered at 0 + 0i and with radius 1. What is the sum of the reciprocals of the roots of the polynomial?

Summer Problem Solving Marathon Question #37

[Value = 6 points]

Let FLOOR(x) be the greatest integer that is less than or equal to x. What is the sum from N = 1 to N = 1024 of FLOOR(log₂ N)?

Summer Problem Solving Marathon Solution #36

The least common multiple of 6 and 8 is 24. Every multiple of 24 thus has a remainder of 0 when divided by 6 and when divided by 8.

For similar reasons, numbers that are 1 to 5 more than a multiple of 24 have remainders of 1 to 5 when divided by either 6 or 8.

We thus find numbers that are within 5 of a multiple of 24, between 101 and 199. There are four multiples of 24 in that range -- 120, 144, 168, and 192. That gives us 24 numbers with the same remainder upon division by 6 or 8. In addition, 101 is 5 more than 96, a multiple of 24. It thus works as well. This gives a total of 25 numbers.

Summer Problem Solving Marathon Solution #35

The volume of a sphere is (4/3)π r³, so the volume of the hemisphere of ice cream is (2/3)π 2³ = (16/3)π.

The volume of a cone is (1/3)hB, where B is the area of the circular base of the cone. Here the height is 8, and the circular base has radius 2 and hence area 4π. So the volume of the cone is (32/3)π.

Thus the total volume of ice cream is (16/3)π + (32/3)π = (48/3)π = 16π.

Summer Problem Solving Marathon Solution #34

When a right triangle is inscribed in a circle, its hypotenuse is a diameter of the circle. Thus the hypotenuse of the triangle is 14. Since the triangle is a 3:4:5 right triangle, its legs are then 42/5 and 56/5. The area of the triangle is then 1/2(42/5)(56/5) = 1176/25 = 47.04.

Summer Problem Solving Marathon Solution #33

For any choice of n, we can write 2ⁿ in the form nk + m, for some k and some m < n. (That is, we write 2ⁿ as some multiple of n, plus a remainder on division by n.)

We then rewrite 2⁽²ⁿ⁾ as 2^{nk + m} = 2^nk2^m.

We are now interested in the remainder of this when divided by 2ⁿ - 1. Since m < n, the remainder of 2^m when divided by 2ⁿ -1 is 2^m. 2^nk can be rewritten as (2ⁿ)^k. But the remainder of 2ⁿ when divided by 2ⁿ -1 is clearly 1, so the remainder of (2ⁿ)^k when divided by 2ⁿ -1 is 1^k = 1.

Thus the remainder of 2⁽²ⁿ⁾ when divided by 2ⁿ -1 is 2^m. We want the remainder not to be a power of 4, which means we want m to be odd.

We are thus looking for n such that 2ⁿ, when divided by n, produces an odd remainder. Examination of cases then shows that 25 is the first value of n that works.

Summer Problem Solving Marathon Question #36

[Value = 3 points]

How many integers n, 100 < n < 200, have the same remainder when divided by 6 and when divided by 8?

Summer Problem Solving Marathon Question #35

[Value = 3 points]

The interior of a right circular cone is 8 inches tall with a 2 inch radius at the opening. The interior of the cone is filled with ice cream, and the cone has a hemisphere of ice cream exactly covering the opening of the cone. What is the volume of the ice cream?

Summer Problem Solving Marathon Solution #32

Given any S meeting the conditions, let s be the number of elements in S, and let N be the sum of all the elements in S.

Given any n∈S, the mean of the elements in S other than n is an integer. There are s-1 other elements in S, and their sum is N - n. So s-1 is a factor of N-n. Thus N and n are congruent mod s-1 (that is, they have the same remainder when divided by s-1).

Since 1∈S, N and 1 are congruent mod s-1, and in fact all elements of S are congruent to 1 mod s-1. Since 2002∈S, we know that 2002 is congruent to 1 mod s-1, so 2001 is a multiple of s-1. Since 2001 = 3 x 23 x 29, the possible values for s are 4, 24, 30, 70, 88, 668, and 2002.

If we list the numbers that are congruent to 1 mod s-1, we get 1, s, 2s-1, 3s-2, ... . The s^th such number is 1 + (s-1)(s-1) = s² - 2s + 2. So the largest element of S must be of size at least s² - 2s + 2. Since the largest element is 2002, this rules out 70, 88, 668, and 2002 as values of s.

Thus the largest possible value of s is 30. It remains to see that this value of s can be made to work. We will need a set of 30 elements, each of which is congruent to 1 mod 29. There are many sets meeting this constraint, including {1, 30, 59, 88, ..., 813, 2002}.

Summer Problem Solving Marathon Question #34

[Value = 2 points]

A right triangle similar to a 3:4:5 right triangle is inscribed in a circle of radius 7. What is the area of the triangle?

Summer Problem Solving Marathon Question #33

[Value = 10 points]

Let N be

(a) the smallest integer n > 1 such that when 2⁽²ⁿ⁾ is divided by 2ⁿ-1, the remainder is not a power of 4

or

(b) 0, if there is no such n.

What is N?

Summer Problem Solving Marathon Question #32

[Value = 9 points]

Let S be a set of positive integers, such that 1 ∈ S, 2002 ∈ S, and for all n ∈ S, n < 2003.

Suppose S has the following feature: for any n ∈ S, the mean of the elements of S - {n} is an integer.

What is the greatest number of elements S could have?

Summer Problem Solving Marathon Solution #31

The length of CE is 2, so CDE has the same length base as ABC. Since D is the midpoint of AC, the perpendicular distance from D to BC is half that of the perpendicular distance from A to BC. Thus the height of CDE is half that of ABC.

CDE thus has the same base as ABC and half the height of ABC. Its area is thus half that of ABC.

The area of an equilateral triangle of side length s is (s²√3)/4. So the area of ABC is √3, and the area of CDE is (√3)/2.

Summer Problem Solving Marathon Solution #30

Diagonal AC is the hypotenuse of right triangle ABC, with legs 18 and 21. So AC is √(18² + 21²) = &radic(324 + 441) = √765.

AD is then the hypotenuse of right triangle ACD, with legs 14 and √765. AD is thus √(14² + 765) = √(196 + 765) = √961 = 31.

The perimeter of ABCD is thus 18 + 21 + 14 + 31 = 84.

Summer Problem Solving Marathon Solution #29

First we determine the length of the longest possible growing path. We do this by enumerating the distances between points. Two points in the grid can be separated by 0 to 3 units horizontally and 0 to 3 units vertically. Using the Pythagorean Theorem, that gives us the following possible distances:

√1: 0,1 and 1,0 separation
√2: 1,1 separation
√4: 0,2 and 2,0 separation
√5: 1,2 and 2,1 separation
√8: 2,2 separation
√9: 0,3 and 3,0 separation
√10: 1,3 and 3,1 separation
√13: 2,3 and 3,2 separation
√18: 3,3 separation

That's a total of 9 different distances, so the longest possible growing path would be of length 10.

Next we check to see if a length 10 path is possible, and (if it is possible) how many such paths there are. Since the movement options are more constrained when the distances are long than when they are short, we construct the path in reverse.

First we need a length √18 gap. This is possible only by moving from one corner dot to the opposite corner dot. For convenience, we label the dots (x,y), where x and y both range from 1 to 4. Then our path can start:

(1,1), (4,4)
(4,4), (1,1)
(1,4), (4,1)
(4,1), (1,4)

So there are four choices for the first pair of elements in the path.

Since the diagram is symmetric, these four choices will proceed in the same way. For convenience, we focus on the case (1,1), (4,4). From (4,4) we need a gap of √13. This requires moving 3 units in one direction and 2 in the other. There are two points that meet this requirement: (2,1) and (1,2). Symmetry is still preserved, so we focus on the case (2,1).

Next we need a gap of √10. This requires a movement of 3 units in one direction and 1 in the other. There are two points meeting this condition: (1,4) and (3,4).

However, (1,4) will not work. We next need a gap of distance √9, which requires a separation of 3 in one direction and 1 in the other. The points meeting this constraint are (1,1) and (4,4), but both of these are already in the path.

So we must proceed with (3,4). The only point of distance √9 from (3,4) is (3,1), so this move is forced. From here, we need a gap of distance √8, which requires a separation of 2 units in both directions. The only point meeting this constraint is (1,3), so this move is also forced.

Next we need a distance of √5, with separations of 2 and 1. (2,1), (3,2), and (3,4) all meet this constraint, but only (3,2) is unused, so this move is also forced.

Next we need a distance of &rdic;4, with separations of 2 and 0. (3,4) and (1,2) both meet this constraint, but only (1,2) is unused, so this move is also forced.

Next we need a distance of √2, with a separation of 1 unit in both directions. (2,1) and (2,3) meet this constraint, but only (2,3) is unused, so this move is also forced.

Finally we need a distance of √1, which requires a move of a single unit horizontally or vertically. There are four points meeting this constraint, but (3,1) is already in the path. The other three are available.

Looking back, we had 4 choices for the first move, 2 for the second, and 3 for the final. All other moves were forced. So there are a total of 4 x 2 x 3 = 24 paths of length 10.

The desired answer is then 10 x 24 = 240.

Summer Problem Solving Marathon Question #31

[Value = 4 points]

ABC is an equilateral triangle with side length 2. D is the midpoint of AC. Point E is chosen so that C is the midpoint of BE. What is the area of the triangle CDE?

Summer Problem Solving Marathon Solution #28

Working from right to left, we find that the first place with an obvious error is 4 + 2 = 1 in the 10,000 place. Assuming the previous parts are correct, we should have carried 1 from the previous place. So we have two possibilities:

(1) The 1 is correct, and the sum should be 11. Then either 4 needs to be replaced by 8, or 2 needs to be replaced by 6. However, replacing 4 with 8 creates an error in the hundreds place, so this possibility can be ignored.

(2) The 1 is incorrect. Then the 4 and 2 are correct, so 1 needs to be replaced with 7. But replacing 1 with 7 creates an error in the tens place.

So the only option is that 2 is replaced with 6. Making this substitution throughout, we find that the result is a correct addition. So the answer is 2 + 6 - 8.

Summer Problem Solving Maraton Question #30

[Value = 4 points]

Let ABCD be a quadrilateral in which AB is of length 18, BC is of length 21, and CD is of length 14. Let angle ABC be right, and let the diagonal AC be perpendicular to CD. What is the perimeter of ABCD?

Summer Problem Solving Marathon Solution #27

There are eight triangles of minimal size in the diagram. For convenience, number these 1 through 8, reading left to right on the top half of the square and then left to right on the bottom half of the square.

There are four triangles each made up of two of the smallest triangles -- one made from triangles 2 and 3, one made from 4 and 8, one made from 6 and 7, and one made from 1 and 5.

There are no triangles of size 3. There are four triangles of size 4 -- one made of triangles -- one made from triangles 1, 2, 3, and 5; one made from triangles 2, 3, 4, and 8; one made from triangles 4, 6, 7, and 8; and one made from triangles 1, 5, 6, and 7.

There are no triangles of any other size. Thus there are a total of 8 + 4 + 4 = 16 triangles in the diagram.

Summer Problem Solving Marathon Question #29

[Value = 6 points]

Consider a 4 x 4 square array of points, in which points immediately adjacent horizontally or vertically are 1 unit apart in distance.

A growing path is a non-repeating sequence of distinct points in the array, in which the distances between consectuive points in the path is strictly increasing.

Let x be the length of the longest possible growing path, and let y be the number of growing paths of that length. What is xy?

Summer Problem Solving Marathon Question #28

[Value = 2 points]

The following addition is incorrect:

742586 + 829430 = 1212016

However, it can be made correct by uniformly one of the digits in the problem throughout with another digit. What is the sum of the digit to be replaced and the digit that replaces it?

Summer Problem Solving Marathon Question #27

[Value = 1 point]

Let ABCD be a square. Draw in the diagonals AC and BD. Let E be the midpoint of AB, F be the midpoint of BC, G be the midpoint of CD, and H be the midpoint of DA. Draw in the line segments EG and FH.

How many triangles are there in the resulting diagram?

Summer Problem Solving Marathon Solution #26

We solve the problem by casework on the size of B. For convenience, let S = {1,2,3,4,5,6}.

If the size of B is zero, then A is definitely a subset of S - B = S. There are 2⁶ = 64 subsets of S, so there is a 1/64 chance that the size of B is zero. This gives us a 1/64 chance of success from this option.

The case when the size of B is six is similar -- in this case, A is definitely a subset of S. And there is a 1/64 chance that the size of B is 6. More generally, we will get the same results for size of B = 1 and size of B = 5, and for size of B = 2 and size of B = 4.

So far we have 1/64 + 1/64 chance of success, from the cases size B = 0 and size B = 6. Now consider the case when the size of B = 1. There are 6 subsets of size 1, so the chance of choosing such a subset is 6/64. If the size of B is 1, then there are two choice of A that are subsets of B. S - B is then of size 5, so there are 32 choices of A that are subsets of S - B. That makes a total of 32 + 2 - 34. However, we have counted the empty set twice, so in fact there are 33 choices of A. So we have a 6/64 chance of picking B of size 1, and then a 33/64 chance of picking A that works. This gives us a (33 x 6)/64² = 198/64² chance of success. Since the case when size B = 5 is the same, we double this to get 396/64².

Next we consider the case when the size of B is 2. There are 6 choose 2 = 15 ways of picking such a subset. There are then 4 choices of A that are subsets of B, and 16 choices of A that are subsets of S - B, for a total of 20. Again, we have double-counted the empty set, so we subtract 1 to get 19. So the chance of success when size B = 2 is (15 x 19)/64² = 285/64². Again we double this for the case size B = 4, to get 570/64².

Finally, we consider the case when the size of B is 3. There are 6 choose 3 = 20 ways of picking such a subset. There are then 8 choices of A that are subsets of B, and 8 choices of A that are subsets of S - B, for a total of 16. Subtracting the double-counted empty set, we get 15 choices of A. So the probability of success here is (20 x 15)/64² = 300/64².

Adding up the cases, we have (64 + 64 + 396 + 570 + 300)/64² = 1394/2¹² = 697/2¹¹ = 697/2048.

Summer Problem Solving Marathon Solution #25

The base of triangle AEF is AE, which is 3/4 the length of AB. Since F is the midpoint of BC, the height of triangle AEF is 1/2 the height of triangle ABC. So the area of AEF is (3/4)(1/2) = 3/8 the area of triangle ABC. Thus the area of AEF is (3/8)96 = 36.

Summer Problem Solving Marathon Solution #24

The first perpendicular is drawn from point C to point D. The second is drawn from point D to point E. Let points F, G, H, and so on be the ending points of the further perpendiculars.

CDE forms a right triangle, as does DEF. Furthermore, CD is parallel to EF (because both meet OB at a right angle). Thus the triangles CDE and DEF are similar. In fact, we form an infinite sequence of similar triangles -- CDE, DEF, EFG, FGH, and so on.

CD is the hypotenuse of CDE, and DE is the hypotenuse of DEF. CD is of length a, and DE is of length b. So the ratio of similarity between DEF and CDE is b/a. Side EF of DEF corresponds to side DE of CDE. So EF is of length b(b/a) = b²/a.

EF is the hypotenuse of EFG, and DE is the hypotenuse of DEF. So the ration of similarity between EFG and DEF is (b²/a)/b = b/a. In fact, we can easily see that the ratio of similarity between consecutive similar triangles in our sequence is always b/a. So side FG is EF(b/a) = (b²/a)(b/a) = b³/a².

Generalizing, we get a sequence of lengths:

a, b, b²/a, b³/a², b⁴/a³, ...

This is an infinite sequence with first term a and ratio b/a. Its sum is then:

a/(1 - b/a) = a((a-b)/a) = a²/(a-b)

Summer Problem Solving Marathon Question #26

[Value = 4 points]

Two subsets A and B are chosen at random from the set {1,2,3,4,5,6}. What is the probability that either A is a subset of B, or A is a subset of {1,2,3,4,5,6} - B?

Summer Problem Solving Marathon Question #25

[Value = 1 point]

The area of triangle ABC is 96. D is the midpoint of AB, E is the midpoint of DB, and F is the midpoint of BC. What is the area of the triangle AEF?

Summer Problem Solving Marathon Solution #23

Let D be the vertex of the rhombus on side AB of the triangle, E be the vertex of the rhombus on side AC, and F be the vertex of the rhombus on side BC.

Opposite sides of a rhombus are parallel, so DF is parallel to AE, and hence also to AC. So the triangles BDF and BAC are similar. Thus their sides are in the same ratio. Let s be the side length of the rhombus. Then BD = 12-s. The ratio of BD to BA is the same as the ratio of DF to AC. So we have:

(12 - s)/12 = s/6

Multiplying both sides by 12, we have:

12 - s = 2s
12 = 3s
s = 4

This gives us the side length of the rhombus, but not yet the area. To find the area of the rhombus, we view it as two congruent triangles -- ADE and FDE. To find the area of ADE, we use the area formula:

Area = (ab sin C)/2

In this case we have the two sides, so we just need the sine of the included angle. That angle is angle BAC.

We now use the law of cosines applied to triangle ABC. Recall that the law of cosines says:

c² = a² + b² - 2ab cos C

We use the particular instance in which c is side BC, a is side AB, b is side AC, and C is angle BAC (which we will call angle A). Then we have:

64 = 144 + 36 - 2(12)(6)cos A
64 = 180 - 144cos A
144cos A = 114
cos A = 114/144

However, we want sin A. So we use the fact that cos² A + sin² A = 1, so sin A = (1 - cos² A)^1/2.

So sin A = (1 - (114/144)²)^1/2 = ((144 - 114)²/144²)^1/2 = ((144 + 114)(144-114))^1/22/144 = (258 x 30)^1/2/144 = (6/144)√215 = (1/24)√215.

The area of triangle ADE is then (1/2)(4)(4)(1/24)√215 = (1/3)√215. The area of the rhombus is twice this, or (2/3)√215.

Summer Problem Solving Marathon Question #24

[Value = 9 points]

Let AOB be an acute angle, and Let C be a point on OA. From C, a perpendicular is drawn to OB, meeting OB at point D. The length of CD is a. From D, a perpendicular is drawn to OA, meeting OA at E. The length of DE is b. From E, a perpendicular is drawn to OB. From the base of this perpendicular, a perpendicular is drawn to OA. This procedure is repeated infinitely many times. What is the sum of the lengths of the perpendiculars dawn?

Summer Problem Solving Marathon Solution #22

In any quadratic equation with a leading coefficient of 1, the coefficient of x is the negative of the sum of the roots, and the constant is the product of the roots. So in our equation x² - ax + b = 0, the sum of the roots is a, and the product of the roots is b.

We are told that a, written in base n, is 18. Thus a = n + 8. Since n is one of the roots, the other root must be n.

The product of the roots is then b. So b = 8n. Written in base n, this is 80.

Summer Problem Solving Marathon Question #23

[Value = 4 points]

In triangle ABC, side AB is of length 12, side BC of length 8, and side AC of length 6. A rhombus is inscribed in ABC, with one vertex of the rhombus being at point A and two sides of the rhombus lying along AB and AC. What is the area of the rhombus?

Week Four Standings

Points through the end of the fourth week of the summer problem solving marathon:

Matthew: 122
Amelia: 62
Michael: 27
Sophia: 20
Paul: 3
Lily: 2
Ben: 2
Sharon: 2

Summer Problem Solving Marathon Question #22

[Value = 3 points]

The equation x² - ax + b = 0 has an integer solution n, where n is greater than 8. The coefficient a, written in base n, is 18. What is coefficient b written in base n?

Summer Problem Solving Marathon Solution #21

Let a be Ann's current age, and let b be Barbara's age. Then we are told a + b = 44.

Now consider the constraint that "Barbara is as old as Ann was when Barbara was as old as Ann had been when Barbara was half as old as Ann is". We will work out way backward through this.

When Barbara was half as old as Ann is, she was a/2. Since she is now b, that was b - a/2 years ago.

b - a/2 years ago, Ann was a - (b - a/2) = 3a/2 - b years old.

So Barbara was that age b - (3a/2 - b) = 2b - 3a2 years ago.

2b - 3a/2 years ago, Ann was a - (2b - 3a/2) = 5a/2 - 2b years old.

That's the age Barbara is now. So b - 5a/2 - 2b, or 3b = 5a/2, or b = 5a/6.

We now substitute this in a + b = 44. So a + 5a/6 = 44, or 11a/6 = 44, or a = 24.

Summer Problem Solving Marathon Solution #20

Let A be the vertex opposite the side of length 39, B be the vertex opposite the side of length 40, and C be the vertex opposite the side of length 25. Drop a perpendicular from A to BC. Let D be the point at which the perpendicular intersects BC.

Then BDA and CDA are both right triangles. Let BD be of length x. Then CD is of length 39 - x. Let y be the length of AD. Then, using the Pythagorean Theorem on the two right triangles, we get the following two equations:

x² + y² = 25²
(39 - x)² + y² = 40²

Subtracting the first equation from the second, we have:

(39 - x)² - x² = 40² - 25²

We factor each side using the difference of squares:

(39 - x + x)(39 - x - x) = (40 - 25)(40 + 25)
39(39 - 2x) = 15(65)
1521 - 78x = 975
78x = 546
x = 7

We then determine y:

7² + y² = 25²
49 + y² = 625
y² = 576
y = 24

Now let O be the center of the circle. Draw the two radii OA and OC. Angle CBA and angle COA both cut off the same arc of the circle. CBA is an inscribed angle and COA is a central angle, so COA is twice the size of CBA. Now drop a perpendicular from O to AC, intersecting AC at E. This perpendicular bisects AC, so AEO and CEO are congruent triangles, and the two angles AEO and CEO are both congruent to angle CBA.

Angle CBA is also an angle in the right triangle DBA. Its sine is thus AD/AB = 24/25. So the sine of angle AEO is also 24/25. Now side AE is of length 20, so OA is 20/(24/25) = 500/24 = 125/6.

This gives us the length of a radius of the circle. The length of the diameter is then twice that, or 125/3.

Summer Problem Solving Marathon Solution #19

Let the amount of money paid by the four kids be a, b, c, and d, respectively. Then we are given a + b + c + d = 60.

The first kid paid half as much as the other three combined. So a = (b + c + d)/2. Substituting this into the original equation, we have:

(b + c + d)/2 + b + c + d = 60

Or:

(b + c + d)3/2 = 60

b + c + d = 40.

Hence a = 20.

Next we are told that the second kid paid one-third as much as the other kids combined. So b = (a + c + d)/3 = (20 + c + d)/3. Substituting this into b + c + d = 40, we have:

(20 + c + d)/3 + c + d = 40

Or (multiplying both sides by 3):

20 + c + d + 3c + 3d = 120

4c + 4d = 100

c + d = 25.

Hence b = 15.

Next, we are told that the third kid paid one-fourth as much as the other kids combined. So c = (a + b + d)/4 = (35 + d)/4. Substituting this into c + d = 25, we have:

(35 + d)/4 + d = 25

35 + d + 4d = 100

35 + 5d = 100

5d = 65

d = 13

So the amount of money paid by the fourth kid is $13.

Summer Problem Solving Marathon Question #21

[Value = 5 points]

Barbara is as old as Ann was when Barbara was as old as Ann had been when Barbara was half as old as Ann is. The sum of their current ages is 44. How old is Ann?

Summer Problem Solving Marathon Solution #18

Call the triangle ABC, and the center of the circumscribed circle O. Draw the lines OA and OB. The result is the triangle OAB. Angle O is a 120 degree angle (because it cuts off a third of the circle), and side AB is length p/3.

Now drop a perpendicular from O to AB. Call the point of intersection D. Consider triangle ODA. This is a 30-60-90 right triangle. (It is a right triangle because OD is a perpendicular to AB, and since OAB is isoceles, the perpendicular bisects angle O, so DOA is a 60 degree angle).

In this 30-60-90 right triangle, side AD is of length p/6. This side corresponds to the 60 degree angle. So side OA is (p/6)(2/√3) = p/(3√3).

But OA is a radius of the circumscribed circle. So the area of that circle is π(p/3√3)² = (πp²)/27.

Summer Problem Solving Marathon Question #20

[Value = 7 points]

A triangle with side lengths 25, 39, and 40 has a circle circumscribed about it. What is the diameter of the circle?

Summer Problem Solving Marathon Question #19

[Value = 2 points]

Four kids bought a boat for $60. The first paid half what the other kids combined paid. The second paid a third what the other kids combined paid. The third paid a fourth what the other kids combined paid. How much did the fourth pay?

Summer Problem Solving Marathon Solution #17

Let x be the number that is added. Then 20 + x, 50 + x, and 100 + x form a geometric sequence. Let r be the common ratio of that sequence. Then:

50 + x = (20 + x)r
100 + x = (50 + x)r

From the first equation, we have r = ^{(50 + x)}⁄_{(20 + x)}. From the second, we have r = ^{(100 + x)}⁄_{(50 + x)}. So we have:

^{(50 + x)}⁄_{(20 + x)} = ^{(100 + x)}⁄_{(50 + x)}

Cross-multiplying:

(50 + x)(50 + x) = (20 + x)(100 + x)

Or:

x² + 100x + 2500 = x² + 120x + 2000

20x = 500

x = 25

So r = (50 + 25)/(20 + 25) = 75/45 = 5/3.

Summer Problem Solving Marathon Question #18

[Value = 4 points]

When an equilateral triangle of perimeter p is inscribed in a circle, what is the area of the circle?

Week Three Standings

Points through the end of the third week of the summer problem solving marathon:

Matthew: 98
Amelia: 41
Michael: 23
Sophia: 20
Paul: 3
Lily: 2
Ben: 2

Summer Problem Solving Marathon Question #17

[Value = 2 points]

Some particular number is added to each of 20, 50, and 100. The result is a geometric sequence. What is the ratio of the sequence?

Summer Problem Solving Marathon Solution #16

Because the sides of the triangle are consecutive integers, we can call them x, x+1, and x+2 for some integer x.

We start with the law of sines. Call the three angles of the triangle A, B, and C, and let a, b, and c be the lengths of the sides opposite angles A, B, and C respectively. Then the law of sines says:

^a⁄_{sin A} = ^b⁄_{sin B} = ^c⁄_{sin C}

In our triangle, the largest angle is twice the smallest angle. Call the smallest angle θ. Then the largest angle is 2θ. The smallest angle is opposite the shortest side, and the largest angle is opposite the longest side. So from the law of sines, we have:

^x⁄_{sin θ} = ^x+2⁄_{sin 2θ}

Next we use the fact that sin 2θ = 2sinθcosθ. Making this substitution, we get:

^x⁄_{sin θ} = ^x+2⁄ _2sinθcosθ

Then multiplying both sides by sinθ, we have:

x = ^x+2⁄ _2cosθ

So cosθ = ^x+2⁄_2x

Now we appeal to the law of cosines. The law of cosines says:

c² = a² + b² - (2ab)cos C

We'll take the instance in which C is the smallest angle (θ). Then c = x, and a and b are x+1 and x+2. So we have:

x² = (x+1)² + (x+2)² - 2(x+1)(x+2)cos θ

But we already know that cosθ = ^x+2⁄_2x. Making this substitution, we have:

x² = (x+1)² + (x+2)² - 2(x+1)(x+2)(^x+2⁄_2x)

Or:

x² = x² + 2x + 1 + x² + 4x + 4 - (x³ + 5x² + 8x + 4)/x

Or (multiplying both sides by x):

x³ = x³ + x² - 3x - 4

Or:

x² - 3x - 4 = 0

Or:

(x - 4)(x + 1) = 0

So x = 4 or x = -1. But since x is the length of the shortest side, it cannot be negative. So x = 4. Thus the cosine of the shortest side is (4 + 2)/8 = 3/4.

Summer Problem Solving Marathon Solution #15

[No correct answers were submitted for this problem.]

We will use the method of partial fractions. We take the fraction ¹⁄_(n²-4) and note that the denominator factors as (n+2)(n-2). We will thus try to rewrite the fraction as the sum of a fraction with denominator n+2 and a fraction with denominator n-2.

So we want to find numerators A and B such that:

¹⁄_(n²-4) = ^A⁄_(n+2) + ^B⁄_(n-2)

Multiplying both sides by (n+2)(n-2), we obtain:

1 = A(n-2) + B(n+2)

Rearranging terms, we have:

1 = (A+B)n + 2(B-A)

This identity needs to hold for all values of n. So we need to have:

A + B = 0
2(B - A) = 1

Solving this pair of equations, we get A = -1/4 and B = 1/4.

So ¹⁄_(n²-4) = (¹⁄_(n-2) - ¹⁄_(n+2))/4.

Now, the sum we are trying to approximate is:

1000(¹⁄_(3²-4) + ¹⁄_(4²-4) + ¹⁄_(5²-4) + ... + ¹⁄_(10000²-4))

Using our partial fraction decomposition, we can rewrite this as:

1000((¹⁄_(3-2) - ¹⁄₍₃₊₂₎)/4 + (¹⁄_(4-2) - ¹⁄₍₄₊₂₎)/4 + (¹⁄_(5-2) - ¹⁄₍₅₊₂₎)/4 + ... + (¹⁄_(10000-2) - ¹⁄_(10000+2))/4)

Factoring out the common 1/4 factor, we obtain:

250(¹⁄_(3-2) - ¹⁄₍₃₊₂₎ + ¹⁄_(4-2) - ¹⁄₍₄₊₂₎ + ¹⁄_(5-2) - ¹⁄₍₅₊₂₎ + ... + ¹⁄_(10000-2) - ¹⁄_(10000+2))

= 250(1 - ¹⁄₅ + ¹⁄₂ - ¹⁄₆ + ¹⁄₃ - ¹⁄₇ + ... + ¹⁄₉₉₉₈ - ¹⁄₁₀₀₀₂)

The series inside the parentheses is an example of a telescoping series. The same terms show up both positive and negative, and cancel out. Examining the series, we see that the positive terms 1, 1/2, 1/3, and 1/4 do not cancel out, and the negative terms -1/9999, -1/10000, -1/10001, and -1/10002 do not cancel out. All other terms do. So our expression reduces to:

250(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002)

1 + 1/2 + 1/3 + 1/4 = 2.083. Multiplying this by 250 gives a bit more than 520.8. From this, we have to subtract 250 times (1/9999 + 1/10000 + 1/10001 + 1/10002). But this is approximately 1000(1/10000) = .1, so subtracting it will leave roughly 520.7. We want the nearest integer, which will be 521.

Summer Problem Solving Marathon Solution #14

Dividing by the leading coefficient, we have x² - ⁴⁄₃x + k/3 = 0.

In a quadratic equation in which the x² term has a coefficient of 1, the sum of the roots is the negative of the x coefficient. So the sum of the two roots in our equation is 4/3.

We thus want two real numbers whose sum is 4/3 and whose product is as large as possible. Clearly both must be positive. In fact, we maximize the product by taking each to be half of 4/3. (This is equivalent to the problem of maximizing the area of a rectangle with fixed perimeter, which is done by making the rectangle a square.)

So the two roots are both 2/3. Now we need to find k. In a quadratic equation in which the x² term has a coefficient of 1, the constant term is the product of the roots. So k/3 = (2/3)(2/3), or k/3 = 4/9, or k = 4/3.

Summer Problem Solving Marathon Question #16

[Value = 8 points]

In a certain triangle, the sides are consecutive integers, and the largest angle is twice the smallest angle. What is the cosine of the smallest angle?

Summer Problem Solving Marathon Question #15

[Value = 6 points]

What integer is closest to the value of 1000 times the sum of the expression ¹⁄_n²-4 evaluated for every integer value of n from n=3 to n = 10000 inclusive.

(That is, what is the closest integer to 1000(¹⁄_3²-4 + ¹⁄_4²-4 + ... + ¹⁄_10000²-4)?)

Summer Problem Solving Marathon Solution #13

If we drop perpendiculars from A and B to side DC, hitting that side at E and F respectively, then AEFB is a rectangle of length 5 and unknown width.

Triangle BFC is then a 45-45-90 right triangle. The sides of a 45-45-90 right triangle are in the ratio 1-1-√2. In this case, the hypotenuse BC is 3√2, so the two legs are each of length 3. Thus BF is 3, and so AF and FB must also be 3.

Triangle AED is a 30-60-90 right triangle. The sides of a 30-60-90 right triangle are in the ratio 1-√3-2. In this case, the side AE, which is opposite the 60 degree angle, is of length 3. So side DE, opposite the 30 degree angle, is of length 3/√3, which simplifies to √3.

Side DC is then the sum of DE, EF, and FC. That sum is 5 + 3 + √3 = 8 + √3.

Summer Problem Solving Marathon Solution #12

Note that since 4² = 16, and since 554 ends in a 4, b must be such that 16 in base b ends in a 4. That means that the remainder when 16 is divided by b is 4. There are then only two possibilities: b = 6 and b = 12.

We then note that in base 10, 24² = 576. Since in base b, 24² = 554, b must be larger than 10. (Try some cases if you don't see why this works.) It thus follows than b = 12. Testing with the actual values, we see that this is correct.

(This can also be solved algebraically, by saying that (2b² + 4)² = 5b³ + 5b² + 4 and solving for b. But the above method is a lot quicker.)

Summer Problem Solving Marathon Question #14

[Value = 5 points]

Find the value of k such that both roots of 3x² - 4x +k = 0 are real, and the product of the two roots is maximized.

Week Two Standings

Points through the end of the second week of the summer problem solving marathon:

Matthew: 66
Amelia: 29
Sophia: 20
Michael: 14
Paul: 3
Lily: 2
Ben: 2

Summer Problem Solving Marathon Question #13

[Value = 3 points]

ABCD is a trapezoid with AB||DC, AB=5, and BC = 3√2. Angle BCD is 45 degrees, and angle CDA is 60 degrees. What is the length of DC?

Summer Problem Solving Marathon Solution #10

We start by considering the non-empty subsets of {1,2,...,9}. We can divide these into two groups:

(i) Those containing 9.
(ii) Those not containing 9.

Note that, if we set aside the set {9}, there is a perfect correlation between the two groups -- for each subset containing 9, there is a unique subset not containing 9, obtained by removing 9 from it.

Furthermore, compare the power sum of two correlated sets, such as {1,3,7} and {1,3,7,9}. The two power sums are:

(i) 7i³ + 3i² + i¹
(ii) 9i⁴ + 7i³ + 3i² + i¹

That is, the two power sums are the same, except that the power sum of the 9-containing set has one further term which is 9 times some power of i. This is because in the power sum the elements of the set are listed in decreasing order, and 9 is always the largest element in any set containing it.

As a result, S₉ splits into three parts:

(i) The power sums of the sets not containing 9.
(ii) Those same power sums again, appearing as parts of the power sums of the 9-containing sets (the 9-free parts).
(iii) The terms generated by the 9s, which all have the form of 9 times a power of i, where the power of i is equal to the size of the 9-containing set. (In this part we will include the power sum of the set {9}.)

Now, part (i) is just S₈. Part (ii) is S₈ again. We are given S₈, so these two parts can be quickly calculated to be twice -176 - 64i, or -352 - 128i. So all that remains is to calculate part (iii).

Any 9-containing set of size n will generate a term of the form 9iⁿ. How many 9-containing sets of size n are there? Well, to form such a set, we select n-1 elements from {1,2,...,8}. There are C(8,n-1) ways to do this (8 choose n-1). So we have:

(i) One term of the form 9i, for a total of 9i.
(ii) 8 terms of the form 9i², for a total of -72.
(iii) 28 terms of the form 9i³, for a total of -252i.
(iv) 56 terms of the form 9i⁴, for a total of 504.
(v) 70 terms of the form 9i⁵, for a total of 630i.
(vi) 56 terms of the form 9i⁶, for a total of -504.
(vii) 28 terms of the form 9i⁷, for a total of -252i.
(viii) 8 terms of the form 9i⁸, for a total of 72.
(ix) One term of the form 9i⁹, for a total of 9i.

Summing this up, we have 144i.

So S₉ = -352 - 128i + 144i = -352 + 16i.

Summer Problem Solving Marathon Question #12

[value = 1 point]

When numbers are written in base b, 24² = 554. What is b?

Summer Problem Solving Marathon Solution #11

Let r be the solution of x² - 3x + c = 0 whose negative -r is a solution of x² + 3x - c = 0. Then we can substitute r for x in the first equation, and -r for x in the second. This yields:

r² - 3r + c = 0
r² - 3r - c = 0

Subtracting the second equation from the first, we obtain 2c = 0, so c = 0.

We want the solutions of x² - 3x + c = 0. Since c = 0, we thus want the solutions of x² - 3x = 0.

So we have x(x-3) = 0, and the solutions are x = 0 and x = 3.

Summer Problem Solving Marathon Solution #9

Let P_n be the probability of winning the game when it is played with n pairs of cards in the deck. Clearly P₁ = P₂ = 1, because the game cannot be lost with either 1 or 2 pairs in the deck.

Suppose we already know the value of P_n-1, and we are thinking about what happens in the game with n pairs. Here are four possibilities:

#1: The first two cards we draw form a pair. This pair is then discarded, and from this point on, the game proceeds as if it were a game with n-1 pairs, so the probability of victory from this point is P_n-1.

#2: The first two cards are different, but the third card matches the first. Then the pair formed by the first and third cards is discarded, and play continues as if it were a game with n-1 pairs (with the first card already drawn, but this does not affect the probabilities). So the probability of victory from this point is P_n-1.

#3: The first two cards are different, but the third card matches the second. Then the pair formed by the second and third cards is discarded, and play continues as if it were a game with n-1 pairs. Again, the probability of victory from this point is P_n-1.

#4: The first three cards are all different. The player then loses.

These four options are exclusive and exhaustive, so the sum of their probabilities of occurring must be 1. Consider the probability of the fourth option. Once the first card is drawn, 2n-1 cards remain. 2n-2 of these are different from the first card, so there is a ^2n-2⁄_2n-1 chance that the first two cards will be different.

After the second card has been drawn, 2n-2 cards remain. Of these, 2n-4 are different from both the first and the second card. So the probability that the third card will not match the first or the second is ^2n-4⁄_2n-2.

So the chance that the first three cards are all different is ^2n-2⁄_2n-1 x ^2n-4⁄_2n-2 = ^2n-4⁄_2n-1.

So the chance that one of the first three options occurs is 1 - ^2n-4⁄_2n-1 = ^{2n-1 - (2n-4)}⁄_2n-1 = ³⁄_2n-1.

Given that one of the first three options occurs, the probability of victory is P_n-1>. So P_n, the overall probability of victory when there are n pairs, is ^3P_n-1⁄_2n-1.

We already know that P₂ = 1. So using this formula, we have:

P₃ = ^3x1⁄₅ = ³⁄₅.
P₄ = ^{3 x 3⁄₅}⁄₇ = ⁹⁄₃₅.
P₅ = ^{3 x 9⁄₃₅}⁄₉ = ³⁄₃₅.
P₆ = ^{3 x 3⁄₃₅}⁄₁₁ = ⁹⁄₃₈₅.

So the probability of victory is ⁹⁄₃₈₅.

Summer Problem Solving Marathon Question #11

[Value = 3 points]

The negative of one of the solutions of x² - 3x + c = 0 is a solution of x² + 3x - c = 0.

What are the solutions of x² - 3x + c = 0?

Summer Problem Solving Marathon Solution #8

The equation x² + y² = 4 has as its graph a circle of radius 2 centered at the origin. The equation y = |x| then cuts off a 90 degree sector of that circle. The area of the whole circle is 4π, so the area of the 90 degree sector is ¹⁄₄ of that area, or π.

Summer Problem Solving Marathon Question #10

[Value = 9 points]

Let {a₁, a₂, ..., a_n} be a set of real numbers, such that a₁ < a₂ < ... < a_n. We define the power sum of {a₁, a₂, ..., a_n} to be a₁i¹ + a₂i² + ... + a_niⁿ.

Given any n, let S_n be the sum of the power sums of all non-empty subsets of the set {1,2,...,n}.

S₈ is -176 - 64i. What is S₉?

Summer Problem Solving Marathon Solution #7

The averages of X and Y are 14 apart. The average of X ∪ Y is 6 above that of Y and 8 below that of X. So the sizes of X and Y must be in the ratio of 6:8.

The averages of Y and Z are 18 apart, with the average of Y ∪ Z 10 above that of Y and 8 below that of Z. So the sizes of Y and Z must be in the ratio of 8:10.

Thus X, Y, and Z are in the ratio of 6:8:10. For convenience, we can take them to in fact be of sizes 6, 8, and 10.

Then X has 6*37 = 222 years of age in it. Y has 8*23 = 184 years of age in it. And Z has 10*41 = 410 years of age in it. That's a total of 222 + 184 + 410 = 816 years of age. There are a total of 24 people in X, Y, and Z combined, so the average age is 816/24 = 34.

Summer Problem Solving Marathon Question #9

[Value = 8 points]

A deck of cards contains twelve cards: two marked with the letter A, two with the letter B, two with the letter C, two with the letter D, two with the letter E, and two with the letter F.

A person plays a game with these cards in the following way. The deck is randomized. The player then draws cards one at a time from the deck. Any time the player has two matching cards in his hand, he discards them. If the player ever has three unmatched cards in his hand, he loses. If he draws every card from the deck without losing, he wins.

What is the probability that the player wins the game?

Summer Problem Solving Marathon Question #8

[Value = 3 points]

What is the area of the smallest region bounded by the graphs of y = |x| and x² + y² = 4?

Week 1 Standings

Points through the end of the first week of the summer problem solving marathon:

Matthew: 28
Amelia: 16
Sophia: 13
Michael: 5
Paul: 3
Lily: 2
Ben: 2

Summer Problem Solving Marathon Question #7

[Value = 6 points]

X, Y, and Z are pairwise disjoint sets of people. The average age of people in X is 37. The average age of people in Y is 23. The average age of people in Z is 41. The average age of people in X ∪ Y is 29. The average age of people in X ∪ Z is 39.5. The average age of people in Y ∪ Z is 33.

What is the average age of people in X ∪ Y ∪ Z?

Summer Problem Solving Marathon Solution #6

Since n + 125 and n + 199 are both perfect squares, we have n + 125 = a² and n + 199 = b² for some a and b. Subtracting the first equation from the second, we have:

b² - a² = 76

or:

(b + a)(b - a) = 76

This yields three possibilities:

#1: b + a = 76, b - a = 1
#2: b + a = 38, b - a = 2
#3: b + a = 19, b - a = 4

But only the second yields an integer value for b. So b = 20, and a = 18. So n + 201 = 20² = 400, and n = 199.

Summer Problem Solving Marathon Solution #5

We have d = ⁴⁄_c. c = πd, so by substitution d = ⁴⁄_πd. Cross-multiplying, πd²=4, and d²=⁴⁄_π.

Since r = ^d⁄₂, r² = ^d2⁄₄. So the area of the circle is πr² = π(^d2⁄₄) = π(¹⁄_π) = 1.

Summer Problem Solving Marathon Question #6

[Value = 2 points]

What is the smallest positive integer n such that n + 125 and n + 201 are both perfect squares?

Summer Problem Solving Marathon Solution #4

Consider first the general form of the binomial theorem. If we have (x + y)ⁿ, the full product will have terms of the form xⁿy⁰, x^n-1y¹, x^n-2y², ..., x¹y^n-1, x⁰yⁿ.

The coefficients of these terms will then be C(n,0), C(n,1), C(n,2), ..., C(n,n-1), C(n,n), where C is the "choose" function.

In our case, y is -(x^-1⁄₂). If we number the terms in the expansion from 0 to n=7, the k^th term will be C(7,k) x^7-k(-(x^-1⁄₂))^k. This simplifies to (-1)^kC(7,k) x^{(7-k) - k⁄₂} = (-1)^kC(7,k) x^{7- 3k⁄₂}.

We want the term in which the exponent of x is ^-1&frasl₂. So we want 7 - ^3k⁄₂ = ^-1&frasl₂, or 14 - 3k = -1, or 3k = 15, or k = 5.

The coefficient is then (-1)⁵C(7,5) = -^7!⁄_(5!2!) = -7*6/2 = -21.

Summer Problem Solving Marathon Question #5

[Value = 1 point]

In a certain circle, the diameter is four times the reciprocal of the circumference. What is the area of the circle?

Summer Problem Solving Marathon Solution #3

From a + d = b + c, we have a = b + c - d. We substitute this for a in bc - ad = 93 to get:

bc - (b + c - d)d = 93

or:

bc - bd - cd + d² = 93

Rearranging terms, we get:

bc - cd + d² - bd = 93

Factoring out c from the first two terms and d from the second two terms, we have:

c(b - d) + d(d - b) = 93

or:

d(d - b) - c(d - b) = 93

Then factoring out d - b, we have:

(d - c)(d - b) = 93

Now d - b and d - c must be integers, and d - b > d - c (since c > b). Since 93 = 3 * 31, there are two possibilities:

#1: d - b = 93, d - c = 1.
#2: d - b = 31, d - c = 3.

On the first possibility, b = d - 93, and c = d - 1. Since a = b + c - d, we have a = d - 93 + d - 1 - d = d - 94.

Since d < 500, the greatest possible value for d is 499. Since a = d - 94 > 0, the smallest possible value for d is 95. Once we pick d, the values of a, b, and c are fixed. There are 499 - 95 + 1 = 405 possible ordered 4-tuples under this possibility.

On the second possibility, b = d - 31 and c = d - 3. So a = b + c - d = d - 31 + d - 3 - d = d - 34.

So d is at most 499 as before, and at smallest 35 (so that a is at least 1). So there are 499 - 35 + 1 = 465 possible value for d on this possibility, and hence 465 possible ordered 4-tuples.

So between the two possibilities, there are 405 + 465 = 870 possible ordered 4-tuples meeting the conditions.

Summer Problem Solving Marathon Solution #2

The perpendicular from A to BC creates two 30-60-90 right triangles:



Because the two triangles ABD and ACD are congruent, BD and CD must both be 6:



In a 30-60-90 right triangle, the sides are in length ratios of 1 - √3 - 2. In the 30-60-90 right triangle ABD, the three sides in this ratio are BD, AD, and AB. Since BD is of length 6, AD must be of length 6√3 and AB of length 6*2 = 12 (which it is):



We now add point E at the midpoint of AD:



Since AD is of length 6√3 and E is its midpoint, ED is of length 3√3.

Now we add the line BE, which we want to find the length of:



This creates the right triangle BDE. We know that BD is of length 6, and DE of length 3√3. Using the Pythagorean Theorem, we then have:

BE =
√ 6^2 + (3√3)^2 

=
√ 36+27 

=
√ 63 

=3√7

Summer Problem Solving Marathon Question #4

[Value = 4 points]

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

Summer Problem Solving Marathon Question #3

[Value = 7 points]

How many ordered four-tuples of integers (a,b,c,d) are there such that:
(i) 0 < a < b < c < d < 500
(ii) a + d = b + c
(iii) bc - ad = 93

Summer Problem Solving Marathon Question #2

[Value = 3 points]

Each side of triangle ABC is length 12. D is the foot of the perpendicular dropped from A on BC, and E is the midpoint of AD. What is the length of BE?

Summer Problem Solving Marathon Solution #1

Let x and y be the two numbers. Then we are told x + y = 10, and xy = 20. We want to find ¹⁄_x + ¹⁄_y.

Converting both fractions to the common denominator xy, we get ^y⁄_xy + ^x⁄_xy, or ^(x+y)⁄_xy.

Substituting our starting values for x + y and xy, we have ¹⁰⁄₂₀, or ¹⁄₂.

[Put any questions, objections, or alternative solutions in the comments.]

Summer Problem Solving Marathon Question #1

[Value = 2 points]

The sum of two numbers is ten, and their product is twenty. What is the sum of their reciprocals?