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!