twenty-third annual unc math contest first round november ... · 14-10-2011 · whole way, he is 3...

Twenty-third Annual UNC Math Contest First Round November, 2014

Rules: 90 minutes; no electronic devices.

The positive integers are 1, 2, 3, 4, . . .

1. A rectangle 20 feet by 100 feet has a fence around its perimeter. There are posts every 5feet along each of the four sides, arranged so that there is one post at each corner. How manyposts are there in all around the perimeter of the rectangle?

2. The bottom edge of the region at the left is a halfcircle whose diameter has length six. The top edge ismade of three smaller, congruent half circles whose di-ameters lie on the diameter of the bigger half circle.What is the area inside the region?

B

A

C

3. The square in the figure has area 36. Points A and Bare midpoints of sides. What is the area of the triangleABC?

4. How many pairs of consecutive positive integers have product less than 300? For example,3x4=12 is less than 300 and 10x11=110 is less than 300. How many such pairs are there? (Count10x11 and 11x10 as the same pair.)

5. The large circle at left is split into two congruentregions by two half circles that meet each other at thecenter of the large circle and meet the large circle atpoints directly above and below the center. There isa straight line that simultaneously bisects (i.e. cuts inhalf) the area of both the regions. What is the slope ofthe line?

TURN PAGE OVER

6. (a) If two different integers are taken at random from among the integers 1, 2, . . ., 1000,what is the probability that the sum of the two integers is odd?(b) If three different integers are taken at random from among the integers 1, 2, . . ., 1000, whatis the probability that the sum of the three integers is odd?

7. Earl E. Bird drives the same route to work each day and leaves at 8 am. He finds that if hegoes 40 miles per hour the whole way he is 3 minutes late. If he goes 60 miles per hour thewhole way, he is 3 minutes early. What speed must he go in order to arrive exactly on time?

8. Suppose p(x) = x

2 + 3x� 7 and q(x) = x

2 � 2x� 99. If a value of x is chosen at randomfrom the interval �100 x 100, then what is the probability that q(p(x)) is negative?

9. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, . . . (After the first two Fibonacci numbers, eachFibonacci number is equal to the sum of the two before it.) If F is the 2014th Fibonacci number,find the remainder when F

F is divided by 7.

IHG

FED

CBA

10. A code is typed by always jumping from a positionon the keypad to another adjacent position, diagonally,vertically, or horizontally. The moving finger, havingwrit, must move on: it cannot stay frozen in position topick a letter twice in a row. The first letter can be anyletter on the keypad. Examples. The code EIE can becreated, but AAB, ACF, and EII cannot. Also, ABC andCBA are different codes.

(a) How many different three-letter codes can be created?(b) How many different four-letter codes can be created?

END OF CONTEST

University of Northern Colorado Mathematics Contest

Problems are duplicated and solved by Ming Song ([email protected]) 1

University of Northern Colorado Mathematics Contest 2014-2015

Problems of First Round with Solutions

1. A rectangle 20 feet by 100 feet has a fence around its perimeter. There are posts every 5

feet along each of the four sides, arranged so that there is one post at each corner. How

many posts are there in all around the perimeter of the rectangle?

Answer: 48

Solution:

The perimeter is

.

The number of posts is

.

2. The bottom edge of the region below is a half circle whose diameter has length six. The top

edge is made of three smaller, congruent half circles whose diameters lie on the diameter of

the bigger half circle. What is the area inside the region?

Answer:

Solution:

The area is

.

3. The square in the figure has area 36. Points A and B are midpoints of sides. What is the area

of the triangle ABC?

Answer:

A

B

C



Solution:

The red-shaded part, the green-shaded part and the blue-shaded part have areas equal to , ,

and of the area of the square respectively.

So triangle ABC’s area is of the area of the square.

The answer is .

4. How many pairs of consecutive positive integers have product less than 300? For example,

is less than 300 and is less than 300. How many such pairs are there?

(Count and as the same pair.)

Answer: 16

Solution:

Note that

and .

So there are 16 pairs, 1×2, 2×3, 3×4, …, 16×17.

5. The large circle at left is split into two congruent regions by two half circles that meet each

other at the center of the large circle and meet the large circle at points directly above and

below the center. There is a straight line that simultaneously bisects (i.e. cuts in half) the

area of both the regions. What is the slope of the line?

Answer:

A

B

C



Solution:

If we draw the other two small semicircles, we see the four regions of the same area:

So each of two original regions (green and yellow parts) is equal to two small circles in area. The

half of one region is equal to one small circle in area.

Let O be the center. Let A be the lowest point of the large circle.

Draw OA. Draw radius OB cutting the right original region into equal parts in area. Then the

sector OAB of the large circle must equal half the area of one small circle. Since the large circle

has an area equal to 4 times the area of one small circle. Therefore, sector OAB must be of the

circle. That is, .

Draw line along OB obviously dividing both original regions into two parts of equal area.

The slope of OB is .

O

A

B

B

O

45

º

A



6. (a) If two different integers are taken at random from among the integers 1, 2, , 1000,

what is the probability that the sum of the two integers is odd?

(b) If three different integers are taken at random from among the integers 1, 2, , 1000,

what is the probability that the sum of the three integers is odd?

Answer: (a) , (b)

Solution 1 to (a):

Take the first number. If it is odd, we need the next to be even for the sum to be odd. If it is even,

we need the next to be odd for the sum to be odd.

For whatever it is there are 500 numbers to match it such that the sum is odd. There are 999

numbers remaining. So the desired probability is .

Solution 2 to (a):

There are ways to choose two numbers.

For the sum to be odd, one must be odd and the other must be even. There are 500 ways to choose

an odd number and 500 ways to choose an even number. So there are ways to choose

an odd number and an even number.

The desired probability is

.

Solution to (b):

There are 500 odd numbers and 500 even numbers. An odd number + 1 is en even number, and

an even number – 1 is an odd number.

There is a one-to-one correspondence between the odd sums and the even sums. The probability

is exactly a half for each.

The answer is .

Or we can understand it as follows.

For the sum to be odd, we have two cases.

Case 1: all numbers are odd.

Case 2: one is odd, and the other two are even.

For the sum to be even, we have two cases as well.

Case 1: all numbers are even.

Case 2: one is even, and the other two are odd.

Since there are 500 odd numbers and 500 even numbers, the chances for the sum to be odd and to

be even are equal.



7. Earl E. Bird drives the same route to work each day and leaves at 8 am. He finds that if he

goes 40 miles per hour the whole way he is 3 minutes late. If he goes 60 miles per hour the

whole way, he is 3 minutes early. What speed must he go in order to arrive exactly on time?

Answer: 48 miles per hour.

Solution 1:

Let x in miles be the distance one way. The time difference for the two trips is minutes. So

.

We obtain .

Driving at 60 miles per hour he needs minutes. To be on time, he should drive for 15

minutes. So driving speed should be 12/ (15/60) = 48 mph.

Solution 2:

The time for coming on time is the average of the times needed for the two trips at 40 miles per

hour and 60 miles per hour respectively. So the speed for coming on time is a kind of average

speed of these two trips. The answer to this kind of question is not the usual arithmetic mean

average. In this problem the answer is not the arithmetic mean, 50. It is instead the reciprocal of

the average of the reciprocals of the two speeds. This is called the harmonic mean of the speeds.

Let v be the average speed in miles per hour.

We have

.

We get . This is the answer.

8. Suppose and . If a value of x is chosen at random

from the interval , then what is the probability that is negative?

Answer:

Solution:

Note that . For , we must have .

Let .

Then . That is, . We have .

Let .

Then . That is, . We have or .



Therefore, for , we must have or . This can be seen in the

figure:

The total interval length for is .

The length of interval is .

Therefore, the desired probability is

= .04

9. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, (After the first two Fibonacci numbers, each

Fibonacci number is equal to the sum of the two before it.) If F is the 2014th

Fibonacci

number, find the remainder when is divided by 7.

Answer:

Solution:

Let be the ith

Fibonacci number for .

Let us write the Fibonacci sequence in mod 7:

1, 1, 2, 3, 5, 1, , 0, , . , . , , 1, 0, 1, 1,

We see the period of length 16 in mod 7.

Note that mod 16.

So mod 7.

Then is either or 1 mod 7 depending on being odd or even.

Look at the Fibonacci numbers again:

1, 1, 2, 3, 5, 8,

Every third number is even. That is, mod 2 if and only if 0 mod 3.

Note that mod 3. So is odd.

Then mod 7. The answer is 6.

0 –1 –2 100 –100 3 –6

x



10. A code is typed by always jumping from a position on the keypad to another adjacent

position, diagonally, vertically, or horizontally. The moving finger, having writ, must move

on: it cannot stay frozen in position to pick a letter twice in a row. The first letter can be

any letter on the keypad. Examples: the code EIE can be created, but AAB, ACF, and EII

cannot. Also, ABC and CBA are different codes.

(a) How many different three-letter codes can be created?

(b) How many different four-letter codes can be created?

Answer: (a) 200, (b) 952

Solution:

Let be the number of codes of length n starting from a corner letter (one of A, C, I, and G).

Let be the number of codes of length n starting from an edge letter (one of B, F, H, and D).

Let be the number of codes of length n starting from the center letter E.

Let . Then is the total number of codes of length n.

We need to find and .

From a corner letter, we may go to one of two edge letters or to the center letter. So we have the

following recursion:

(1)

Similarly, we have

(2)

(3)

Obviously, , , and .

Calculate the recursions with the following table:

n 1 2 3 4

1 3 18 80

1 5 24 116

1 8 32 168

9 40 200 952

We obtain and .

A B C

D E F

G H I



Although it was not asked for, we will find the general formula for . If you have questions

about the technique below, please email us or look some things up. These techniques are beyond

what it is expected participants would know for this contest.

Plug (1) into (2):

(4)

Plug (4) into (3):

That is,

(5)

Plug (4) and (5) into (1)

That is,

(6)

The characteristic equation of (6) is

(7)

By testing, is a root. So we can factor out from the left side of (7):

.

We have or .

From the second equation we have .

That is, . So , or .

We obtain three characteristic values of (7): , , and .

So can be expressed as

(8)

With three initial values , , and , we can determine A, B, and C.

However, I would like to introduce .

Obviously, .

Plugging , 1, 2 into (8), we have respectively.

,

,

.

Solving for A, B, and C we obtain



, , and .

Plugging these into (8) we have

.

That is,

,

or

.

From (4) and (5)

.

We get the formula for in the closed form:

.

Do you believe it?

If not, plug , 2, 3, and 4 and see what you get.

The Moving Finger writes and, having writ,

Moves on: nor all your Piety nor Wit

Shall lure it back to cancel half a Line,

Nor all your Tears wash out a Word of it.

- Omar Khayyam

Twenty-third Annual UNC Math Contest Final Round January 31, 2015

Three hours; no electronic devices. Show your work. Answers must be justified to receive

full credit. We hope you enjoy thinking about these problems, but you are not expected to

do them all.

The positive integers are 1, 2, 3, 4, . . .A polynomial is quadratic if its highest power term has power two.

1. The sum of three consecutive integers is 54. What is the smallest of the three integers?

2. Find the area of the shaded region. The outer circlehas radius 3. The shaded region is outlined by half cir-cles whose radii are 1 and 2 and whose centers lie onthe dashed diameter of the big circle.

3. If P is a polynomial that satisfies P(x

2 + 1) = 5x

4 + 7x

2 + 19, then what is P(x)? (Hint: P isquadratic.)

4. Tarantulas A, B, and C start together at the same time and race straight along a 100 footpath, each running at a constant speed the whole distance. When A reaches the end, B stillhas 10 feet more to run. When B reaches the end, C has 20 feet more to run. How many morefeet does Tarantula C have to run when Tarantula A reaches the end?

5. A termite nest has the shape of an irregular polyhe-dron. The bottom face is a quadrilateral. The top faceis another polygon. The sides comprise 9 triangles, 6quadrilaterals, and 1 pentagon. The nest has 10 ver-tices on its sides and bottom, not counting the severalaround the top face. How many edges does the top facehave?

You may use Euler’s polyhedral identity, which says that on a convex polyhedron the numberof faces plus the number of vertices is two more than the number of edges. (A vertex is a cornerpoint and an edge is a line segment along which two faces meet.)

6. How many ordered pairs (n, m) of positive integers satisfying m < n 50 have the prop-erty that their product mn is less than 2015?

TURN PAGE OVER

7. (a) Give an example of a polyhedron whose faces can be colored in such a way that eachface is either blue or gold, no two gold faces meet along an edge, and the total area of allthe blue faces is half the total area of all the gold faces. A blue face may meet another blueface along an edge, and any colors may meet at vertices. Describe your polyhedron and alsodescribe how to assign colors to the faces.(b) Show that if the faces of a polyhedron are colored in such a way that each face is either blueor gold and no two gold faces meet along an edge, and if the polyhedron contains a sphereinside it that is tangent to each face, then the total area of all the blue faces is at least as largeas the total area of all the gold faces.

8. A garden urn contains 18 colored beetles: 6 red beetles, numbered from 1 to 6, and 12yellow beetles, numbered from 1 to 12. Beetles wander out of the urn in random order, oneat a time, without any going back in. What is the probability that the sequence of numberson the first four beetles to wander out is steadily increasing, that is, that the number on eachbeetle to wander out is larger than the number on the beetle before and that no number isrepeated? Give your answer as a fraction in lowest terms. You may leave the numerator anddenominator in a factored form.

9. Starting at the node in the center of the diagram,an orb spider moves along its web. It is permissiblefor the spider to backtrack as often as it likes, in eitherdirection, on segments it has previously travelled. Oneach move, the spider moves along one of the segments(curved or straight) to some adjacent node that is dif-ferent from the node that it currently occupies.

(a) How many different five-move paths start at the center node and end at the center node?(b) How many different seven-move paths start at the center node and end at the center node?

10. (a) You want to arrange 8 biologists of 8 differentheights in two rows for a photograph. Each row musthave 4 biologists. Height must increase from left toright in each row. Each person in back must be tallerthan the person directly in front of him. How manydifferent arrangements are possible?

(b) You arrange 12 biologists of 12 different heights in two rows of 6, with the same conditionson height as in part (a). How many different arrangements are possible? Remember to justifyyour answers.(c) You arrange 2n biologists of 2n different heights in two rows of n, with the same conditionson height as in part (a). Give a formula in terms of n for the number of possible arrangements.

BONUS: You arrange 12 biologists of 12 different heights in three rows of 4, with the sameconditions on height as in part 10(a) for all three rows. How many different arrangements arepossible?

END OF CONTEST

Twenty-third Annual UNC Math Contest Final Round January 31, 2015

SOLUTIONS

1. The sum of three consecutive integers is 54. What is the smallest of the three integers?ANSWER: 17SOLUTION: The sum of three consecutive integers is three times the average, or middle, inte-ger, so the middle integer is 54/3 = 18 and the smallest is 17. Another approach is to write thethree integers as n, n+1, and n+2 and compute 54=n+(n+1)+(n+2)=3n+3 so 51=3n and 17=n.

2. Find the area of the shaded region. The outer circlehas radius 3. The shaded region is outlined by half cir-cles whose radii are 1 and 2 and whose centers lie onthe dashed diameter of the big circle.

ANSWER: 3pSOLUTION: The area of the shaded region is the areaof a circle of diameter radius 2 minus the area of a circleof radius 1 = 4p � p = 3p. One elegant way to see thisis to reflect the top half of the diagram left to right.

3. If P is a polynomial that satisfies P(x

2 + 1) = 5x

4 + 7x

2 + 19, then what is P(x)?

ANSWER: P(x) = 5x

2 � 3x + 17SOLUTION: P is clearly quadratic. Let P(x) = Ax

2 + Bx + C. Then P(x

2 + 1) = A(x

2 +1)2 + B(x

2 + 1) + C = Ax

4 + 2Ax

2 + A + Bx

2 + B + C = Ax

4 + (2A + B)x

2 + (A + B + C).Therefore, A = 5, 2A + B = 7, and A + B + C = 19. Solve for A, B, and C.

4. Tarantulas A, B, and C start together at the same time and race straight along a 100 footpath, each running at a constant speed the whole distance. When A reaches the end, B stillhas 10 feet more to run. When B reaches the end, C has 20 feet more to run. How many morefeet does Tarantula C have to run when Tarantula A reaches the end?ANSWER: 28SOLUTION: In the time A travels 100 feet, B travels 90 feet. Thus B travels at a speed 9/10 thespeed of A. Similarly, C travels at a speed 8/10 the speed of B and (8/10)(9/10)=72/100 thespeed of A. When A has gone 100 feet, C has gone 72 feet and has 100-72=28 feet to go.

5. A termite nest has the shape of an irregular polyhe-dron. The bottom face is a quadrilateral. The top faceis another polygon. The sides comprise 9 triangles, 6quadrilaterals, and 1 pentagon. The nest has 10 ver-tices on its sides and bottom, not counting the severalaround the top face. How many edges does the top facehave?

You may use Euler’s polyhedral identity, which says that on a convex polyhedron the numberof faces plus the number of vertices is two more than the number of edges. (A vertex is a cornerpoint and an edge is a line segment along which two faces meet.)ANSWER: 8SOLUTION: F = 1 bottom + 1 top + 9 triangles + 6 quadrilaterals +1 pentagon= 18 faces. Letn be the number of edges around the open top. Then there are also n vertices on the rim andthe total number of vertices is V = 10 + n. Count 9⇥ 3 edges on triangles, 7⇥ 4 edges onquadrilaterals (including the bottom), and so on. Observe that this will count each edge twotimes. So 2E = 9⇥ 3 + 7⇥ 4 + 1⇥ 5 + n = 27 + 28 + 5 + n = 60 + n. Using the Euler identity,we obtain 18 + 10 + n = (1/2)(60 + n) + 2 or n = 8.

6. How many ordered pairs (n, m) of positive integers satisfying m < n 50 have the prop-erty that their product mn is less than 2015?ANSWER: 1197SOLUTION: Note that 50⇥ 40 = 2000 so for n = 50, m = 1, 2, 3, . . . , 40. There are 40 suchpairs.For n = 49 clearly m can be 40. Try 49⇥ 41 = 2009. For n = 49, m = 1, 2, 3, . . . , 41.For n = 48 clearly m can be 41. Try 48⇥ 42 = 2016. No! For n = 48, m = 1, 2, 3, . . . , 41.For n = 47 m = 1, 2, 3, . . . , 42.For n = 46 m = 1, 2, 3, . . . , 43.For n = 45 m = 1, 2, 3, . . . , 44.From now on, all m n will work. Adding up, we get 1 + 2 + . . . + 44 plus 43 + 42 + 41 + 41 +40. The first sum is 44⇥ 45/2 = 22⇥ 45 = 990 and the second is 200 + 3 + 2 + 1 + 1 = 207.The total is 1197.

Alternative way of looking at this one:

There are 50⇥492 = 1225 ordered pairs (n, m) that satisfy m < n 50. Because most of them

will satisfy the inequality mn < 2015, it is easier to concentrate on counting the exceptions(shown in the shaded region ) that satisfy mn > 2015 but n 50, hence m > 40.

Out[299]=

42 44 46 48 50

42

44

46

48

The bounding curve nm = 2015 is a hyperbola, mirror-symmetric across the line m = n. By mirror sym-metry, the tangent line through (

p2015),

p2015) has

slope �1. The shaded region is convex, so it lies abovethis tangent line, inside the triangular region definedby m + n > 2

p45 = 89.7 . . .. Since we seek only in-

teger solutions, this narrows the field of candidates tom + n � 90. There are 30 lattice points in the triangular

region that lie on or above this border-line. It seems probable that most of these candidateswill lie in the shaded region, but now we must check carefully if any of the points on or abovethe border-line are false solutions (not actually on the shaded region). Points (m, n) on this linecan be written in parametric form as n = 45 + t, m = 45� t. Since (45 + t)(45� t) = 2025� t

2

it is easy to create this table of values for (n, m, n ⇤m) for the small values of t of interest:n = 45 46 47 48 49 50m = 45 44 43 42 41 40

n ⇤m = 2025 2024 2021 2016 2009 2000Note that the last two points (49, 41) and (50, 20) lie outside the shaded region. Thus thereare 30� 2 = 28 candidates that remain. One can check easily (numerically) that the latticepoints immediately above these two discarded false solutions are true solutions; that is, allcandidates have been classified, and all lattice point solutions in the shaded region are nowaccounted for. Take 1225-28= 1197.

7. (a) Give an example of a polyhedron whose faces can be colored in such a way that eachface is either blue or gold, no two gold faces meet along an edge, and the total area of allthe blue faces is half the total area of all the gold faces. A blue face may meet another blueface along an edge, and any colors may meet at vertices. Describe your polyhedron and alsodescribe how to assign colors to the faces.(b) Show that if the faces of a polyhedron are colored in such a way that each face is either blueor gold and no two gold faces meet along an edge, and if the polyhedron contains a sphereinside it that is tangent to each face, then the total area of all the blue faces is at least as largeas the total area of all the gold faces.ANSWER: (a) One possible example is a square box with large top and bottom and smallheight, with the top and bottom colored gold and the four sides blue. If the top and bottomare squares of side s, then the total area colored gold is 2⇥ s

2. If the height of the box is h, thenthe total area colored blue is 4⇥ h⇥ s. It is easy to select s and h to make 4⇥ h⇥ s equal tohalf of 2⇥ s

2. Set 4⇥ h⇥ s = s

2 or 4⇥ h = s. Choose, say, h = 1 and s = 4.(b) Triangulate each face by connecting the point of tangency on the face to each vertex of theface. Look at a gold face and one of the triangles into which it has been cut. The gold trianglemeets a blue triangle along the edge where the two faces meet. Consider the two trianglesand the radii that go from the tangent points on the two faces (=the apexes of the triangles)to the center of the sphere. By symmetry, the two triangles are congruent. This means that foreach triangle colored gold, there is a congruent triangle colored blue. Therefore, the total areacolored blue is at least as large as the total area colored gold.

F

G More detail: We are told that each face has a point oftangency to the sphere. Each face is perpendicular tothe radius that connects the center of the sphere to thepoint of tangency. Consider a pair of faces that meetalong an edge, as in the diagram.P

C

F

G Using the radii, which are perpendicular to the respec-tive faces, we can find some congruent triangles. Callthe far end of the edge in the diagram P and the centerof the sphere C. Call the point of tangency of one faceF and the point of tangency of the other face G.

The segments FC and GC are radii and so have the same length. The segments FP and GP liein the faces so they are perpendicular to the radii to those two faces. This makes triangles CFPand CGP both right with right angles at F and G. They have the common hypotenuse PC and

they each have legs the length of the radius of the sphere. This makes them congruent (by theHypotenuse-Leg rule for right triangles) and tells us that FP and GP have the same length.

P

C

F

G

Q

There is another pair of congruent right triangles: Justreplace P with the other end of the edge, Q: TrianglesCFQ and CGQ are also congruent. This will give us theresult about color. Consider a face colored gold- thepentagon in the example, say. Triangulate it by

connecting the point of tangency to each vertex. Consider one of the gold triangles: we knowit meets a blue face and that the blue face contains a triangle that is congruent to the goldone. Therefore, the amount of surface area that is blue is at least as large as the amount that iscolored gold.

8. A garden urn contains 18 colored beetles: 6 red beetles, numbered from 1 to 6, and 12yellow beetles, numbered from 1 to 12. In random order, beetles wander out of the urn, oneat a time, without any going back in. What is the probability that the sequence of numberson the first four beetles to wander out is steadily increasing, that is, that the number on eachbeetle to wander out is larger than the number on the beetle before and that no number isrepeated? Give your answer as a fraction in lowest terms. You may leave the numerator anddenominator in a factored form.ANSWER: 157/(18⇥ 17⇥ 16) or 157/4896SOLUTION: There are 18 numbered beetles that can wander out of the urn and therefore18⇥ 17⇥ 16⇥ 15 ways for a sequence of four beetles to emerge. There are two copies of eachnumber 6 and a single copy of each of the numbers � 7.Categorize the possibilities in which the numbers increase into disjoint cases.Case 1: All 4 beetles have numbers 6. There are C(6, 4) combinations of 4 distinct numbers 6. Once we have selected the 4 numbers, there is only one way to list them in increasingorder. Each number 6 occurs on 2 different beetles in the urn. There are 24

C(6, 4) ways tocreate strictly increasing sequences of 4 distinct numbers 6.Case 2. The first 3 beetles have numbers 6 and the last one out has a number � 7. Thereare C(6, 3) combinations of 3 distinct numbers 6. Each of the 3 numbers occurs on 2 beetles.There are 6 possibilities for the fourth number and the fourth number chosen is on only onebeetle: 23

C(6, 3)C(6, 1)Case 3. The first 2 beetles have numbers 6 and the last 2 have numbers� 7: 22

C(6, 2)C(6, 2)Case 4. The first beetle has a number 6 and the last three have numbers� 7: 2C(6, 1)C(6, 3)Case 5. All four beetles have numbers � 7: C(6, 4)Total number of possibilities that give increasing sequences:

= 24(64) + 23(6

3)(61) + 22(6

2)(62) + 2(6

1)(63) + (6

4)

= (24) 6⇥5⇥4⇥34⇥3⇥2 + (23)( 6⇥5⇥4

3⇥2 )(6) + (22) 6⇥52

6⇥52 + 2(6)( 6⇥5⇥4

3⇥2 ) + 6⇥5⇥4⇥34⇥3⇥2 = 3⇥ 785.

Total number of random sequences: 18⇥ 17⇥ 16⇥ 15.Probability of the sequence being increasing is the quotient:(3⇥ 785)/(18⇥ 17⇥ 16⇥ 15) = 785/(18⇥ 17⇥ 16⇥ 5) = 157/(18⇥ 17⇥ 16) = 157/4896.Note on generating functions:

Those students familiar with the binomial theorem may recognize the sum

24(64) + 23(6

3)(61) + 22(6

2)(62) + 2(6

1)(63) + (6

4)

as the coefficient of x

4 in the product (1 + 2x)6(1 + x)6. In fact, the problem can be worked bythe method of generating functions, as follows.Think of picking an increasing sequence of numbers on the emerging beetles as selecting either0 or 1 beetle with the number 1, then selecting either 0 or 1 beetle with the number 2, and soon. Since there are two beetles with each of the numbers 6, the generating function foreach of those numbers is 1 + 2x. There is just one beetle with each of the numbers � 7, so thegenerating function for each of those numbers is 1 + x. The combined generating function forincreasing sequences is thus (1 + 2x)6(1 + x)6. The number of increasing sequences of length4 is the coefficient of x

4. Finish the problem by finding that coefficient and taking the quotientwith 18⇥ 17⇥ 16⇥ 15. (Finding the coefficient is a bit of work.)

9. Starting at the node in the center of the diagram,an orb spider moves along its web. It is permissiblefor the spider to backtrack as often as it likes, in eitherdirection, on segments it has previously travelled. Oneach move, the spider moves along one of the segments(curved or straight) to some adjacent node that is dif-ferent from the node that it currently occupies.

(a) How many different five-move paths start at the center node and end at the center node?(b) How many different seven-move paths start at the center node and end at the center node?ANSWER: (a) 264SOLUTION: (a) The various types of paths can be listed and counted. Let O stand for a moveto any outside point from the inside, I stand for a move to the inside (center) point, and Xstand for a move from any outside point to any other outside point along any path. There arethree cases: OXXXI, OXIOI, OIOXI with the following numbers of different paths, respectively,3⇥ 4⇥ 4⇥ 4⇥ 1 = 192, 3⇥ 4⇥ 1⇥ 3⇥ 1 = 36, 3⇥ 1⇥ 3⇥ 4⇥ 1 = 36. Total=264ANSWER: (b) 5700SOLUTION: (b) The various types of paths can be listed and counted, as before. This timethe possibilities are OIOIOXI, OIOXIOI, OIOXXXI, OXIOIOI, OXIOXXI, OXXIOXI, OXXXXOI,OXXXXXI. Count as before.Alternatively, work as in the keypad question on the First Round. Label the center node x andthe peripheral nodes as a, b, c.Let A

n

, B

n

, C

n

, and X

n

represent the number of paths of length n that start at the center x

and end at the nodes a, b, c, x respectively. Now extend the paths one extra step, and use thediagram to deduce this highly symmetrical system of recurrence equations:A

n+1 = 2B

n

+ 2C

n

+ X

n

B

n+1 = 2A

n

+ 2C

n

+ X

n

C

n+1 = 2A

n

+ 2B

n

+ X

n

X

n+1 = A

n

+ B

n

+ C

n

;hence X

n+2 = A

n+1 + B

n+1 + C

n+1.Add the top three equations to deduce that the X values satisfyX

n+2 = A

n+1 + B

n+1 + C

n+1 = 4X

n+1 + 3X

n

Next use this recurrence equation to eventually solve for X7, as follows.Note that by inspection of the network, X1 = 0, X2 = 3.Insert these into the recurrence formula to deduce X3 = 4(3) + 3(0) = 12. Likewise X4 =4(12) + 3(3) = 57, X5 = 4(57) + 3(12) = 228 + 36 = 264,X6 = 1227 and finally X7 = 5700.

10. (a) You want to arrange 8 biologists of 8 differentheights in two rows for a photograph. Each row musthave 4 biologists. Height must increase from left toright in each row. Each person in back must be tallerthan the person directly in front of him. How manydifferent arrangements are possible?

(b) You arrange 12 biologists of 12 different heights in two rows of 6, with the same conditionson height as in part (a). How many different arrangements are possible? Remember to justifyyour answers.(c) You arrange 2n biologists of 2n different heights in two rows of n, with the same conditionson height as in part (a). Give a formula in terms of n for the number of possible arrangements.

ANSWER: (a) 14 (b) 132

(c)1

n + 1

✓2n

n

◆=

✓2n

n

◆�

✓2n

n� 1

◆=

✓2n

n

◆�

✓2n

n + 1

◆=

12n + 1

✓2n + 1

n

◆=

(2n)!(n + 1)!n!

These are the Catalan numbers and there are many acceptable forms for them. Any of theforms listed here, for instance, are acceptable answers.SOLUTION: (a) Either list the 14 arrangements or refer to the reasoning shown for eitherof parts (b) or (c). Here is a systematic list in a lexicographic order: FFFFBBBB; FFFBFBBB;FFFBBFBB; FFFBBBFB; FFBFFBBB; FFBFBFBB; FFBFBBFB; FFBBFFBB; FFBBFBFB; FBFFFBBB;FBFFBFBB; FBFFBBFB; FBFBFFBB; FBFBFBFB. Keep in mind as you list that acceptable stringshave the 4 Fs and 4 Bs and the Bs never outnumber the Fs as you move left to right.

SOLUTION:(b) We can use a better system than listing. Line up the biologists by height. Startfilling in seats. Each new person must go in the next empty seat in either the front row or theback row. Look at the patterns and keep track of how many of the seating arrangements beginwith j Fs and k Bs. We will construct a table. (See below.)When j=k we have two full rows: this is what (a) and (b) ask about, j=k=4 and j=k=6.There will be no arrangements with k bigger than j (more B’s than F’s so far) so all entriesabove the diagonal are 0.1F & 0B or 2F & 0B, etc. are easy. There is one arrangement with each of these, so the firstcolumn is all 1s.Next column, the column with 1 in the back row: 1F & 1B one way. 2F & 1B two ways.Notice that to get 2F & 1B then just before you added the most recent biologist, you had either

1F & 1B and added an F or you had2F & 0B and added a B.

This means each entry is the sum of entry to left + entry above. Fill in the table.0B 1B 2B 3B 4B 5B 6B

1F 1 1 0 0 0 0 02F 1 2 2 0 0 0 03F 1 3 5 5 0 0 04F 1 4 9 14 14 0 05F 1 5 14 28 42 42 06F 1 6 20 48 90 132 132

The 4F & 4B entry is the 14 from part (a). For part (b) we want the 6F & 6B entry, 132.

SOLUTION: (c) First select n of the 2n biologists to go in the front row. There are (2n

n

) ways tomake this selection. Given a selection, the front row people and the back row people can beput in order so that height increases left to right. Some of these seating arrangements will sat-isfy the condition that every person in the back row is taller than the person in front and someof the arrangements will not. We will count how many violate this condition and subtract thatfrom (2n

n

). Number the biologists in order of height and then write down FFBBFB.... to recordwhether each person is seated in the front row or the back row. All (2n

n

) of our arrangementshave equal numbers of F and B, because we selected n for the front row. The strings that sat-isfy the last requirement are exactly the strings that have the property that as you count fromleft to right, the total number of F’s never exceeds the total number of B’s. We want to countthe strings that violate this. We will associate to each of the "bad" strings one unique stringof n + 1 F’s and n� 1 B’s in such a way that each "bad" string has such an unbalanced stringand that each such n + 1/n� 1 string has exactly one "bad" string associated with it. Then allwe have to do is count how many of these unbalanced strings there are, the ones with n + 1F’s and n� 1 B’s. That is, of course, ( 2n

n�1). Here is the association: given a "bad" string, movefrom left to right until you first have more F’s than B’s. Leave that first excess F in place. Tothe left of that F you will have equal numbers of F’s and B’s. To the right of that F you willhave one more B than F. In every place to the right of the F, change every F to a B and everyB to an F. The new string will have n + 1 F’s and n� 1 B’s. To check that the assignment isreversible, consider a string with n + 1 F’s and n� 1 B’s. As you move from left to right, thereis a first place the number of F’s exceeds the number of B’s. To the left of that F, the string isbalanced. Switch all the F’s and B’s to the right of this F. The resulting string will be the "bad"balanced string that went with the unbalanced string. Therefore, the number of good stringsis (2n

n

)� ( 2n

n�1). This expression is an acceptable answer. It is also easily simplified to 1n+1(2n

n

).

BONUS: You arrange 12 biologists of 12 different heights in three rows of 4, with the sameconditions on height as in part 10(a) for all three rows. How many different arrangements arepossible?ANSWER. 462

SOLUTION: Work as in part (b), but with a third index for the third row. We will call the rowsA, B, and C. The recursive rule for filling in the table is similar to the rule in part (b), but fora three-dimensional array, displayed below in a compressed format as follows: the upper leftnumber in each box is for 0C; the next one is for 1C, and so on. The last entry in the tablerepresents (4A, 4B, 3C) =462. However, this will also equal the next recursive entry: (4A, 4B,4C) =462.

0B 1B 2B 3B 4B1 1 1c 2c 3c 0 0 0

0 1 0 0 0!A 0 0 0 0 0

0 0 0 0 01 2 2 0 0

0 3 5 0 02A 0 0 5 0 0

0 0 0 0 01 3 5 1c 2c 3c 5 0

0 6 16 21 03A 0 0 21 42 0

0 0 0 42 01 4 9 14 14

0 10 35 70 844A 0 0 56 168 252

0 0 0 210 462

Solution to 2015 UNC Final by Ming Song

Copyright © 2015 Ming Song 719-266 8036, [email protected]. Don’t use or copy without permission 1

1.

Answer: 17

Solution:

The number at the middle is 18354 .

The smallest number is 17.

2.

Answer: S3

Solution:

The shaded area is SSS 312 22 � .

3.

Answer: � � 1735 2 �� xxxp .

Solution:

Let 12 � xy . Then 12 � yx .

So � � � � � � 1735191715 22 �� yyyyyp .

Therefore, � � 1735 2 �� xxxp .

4.

The speed ratio of A to B is 90:100 .

The speed ratio of B to C is 80:100 , 8:10 , or 72:90 .

Then the speed ratio of A to C is 72:100 .

When A runs 100 feet, C. runs 72 feet. C has to run 28 feet more.

The answer is 28.

5.

Answer: 8

Solution:

Let x to the number of vertices on the top face. Then the top face has x edges.

Let V be the number of faces of the polyhedron, F be the number of faces, and E be the number of edges.



According to the Euler Theorem we have 2� � EFV .

In this problem,

xV � 10 ,

1811691 �� F ,

230

24514639 xxE �

�� .

We have the equation:

22

301810 �� xx .

Solving for x we obtain 8 x .

The answer is 8.

6.

Answer: 1197

Solution:

Note that 22 45201544 �� and 201519804544 � �

If 45d�nm , we always have 2015�mn .

There are 990245

¸̧¹

·¨̈©

§ ways to assign values for m and n in this range.

If 46 n , we must have 43dm by noticing 201520241454644 2 ! � � .



If 49 n , we must have 41dm by noticing 201520094454941 22 � � � .

If 50 n , we must have 40dm by noticing 201520005040 � � .

Therefore, the answer is

11974041414243990 �� .



7.

Solution to (a):

Color a cube with two opposite faces being golden and the other faces to be blue.

Solution to (2):

In 2D-geometry we have PBPA is P is a point outside a circle and PA and PB are tangent to the circle at A and B respectively.

We will prove a lemma, a similar conclusion, in 3D-geometry.

Lemma:

Plane m and n intersect at line l. A sphere is tangent to both planes at A and B respectively.

Let C and D are two points on line l. Then BCDACD '#' .

Proof:

Draw the plane through C, A, and B. The plane cuts the sphere with a circle. CA and CB are tangent to the circle at A and B respectively. According to the conclusion in 2D-Geometry, we have CBCA .

Similarly, we have .DBDA

A Bl

C

m

n

D

A Bl

D

C

m

n

A

B

P



With CD is common, we have BCDACD '#' .

Particularly, triangle ACD and BCD have the same area.

Now the solution to (b) follows.

Consider all golden-blue edges. The blue area is at least as much as the golden area.

8.

Answer: 4896157

Solution 1:

Put beetles in a line with numbers from small to large. With the same number the red beetle is to the left of the yellow beetle.

There are 15161718 �� ways for 18 beetles to wander out.

There are ¸̧¹

·¨̈©

§418

ways to choose four beetles.

There are ¸̧¹

·¨̈©

§¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§15

216

16

ways to choose four beetles with a pair which have the same

number.

There are ¸̧¹

·¨̈©

§26

ways to choose four beetles with two pairs each of which have the same

number.

So there are ¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§26

15

216

16

418

ways to choose four beetles all of which have

different numbers.

Put this four beetles in with right order with the numbers increasing.

Therefore, the answer to the problem is

� �

4896157

161718474173

24151617182470515161718

15161718

155120624

15161718

1516171826

15

216

16

418

��

��

��

��

��

¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§

.

Solution 2:

The denominator is 15161718 �� , obviously.

The numerator is the coefficient of 4x in � � � �66 121 xx �� . That is,



235546

36

16

226

26

216

36

246

2 234 ¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§�¸̧¹

·¨̈©

§.

Therefore, the desired probability is

4896157

161718157

151617182355

��

��

.

9.

Solution 1 (Recursion):

Let na be the number of ways of paths of length n such that the spider comes back to A when the spider is at A right now.

Let nb be the number of ways of paths of length n such that the spider comes back to A when the spider is at B right now.

Define nc and nd likewise.

By the symmetry, we have nnn dcb .

We can establish the following recursion:

111 �� nnnn dcba ,

111 22 �� nnnn dcab .

So we have

13 � nn ba (1)

11 4 �� nnn bab (2)

Plugging (2) into (2) we obtain

21 34 �� nnn bbb (3)

With (1) we have the same recursion for na :

21 34 �� nnn aaa

With 01 a and 32 a , we calculate with the recursion:

1203343 �� a ,

A

B

DC



57331244 �� a ,

2641235745 �� a ,

122757326445 �� a ,

57002643122745 �� a .

The answer to (a) is 264, and the answer to (b) is 5700.

Solution 2:

Let x be the number of moves that the spider moves along a radius, and let y be the number of moves that the spider moves among B, C, and D.

Problem (a):

We have 5 �yx .

Case 1: 2 x and 3 y .

In the first step, the spider must go out along a radius. There are 3 choices.

In the last step, the spider must go in along the radius where it is. There is one choice.

In any of three other moves, there are 4 choices.

The number of ways in this case is 192143 3 �� .




In between there are two chances for the spider go in and then immediately go out with 3 choices.

The spider has one move among points B, C, and D. There are 4 choices in this move.

The number of ways in this case is 724323 �� .

The total number of ways is 26472192 � .

Problem (b):

We have 7 � yx .



A

B

DC




In any of five other moves, there are 4 choices.





In between there are four chances for the spider go in and then immediately go out with 3 choices.

The spider has three moves among points B, C, and D. There are 4 choices in each move.





In between there are three ways for the spider go in and then immediately go out with 3 choices twice.

The spider has one move among points B, C, and D. There are 4 choices in this move.


The total number of ways is 570032423043072 �� .

Solution 3 (Maxtrix):

We define the matrix:

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

0221202122011110

A .

The entry ija ( 4,1 dd ji ) is the number of ways of the spider to move from iA to jA .

A1

A2

A3 A4



Then the entry ijb ( 4,1 dd ji ) in matrix nA is the number of ways of the spider to reach jA

from iA in m moves.

Let us calculate:

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

�

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

9554595455944443

0221202122011110

0221202122011110

2A ,

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

�

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

24323219322432193232241919191912

9554595455944443

0221202122011110

23 AAA ,

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

�

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

???409???409???409???264

24323219322432193232241919191912

9554595455944443

325 AAA ,

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

�

¸̧¸¸¸

¹

·

¨̈¨¨¨

©

§

???????????????5700

???409???409???409???264

9554595455944443

527 AAA .

The answer to (a) is 264, and the answer to (b) is 5700.



10.

Let us consider to arrange 1 to 6.

Look at the one-to-one correspondence for the five arrangements:

The number of arrangements for numbers 1 to 2n satisfying the given conditions is equal to the number of shortest routes in the n by n gird from the left-bottom corner to the right-top corner without passing the diagonal as shown.

2 4 6

1 3 5

6

2

4

1

3

5

2 5 6

1 3 4

6

2 4

1

3

5

3 4 61 2 5

6

2

4

1

3

5

3 5 61 2 4

6

2

4

1

3

5

4 5 6

1 2 3

6

2

4

1 3

5



So we can easily get the answer by marking numbers:

The answer to Problem (a) is 14, and the answer to problem (b) is 132.

If you know the Catalan numbers, the general formula is

¸̧¹

·¨̈©

§� n

nn

21

1.

For 4 n ,

1448

51

¸̧¹

·¨̈©

§.

This is the answer to problem (a).

For 6 n ,

1326

1271

¸̧¹

·¨̈©

§.

This is the answer to problem (b).

1 1 1 1 111

1 2 5 643

2 5 2014 9

5 14 4828

14 9042

13242

132

twenty-third annual unc math contest first round november ... · 14-10-2011 · whole way, he is 3...

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