Gator Mathematics Competition 2013 Tests and Solutions

Gator Mathematics Competition 2013

Mitchell HarrisSalim Hyder

Sitharthan SekarEndrit Fejzullahu

Jon ClausDominic Guzzo

Kurtis LeeSergei ShabanovMike Goodman

Eli Ross

January 26, 2013

1

Contents

1 Individual Test 3

2 Individual Solutions 5

3 Mathathon 8

4 Mathathon Solutions 11

5 Relay 18

6 Relay Solutions 21

7 Tiebreaker 23

8 Tiebreaker Solutions 24

2

1 Individual Test

Individual Round One GaMaCo 2013

1. Compute the area of the triangle with vertices at the coordinates (1337, 1337), (1337, 1344), and(1349, 1337).

2. Three points are selected randomly (and uniformly) along the circumference of a circle with radius 1.Compute the probability that they are the vertices of an acute triangle.

Individual Round Two GaMaCo 2013

3. Circle O (with positive area) has diameter AB, and C lies on O. Compute the maximum possibleratio of the area of ∆ABC to the area of circle O.

4. Compute the numerical value ofV

x3for some nonzero V in the system of equations:

y + 2z = λyz

x+ 2z = λxz

2x+ 2y = λxy

xyz = V

Individual Round Three GaMaCo 2013

5. Compute the largest real x such that

x− 3√x− 3 =

1√x− 3

.

6. The sum 712 + 9360

d is an integer. Compute the smallest positive integral value for d.

3

Individual Round Four GaMaCo 2013

7. An ant is at the origin of a coordinate plane. The first day, the ant decides to move 4 units in somedirection parallel to the x axis, chosen at random. The next day, the ant randomly chooses to turn rightor left and then walk 4 units in that direction (now moving parallel to the y axis). Every day, the antcontinues moving in this fashion – turning left or right from the direction of the previous day and thenmoving 4 units. The probability he returns to the origin at the end of the 16th day is k

215. Compute k.

8. A spy-aircraft is flying at constant altitude along a circle of radius 12 miles with center at a point A.The speed of the aircraft is v. At some moment in time, a missile is fired at the aircfraft from the pointA, which has speed v and is guided so that it always points towards the aircraft. How far does the missiletravel before colliding with the aircraft?

Individual Round Five GaMaCo 2013

9. Compute the minimum value of f(x) =√x2 − 8x+ 21.

10. Compute the sum of the solutions to the equation sin(x) + 8 cos2(x) sin(x) + 3√

3 cos(x) = 0 on theinterval 0 < x < 2π.

4

2 Individual Solutions

Individual Round One GaMaCo 2013

1. Compute the area of the triangle with vertices at the coordinates (1337, 1337), (1337, 1344), and(1349, 1337).

Answer: 42

Solution: The triangle is right and has side lengths of 12 and 7. So A = 12(12)(7) = 42.

2. Three points are selected randomly (and uniformly) along the circumference of a circle with radius 1.Compute the probability that they are the vertices of an acute triangle.

Answer: 14

Solution: We calculate the complementary probability – that the triangle is obtuse. Note that theprobability of a right triangle is zero. The triangle will be obtuse if and only if all three points lie on thesame side of some diameter. Let the random variable Xk be associated with the point on the circle Pk,for k = 1, 2, 3, and define it as 1 if the other two points are within π clockwise from it and 0 otherwise.We are looking for P (X1 +X2 +X3 = 1) = E(X1 +X2 +X3) since at most one of these variables can be1. Because the points are chosen uniformly, their probability distributions are identical. So the aboveequals 3E(X1) = 3

4 because the probability that each of the other two points lie in the semicircle startingat X1 is 1

2 ·12 . Our desired probability is then 1− 3

4 .

Individual Round Two GaMaCo 2013

3. Circle O (with positive area) has diameter AB, and C lies on O. Compute the maximum possibleratio of the area of ∆ABC to the area of circle O.

Answer: 1π

Solution: Let the legs of the right triangle be p and q. Then

(p− q)2 ≥ 0⇒ p2 + q2 ≥ 2pq ⇒ p2+q2

4 ≥ pq2 ⇒

(p2+q2)/4pq/2 ≥ 1⇒ π(p2+q2)/4

pq/2 ≥ π, where the left hand side of

the inequality is precisely the inverse of the ratio in question. So the answer is 1π .

4. Compute the numerical value ofV

x3for some nonzero V in the system of equations:

y + 2z = λyz

x+ 2z = λxz

2x+ 2y = λxy

xyz = V

Answer: 12

Solution: Multiply each of the equations by x, y, and z respectively. Noting that xyz = V , the system is

5

transformed into

xy + 2xz = λV

xy + 2yz = λV

2xz + 2yz = λV

xyz = V

Subtracting the first two equations gives that either z = 0 (which is impossible if V is nonzero) or y = x.Using the latter in the next two equations gives that x = 2z (since x 6= 0). By the last equation

(2z) · (2z) · z = V ⇒ 4z3 = V . But x = 2z, so Vx3

= 4z3

8z3= 1

2 .

Individual Round Three GaMaCo 2013

5. Compute the largest real x such that

x− 3√x− 3 =

1√x− 3

.

Answer: 4

Solution: Add and subtract 3 to the left hand side:(x−3)−3

√x− 3+3 = 1√

x−3 ⇒ (x−3)32 −3(x−3)

22 +3(x−3)

12 −1 = 0⇒ ((x−3)

12 −1)3 = 0⇒ (x−3) = 1.

Therefore x = 4 is the only real solution.

6. The sum 712 + 9360

d is an integer. Compute the smallest positive integral value for d.

Answer: 1728

Solution: Write the second fraction as c·mc·n where the greatest common divisor of m and n is 1. If 7

12 + mn

is to be some integer k, then 7n+ 12m = 12nk ⇒ 12m = 12nk − 7n⇒ m = nk − 7 · n12 . Therefore 12divides n. We also have that 7n = 12nk − 12m⇒ 7 = 12k − 12mn . But since m and n are relativelyprime, we have that n divides 12. So n = 12. Therefore our goal is to have that when the fraction isreduced the denominator is 12. We need to find the smallest c so that the numerator 9360 has no factorsof 12. Dividing 9360 by 12 two times shows that 9360 = 122 · 65. So c = 122 for the smallest value of d,and d = 122 · 12 = 1728. We must check that this is indeed an integer, and it is because 7

12 + 6512 = 72

12 = 6.So d = 1728.Individual Round Four GaMaCo 2013

7. An ant is at the origin of a coordinate plane. The first day, the ant decides to move 4 units in somedirection parallel to the x axis, chosen at random. The next day, the ant randomly chooses to turn rightor left and then walk 4 units in that direction (now moving parallel to the y axis). Every day, the antcontinues moving in this fashion – turning left or right from the direction of the previous day and thenmoving 4 units. The probability he returns to the origin at the end of the 16th day is k

215. Compute k.

Answer: 2450

Solution: On odd days he moves in the East/West direction and on even days in the North/Southdirection. The even days and odd days are independent of each other, and so for all the even days he

6

must move up as many times as he moves down (and similarly for the odd days). Therefore the

probability is(84

)2 · 1216

=(8·7·6·54·3·2·1

)2 · 1216

= 4900216

= 2450215

.

8. A spy-aircraft is flying at constant altitude along a circle of radius 12 miles with center at a point A.The speed of the aircraft is v. At some moment in time, a missile is fired at the aircfraft from the pointA, which has speed v and is guided so that its velocity vector always points towards the aircraft. How fardoes the missile travel before colliding with the aircraft?

Answer: 6π

Solution: Call the point B the point on the original circle at which the aircraft is positioned when themissile is fired. We claim that the path of the missile is a circle with radius 6 that has AB tangent to it.Let P be some arbitrary point along the path of the aircraft. Call the intersection of PA with the newcircle be point M . Then ∠PAB is half the measure of the arc MA. Since the missile and aircraft is thesame speed, they should each travel equal distances in equal times, so PB = AM . Since the measure ofPB is the measure of ∠PAB, the radius of the smaller circle is half the radius of the larger. Hence themissile travels half the circumference of the circle, or 6π.Individual Round Five GaMaCo 2013

9. Compute the minimum value of f(x) =√x2 − 8x+ 21.

Answer:√

5

Solution: Note that completing the square inside the radical yields f(x) =√

(x− 4)2 + 5. Notice that(x− 4)2 has a minimum value at x = 4, leaving 0. Hence, the minimum value of f(x) is

√5.

10. Compute the sum of the solutions to the equation sin(x) + 8 cos2(x) sin(x) + 3√

3 cos(x) = 0 on theinterval 0 < x < 2π.

Answer: 7π3

Solution: Our motivation for the solutions comes from factoring the first two terms and trying to writethe summation as (close to) symmetric with both sin3(x) and cos3(x) terms. Making use of the identitysin2(x) + cos2(x) = 1 we see that

sin(x) + 8 cos2(x) sin(x) + 3√

3 cos(x) = sin(x)(1 + 8 cos2(x)) + 3√

3 cos(x)

= sin(x)((sin2(x) + cos2(x)) + 8 cos2(x)) + 3√

3 cos(x) · 1= sin(x)(sin2(x) + 9 cos2(x)) + 3

√3 cos(x)(cos2(x) + sin2(x))

= sin3(x) + 9 sin(x) cos2(x) + 3√

3 cos3(x) + 3√

3 sin2(x) cos(x)

= sin3(x) + 3√

3 sin2(x) cos(x) + 9 sin(x) cos2(x) + 3√

3 cos3(x)

= (sin(x) +√

3 cos(x))3

This is only zero when sin(x) +√

3 cos(x) = 0⇒ tan(x) = −√

3, which happens when x = 2π3 or 5π

3 . Thesum is 7π

3 .

7

3 Mathathon

GaMaCo Mathathon Round 1 – 2 pts each

1. What is the largest distance between any two points on a regular hexagon with a side length of one?

2. If z = 2 + 3i find the magnitude of: (−2z − 1

z − 3

)+ 2

3

(−2z − 1

z − 3

)− 1

3. For how many integers n ≥ 1 is10n − 1

9the square of an integer?


4. Two sides of a triangle have lengths 8 and 13. Given that the triangle is acute, how many integers maybe the length of the third side?

5. The sum of two positive integers is 2431. What is the greatest possible value of their greatest commondivisor?

6. Set S contains the first n positive integers. When one element is removed, the average of the

remaining terms is 4610

13. What number was removed?


7. Megan is flipping 17 coins to decide whether she should bake cookies and bring puppies for her class.If a majority of the coins are heads she will bake cookies and bring in puppies. The probability that shebakes cookies and brings in puppies, given that at least one coin is heads, is m

n for relatively primepositive integers m and n. What is blog4 (m+ n)c?

8. A vector in 3D space that in standard position in the first octant makes an angle of π3 with the x axis

and π4 with the y axis. What angle does it make with the z axis?

9. Compute√

20122 + 20122 · 20132 + 20132 − 20122.

8


10. Let f(x) = x3 + 6x2 + 11x+ 6, and S = {1, 2, 3, ..., 99, 100}. For how many s ∈ S is it true that f(s)is divisible by 12?

11. Find a closed form for 1 + 11 + 111 + · · ·+ 111 · · · 111︸︷︷︸n 1’s

in terms of n (not using sigma notation).

12. Find the area of the region enclosed by the graph of |3x+ 12|+ |5y + 10| = 45.


13. For how many real numbers a does the quadratic equation x2 + ax+ 5a = 0 have only integer roots?

14. Find the sum of the absolute values of the real roots of the equation x4 − 4x− 1 = 0.

15. Round log2

(32∑k=0

(32

k

)· 3k · 5k

)to the nearest integer.


16. Let P be a point inside a ball. Consider three mutually perpendicular planes through P . Theseplanes intersect the ball along three disks. If the radius of the ball is 2 and 1

2 is the distance between thecenter of the ball and P , compute the sum of the areas of the three disks of intersection.

17. Let ~v be an n-dimensional vector such that the nth compononent, vn, is given by vn =

√√√√√ 1(n+ 1

n− 1

) .

Compute the real number that the magnitude of ~v approaches as n grows without bound.

18. The numbers 1, 2, 3, ..., 2013 are written on a board. A student erases three numbers a, b, c andinstead writes the number

1

2(a+ b+ c)

((a− b)2 + (b− c)2 + (c− a)2

)She repeats this process until there is only one number left on the board. List all possible values of theremainder when the last number is divided by 3.

9


19. Let x, y, z be positive real numbers such that x+ y + z + 2 = xyz. Find the maximum value of

1√xy

+1√yz

+1√zx

20. Let f(n) be the sum of digits of the natural number n, e.g. f(35) = 3 + 5 = 8. Find f(f(f(55555))

)

21. The sequences {an} and {bn} both satisfy the same recurrence relation,

ak = kk · ak−1 + ak−2,

andbk = kk · bk−1 + bk−2.

The initial conditions are, however, a1 = b2 = 0 and a2 = b1 = 1. Compute a2013 · b2012 − a2012 · b2013.


22. Let ~a be a vector in R3. Let α, β and γ be the angles that ~a forms with the x, y and z axis,respectively. Find the minimum value of

1

sin2 α+

1

sin2 β+

1

sin2 γ

23. How many ordered triples of integers (a, b, c), where 1 ≤ a, b, c ≤ 10, are such that for every naturalnumber n, the equation (a+ n)x2 + (b+ 2n)x+ c+ n = 0 has at least one real root?

24. Compute b1010 ·√ec. If the first n digits of your answer are correct, your score for this question will

be

⌊n(n+ 1)

4

⌋.

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4 Mathathon Solutions


1. What is the largest distance between any two points on a regular hexagon with a side length of one?

Answer: 2

Solution: Note that the vertices of a regular hexagon all lie on the circumcircle of the hexagon. Thenthe furthest distance from one vertex is to another point on the circle, or the opposite vertex. Since aradius of the circle is the length of a side, the total distance is 2.

2. If z = 2 + 3i find the magnitude of: (−2z − 1

z − 3

)+ 2

3

(−2z − 1

z − 3

)− 1

Answer: 1√13

Solution: While the result can be computed directly it is interesting to consider the composition oftransformations of the complex plane. Writing these transformations as matrices and multiplying gives(

1 23 −1

)·(−2 −11 −3

)= −1

7

(0 11 0

), which is effectively the inversion f(z) = 1

z . And

| 12+3i | =

1|2+3i| = 1√

13.

3. For how many integers n ≥ 1 is10n − 1

9the square of an integer?

Answer: 1

Solution: Since10n − 1

10− 1is the sum of a geometric series with first term 1 and common ratio 10, the

numbers of this form are exactly the numbers 1, 11, 111, etc. Clearly 1 is a square (corresponding ton = 1). But modulo 8, this sequence is 1, 3, 7, 7, 7, . . . , and it will remaind a constant of 7 because everypower of 10 that is added from then on is divisible by 8. On the other hand, the squares modulo 8 are 0,1, and 4. So there are no other perfect squares mod 8, which means there are no other perfect squares inthe original sequence.


4. Two sides of a triangle have lengths 8 and 13. Given that the triangle is acute, how many integers maybe the length of the third side?

Answer: 5

Solution: Let x be the length of the third side. By triangle inequality, 5 < x < 21. Because the triangleis acute, the sum of the squares of any two sides must be greater than the square of the third side. So,

11

64 + x2 > 169→ x ≥ 11. Further, 64 + 169 > x2 → x ≤ 15. So, the integers 11, 12, 13, 14, and 15 work,for a total of 5 possible values.

5. The sum of two positive integers is 2431. What is the greatest possible value of their greatest commondivisor?

Answer: 221

Solution: Recall that by definition, the greatest common divisor divides both integers, and thus dividestheir sum, 2431. Prime factorizing, 2431 = 11 · 13 · 17. It cannot be the case that the greatest commondivisor is 2431 itself, so it is the second largest factor of 2431 which is 17 · 13 = 221

6. Set S contains the first n positive integers. When one element is removed, the average of the

remaining terms is 4610

13. What number was removed?

Answer: 22

Solution: We can create a range for the possible averages. The smallest possible average would be when

the largest term, n, is removed, in which case the average isn

2. The greatest possible average is when the

smallest term, 1, is removed, in which case the average isn+ 2

2. So,

n

2≤ 46

10

13≤ n+ 2

2, and multiplying

by two yields n ≤ 937

13≤ n+ 2. We see that n must either be 92 or 93, and also that13 | n− 1, so

n = 92. Thus, letting x be the number that was removed,12(92)(93)− x

91= 46

10

13→ x = 22


7. Megan is flipping 17 coins to decide whether she should bake cookies and bring puppies for her class.If a majority of the coins are heads she will bake cookies and bring in puppies. The probability that shebakes cookies and brings in puppies, given that at least one coin is heads, is m

n for relatively primepositive integers m and n. What is blog4 (m+ n)c?

Answer: 8

Solution: First find the general formula for the case of n coins where n is odd. The number of differentcombinations is 2n normally, however with the condition imposed it becomes 2n − 1 the number ofpossiblities for heads to be a majority is 2(n− 1) and the condition does not affect this. So the general

case is 2n−1

2n−1 . Plugging in 17 gives 216

217−1 . This become log4 216 + 217 − 1 we can ignore the −1 and then

factor out a 216 giving us log4 216(1 + 2) which then becomes log4 216 + log4 3. The first part is equal to 8and the second part is some positive number less than 1. Thus the final answer is 8 .

8. A vector in 3D space that in standard position in the first octant makes an angle of π3 with the x axis

and π4 with the y axis. What angle does it make with the z axis?

Answer: π3

12

Solution: Note that for an axis ei and a vector ~x, where ωi is the angle between them, thencos(ωi) = 1

|~x|(0, 0, 0, . . . , 1, . . . 0) · ~x. Summing the squares of these must give 1. So

cos2(π3 ) + cos2(π4 ) + cos2(θ) = 1⇒ θ = π3 .

9. Compute√

20122 + 20122 · 20132 + 20132 − 20122.

Answer: 2013

Solution: Set a = 2012 so the radicand is a2 + a2 · (a+ 1)2 + (a+ 1)2 = (a2 + a+ 1)2. So we’re lookingfor 20122 − 20122 + 2013 = 2013.


10. Let f(x) = x3 + 6x2 + 11x+ 6, and S = {1, 2, 3, ..., 99, 100}. For how many s ∈ S is it true that f(s)is divisible by 12?

Answer: 75

Solution: Finding the zeros and factoring, we see f(x) = (x+ 1)(x+ 2)(x+ 3). For a number to bedivisible by 12, it must be divisible by both 3 and 4. Because f(x) is the product of three consecutiveintegers, it will always be divisible by 3. We also see that f(x) will always be divisible by 4 except when

4 | s. Hence, there are3

4(100) = 75 elements in the set that work.

11. Find a closed form for the sum 1 + 11 + 111 + · · ·+ 111 · · · 111︸︷︷︸n 1’s

in terms of n.

Answer: 181(10n+1 − 9n− 10)

Solution: Note that 1 = 10−19 , 11 = 102−1

9 , etc. Therefore we are looking for19

(10− 1 + 102 − 1 + 103 − 1 + · · ·+ 10n − 1

)= 1

9

(10 + 102 + 103 + · · ·+ 10n − n

). The geometric

progression has sum 10 · 10n−110−1 . So the final sum is 181(10n+1 − 9n− 10).

12. Find the area of the region enclosed by the graph of |3x+ 12|+ |5y + 10| = 45.

Answer: 270

Solution: By properties of the absolute value, we may write this as 3|x+ 4| − 5|y + 2| = 45. Now, let’suse a transformation u = x+ 4, v = y + 2, so that the graph is now centered at the origin and symmetric

around all four quadrants. Thenu

15+v

9= 1 is the area in the first quadrant, and represents a right

triangle with legs of length 15 and 9. Thus, the total area of the graph is 4

(1

2

)(15)(9) = 270 .


13. For how many real numbers a does the quadratic equation x2 + ax+ 5a = 0 have only integer roots?

Answer: 4

13

Solution: Let the roots of the quadratic be m andn. Then m+ n = −a, and mn = 5a. So,mn+ 5m+ 5n = 0. This is almost factorable; in fact, adding 25 to both sides makes it so.(m+ 5)(n+ 5) = 25. There are four pairs of factors of 25, (±5,±5), and (±1,±25). Each of these pairscorresponds to a unique real value of a, and hence there are 4 such a

14. Find the sum of the absolute values of the real roots of the equation x4 − 4x− 1 = 0.

Answer:√

4√

2− 2

Solution: Note that x4 − 4x− 1 = (x2 + 1)2 − 2(x+ 1)2 = 0⇒ x2 + 1 =√

2(x+ 1). Hence,

x = 1±√

2√2−1√

2. By Descarte’s rule of signs, these are the only two roots.

15. Round log2

(32∑k=0

(32

k

)· 3k · 5k

)to the nearest integer.

Answer: 128

Solution: We can write the summation as(32k

)· 15k, which counts, for every value of k, the number of

ways to choose k balls out of our starting 32 balls in one bucket and place them into 15 other buckets.Summing over all k counts the total number of ways to distribute the 32 balls among 16 buckets, which isequal to 1632 = 2128. (Note that this expression is also the expansion of (1 + 15)32).


16. Let P be a point inside a ball. Consider three mutually perpendicular planes through P . Theseplanes intersect the ball along three disks. If the radius of the ball is 2 and 1

2 is the distance between thecenter of the ball and P , compute the sum of the areas of the three disks of intersection.

Answer: 47π4

Solution: Let d1, d2, d3 be coordinates of P in some rectangular coordinate system with the origin at thecenter of the ball and whose coordinates are parallel to the three mutually orthogonal planes through P .Then d2 = d21 + d22 + d23. Moreover, the radii of the disks of intersection are

√R2 − d21,

√R2 − d22, and√

R2 − d23. Hence the sum of their areas are π(22 − d21 + 22 − d22 + 22 − d23) = π(12− 14) = 47π

4 .Remark: The result does not depend on the choice of mutually perpendicular planes through P .

17. Let ~v be an n-dimensional vector such that the nth compononent, vn, is given by vn =

√√√√√ 1(n+ 1

n− 1

) .

Compute the real number that the magnitude of ~v approaches as n grows without bound.

Answer:√

2

Solution: Note that |~v| =

√√√√√ ∞∑n=1

1(n+ 1

n− 1

) . Also,

(n+ 1

n− 1

)=n (n+ 1)

2, so

∞∑n=1

1(n+ 1

n− 1

) =

∞∑n=1

2

n(n+ 1).

Using partial fraction decomposition, we determine that the sum telescopes and converges to 2. Hence,

|~v| =√

2

14

18. The numbers 1, 2, 3, ..., 2013 are written on a board. A student erases three numbers a, b, c andinstead writes the number

1

2(a+ b+ c)

((a− b)2 + (b− c)2 + (c− a)2

)She repeats this process until there is only one number left on the board. List all possible values of theremainder when the last number is divided by 3.

Answer: 0

Solution: Observe that

1

2(a+ b+ c)

((a− b)2 + (b− c)2 + (c− a)2

)= a3 + b3 + c3 − 3abc ≡ a+ b+ c (mod 3)

So the sum of the numbers on the board is constant viewed modulo 3. The initial sum is 2013·20142 ≡ 0

(mod 3), so the remaining number in the end must be divisible by 3.


19. Let x, y, z be positive real numbers such that x+ y + z + 2 = xyz. Find the maximum value of

1√xy

+1√yz

+1√zx

Answer: 32

Solution:We will prove that 1√

xy + 1√yz + 1√

zx≤ 3

2 . Observe that x+ y + z + 2 = xyz ⇐⇒ 1x+1 + 1

y+1 + 1z+1 = 1.

Now substitute a = 1x+1 , b = 1

y+1 and c = 1z+1 , from where x = 1−a

a , y = 1−bb and c = 1−c

c .

2

(√ab

(1− a)(1− b)+

√bc

(1− b)(1− c)+

√ca

(1− c)(1− a)

)≤ 3

Since 1−a = b+ c, 1− b = c+a and 1− c = a+ b, by the AM −GM inequality we have (2AB ≤ A2 +B2)

2

(√ab

(b+ c)(c+ a)+

√bc

(c+ a)(a+ b)+

√ca

(a+ b)(b+ c)

)≤∑cyc

a

c+ a+∑cyc

a

a+ b= 3

20. Let f(n) be the sum of digits of the natural number n, e.g. f(35) = 3 + 5 = 8. Find f(f(f(55555))

)Answer: 9

Solution: Observe that f(55555) ≤ 9(log 55555 + 1) = 9 · 55 log 555 + 9 < 9 · 55 · 3 + 9 = 1494. Thereforef(f(55555

))≤ 27. Then f

(f(f(55555

)))≤ 10. However f

(f(f(55555

)))≡ 55555 ≡ (−3)55 ≡ 0

(mod 9). Hence f(f(f(55555

)))= 9.

21. The sequences {an} and {bn} both satisfy the same recurrence relation,

ak = kk · ak−1 + ak−2,

15

andbk = kk · bk−1 + bk−2.

The initial conditions are, however, a1 = b2 = 0 and a2 = b1 = 1. Compute a2013 · b2012 − a2012 · b2013.

Answer: −1

Solution: Define Mk−2 as =

(ak−2 ak−1bk−2 bk−1

). Then matrix

Mk−1 =

(ak−2 ak−1bk−2 bk−1

)·(

0 11 kk

)=

(ak−1 akbk−1 bk

).

Our goal is to find det(M2012), where

M2012 =

(0 11 0

)·2013∏k=3

(0 11 kk

).

Note that the determinant of all of these matrices are −1, and the determinant is multiplicative. Sincethere are (2013− 3 + 1) + 1 = 2012 matrices including the original matrix, the determinant will be(−1)2012 = 1. But we want the opposite of this for the expression, so the answer is−1.


22. Let ~a be a vector in R3. Let α, β and γ be the angles that ~a forms with the x, y and z axis,respectively. Find the minimum value of

1

sin2 α+

1

sin2 β+

1

sin2 γ

Answer: 92

Solution: We begin with a lemma.Lemma(Nesbitt’s inequality) Let x, y, z be positive real numbers. Then the following inequality is true

x

y + z+

y

z + x+

z

x+ y≥ 3

2

Proof. Observe that xy+z = x2

xy+xz . Then using Cauchy-Schwarz inequality we have

x2

xy + yz+

y2

zy + xy+

z2

xz + yz≥ (x+ y + z)2

2(xy + yz + zx)

Now it is enough to show that (x+ y + z)2 ≥ 3(xy + yz + zx). Notice that(x+ y + z)2 − 3(xy + yz + zx) = 1

2

((x− y)2 + (y − z)2 + (z − x)2

)≥ 0. This finishes the proof of the

lemma.Now let ~a = (p, q, r). Then cosα = p√

p2+q2+r2and other similar expressions for cosβ and cos γ. Therefore

using the lemma above we have

1

sin2 α+

1

sin2 β+

1

sin2 γ= 3 +

p2

q2 + r2+

q2

p2 + r2+

r2

p2 + q2≥ 9

2

16

23. How many ordered triples of integers (a, b, c), where 1 ≤ a, b, c ≤ 10, are such that for every naturalnumber n, the equation (a+ n)x2 + (b+ 2n)x+ c+ n = 0 has at least one real root?

Answer: 165

Solution: In order for the equation (a+ n)x2 + (b+ 2n)x+ (c+ n) = 0 to have a real root, we need that(b+ 2n)2 − 4(a+ n)(c+ n) ≥ 0, which is equivalent with b2 − 4ac+ 4n(b− a− c) ≥ 0. It is enough thatb− a− c ≥ 0 because b2 − 4ac ≥ 0 follows from the fact that b ≥ a+ c because b2 ≥ (a+ c)2 ≥ 4ac. Thecase b2 − 4ac ≥ 0 and b < a+ c is not possible because when we let n become large enough the number4n(b− a− c) is very small (i.e. a very negative number), whilst b2 − 4ac is a finite positive number and inthis case we would have b2 − 4ac+ 4n(b− a− c) < 0 for all n large enough, a contradiction. Therefore itis enought to find those triples so that b ≥ a+ c.For b = 1 there is no pair (a, c) so that b ≥ a+ c. If b = 2 the only pair is (1, 1). If b = 3 then3 ≥ 1 + 2 = 2 + 1 and 3 > 1 + 1. In general, if x1, x2, ..., xm are natural numbers and m ≤ n then theequation x1 + x2 + ...+ xm = n has

(n−1m−1

)solutions. In our case m = 2. So we find all those cases when

b = a+ c, b− 1 = a+ c, b− 2 = a+ c etc. Also(n−11

)= n− 1, therefore we find that the desired number is:

10∑i=2

i(i− 1)

2= 165

24. Compute b1010 ·√ec. If the first n digits of your answer are correct, your score for this question will

be

⌊n(n+ 1)

4

⌋.

Answer: 16487212707

Solution: While there are many methods to do this question, using the taylor series for e and then ataylor series for

√x would be a possible one.

17

5 Relay

1-1. Compute the largest integral value of k so that exactly one of the roots of the cubic equationx3 + 8x2 + 7x+ k = 0 is real and positive.

1-2. Let T = TNYWR. Compute the minimum value of the function(x− |T |)2 + (x− |T | − 1)2 + (x− |T | − 2)2.

1-3. Let T = TNYWR. Two externally tangent circles have radii T + 2. Point A is on one circle and Bis on the other. What is largest possible length of AB?

2-1. Mitchell and Sitharthan decide to meet at the Reitz for lunch at some time between 10 a.m. and 11a.m. They both forgot what time precisely they decided on meeting and randomly arrive (uniformly)during that interval. Mitchell will wait for 10 minutes after he arrives, and Sitharthan will wait for 10minutes also. Let the probability that meet each other be m

n for relatively prime positive integers m andn. What is m+ n?

2-2. Let T = TNYWR. The smallest integral value of a strictly greater than 31 such that the equation6Tx+ a · y = 1 has solutions (x, y) in the integers is Q. Compute Q− 27.

2-3. Let T = TNYWR. Three circles with radii 3, 4, and (T − 3) are all mutually externally tangent.What is the area of the triangle with vertices at their centers?

18

3-1. Find the sum of the solutions for x in the following equation where 0 < x ≤ 2π

cot(x) + tan(x) = 4

3-2. Let T = TNYWR. Given z = cos(2T9 ) + i sin(2T9 ), compute 21+z and express in the form a+ ib.

3-3. Let T = TNYWR. An infinite number of pitches are thrown to Tony during batting practice. Theprobability that Tony hits exactly n of these pitches is Pn (for n = 0, 1, 2, . . .), where Pn+1 = 1

3Pn forn ≥ 0. What is the probability that Tony hits exactly |T |2 pitches?

4-1. Find the tens digit of 72014.

4-2. Let T = TNYWR. Find the number of zeros at the end of the decimal expansion of ((T + 1)!)!.

4-3. Let T = TNYWR. Compute the sum of the coefficients of all the terms in the expansion of

(Tx− (T − 1)y)3 (Tx− (T − 2)y)3 (Tx− (T − 3)y)2 (Tx− (T − 4)y)1 + Tx

19

5-1. How many 3-digit (base-10) positive integers are odd and do not contain the digit 4?

5-2. Let T = TNYWR. Joanna, the party animal, invited b T60c of her friends to go to a show aboutgrowing old with cats (so, in total, b T60c+ 1 people are going). If they randomly sit in a row, then theprobability that Joanna is in the middle is m

n for relatively prime positive integers m and n. Find m+ n.

5-3. Let T = TNYWR. How many (base-10) positive integers greater than 50 · T can be formed usingonly the digits in the set {1, 2, 3, 4, 5}, where the digits in any given number are all distinct?

5-4. Find the value of q in the following system of equations

(log3 p)2 = log3 p

2 and log3(p+ q) = log3 p+ log3 q

5-5. Let T = TNYWR. Find the area of the triangle in the complex plane with vertices at the complexcube roots of bT c.

5-6. Let the larger of the two numbers you receive be A and the smaller be B and let Z = b A40c and

T = (4B)2

9 . How many ways can T balls be chosen from a bag with Z balls of distinct colors (no balls areidentical to each other all are distinct) with replacement if order does not matter?

20

6 Relay Solutions

1-1: By Descarte’s rule of signs, there will only be a positive, real root if there is at least one sign change.So k must be negative. The largest integer that is negative is −1 .1-2: The function is an uperwards facing parabola with symmetry along the line x = |T |+ 1 thusplugging in x = |T |+ 1 gives 2 . (Note the value of T is irrelevant.)1-3: The largest possible value for AB is when A is placed at a maximum from the point of tangency andB is placed at the reflection of A on the other circle. The length of AB then becomes 2 ∗ diameter or4 ∗ radius = 4 ∗ (T + 2) = 4 ∗ 4 = 16 .

2-1: Draw a square with side length 60 (for 60 minutes), then go up 10 units on the x and y axis onopposing corners. Shade in the region contained by these points, this shaded region is the probabilitythat they will eat lunch together. To calculate that area find the area of the two isoceles triangles thatare left over (50 ∗ 50) and now subtract by the toal area 3600− 2500 = 1100. The probability is 11

36 and

m+ n equals 47 .2-2: In order for the condition to be true 6T and a must be relatively prime 6T factorizes into 2 ∗ 3 ∗ 47for a to be relatively prime it must be a multiple of 5, since 4 is a multiple of 2 it would not be relativelyprime. The smallest multiple of 5 greater than 30 is 35. So Q− 27 = 8 .2-3: The legs of the triangle are 7, T , and T + 1. We use Heron’s formula to get the area. The

semiperimeter is T + 4. So A =√

(T + 4)(T − 3)(4)(3). Since T = 8, A =√

12 · 5 · 12 = 12√

5 .

3-1: The equation simlifies to sin(x) cos(x) = 14 , which further simplifies to sin(2x) = 1

2 Note that due to

the 2x you have to account for 4 solutions which are π12 , 5π

12 , 13π12 , 17π

12 . The solutions add up to 3π .3-2: Let x = T

9 , thus 1 + z = 1 + cos(2x) + i sin(2x) which equals 2 cos2(x) + i sin(2x) changing everything

into tan(x) gives 21+tan2(x)

+ 2i tan(x)1+tan2(x)

adding fractions gives 2[1+i tan(x)]1+tan2(x)

which simplifies to 21−i tan(x)

plugging this into the original equation gives 1− i tan(T9 ). Plugging in T = 3π gives 1− i√

3 .

3-3: The probabilities follow the pattern P0,13P0,

19P0, ... The sum of this series has to equal 1 because it

contains all the possible probabilities. Factoring out the P0 and doing a geometric series on the sum

yields 32 thus 3

2 · P0 = 1 and P0 equals 2/3. Thus P4 is2

243.

4-1: The pattern for the tens digits is 0,4,4,0 which can be found from calculating 71, 72, 73, and 74 Theremainder when you divide 2014 by 4 is 2 and thus the tens digit is 4 .4-2: First calculate 5! as 120. To find the number of zeros calculate b120/25c + b120/5c which comes outto 28 .4-3: To calculate the sum of the coefficents plug in 1 for x and y doing this cancels the T in the productand you get 13 · 23 · 32 · 41 which equals 288 + 28 = 316 .

21

5-1: For a number to be odd it must end in 1,3,5,7, or 9. Thus there is 5 possiblities for the last digit 9for the second and 8 for the first (cannot be 0). 9 ∗ 8 ∗ 5 = 360 .5-2: There are 7! ways to arrange 7 people in a row with no restrictions. With the restriction that Joannahas to be in the middle there are 6 other spots and 6! ways to arrange those spots. The probability is 6!

7!

whch reduces to 17 1 + 7 = 8 .

5-3: 50T=400. For 3 digit numbers greater than 400 the hundreds digit can be 4 or 5. The tens digit has4 possiblities and the ones digit has 3. Thus for 3 digit numbers the number of possiblities is 2 ∗ 4 ∗ 3. For4 digit and 5 digit numbers there are 5! numbers for each. The sum of possbilities is equal to4! + 2 ∗ 5! = 264 . 5-4: Let x = log3 p. The first equation then reduces to x=2, which makes p = 9.Plugging p = 9 into the second equation gives log3(9 + q) = log3 9 + log3 q and using log rules that

simplifies to log3(9 + q) = log3 9q further simplifying gives 9 + q = 9q solving for q gives q = 9/8

5-5: To solve for the cube roots of 1, divide the unit circle into 3 sectors with the first point at (1, 0), the

other two points will be at (−12 ,√32 ) and (−12 ,−

√3

2 ) these points are already on the complex plane so youcan just calculate the area of the triangle formed by the points using a variety of methods. The

calculated area is3√

3

4

5-6: The generalized solution to this problem is(n+r−1

r

). The problem is the equivalent of asking how

many ways can you arrange r bugs on n leaves. To solve consider | as the border of a leaf and o as a bugso if we had 4 leaves and 2 bugs one combination would be |o||o|| where the 1st and 3rd levaes have bugson them. To find the number of combinations we need to see how many different way we can arrange theinside because the outside must always remain bars (the border of the leaves). So on the inside we haven+ r − 1 items which can be arranged in n+ r − 1! ways but since the bugs and leaves areindistinguishable you have to divide this by r! and n− 1! giving us the generalized solution of

(n+r−1

r

).

Plugging in the values given yields a final answer of 56 .

22

7 Tiebreaker

Tiebreaker Round One GaMaCo 2013

1. Let P = (x, y) be a point in the plane, and define f(P ) = ax+ by for constant real numbers a and b.Two distinct points in the plane, A and B, satisfy that f(A) = f(B) = 2013. If C ′ is the reflection of Cabout the line connecting A and B, and f(C) = 1337, then compute f(C ′).

Tiebreaker Round Two GaMaCo 2013

2. When the least common multiple of two positive integers is divided by their greatest common divisor,the result is 21. One of the integers is 161. Compute the smallest possible value of the other integer.

Tiebreaker Round Three GaMaCo 2013

3. Compute the probability that if the letters in GAMACO are randomly arranged then the letters are inalphabetical order.

23

8 Tiebreaker Solutions

Tiebreaker Round One GaMaCo 2013

1. Let P = (x, y) be a point in the plane, and define f(P ) = ax+ by for constant real numbers a and b.Two distinct points in the plane, A and B, satisfy that f(A) = f(B) = 2013. If C ′ is the reflection of Cabout the line connecting A and B, and f(C) = 1337, then compute f(C ′).

Answer: 2689

Solution: Points A and B satisfy that ax+ by = 2013. So the distance between f(C) = ax+ by and thisline is 676. Since C ′ is the reflection about this line, it is located on the line with constant term2013 + 676 = 2689.

Tiebreaker Round Two GaMaCo 2013

2. When the least common multiple of two positive integers is divided by their greatest common divisor,the result is 21. One of the integers is 161. Compute the smallest possible value of the other integer.

Answer: 69

Solution: Let the other number be n. Thenlcm(161, n)

gcd(161, n)= 21, and we also know that

lcm(161, n) · gcd(161, n) = 161n. Dividing this equation by the previous, we obtain

(gcd(161, n))2 =161n

21=

23n

3. Thus, 3(gcd(161, n))2 = 23n. As gcd(3, 23) = 1, we see that it must be the

case that both 3 | n and 23 | n, so the smallest possible value is 23 · 3 = 69 .

Tiebreaker Round Three GaMaCo 2013

3. Compute the probability that if the letters in GAMACO are randomly arranged then the letters are inalphabetical order.

Answer: 1360

Solution: The only way for the letters to be in alphabetical orer is AACMG, which can only be done inone distinct way. The number of distinct permutations of the letters is 6!

2! = 360. So the answer is 1360 .

24

Documents

Gator Mathematics Competition 2013 Tests and Solutions