Twenty-eighth Annual UNC Math Contest First Round Fall, 2019 · 19.05.2020 · Now ask Uncle...

.

Twenty-eighth Annual UNC Math Contest First Round Fall, 2019

Rules: 90 minutes; no electronic devices.

1. This floor plan of the portal to the Addams castle shows twinchambers and a wedge-shaped pit (shaded) where a trap doorwill be placed. Find the degree measure of the acute angle at thetip of the wedge. The chambers are congruent regular ten-sidedpolygons.

2. Find the value of F(13.333) for the function F defined by F(x)= 2x for x <1 and F(x)=F(x-1) for x � 1.

3. Grandmama starts with identical cauldrons, one on her left that is full of liquid and one on her right that isempty. She pours three-quarters of the liquid from the left one into the right one and casts a spell. Then shepours one third of the liquid in the right one back into the left one. If she now pours one third of the contentsof the left cauldron back into the right cauldron, what fraction of the liquid is now in the cauldron on the left?

4. Find the smallest number of nightshade seedlings Uncle Fester could have if when he arranges them witheight in each row, he has three left over and when he puts 9 in each row, he has 4 left over.

5. Pugsley tries to fit a round circle into a square hole, but partof the circle sticks out. What is the total area of the two excessregions that Pugsley will carve from the circle? The circle hasradius 10, and Pugsley has it placed so that it passes through onecorner of the square and is tangent to two sides of the square.

Length=26Length=26Length=26Length=26Length=26

6. The Mummy’s tomb is a regular pyramid whose base is squareand whose four upright faces are congruent isosceles triangles. Ascarab beetle crawls along the outer surface of the pyramid fromone corner of the base to the midpoint of the opposite uprightedge. Find the length of the shortest path the beetle can take.Each triangular face has top angle 30 degrees; its upright sideseach have length 26.

TURN PAGE OVER

.

7. Wednesday has a lie detector machine that is correct most of the time, but labels a truthful person as a "liar"with probability 0.06 and labels a liar as "truthful" with probability 0.08. She interrogates three children, whoall claim they did not frighten Wednesday’s pet iguana. In actuality, exactly two are guilty, and one is innocent.She tests all three suspects. The machine labels two as liars and one as innocent. What is the probability thatthe machine has correctly identified the guilty pair?

8. Morticia gives a number between 1 and 65 to Lurch, and then Pugsley tries to guess what it is by askingLurch questions. Pugsley asks first "Is your number greater than 20?" and Lurch answers. Then Pugsley asks"Is your number even or odd?" and Lurch answers. "Hmm," says Pugsley, "Is your number a perfect square?"When Lurch answers, Pugsley says "I think I almost have it. I will know if you tell me whether one of the digitsin your number is a 4." Lurch answers and Pugsley says "I have it!" Just then, Morticia looks up and says "OhPugsley-I forgot to tell you! Lurch is having a backwards day today. Every answer he gives is wrong." Pugsleyfurrows his brow. After a minute he says, "Thanks. That is ok. I can still tell what his number is!" What doesPugsley think is Lurch’s number?

Out[735]=

1

23

4567

8

9

10

1112

13 14 1516

17

18

19

9. Lurch plays Whack-a-Thing on a circular table that has holesnumbered 1 through 19 counter-clockwise. If Thing is currentlyat hole numbered x, Thing will next travel x(x � 1) steps counter-clockwise to reach its new hole. Play begins with Thing selectinga starting hole, and a round of play ends when Thing returns toa hole previously visited in that round. What is the maximumnumber of distinct holes that Thing can occupy in a single roundof play?Example. If Thing starts at x = 2, it will then travel 2(2 � 1) = 2steps counterclockwise to hole 2 + 2 = 4. Next Thing will travel4(4 � 1) = 12 steps to hole number 4 + 12 = 16. Next, from hole16, Thing will travel (16)(15) steps, completing several circuitsaround the table and ultimately appearing at hole . . .

10. Gomez, Morticia, Pugsley, Wednesday, Uncle Fester, and Cousin Itt sit at a round table. If each casts a hexon one of the others, chosen at random, what is the probability that everyone gets hexed?

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 2019-2020 Solutions of First Round

1. This floor plan of the portal to the Addams castle shows twin chambers and a wedge-shaped pit (shaded) where a trap door will be placed. Find the degree measure of the acute angle at the tip of the wedge. The chambers are congruent regular ten-sided polygons.

Answer: 72

Solution 1:

One exterior angle of a regular n-sided polygon measures

.

One exterior angle of a regular decagon (10-sided polygon) measures

.

The marked angle contains two exterior angles of regular decagons as shown.

The marked angle in the original diagram measures

.

Solution 2:

One interior angle of a regular n-sided polygon measures

.

360n°

360 3610°= °

2 36 72× ° = °

( )2 180nn

- × °



One interior angle of a regular decagon measures

.

The marked angle in the original diagram measures

.

2. Find the value of for the function F defined by for and for .

Answer: 0.666

Solution:

.

3. Grandmama starts with identical cauldrons, one on her left that is full of liquid and one on her right that is empty. She pours three-quarters of the liquid from the left one into the right one and casts a spell. Then she pours one third of the liquid in the right one back into the left one. If she now pours one third of the contents of the left cauldron back into the right cauldron, what fraction of the liquid is now in the cauldron on the left?

Answer:

Solution 1:

Let us do the operations step by step:

Left Right

1 0

8 180 14410× °

= °

360 2 144 72° - × ° = °

( )13.333F ( ) 2F x x= 1x <

( ) ( )1F x F x= - 1x ³

( ) ( ) ( ) ( )13.333 12.333 1.333 0.333 2 0.333 0.666F F F F= = = = = × =!

13

14

34

1 112 2

- =3 2 14 3 2× =

1 2 12 3 3× =



The answer is .

Solution 2:

We may assume a number for the capacity of the left cauldron. We use 12.

Now we do the operations step by step:

Left Right

12 0

3 9

6 6

4

The desired fraction is .

4. Find the smallest number of nightshade seedlings Uncle Fester could have if when he arranges them with 8 in each row, he has three left over and when he puts 9 in each row, he has 4 left over.

Answer: 67

Solution 1:

We are searching the number n:

mod 8,

mod 9.

The smallest positive value of n is

.

Solution 2:

We are searching the number n:

mod 8,

mod 9.

If you don’t see the same “negative” remainders in solution 1, we may do the following.

Let for some nonnegative integer k. We have

mod 9.

Do mod calculations:

.

The smallest nonnegative value of k is 8.

The smallest positive value of n is .

The answer is 67.

13

4 112 3

=

3 5n = = -

4 5n = = -

( )lcm 8, 9 5 72 5 67- = - =

3n =

4n =

8 3n k= +

8 3 4k + =

8 1mod 9 1mod 9 1 8mod 9k k k= ® - = ® = - =

8 8 3 67n = × + =



Solution 3:

If you don’t know the “mod” sign, we may think this way.

Let Uncle Fester borrow 5 nightshade seedlings.

When he arranges the original nightshade seedlings + 5 borrowed with 8 in each row, he has none left over and when he puts 9 in each row, he has none left over.

Then the least number of nightshade seedlings is the least common multiple of 8 and 9. It is .

Now ask Uncle Fester to return 5 nightshade seedlings.

The answer is .

Solution 4:

We list the numbers that yield remainder 3 upon division by 8 until we hit a number that yields remainder 4 upon division by 9:

3, 11, 19, 27, 35, 43, 51, 59, 67, .

We see that 67 yields remainder 4 upon division by 9.

The answer is 67.

5. Pugsley tries to fit a round circle into a square hole, but part of the circle sticks out. What is the total area of the two excess regions that Pugsley will carve from the circle? The circle has radius 10, and Pugsley has it placed so that it passes through one corner of the square and is tangent to two sides of the square.

Answer:

Solution:

Let ABCD be the square with A on the circle.

Let O be the center of the circle. Let AB intersect the circle at E.

8 9 72× =

72 5 67- =

!

50 100p -

A B

C D

O

E



Draw the diagonal AC of the square. Obviously, O is on AC.

With and , is an isosceles right triangle.

Then we have .

The area of the shaded circular segment AE is

.

The total area of the shaded regions is

.

6. The Mummy’s tomb is a regular pyramid whose base is square and whose four upright faces are congruent isosceles triangles. A scarab beetle crawls along the outer surface of the pyramid from one corner of the base to the midpoint of the opposite upright edge. Find the length of the shortest path the beetle can take. Each triangular face has top angle 30 degrees; its upright sides each have length 26.

Answer:

Solution:

Let PABCD be the pyramid as shown.

We display the lateral faces on a plane.

The shortest path is the straight segment AM on the plane.

With , , and , AMP is a 30°-60°-90° triangle.

Therefore, .

45EAOÐ = ° OA OE= AOE

90AOEÐ = °

2 21 110 10 25 504 2

p p× - × = -

( )2 25 50 50 100p p× - = -

13 3

26PA = 13PM = 2 30 60APMÐ = × ° = °

13 3AM =

30º

26

60º

26

A

B C

D

P

M

P

A B

C

M



7. Wednesday has a lie detector machine that is correct most of the time, but labels a truthful person as a “liar” with probability 0.06 and labels a liar as “truthful” with probability 0.08. She interrogates three children, who all claim they did not frighten Wednesday’s pet iguana. In actuality, exactly two are guilty, and one is innocent. She tests all three suspects. The machine labels two as liars and one as innocent. What is the probability that the machine has correctly identified the guilty pair?

Answer:

Solution:

The probability that the machine labels two as liars and one as innocent is

.

The probability that the machine correctly identifies the guilty pair is

.

The probability that the machine correctly identifies the guilty pair under that the machine labels two as liars and one as innocent is

.

This is the answer.

8. Morticia gives a number between 1 and 65 to Lurch, and then Pugsley tries to guess what it is by asking Lurch questions. Pugsley asks first “Is your number greater than 20?” and Lurch answers. Then Pugsley asks “Is your number even or odd?” and Lurch answers. “Hmm,” says Pugsley, “Is your number a perfect square?” When Lurch answers, Pugsley says “I think I almost have it. I will know if you tell me whether one of the digits in your number is a 4.” Lurch answers and Pugsley says “I have it!” Just then, Morticia looks up and says “Oh Pugsley-I forgot to tell you! Lurch is having a backwards day today. Every answer he gives is wrong.” Pugsley furrows his brow. After a minute he says, “Thanks. That is ok. I can still tell what his number is!” What does Pugsley think is Lurch’s number?

Answer: 14

Solution:

We analyze in two cases.

Case 1: the answer is “no” to “is your number greater than 20?”

The numbers are from 2 to 20 inclusive.

We divide the numbers into two groups: even and odd

Even: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Odd: 3, 5, 7, 9, 11, 13, 15, 17, 19

10811093

20.92 0.92 0.94 0.92 0.08 0.06

1æ ö

´ ´ + ´ ´ ´ç ÷è ø

0.92 0.92 0.94´ ´

0.92 0.92 0.94 92 94 23 47 10812 92 94 2 8 6 23 47 12 1093

0.92 0.92 0.94 0.92 0.08 0.061

´ ´ ´ ´= = =

´ + ´ ´ ´ +æ ö´ ´ + ´ ´ ´ç ÷

è ø



There is a digit 4 only in some even numbers. Hence the answer to “is your number even or odd?” is “even”.

In the even numbers there are 2 square numbers: 4 and 16. One number contains a digit 4 and the other doesn’t.

In the even numbers there are 8 non-square numbers: 2, 6, 8, 10, 12, 14, 18, 20. There are 7 numbers not containing a digit 4.

Since Pugsley can know the number by knowing whether of one of digits of the number is a 4, the answer to “is your number a perfect square?” is “yes”.

If the answer is “yes” to “whether one of the digits is a 4, Lurch’s number is 4.

If the answer is “no” to “whether one of the digits is a 4, Lurch’s number is 16.

Now we examine which situation with the opposite answers still allows Pugsley to determine Lurch’s number.

We use the opposite answers for all.

The numbers are greater than 20: 21 to 63 inclusive.

The numbers are odd: 21, 23, 25, 27, 29, , 47, 49, 51, 53, , 63.

The numbers are non-squares: 21, 23, 27, 29, , 47, 51, 53, , 63.

If the answer is “no” to “whether one of the digits is a 4, the number can be one of 21, 23, 27, 29, , 39, 51, 53, , 63. Pugsley cannot determine the number.

If the answer is “yes” to “whether one of the digits is a 4, the number can be one of 41, 43, 45, 47. Pugsley cannot determine the number either.

Therefore, there is no solution in this case.

Case 2: the answer is “yes” to “is your number greater than 20?”

The numbers are from 21 to 63 inclusive.

We divide the numbers into two groups: even and odd

Even: 22, 24, 26, 28, , 62

Odd: 21, 23, 25, 27, , 63

Subcase a: to “is your number even or odd?”, the answer is “even”

Since there is only one square number: 36, the answer to “is your number a perfect square?” is “no”.

Then Pugsley cannot determine Lurch’s number because there are two or more numbers containing a digit 4 and there are two or more numbers not containing a digit 4 in the non-squares.

Subcase b: to “is your number even or odd?”, the answer is “odd”

There are two squares 25 and 49. One contains a digit 4, and the other doesn’t. Therefore, the answer to “is your number a perfect square?” is “yes”.

If the answer is “yes” to “whether one of the digits is a 4?”, the number is 49.

If the answer is “no” to “whether one of the digits is a 4?”, the number is 25.

! !

! !

! !

!

!



Now we examine which situation with the opposite answers still allows Pugsley to determine Lurch’s number.

We use the opposite answers for all.

The numbers are less than or equal to 20: 2 to 20 inclusive.

The numbers are even: 2, 4, 6, 8, 10, 12, 14, 16, 18.

The numbers are non-squares: 2, 6, 8, 10, 12, 14, 18.

If the answer is “no” to “whether one of the digits is a 4?”, the number cannot be determined.

If the answer is “yes” to “whether one of the digits is a 4?”, the number is 14. Pugsley can still determine Lurch’s number.

Therefore, Lurch’s number is 14.

Let us make a summary:

Lurch’s number is 14.

To “is your number greater than 20?”, the liar’s answer is “yes” (the true answer is “no”).

To “is your number even or odd?”, the liar’s answer is “odd” (the true answer is “even”).

To “is your number a perfect square?”, the liar’s answer is “yes” (the true answer is “no”).

To “whether one of the digits is a 4?”, the liar’s answer is “no” (the true answer is “yes”).

The answer to the problem is 14.

9. Lurch plays Whack-a-Thing on a circular table that has holes numbered 1 through 19 counter-clockwise. If Thing is currently at hole numbered x, Thing will next travel steps counterclockwise to reach its new hole. Play begins with Thing selecting a starting hole, and a round of play ends when Thing returns to a hole previously visited in that round. What is the maximum number of distinct holes that Thing can occupy in a single round of play?

Example. If Thing starts at , it will then travel steps counterclockwise to

hole . Next Thing will travel steps to hole number . Next, from hole 16, Thing will travel steps, completing several circuits around the table and ultimately appearing at hole .

( )1x x -

2x = ( )2 2 1 2- =

2 2 4+ = ( )4 4 1 12- = 4 12 16+ =16 15×!



Answer: 7

Solution:

Note that

.

The next number is the square of the previous number in mod 19.

Let us list.

1 distinct hole

7 distinct holes

7 distinct holes

(see the 2nd line) 6 distinct holes

6 distinct holes

6 distinct holes

2 distinct holes

(see the 7th line) 3 distinct holes

(see the 3rd line) 6 distinct holes







( ) 21x x x x+ - =

1 1®

81 25 362 16 9 5 6 1

247

34® ® ® ® ® ® ®

- -

93 5 6 179 4 16® ® ® ® ® ® ®

4 ®

56 17 45 16 9® ® ® ® ® ®

617 4 16 6 9 5® ® ® ® ® ®

49 6417 718

® ®-

648 7®

9 5®

10 59

®-

11 78

®-

12 77

11 11® ® ®-

13 176

® ®-

14 65

® ®-

15 164

®-



(see the 3rd line) 6 distinct holes

(see the 2nd line) 6 distinct holes

2 distinct holes

1 distinct hole

We see the maximum number of distinct holes to be 7. It can be achieved when is 2, 3, 10, 13, 14, or 15.

The answer is 7.

The following is Professor Diaz’s approach:

You might find it helpful on this problem to organize information in a graph like this:

16 93

®-

17 42

®-

18 11

®-

190

19®

0x



10. Gomez, Morticia, Pugsley, Wednesday, Uncle Fester, and Cousin Itt sit at a round table. If each casts a hex on one of the others, chosen at random, what is the probability that everyone gets hexed?

Answer: 53/3125 When everyone gets hexed, you have each person matched to one of the others. This is a rearrangement of six items in which no item stays in its original place. This is called a “derangement.” There are many ways to compute the number of these. You may want to look up the term “derangement.”

Solution 1:

We first calculate the probability that at least one doesn’t get hexed.

Denote the six people by G, M, P, W, F, and I.

What is the probability that one specified person (say G) doesn’t get hexed?

For G not to get hexed, any of M, P, W, F, and I can cast a hex on one of the four other people (not G), and G can cast a hex on one of M, P, W, F, and I.

The probability that one specified person doesn’t get hexed is .

What is the probability that two specified people (say G and M) don’t get hexed?

For G and M not to get hexed, any of P, W, F, and I can cast a hex on one of the three other people (not G and not M), and any of G and M can cast a hex on one of P, W, F, and I.

The probability two specified people don’t get hexed is .

5 14 55 5æ ö æ ö×ç ÷ ç ÷è ø è ø

4 23 45 5æ ö æ ö×ç ÷ ç ÷è ø è ø

G

M

P

W F

I

G

M P

W F

I



What is the probability that three specified people (say G, M, and P) don’t get hexed?

For G, M, and P not to get hexed, any of W, F, and I can cast a hex on one of the two other people (not G, not M, and not P), and any of G, M, and P can cast a hex on one of W, F, and I.

The probability that three specified people don’t get hexed is .

What is the probability that four specified people (say G, M, P, and W) don’t get hexed?

For G, M, P, and W not to get hexed, F and I must cast a hex on each other, and any of G, M, P, and W can cast a hex on one of F and I.

The probability that four specified people don’t get hexed is .

With the inclusive-exclusive principle, the probability that at least one doesn’t get hexed is

3 32 35 5æ ö æ ö×ç ÷ ç ÷è ø è ø

2 41 25 5æ ö æ ö×ç ÷ ç ÷è ø è ø

( )

( ) ( )

5 1 4 2 3 3 2 4

5 4 2 3 3 2 46

410 5 4 5 3 4 7 4 2

5 5

4

6 6 6 64 5 3 4 2 3 1 21 2 3 45 5 5 5 5 5 5 51 6 4 5 15 3 4 20 2 3 15 1 251 2 36 2 3 2 2 3 3 2 2 3 2 3 15 52

æ ö æ ö æ ö æ öæ ö æ ö æ ö æ ö æ ö æ ö æ ö æ ö× × - × × + × × - × ×ç ÷ ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø è ø è ø è ø è øè ø è ø è ø è ø

= × × - × × + × × - × ×

×= × - × + × - × = - + × -

=10

5 53 2 3 307264

5 5 3125× ×× = =

G

P W F

I M

G

M

P

W

F

I



The probability that everyone gets hexed is

.

Solution 2:

Know that

the number of the circular permutations of n different items is .

The total number of ways that everyone casts a hex is .

For everyone to get hexed, there are four cases.

Case 1: 6 = 6 meaning that six people make one circle.

Let each one cast a hex on his/her left person.

In the diagram A, B, C, D, E, F are Gomez, Morticia, Pugsley, Wednesday, Uncle Fester, and Cousin Itt in some order

The number of ways for this to happen is that is the number of the circular permutations of 6 peoples.

Case 2: 6 = 2 + 4 meaning that six people make two circles with one circle of two people and the other of 4 people


There are ways to choose two people to the first circle. The remaining four people go to the

second circle.

There is only one arrangement in the first circle, and there are ways to arrange 4 people in the second circle.

3072 5313125 3125

- =

( )1 !n -

65

5! 120=

62æ öç ÷è ø

3!

A B

D

C

E

F

A

B

C

E

D F



The number of ways in this case is .

Case 3: 6 = 3 + 3 meaning that six people make two circles with each circle of three people


There are ways to divide 6 people into two groups with each group of three people.

There are ways to arrange 3 people in each circle.


Case 4: 6 = 2 + 2 + 2 meaning that six people make three circles with each circle of two people

Let two people in one circle cast a hex on each other.

There are ways to divide 6 people into three groups with each group of two

people.

There is only one way to arrange 2 people in each circle.


63! 15 6 90

2æ ö

× = × =ç ÷è ø

6 33 32!

æ ö æ ö×ç ÷ ç ÷

è ø è ø

2!

6 33 3 20 1 2 22! 2! 402! 2

æ ö æ ö×ç ÷ ç ÷ × × ×è ø è ø × × = =

6 4 22 2 2

3!

æ ö æ ö æ ö× ×ç ÷ ç ÷ ç ÷

è ø è ø è ø

6 4 22 2 2 15 6 1 15

3! 6

æ ö æ ö æ ö× ×ç ÷ ç ÷ ç ÷ × ×è ø è ø è ø = =

D

E

F

A

B

C

A

B

C

D

E F



The total number of ways that everyone gets hexed is

.

The probability that everyone gets hexed is

.

Solution 3:

We will build a recursion to solve the general problem with n people.

Let n people be A, B, C, .

Let be the number of ways that everyone gets hexed.

A casts a hex on somebody. There are ways to choose one on whom A casts a hex.

Without loss of generality let A cast a hex on B.

There are two cases now.

Case 1: B casts a hex on A.

Then there are ways for the rest people to cast hexes.

Case 2: B casts a hex not on A.

120 90 40 15 265+ + + =

6265 535 3125

=

!

na

( )1n -

2na - 2n -

A C B

D

E F

A

C B

D

E F

n–2 people

A C

B

E F

n–1 people

D



Now look at the people without A.

There are ways for the people to cast hexes.

Somebody ? casts a hex on B.

Now we place A back and let A cast a hex on B. And we let the hex on B by ? be moved to on A.

Therefore, we have

.

That is,

.

The total number of ways that everyone casts a hex is .

The probability that everyone gets hexed is .

In our problem the probability is .

Let us calculate by the recursion.

,

,

,

,

,

1n -

1na - 1n -

( ) ( )2 11n n na n a a- -= - +

( ) ( )1 21n n na n a a- -= - +

( )1 nn -

( )1n

n

an -

665a

6a

0 1a =

1 0a =

( ) ( )2 1 01 1 0 1 1a a a= × + = × + =

( ) ( )3 2 12 2 1 0 2a a a= × + = × + =

( ) ( )4 3 23 3 2 1 9a a a= × + = × + =

C B

E F

n–1 people

?

D

A C B

E F

n–1 people

?

D



.

The desired probability is

.

( ) ( )5 4 34 4 9 2 44a a a= × + = × + =

( ) ( )6 5 45 5 44 9 5 53a a a= × + = × + = ×

65 53 535 3125×

=

Solution to #10 contributed by Jon Meilstrup Problem: If 6 people cast a hex on one of the others, chosen at random, what is the probability that everyone gets hexed? Call Hex(n) a function that computes the number of ways n people can cast on others in the group, with everyone getting hexed, and nobody casting on themselves. We need to compute Hex(6). Choose a person at random to be number 1. They cast a hex on one of the remaining 5 people. Call the person they choose to be number 2. Case 1) person 2 casts on person 1. Number of possibilities here is Hex(4), computed below. Case 2) person 2 casts on person 3. There are 4 people to choose from to be person 3. Case 2 has two sub cases: Case 2.1) person 3 casts on person 1. Number of possibilities here is Hex(3). Case 2.2) person 3 casts on person 4. There are 3 people to choose from to be person 4. Case 2.2 has two sub cases: Case 2.2.1) person 4 casts on person 1. Number of possibilities here is Hex(2) Case 2.2.2) person 4 casts on person 5. There are 2 people to choose from to be person 5. After 5 is chosen, there is only one choice for person 6. Hex(2) has only 2 people, each casts on each other, so Hex(2) =1 Hex(3) has 3 people, the first person can choose one of 2 people, so Hex(3)=2 Hex(4) has 4 people. We can follow similar reasoning as above, or note that the possibilities are: (1-2-3-4) or (1-2)(3-4) For (1-2-3-4) there are 3 choices for the second person, and 2 for the third, so there are 6 possibilities. For (1-2)(3-4) there are 3 choices for person 2, and the rest are fixed, so only 3 possibilities. So, Hex(4) = 6+3 = 9 Now, to total up the answer, in reverse order, as in each step, we need to sum the cases, then multiply by the number of ways that case could be used: Case 2.2.2) 2 choices Case 2.2.1) Hex(2) = 1 choice Case 2.2) sum of 2.2.2 and 2.2.1, or 2+1, then times 3 (number of ways to choose person 4) = 3*(2+1) = 9 Case 2.1) Hex(3) = 2

Case 2) sum of 2.2 and 2.1, or 9+2, times 4 (number of ways to choose person 3) = 4*(9+2)=44 Case 1) Hex(4) = 9 So, Hex(6) is the sum of Case 1 and 2, times the number of ways to choose person 2 = (44+9)*5 The probability is this number, divided by the total possible, so (53*5)/5^6 or 53/5^5

Twenty-eighth Annual UNC Math Contest Final Round January 26, 2020

Three hours; no electronic devices. Justify your answers and show your work clearly.The positive integers are 1, 2, 3, 4, ...

1. Show how to draw two lines on the face of a clock sothat the circle is divided into three regions in such a waythat the sum of the numbers in each of the three regionsis the same. On your answer form, list the numbers ineach of the three groups.

2. A box contains colored balls, each marked with a positive integer. Half of the balls are blueand the others are gold. One fifth of the balls are marked with prime numbers. Four dozen aremarked with numbers that are not prime. One third of the blue balls are marked with primenumbers.(a) How many of the gold balls are marked with numbers that are not prime?(b) How many balls are not blue or are marked with numbers that are not prime or are both notblue and marked with a number that is not prime?

3. Compute 1 + 2 – 3 + 4 + 5 – 6 + 7 + 8 – 9 + ... + 2017 + 2018 – 2019 + 2020

4. Find the length of a side of a square if a circle thatpasses through one corner of the square and is tangentto two sides of the square has radius ten.

5. At what time will a bell be rung for the last time if it is rung once every five hours, beginningat 1 pm, and rung 2020 times in all, so that it is rung at 1pm, 6 pm, 11 pm, 4 am, 9 am, ... ?(Assume there are no daylight savings time changes.)

27 26 25 24 23 22 28 9 8 7 6 21 29 10 5 20 30 11 4 19 31 12 1 2 3 18 32 13 14 15 16 17 33 34 35 36 37 · · ·

6. How many rows above or below the 1 and how manycolumns to the left or right of the 1 will 213 fall if thepositive integers are arranged in a ring as shown at left?(For example, 34 falls two rows below and one columnto the left of the 1.)

TURN PAGE OVER

7. There is one spider sitting at each of the four verticesof a tetrahedron. Simultaneously, each spider chooses atrandom one of the edges at its vertex and each spiderruns along its chosen edge to the next vertex. Whatis the probability that no two of the spiders collide,either while running along the edge or at the destinationvertex? A tetrahedron is a solid with four triangles forfaces, as shown in the figure. Give your answer as afraction in lowest terms.

8. The notation [2, 3, 5, 13] is an abbreviation for the fraction

2 +1

3 + 15+ 1

13

Solve for the positive integer a that satisfies this equation:[a + 1, a, a � 1]� [a � 1, a, a + 1] = [1, 1, 665, 2]

9. A right triangle has integer side lengths a b c and one of its three altitudes has length 12.Find all possible integer solutions (a, b, c). Be sure to consider all possible choices of the altitude.(An altitude is a perpendicular drawn from a vertex to the opposite side.)

10. (a) How many DISTINCT values occur for the expression c = ab, if (a, b) can be any pair ofdistinct roots of the equation x4 � x � 1 = 0? Justify your answer by explaining how you knowthey are distinct. The roots may be real or complex.(b) Find a polynomial P(x) that has each of these values of c as one of its roots. That is, P(ab) = 0whenever a 6= b are both roots of the fourth-degree equation given above.Hint. Let c = ab. Look for a pattern in the sequence of expressions(a � b), c(a � b), c2(a � b), c3(a � b), . . ..

END OF CONTEST

Thanks to Richard Grassl, founder of the UNC Contest, for his assistance in preparing this year’s contest.



University of Northern Colorado Mathematics Contest 2019-2020 Solutions of Final Round

1. Show how to draw two lines on the face of a clock so that the circle is divided into three regions in such a way that the sum of the numbers in each of the three regions is the same. On your answer form, list the numbers in each of the three groups.

Answer: , ,

Solution:

The sum of all 12 numbers on the face of a clock is .

The sum of all numbers in one region should be .

Note that we can make 6 pairs of two numbers whose sum is 13.

We may draw the lines as shown:

The numbers in the three regions are , , .

2. A box contains colored balls, each marked with a positive integer. Half of the balls are blue and the others are gold. One fifth of the balls are marked with prime numbers. Four dozen are marked with numbers that are not prime. One third of the blue balls are marked with prime numbers.

(a) How many of the gold balls are marked with numbers that are not prime?

(b) How many balls are not blue or are marked with numbers that are not prime or are both not blue and marked with a number that is not prime?

Answer:

(a) 28

(b) 50

( )1, 2,11,12 ( )3, 4, 9,10 ( )5, 6, 7, 8

12 131 2 3 12 782×

+ + + + = =!

78 263=

( )1, 2,11,12 ( )3, 4, 9,10 ( )5, 6, 7, 8

12 1 2

3

4 5 6 7

8

9

11 10



Solution:

Four dozen is of the number of balls. The total number of balls is .

There are balls marked with prime numbers and 48 balls are marked with non-prime numbers.

There are 30 blue balls and 30 gold balls.

There are blue balls marked with prime numbers. Then there are gold balls

marked with prime numbers.

Therefore, there are gold balls marked with non-prime numbers.

This is the answer to (a).

There are 10 balls are blue and are marked with prime numbers.

The answer to (b) is .

We may use a table to sort information:

Blue Gold Total

Prime 10 2 12

Non-prime 20 28 48

Total 30 30 60

3. Compute

.

Answer:

Solution 1:

.

Solution 2:

.

1 415 5

- =448 605

÷ =

60 48 12- =

1 30 103× = 12 10 2- =

30 22 8- =

60 10 50- =

1 2 3 4 5 6 7 8 9 2017 2018 2019 2020+ - + + - + + - + + + - +!

680,404

( ) ( ) ( ) ( )( )

1 2 3 4 5 6 7 8 9 2017 2018 2019 20201 2 3 4 5 6 7 8 9 2017 2018 20190 3 6 2016 3 1 2 672672 6733 2020 680,404

20202020 2020

2

+ - + + - + + - + + + - +

= + - + + - + + - + + + - +

= + + + + + = × + + + +

×= × + =

!

!

! !

( )( )

1 2 3 4 5 6 7 8 9 2017 2018 2019 20201 2 3 4 5 6 2017 2018 2019 2020 2 3 6 9 20191 2 3 2020 6 1 2 3 6732020 2021 673 6746 2,041,210 1,360,806 680,404

2 2

+ - + + - + + - + + + - +

= + + + + + + + + + + - × + + + +

= + + + + - × + + + +

× ×= - × = - =

!

! !

! !



4. Find the length of a side of a square if a circle that passes through one corner of the square and is tangent to two sides of the square has radius ten.

Answer:

Solution:

Let ABCD be the square with A on the circle as shown.

Let O be the center of the circle. Let BC be tangent to the circle at E. Draw the diagonal AC passing through O.

Draw OE. Extend EO intersecting AD at F. We have .

AFO is an isosceles right triangle. Then

Therefore, . This is the side length of square ABCD.

5. At what time will a bell be rung for the last time if it is rung once every five hours, beginning at 1 pm, and rung 2020 times in all, so that it is rung at 1pm, 6 pm, 11 pm, 4 am, 9 am, ?

Answer: 4 am.

Solution:

The number of hours elapsed from the first ring to the last (2020th) ring is

.

with remainder 15.

The 2020th ring happens in 420 days and 15 hours.

The time that is 15 hours after 1 pm is am.

The bell will be rung at 4 am.

10 5 2+

10OE OA= =

10 5 22

OF = =

10 5 2EF = +

!

( )2020 1 5 10095- × =

10095 24 420÷ =

1 15 12 4+ - =

A B

C D

O E F



6. How many rows above or below the 1 and how many columns to the left or right of the 1 will 213 fall if the positive integers are arranged in a ring as shown? (For example, 34 falls two rows below and one column to the left of the 1.)

Answer: 213 falls on the same row as the 1 and 8 columns to the right of the 1

Solution:

Let us add 4 to all numbers and fill four numbers 1, 2, 3, 4 into the four empty positions at the center.

We will find the position of number relative to the number 5 in the above diagram.

Now look at the square numbers. They are on the two marked half diagonals.

The closest square to 217 is . Note hat .

So 225 is 7 rows above the red 1 and 7 columns to the right of the red 1 in the above diagram.

Note that .

Then the 217 is one row below the red 1 and 7 columns to the right of the red 1.

That is, the 217 falls on the same row as the 5 and 8 columns to the right of the 5 in the above diagram.

Therefore, the 213 falls on the same row as the 1 and 8 columns to the right of the 1 in the original diagram.

213 4 217+ =

215 225= 15 1 2 7= + ×

225 217 8- =

31 12 1 2 3 18

30 11 4 19

28 10 5 20

28 9 8 7 6 21

32 13 14 15 16 17

27 26 25 24 23 22

33 34 35 36 37

35 16 5 6 7 22

34 15 4 1 8 23

33 14 3 2 9 24

32 13 12 11 10 25

36 17 18 19 20 21

31 30 29 28 27 26

37 38 39 40 41

225

7 positions

7 positions

217

8 positions



7. There is one spider sitting at each of the four vertices of a tetrahedron. Simultaneously, each spider chooses at random one of the edges at its vertex and each spider runs along its chosen edge to the next vertex. What is the probability that no two of the spiders collide, either while running along the edge or at the destination vertex? A tetrahedron is a solid with four triangles for faces, as shown in the figure. Give your answer as a fraction in lowest terms.

Answer:

Solution:

Let ABCD be the tetrahedron. Let a, b, c, and d be the spiders originally at vertices A, B, C, and D respectively.

Without loss of generality let spider a go to vertex B.

227

A

B C

D

a

b

c

d

A

B C

D

a

b

c

d



To avoid a collision spider b must go to vertices C or D. The probability for this to happen is .

Without loss of generality let spider b go to vertex C.

To avoid a collision spider c must go to vertices A or D.

We have two cases.

Case 1: c goes to vertex A

For this to happen, the probability is .

Then spider d cannot go anywhere to avoid a collision.

Case 2: c goes to vertex D


Then spider d can go only to vertex A to avoid a collision.


Therefore, the probability that there is no collision is

.

23

13

13

13

2 1 1 23 3 3 27× × =

A

B C

D

a

b

c

d

A

B C

D

a

b

c

d



8. The notation is an abbreviation for the fraction

Solve for the positive integer a that satisfies this equation:

.

Answer: 6

Solution:

The equation is

.

Let us simplify:

,

,

,

.

Note that

and .

We obtain .

9. A right triangle has integer side lengths and one of its three altitudes has length 12. Find all possible integer solutions . Be sure to consider all possible choices of the altitude. (An altitude is a perpendicular drawn from a vertex to the opposite side.)

Answer:

, , , ,

Solution:

Case 1: one leg is 12

Let the other leg be x ( ). We have .

[ ]2, 3, 5,13

12 13 1513

++

+

[ ] [ ] [ ]1, , 1 1, , 1 1,1, 665, 2a a a a a a+ - - - + =

1 1 11 1 11 1 11 11 1 6652

a aa aa a

+ + - + - = ++ + +

- + +

2 21 1 21 1 1333

a aa a a a

- +- = -

- + + +

( )( )( )

3 3

2 2

1 1 213331 1

a a

a a a a

- - += -

- + + +

( )( )2 2

2 213331 1a a a a

-= -

- + + +

( )( )2 21 1 1333 31 43a a a a- + + + = = ×

26 6 1 31- + = 26 6 1 43+ + =

6a =

a b c£ £( ), ,a b c

( ) ( ), , 12, 35, 37a b c = ( )12,16, 20 ( )9,12,15 ( )5,12,13 ( )15, 20, 25

0x > 2 2 212c x- =



That is,

Note that and are of the same parity. We have the following table:

x c

2 72 35 37

4 36 16 20

6 24 9 15

8 18 5 13

We have four solutions: , , , and .

Case 2: the height to the hypotenuse is 12

Then .

With we have .

That is,

.

We have

.

We have the following table:

c a b Solution?

2 72 35 37 25 15 20 yes

4 36 16 20 8 no

no

We have the fifth solution: . We found all five solutions:

, , , , .

( )( ) 4 22 3c x c x- + = ×

c x- c x+

c x- c x+ { },a b

{ }12, 35

{ }12,16

{ }9,12

{ }5,12

( ) ( ), , 12, 35, 37a b c = ( )12,16, 20 ( )9,12,15 ( )5,12,13

12ab c=2 2 2a b c+ = 2 2 22 24a b ab c c+ + = +

( ) ( )2 2 212 12c a b+ - + =

( ) ( ) 212 12 12c a b c a bé + - + ù × é + + + ù =ë û ë û

( )12c a b+ - + ( )12c a b+ + + a b+ 12c +

! ! ! !

( ) ( ), , 15, 20, 25a b c =

( ) ( ), , 12, 35, 37a b c = ( )12,16, 20 ( )9,12,15 ( )5,12,13 ( )15, 20, 25



10. (a) How many DISTINCT values occur for the expression , if can be any

pair of distinct roots of the equation ? Justify your answer by explaining how you know they are distinct. The roots may be real or complex.

(b) Find a polynomial that has each of these values of c as one of its roots. That is, whenever are both roots of the fourth-degree equation given above.

Hint. Let . Look for a pattern in the sequence of expressions , , , , .

Answer:

(a) 6

(b) various, e.g.

Solution:

(a)

Conclusion 1: has two real roots

To see this, we may draw graphs of and in the coordinate plane. There are two intersections. Therefore, has two real roots, one greater than one (it is even easy to see that it is between one and two) and the other between minus one and zero.

Alternatively, we may use the Descartes’ rule of signs.

The coefficients in are 1, , and . The signs are changed once. Therefore, there is one positive real root.

Plug into . We have polynomial . The coefficients in

are 1, 1, and . The signs are changed once. Therefore, has one negative real root.

Altogether has two real roots, a positive root greater than one and a negative root between minus one and zero.

Conclusion 2: has no double root.

Let . Then .

Use Euclidean Algorithm, they don’t have a common factor.

Therefore, has no double root.

This can also be checked with more elementary tools. If there is a double root (or a root of higher multiplicity), then call that root r and check to see whether there can be an r for which (x-r)2 divides into without remainder. Do long division and see that there can be no such r.

Conclusion 3: has 4 different roots.

c ab= ( ),a b4 1 0x x- - =

( )P x

( ) 0P ab = a b¹c ab= ( )a b-

( )c a b- ( )2c a b- ( )3c a b- !

( ) 6 4 3 2 1P x x x x x= + - - -

( ) 12 9 8 6 5 4 3 22 2 3 4 3 2P x x x x x x x x x x= - - + + + - - - -

4 1x x- -4y x= 1y x= +

4 1 0x x- - =

4 1x x- - 1- 1-

x- 4 1x x- - ( ) ( )4 41 1x x x x- - - - = + -4 1x x+ - 1- 4 1x x- -

4 1 0x x- - =

4 1x x- -

( ) 4 1f x x x= - - ( ) 34 1f x x¢ = -

4 1x x- -

4 1x x- -4 1x x- -



has four roots. Two are real, and the other two must be two complex numbers in conjugates: Suppose a complex number u+iv is a root of a polynomial P that has real coefficients. Then P(u+iv)=0, which is real. The conjugate of P(u+iv) is also zero, but that conjugate is equal to P(u-iv). Therefore u-iv is also a root. Complex roots of polynomials with real coefficients come in conjugate pairs.

Therefore, has 4 different roots.

Then there are possible values for .

Let us prove that all these 6 values are different.

Let two real roots be r and s and two complex roots be and where u and v are real.

Obviously, , , , and are different.

We just need to prove that . That is, .

Note that rs is negative and is positive. They cannot be equal.

It can also be helpful to roughly sketch the locations of the products in the complex plane:

Consider first the complex roots u+iv and u-iv. They do not lie on the imaginary axis, they lie in

the interior of two distinct quadrants. Multiplied by the real positive root, these products lie in

those same quadrants. Multiplied by the negative root, the products lie in two more opposite

quadrants. Those four products are all distinct. The other two products are real, but of opposite

sign, as shown above.

4 1x x- -

4 1x x- -

46

2æ ö

=ç ÷è ø

c ab=

u iv+ u iv-

( )r u iv+ ( )s u iv+ ( )r u iv- ( )s u iv-

( )( )rs u iv u iv¹ + - 2 2rs u v¹ +

2 2u v+

–2–2–2–2–2 –1–1–1–1–1 11111 22222 33333 44444 55555

–3–3–3–3–3

–2–2–2–2–2

–1–1–1–1–1

11111

22222

33333

44444

00000



(b)

Solution 1:

Let a and b be two different roots of .

We have

(1)

(2)

(1) × (2) yields

(3)

(1) × :

(4)

(2) × :

(5)

(4) – (5) yields

.

That is,

.

Since , we have

(6)

Let .

(3) becomes

(7)

(6) becomes

(8)

Plug (7) into (8)

Simplify it. We obtain

.

That is, the polynomial

.

has each value of as a root. This is the monic polynomial of the lowest degree.

4 1 0x x- - =

4 1a a= +4 1b b= +

( )4 1ab ab a b= + + +

2b4 2 2 2a b ab b= +

2a4 2 2 2b a ba a= +

( ) ( ) ( )2 2 2 2 2 2a b a b ab a b a b- = - - - -

( )( )( ) ( )2 2 1a b a b a b ab a b+ - + = - -

0a b- ¹

( ) ( )2 2 1a b a b ab+ + = -

c ab=

4 1a b c c+ = - -

( )( )2 1c a b c+ + = -

( )( )2 41 1c c c c+ - - = -

6 4 3 2 1 0c c c c+ - - - =

( ) 6 4 3 2 1P x x x x x= + - - -

c ab=



Solution 2:

In fact, I had the following polynomial of degree 12 as the very first solution to claim the full credit because the problem asks A POLYNOMIAL.


We have

(1)

(2)

(1) × (2) yields

(3)

(1) – (2) yields

.

That is,

.

Since , we have

(4)

Let .

(3) becomes

(5)

(4) becomes

(6)

Plug (5) into (6)

Simply it. We obtain

.


has each value of as a root.

In fact,

.

4 1 0x x- - =

4 1a a= +4 1b b= +

( )4 1ab ab a b= + + +

4 4a b a b- = -

( )( )( )2 2a b a b a b a b- + + = -

0a b- ¹

( )( )2 2 1a b a b+ + =

c ab=

4 1a b c c+ = - -

( ) ( )2 2 1a b a b cé ù+ + - =ë û

( ) ( )24 41 1 2 1c c c c cé ù- - - - - =ê úë û

12 9 8 6 5 4 3 23 3 3 4 3 2 0c c c c c c c c c- - + + + - - - - =

( ) 12 9 8 6 5 4 3 23 3 3 4 3 2P x x x x x x x x x x= - - + + + - - - -

c ab=

( )( )( )

12 9 8 6 5 4 3 2

6 4 3 2 6 4 3 2

3 3 3 4 3 2

2 2 1

P x x x x x x x x x x

x x x x x x x x x

= - - + + + - - - -

= - - - + + + - - -



Solution 3:

We have the following solution with the hint.


We have

(1)

(2)

Observe

We have

Then

That is,

.

Since , we have

.

We obtain

.


.

has each value of as a root.

Solution 4:

Let 4 roots of be a, b, c, and d.

By Vieta’s Formulas we have

, , , .

4 1 0x x- - =

4 1a a= +4 1b b= +

( ) 2 2c a b a b ab- = -

( )2 3 2 2 3c a b a b a b- = -

( )3 4 3 3 4 3 3 3 3 3 3 3 3c a b a b a b ab b a b a ab a b b a- = - = + - - = - + -

( ) ( )( ) ( )( )( )

4 5 4 4 5 2 2 2 2 2 2

2 2

1 1c a b a b a b a a b a b b a b ab a b a b

c a b a b a b

- = - = + + - + + = - + - + -

= - + - + -

( )( )4 2 21c c a b a b- - - = -

( )( ) ( ) ( )6 3 2 4 2 2 4 2 2 2 2 2 2c c c a b a b a b ab b a b a ab a b a b- - - = - = + - - = - - - -

( )( ) ( ) ( )( )6 3 2 4 1c c c a b c a b c c a b- - - = - - - - - -

0a b- ¹

( )6 3 2 4 1c c c c c c- - = - - - -

6 4 3 2 1 0c c c c+ - - - =

( ) 6 4 3 2 1P x x x x x= + - - -

c ab=

4 1 0x x- - =

0cyca =å 0

sysab =å 1

cycabc =å 1abcd = -



We want to build a polynomial with roots ab, ac, ad, bc, bd, and cd (call them the six numbers).

Let the polynomial be

.

A: the sum of the six numbers

.

B: the sum of the products of possible two numbers from the six numbers

.

There are terms in . B is the sum of 15 terms.

Multiply and :

.

We have

.

We obtain

.

C: the sum of the products of possible three numbers from the six numbers

.

There are terms in , there are terms in , and there are

terms in . C is the sum of 20 terms.

Square :

.

Square :

( )P x

( ) 6 5 4 3 2P x x Ax Bx Cx Dx Ex F= - + - + - +

0sys

A ab= =å

615

2æ ö

=ç ÷è ø

2 23 3sys sys

B a bc abcd a bc= + = -å å

4 312

1 2æ ö æ ö

× =ç ÷ ç ÷è ø è ø

2

sysa bcå

0cyca =å 1

cycabc =å

2 4 0sysa bc abcd+ =å

2 4 4sysa bc abcd= - =å

4 3 1B = - =

620

3æ ö

=ç ÷è ø

3 2 2 2 2 2 2 2 2 2 2 2 2 22 2cyc cyc sys cyc cyc sys cyc cyc

C a bcd a b c a b cd a a b c ab a a b c= + + = - + - = - +å å å å å å å å

44

1æ ö

=ç ÷è ø

3

cyca bcdå

44

1æ ö

=ç ÷è ø

2 2 2

cyca b cå

42 122æ ö× =ç ÷è ø

2 22sysa b cdå

0cyca =å

2 22 0cyc sys cyca ab a+ = =å å å

1cycabc =å



.

We obtain

.

D: the sum of the products of possible four numbers from the six numbers

.

There are terms in . D is the sum of 15 terms.

E: the sum of the products of possible five numbers from the six numbers

.

F: the product of the six numbers

.

We obtain the polynomial

.

2 2 2 2 2 2 2 2 2 2 22 2 1cyc sys cyc sys cyca b c a b cd a b c ab a b c+ = - = =å å å å å

0 1 1C = - + =

615

4æ ö

=ç ÷è ø

3 2 2 2 2 2 2 23 3 4 3 1sys sys

D a b c d a b c d a bc= + = - + = - + = -å å

4 312

1 2æ ö æ ö

× =ç ÷ ç ÷è ø è ø

3 2 2

sysa b c då

66

5æ ö

=ç ÷è ø

3 3 3 32 2 2 2 0

sys sys sys sys

a b c d abcd abcdE a b c d abab ab ab

= = = = =å å å å

3 3 3 3 1F a b c d= = -

( ) 6 4 3 2 1P x x x x x= + - - -

Twenty-eighth Annual UNC Math Contest First Round Fall, 2019 · 19.05.2020 · Now ask Uncle...

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Transcript of Twenty-eighth Annual UNC Math Contest First Round Fall, 2019 · 19.05.2020 · Now ask Uncle...