Download 11-12 - UNC Math Contest

UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
First Round
For all Colorado Students Grades 7-12
November 3-6, 2011
You have 90 minutes- no calculators allowed
•
A regular hexagon has six sides with equal length and six angles with equal measure.
•
The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …
1. A 4 x 9 cardboard rectangle is cut up and the pieces rearranged, without gaps or overlap, to
form a square. What is the perimeter of that square?
2. Solve for N:
23 x 54 x 72 = 250 x N
C
3. In triangle ABC, side AB has length 6, side BC has length 5, side AC
has length 7. Segment CD is perpendicular to AB and point D divides
segment AB into two pieces. What is the length of the longer piece?
A
DB
4
4. The ones digit in the number 2 = 16 is 6.
a. What is the ones digit in the number 26 ?
b. What is the ones digit in the number 28 ?
c. What is the ones digit in the number 22011 ?
5. The hexagon ABCDEF has one internal angle greater
than 180 degrees, angle BCD. What is the largest number of
internal angles greater than 180 degrees that any single
hexagon can contain?
F
A
B C
D
E
B
6. Find the shortest distance from point A to point B, measured on the curved
surface of the cylinder. Segment PQ is a diameter of the circular base, and
the base has circumference 6 centimeters. Point A is 2 centimeters above
point P. Point B is 6 centimeters above point Q.
A
Q
Over
P
7. A drawer contains 24 utensils: one knife, one fork, and one spoon, each in 8 different colors. If
you pull items at random from the drawer without looking, what is the smallest number of items
you must take to be certain to have pulled out a complete matching table setting, containing a
knife, fork, and spoon of the same color?
8. Find the product of all of the positive integers n that satisfy the following inequality.
n < 12 < n + 17 < 2n + 10 < n2 ! 51
9. Square Meal You want to eat a lump of cookie dough in stages. A cookie
press converts the dough into a square of uniform thickness. On day 1 you
divide the square into 4 equal smaller square pieces, using a 2x2 grid, then
eat one of these 4 pieces. On day 2 you press the remaining dough into a
new square, subdivide it using a 3x3 grid, and eat one of these 9 pieces.
Continue pressing, subdividing, and eating pieces of the remaining dough.
What fraction of the original lump remains immediately after the 100th
meal? Give your answer as a fraction c /d, expressed in lowest terms.
10. Treasure Chest You have a long row of boxes.
The 1st box contains no coin. The next 2 boxes each contain 1 coin. The next 4 boxes each
contain 2 coins, the next 8 boxes each contain 3 coins, the next 16 boxes each contain 4 coins,
and so on. (The number of boxes that contain N coins is twice the number of boxes that contain
(N-1) coins.)
(a) How many coins are in the 100th box?
(b) How many coins are there when the contents of the first 100 boxes are combined?
11. Hex Consider the sequence of honeycomb-shaped figures below. The first figure has
one cell and is made of 6 line segments. The second figure has 7 cells and is made
of 30 line segments. How many line segments are there in the 20th figure? (The next
page is a sheet of paper tiled in hexagons for your use in considering this problem.)
First figure
Second figure
Third figure
Solutions for First Round
Fall 2011
1. The perimeter of the square is 24. The area of the 4 × 9 cardboard rectangle is 36. However
the rectangle is cut up and arranged into a square, that square will also have area 36. For a
square to have area 36, the side length must be 6. The perimeter of a square with side length
6 is 4 × 6 = 24.
2. N = 980. If you factor 250, you get 2 × 53 . Dividing both sides of the equation by 2 × 53 gives
22 × 5 × 72 = N . Multiplying these out gives N = 2 × 10 × 49 = 2 × 490 = 980.
3. The length of the longer piece is 5. Let’s call the length of the longer piece (AD) x. Then the
length of the segment DB is 6 − x since the length of AB is given as 6. We also do not know
the length of the segment CD, so let’s call that y. To solve for x, we can use the Pythagorean
theorem on the right triangle ADC. So x2 +y 2 = 72 (AC is the hypotenuse of the triangle and is
given to have length 7). To find y, we note that CDB is also a right triangle, so (x−6)2 +y 2 = 52
(BC is the hypotenuse and is given to have length 5). We can rewrite this as y 2 = 52 − (x − 6)2
and then substitute in to the other equation and solve:
x2 + (52 − (x − 6)2 ) = 72
x2 + 25 − (x2 − 12x + 36) = 49
12x − 11 = 49
x=5
4. Check the last digit of small powers of 2: 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64,
27 = 128, 28 = 256,. . . . It appears the pattern among last digits is 2, 4, 8, 6, 2, 4, 8, 6, . . .. This
pattern repeats every 4 numbers. When 2n has an n divisible by 4, the last digit is a 6. Now
2008 is divisible by 4, so we expect 22008 to end in a 6, so 22009 will end in a 2, 22010 will end
in a 4, and most importantly 22011 will end in an 8.
(a) The ones digit of the 6th power is 4.
(b) The ones digit of the 8th power is 6
(c) The ones digit of the 2011th power is 8.
5. The maximum number of angles with measure greater than 180 is 3. For example:
You cannot have a larger number of such angles because the sum of the interior angles of any
hexagon is 4 × 180 = 720◦ , and if you had 4 angles greater than 180◦ , that would already be
over 720◦ .
6. The length of the path connecting A to B, in centimeters, is 5. Unroll the cylinder! doing so
with reveal the line connecting A to B to be the hypotenuse of a right triangle with base 3 (half
of the circumference) and height 4 (the difference between the height of B, 6, and the height
of A, 2).
7. The smallest number of items one must select is 17. You could pull out all the 8 knives, all the
8 spoons, and still not have a set. However, the next utensil you pull will need to be a fork,
and will match the colors of the utensils you have already pulled. Alternatively, for each of the
8 colors, you can pull 2 utensils of that color without having a set - that’s 16 items. The next
one will have to give you a matching table setting.
8. The product of the positive integer solutions is 990. We need n < 12 so the possibilities for n
are 1, 2, 3, . . . , 11. We also need n + 17 < 2n + 10, or equivalently 7 < n. So now n can only
be 8, 9, 10, or 11. Now simply plug in these possibilities to the inequality and see which ones
work:
×
n=8:
8 < 12 < 25 < 26 < 13
n=9:
9 < 12 < 26 < 28 < 30
!
n = 10 :
10 < 12 < 27 < 30 < 49
!
n = 11 :
11 < 12 < 28 < 32 < 70
!
Thus the values of n which satisfy the equation are 9, 10, and 11, whose product is 990.
9. c = 51 and d = 101. After day one, 34 of your original lump remains. After day two, 89 of what
you had at the end of day one is left, so 34 89 of your original lump is left. Continuing, after the
100th day, the fraction of what you started with remaining is
"
!
"
! "! "! "
! 2
3
8
15
n −1
1012 − 1
···
···
4
9
16
n2
1012
2
−1
1
We stop at 101
because on day n we eat (n+1)
2 of what we currently have. To simplify this
1012
2
long product, note that n − 1 = (n − 1)(n + 1). So we can rewrite:
(1 · 3) (2 · 4) (3 · 5)
(n − 1)(n + 1)
(100 · 102)
···
···
(2 · 2) (3 · 3) (4 · 4)
n·n
(101 · 101)
The product collapses - there is massive cancellation. All that remains is
1 · 102
51
=
2 · 101
101
10. Here is a table of which boxes have how many coins:
Box numbers: 1 2-3 4-7 8-15 16-31
Coins per box: 0
1
2
3
4
32-63
5
64 - 100
6
(a) The number of coins in the 100th box is 6.
(b) The number of coins in all the first 100 boxes together is 480. We need to compute 2 · 1 +
4 · 2 + 8 · 3 + 16 · 4 + 32 · 5 + 37 · 6.
11. The number of segments in the 20th figure is 3540. One approach is to cut the figures into six
equal “wedges.” Draw lines from the center of the middle hexagon through each of its corners
- these lines will lie on top of edges of hexagons every other ring. Now count the number of
line segments.
First, just consider the line segments parallel to the lines we used to cut up the figures. For
those coinciding with our cutting lines, we will count only the one of the left (so we don’t over
count). The first figure has none of these lines. The second figure has just one. The third has
2 more, the forth has three more (one on the edge of our wedge), and so on. So to count these
lines, we must find 1 + 2 + 3 + · · · + 19 = 190.
Now for the edges in each wedge not parallel to our dividing lines. The first figure has 1, the
second figure has 3 more, the third has 5 more, the fourth has 7 more, and so on. So to count
these line segments, we sum 1 + 3 + 5 + 7 + · · · + 39 = 40 · 20/2 = 400.
So the total number of line segments each wedge contributes is 590. But there are 6 wedges,
so the total number of line segments is 6 · 590 = 3540.
UNIVERSITY OF NORTHERN COLORADO
MATHEMATICS CONTEST
FINAL ROUND
For Colorado Students Grades 7-12
January 21, 2012
You have three hours. No calculators are allowed. Show your work for each problem
on pages behind your answer sheet. Your score will be based on your answers and
your written work, including derivations of formulas you are asked to provide.
•
•
The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …
An ordinary die is a cube whose six faces contain 1, 2, 3, 4, 5, and 6 dots.
1. (a) What is the largest factor of 180 that is not a multiple of 15?
(b) If
satisfies
that is a factor of ?
, then what is the largest perfect square
2. Four ordinary, six-sided, fair dice are tossed. What is the probability that the sum of the
numbers on top is 5?
3. Mrs. Olson begins a journey at the
intersection of Avenue A and First Street in
the upper left on the attached map. She ends
her journey at one of the Starbucks on
Avenue D. There is a Starbucks on Avenue
D at every intersection from First Street
through Sixth Street! If Mrs. Olson walks
only East and South, how many different
paths to a Starbucks on Avenue D can she
take? Note that Mrs. Olson may pass one
Starbucks on her way to another Starbucks
farther to the East.
4. (a) What is the largest integer
for which
is divisible by
(b) For how many positive integer values of
5. What is the remainder when
is
divisible by
is divided by seven?
over
?
?
6. How many 5-digit positive integers have the property that the product of their digits is 600?
7. A circle of radius 1 is externally tangent to a circle of radius 3
and both circles are tangent to a line. Find the area of the shaded
region that lies between the two circles and the line.
8. An ordinary fair die is tossed repeatedly until the face with six dots appears on top. On
average, what is the sum of the numbers that appear on top before the six? For example, if
the numbers 3, 5, 2, 2, 6 are the numbers that appear, then the sum of the numbers before the
six appears is
. Do not include the 6 in the sum.
9. Treas u re C h es t. You have a long row of boxes. The 1st box contains no coin. The next
2 boxes each contain 1 coin. The next 4 boxes each contain 2 coins. The next 8 boxes each
contain 3 coins. And so on, so that there are
boxes containing exactly
coins.
(a) If you combine the coins from all the boxes that contain 1, 2, 3, or 4 coins you get 98
coins. How many coins do you get when you combine the coins from all the boxes that
contain 1, 2, 3, …, or
coins? Give a closed formula in terms of . That is, give a
formula that does not use ellipsis (…) or summation notation.
(b) Combine the coins from the first
boxes. What is the smallest value of
the total number of coins exceeds 20120? (Remember to count the first box.)
for which
10. An integer equiangular hexagon is a six-sided polygon whose side
lengths are all integers and whose internal angles all measure
120 degrees.
(a) How many distinct (i.e., noncongruent) integer equiangular
hexagons have no side length greater than 6? Two such hexagons
are shown.
(b) How many distinct integer equiangular hexagons have no side
greater than ? Give a closed formula in terms of .
(A figure and its mirror image are congruent and are not
considered distinct. Translations and rotations of one another are
also congruent and not distinct.)
11. Construct a 4th degree polynomial
that meets as many of the
following conditions as you can: The sum of the roots is 1, the sum of the squares of the roots
is 2, the sum of the cubes of the roots is 3, and the sum of the 4th powers of the roots is 4.
must approach its destination from a neighboring intersection
that is one block away either north or west. Therefore, the number of paths to any intersection is the sum of the numbers of
paths to these two nearest neighboring intersections. This is exactly the rule for generating Pascal’s triangle, as you can see in
the diagram. (Not all path counting problems will be this regular.) Take the sum of all such path-numbers along the Starbucks
on Avenue D: 1 + 4 + 10 + 20 + 35 + 56 = 126.
UNC MATH CONTEST SOLUTIONS
FINAL ROUND JANUARY 2012
(1.A.) 36
180= 2 x 2 x 3 x 3 x 5. Factors of 180 are produced by selecting
subsets of these prime factors: 2 x 2 = 4 and 2 x 3 x 5 = 30, for
instance. Any factor that is not a multiple of 15 must leave out
either the 5 or both of the 3s. The largest factor is found by
leaving out the 5. That factor is 2 x 2 x 3 x 3= 36. This question
and 1.B extend Problem 2 from the First Round.
(1.B) 196 = 142
Factor 99000 into prime factors, so that the equation becomes
25 × 32 × 54 × 73 × 11 = 23 × 32 × 53 × 11 × N. Cancel common factors and deduce 22 × 5 × 73 = N. The largest perfect
square factor of N is constructed by taking each prime factor
to the largest available even power. You get 22 × 72 = 142 =
196.
F IGURE 1. Path-counting, step by step
(2.) 1/324
Solution (ii). (33) + (43) + (53) + (63) + (73) + (83) = 126. Justification.
Consider first a typical path, described as a list of eastward and
southward steps. The path SSEESEE goes to the Starbucks on
Fifth Street. All paths going to this Starbucks will have seven
letters and exactly four of those letters will be E’s and three will
be S. The number of such paths is (73). In general, to reach the
intersection of Avenue D and Nth Street, going only east and
south, a path must go N − 1 blocks east and 3 blocks south, or
N + 2 blocks in all. Out of these N + 2 blocks, exactly 3 will be
south. The number of choices will be ( N3+2). Compute this for
N = First, Second,. . ., Sixth Streets, and add.
To get a sum of 5 you must get three 1s and one 2. There are four
different ways to get this combination: 2111, 1211, 1121, and
1112. Each of these has probability 1/64 , so the total probability
is 4/64 = 1/324.
(3.) 126
Solution (i). Starting in the upper left corner, begin tabulating
the number of pathways that reach intermediate destinations
on the grid. Each path that ends at an intermediate destination
1
2
FINAL ROUND JANUARY 2012
Solution (iii) Here is a solution that counts the paths to all the
Starbucks on Avenue D at once. Imagine a rope that follows
Mrs. Olson’s path from the starting point at the intersection
of Avenue A and First Street to her destination Starbucks on
Avenue D, and then continues eastward to the lowest right corner of
the diagram: the intersection of Avenue D and Sixth Street. The rope
must always travel eight blocks. Place a knot at each intersection, including the start and end. No matter which path Mrs.
Olson takes, there will be exactly nine knots on the rope. There
are three special knots on the rope where Mrs. Olson makes the
decision to head south to the next Avenue after passing through
an intersection, and there is a final fourth knot where she decides to stop, somewhere on Avenue D. Mark these four special
"#$%&'(&")*&+,-*.&/012&+,-*.&/31
knots X. For any rope that has four of its nine knots marked X,
there will be one and only one path Mrs. Olson can choose that
will correspond to this marking of the knots. Therefore, Mrs.
Olson has exactly (94)=126 paths from which to choose.
.:;&!;
56%&5
56%&7
56%&8
56%&9
+(<&!;
-=<&!;
>;?&!;
@;?&!;
A;?&!;
4
4
4
4
!;BC
F IGURE 2. Nine knots with four X’s
Solution (iv) Here is another solution that counts all the paths
at once. We discuss this solution at length because it will play
a role in the solution of Problem 10. Let W, X, Y and Z be the
street numbers of the locations of the four knots in solution (iii)
with 1 ≤ W ≤ X ≤ Y ≤ Z ≤ 6. We want to count the number
of choices we have for the ordered list W, X, Y, Z. Imagine a
row of rooms 1, 2, 3, 4, 5, and 6 with five walls dividing them:
#1
#2
#3
#4
#5
#6
F IGURE 3. Rooms with walls
Think of the letters as balls that will be dropped into the rooms
corresponding to their values. Several balls can go into a single
room and some rooms may be empty. For the example above
with W = 1, X = 1, Y = 3, and Z = 5 we get
oo
#1
#2
o
#3
#4
o
#5
#6
F IGURE 4. Balls with walls
We can use the shorthand o o | | o | | o |. Each solution
W, X, Y, Z corresponds to a string of nine symbols, four of which
are balls o and five of which are walls |. Therefore, there are
(94)=126 solutions in all. That is, the number of ways to pick an
ordered list of four numbers in nondecreasing order W ≤ X ≤
Y ≤ Z from the set S = {1, 2, 3, 4, 5, 6} is (94). More generally,
the number of ways to choose four numbers in nondecreasing
order W ≤ X ≤ Y ≤ Z from the set S = {1, 2, 3, 4, 5, 6, . . . n}
3
is (n+
4 ). This counting technique is sometimes called ”balls and
walls”, ”stars and bars”, or ”sticks and stones.”
A handful of students interpreted the problem to say that Mrs.
Olson could pass at most one Starbucks on her way to her final destination Starbucks. So interpreted, the problem becomes
somewhat harder; and full credit was given for correct solutions to this variant interpretation, whose answer is 91 paths.
UNC MATH CONTEST SOLUTIONS
(4.A) 289
Use long division to get
(n3 + 1631)
300
= (n2 − 11n + 121) +
.
(n + 11)
(n + 11)
The polynomial n2 − 11n + 121 is an integer for all integer values of n. The largest n for which (n300
is integer is the n that
+11)
makes the denominator 300: n = 300 − 11 = 289.
(4.B) 11
Each factor m of 300 that satisfies m ≥ 12 will produce a positive integer solution n = m − 11. Count all the factors of 300,
and then discard the seven small factors 1, 2, 3, 4, 5, 6, 10 that are
smaller than 12. To find all the factors of 300, use the prime factorization 300 = 22 × 31 × 52 . Each factor of 300 can be written
as 2r × 3s × 5t with r = 0, 1, 2; s = 0, 1; and t = 0, 1, 2. There
are 3 × 2 × 3 = 18 choices for these powers, hence 18 factors of
300. After discarding the seven factors that are too small, eleven
factors remain.
(5.) 0
Strategy: Look for cyclic patterns in the powers. This can be
done by computing powers of 11 and 12, or more efficiently by
first reducing these mod 7 and then computing powers: 12 ≡
5 mod 7, hence122011 ≡ 52011 mod 7; and similarly 112012 ≡
42012 mod 7.
Now tabulate the powers working mod 7, and look for cyclic
patterns:
k = 1 2 3 4 5 6 7 8 9 10 11 12
12k ≡ 5 4 6 2 3 1 5 4 6 2 3 1
11k ≡ 4 2 1 4 2 1 4 2 1 4 2 1
3
The table reveals that powers of 12 and 11 repeat with periods
of length 6 and 3 respectively. Thus each power can be reduced
by removing whole multiples of its period. Reduce the power
2011 ≡ 1 mod 6, and deduce 122011 ≡ 52011 ≡ 51 ≡ 5 mod 7.
Similarly, reduce 2012 ≡ 2 mod 3, and deduce 112012 ≡ 42012 ≡
42 ≡ 2 mod 7 . Therefore 122011 + 112012 ≡ 5 + 2 = 7 ≡ 0 mod 7.
This problem is an extension of Problem 4 on the First Round.
6. 210
Factoring into primes, 600 = 2 × 2 × 2 × 3 × 5 × 5. These six
primes must be placed in five digit places, and some primes
must share a place. Clearly both 5s must be alone, occupying two places. Now use trial and error to place the other four
primes in three places. The possible unordered lists of five digits are found to be
(i) 2,3,4,5,5; (ii) 1,8,3,5,5; (iii) 1,4,6,5,5; and (iv) 2,2,6,5,5.
Now count the possible orderings of the digits in each list. To
count the orderings of list (i)=2,4,3,5,5, first choose two slots for
the 5s: there are 10 ways. Then rearrange the remaining three
digits 3!=6 different ways. Thus there are 10 × 6 = 60 different
5 digit numbers whose digits are 2, 3, 4, 5 and 5. Similarly there
are 60 orderings of list (ii) and 60 orderings of list (iii). List (iv)
has only 10 × 3=30 distinct orderings, because of the repeated
2. The total is 210.
√
7. 4 3 − 11π
6
Draw a trapezoid ADFE by dropping perpendicular feet from
the two centers A and D to the points E and F on the tangent
line. The area sought is found by removing two circular sectors
from the trapezoid. Note that because AD has length 4, DG has
length 2, and the triangle is right, we see that √
triangle ADG is
a 30-60-90 triangle. Therefore AG has length 2 3. The area of
the trapezoid is the average of AE and DF √
times the distance
between two tangent points on the line, or 4 3. Angle GAD is
4
FINAL ROUND JANUARY 2012
!
computable by the methods used in Problem 9 below, but there
is also a shortcut method to do this problem:
Note that the table contains a duplicate of itself after the first
roll. This means that 5/6 of the time the first roll is a win that
adds 3, and then we get to start rolling all over again. Thus
S = (1/6)(0) + (5/6)(3 + S), which implies S = 15.
D
3
A1
G
E
F
2
1
!
F IGURE 5. Trapezoid minus two sectors
30 degrees, so angle EAD is 30+90=120 degrees and the sector
in the smaller circle is one third of that circle. It has area π3 .
Angle GDA is 60 degrees and the sector in the larger circle is
one sixth of that circle, hence has area 3π
2 .
√
√
11π
Desired area= 4 3 − π3 − 3π
2 =4 3 − 6 .
8. 15
On any single roll, 5/6 of the time you win some points. On
average this win W adds (1 + 2 + 3 + 4 + 5)/5 = 3. On the
other hand, 1/6 of the time you roll a losing value L = 6 that
halts the game. Now let S be the average sum of all possible
strings of rolls. Below is a table of all possible strings. Each W
in a string contributes 3 points to its average value.
String Average Value Probability
1
L
0
6
5
1
WL
3
6 × 6
5 2
WWL
6
( 6 ) × 16
WWWL
9
( 56 )3 × 16
...
...
...
Taking the sum of these values, weighted by the probability of
5
k
each case, gives S = (3 × 16 ) ∑∞
k =1 kp where p = 6 . This is
(9.a) ( N − 1)2 N +1 + 2
0
1
1
2
2
2
2
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
F IGURE 6. Coins in boxes
Let TN = 1 · 2 + 2 · 4 + 3 · 8 + 4 · 16 + . . . + N · 2 N be the total
number of coins in all the boxes in all the rows up to the row
with N coins per box. Let S N = 1 + 2 + 4 + 8 + 16 + . . . + 2 N be
the total number of boxes, including the box at the top with zero
coins in it. The question asks for a formula for TN . First get a
formula for S N , which is a geometric series, by using a standard
method: investigate the effect of doubling the sum.
S N = 1 + 2 + 4 + 8 + 16 + . . . + 2 N
2 · S N = 2 + 4 + 8 + 16 . . . + 2 N + 2 N +1
Subtract the first equation from the second one, and cancel pairs
of duplicated terms to deduce that 2S N − S N = 2 N +1 − 1.
Now try a similar trick on the TN : investigate the effect of doubling the sum that defines Tn .
2 · TN = 1 · 4 + 2 · 8 + 3 · 16 + 4 · 32 + . . . + N · 2 N +1
TN = 1 · 2 + 2 · 4 + 3 · 8 + 4 · 16 + . . . + N · 2 N
UNC MATH CONTEST SOLUTIONS
Subtract to get
2TN − TN =
5
3
(10.b) (n+
4 )
!"#$%&'(&")*&+*+,-&./01&+*+,-&.20
(0 − 1)2 + (1 − 2)4 + (2 − 3)8 + . . . + ( N − 1 − N )2 N
+ N2 N +1
c
= N2 N +1 − (2 + 4 + 8 + . . . + 2 N )
c
= N2 N +1 − (2 N +1 − 2) = ( N − 1)2 N +1 + 2
B
= N2 N +1 − (S N − 1)
A
c
b
b
(9.b) 2201
Note that if ( N − 1)2 N +1 + 2 ≈ 20, 000 then ( N − 1)2 N ≈ 10, 000.
Tabulate powers of 2 and estimate N2 N . Guessing N ≈ 10, we
check that indeed with this choice
(N
− 1 ) 2 N +1
+ 2 = 9 × 2048 + 2 = 18434,
which is ≈ 20000. So far we have taken all boxes up to all the
boxes with 10 coins. We still need to get 20121 − 18434 = 1687
more coins using boxes that now have 11 coins. Divide 1687/11
to get 153 and a remainder that forces us to take one more box,
making 154 more in all. (There are 211 boxes with 11 coins each,
so there are plenty available.) Thus we need S10 + 154 = 2047 +
154 = 2201 boxes. It is possible to do part (b) without getting
the formula in (a). Compute explicitly how many coins are in
the row with one coin per box, two coins per box, and so on.
A number of students did this. This problem is an extension of
Problem 10 from the First Round.
(10.a) 126
One approach is to enumerate the possibilities. In addition to
understanding the geometric properties of equiangular hexagons,
one must have a careful and systematic method to do the enumeration (see appendix). One student did this correctly. Obviously, another approach is to complete part (b) for the explicit
case n=6.
b
a
C
a
a
F IGURE 7. Triangles erected exterior to hexagon
Label the sides of the equiangular hexagon so that A is longest,
and a is opposite it. Let B be the longer side adjacent to a, and
C be the shorter side adjacent to a. Use lower-case letters to label their opposite sides. Erect exterior equilateral triangles to
the lower-case sides. This frames the hexagon inside an equilateral triangle. Equate the lengths of the three sides of the
equilateral frame: c + A + b = c + B + a = b + C + a; hence
A − a = B − b = C − c. Call this common difference d. It satisfies the inequality 0 ≤ d ≤ C − 1. The side length c = C − d satisfies 1 ≤ c ≤ C. Every equiangular hexagon determines a list
of four such whole numbers A ≥ B ≥ C ≥ c ≥ 1. Conversely,
by reversing the steps in the construction, we see that for every
list of four such whole numbers there exists exactly one equiangular hexagon that has these given side lengths. The reversed
construction starts by computing d = C − c, b = B − d, and
a = A − d; then drawing an equilateral triangle with side length
d (possibly zero); then assembling the equiangular hexagon by
adjoining parallelograms as drawn in the figure below.
6
FINAL ROUND JANUARY 2012
!"#$%&'(&")*&+*+,-.&/012&+*+,-.&/31
c
x is c = −(rst + rsu + rtu + stu), and the constant term d is
d = rstu.
A
B
d
d
b
d
C
a
F IGURE 8. Parallelograms around a triangle
We now count all such lists that satisfy n ≥ A ≥ B ≥ C ≥ c ≥ 1
by recognizing that this is version (iv) of Problem 3 above! The
3
number of such lists is (n+
4 ).
The appendix shows all 126 of the hexagons for n=6. They
are ordered and color-coded first by the length of the longest
side. Then they are ordered by the lengths of the three other
sides.
(11) x4 − x3 − 12 x2 − 16 x +
1
24
The polynomial P( x ) = ( x − r )( x − s)( x − t)( x − u) has roots
r, s, t, and u. Multiply out:
P( x ) = x4 − (r + s + t + u) x3 + (rs + rt + ru + st + su + tu) x2
−(rst + rsu + rtu + stu) x + rstu.
The coefficient a of x3 is a = −(r + s + t + u), the coefficient
b of x2 is b = (rs + rt + ru + st + su + tu), the coefficient c of
We are told that r + s + t + u = 1 and r2 + s2 + t2 + u2 = 2 and
r3 + s3 + t3 + u3 = 3 and r4 + s4 + t4 + u4 = 4. Therefore a =
−1. Next observe that (r + s + t + u)2 − (r2 + s2 + t2 + u2 ) =
2(rs + rt + ru + st + su + tu), which is 2b. Since we know that
(r + s + t + u) = 1 and the sum of the squares is 2, 2b = 1 − 2 =
−1 or b = − 21 . That is, the coefficient of x2 is b = − 21 . We use
similar observations to find c and d. It will be convenient to
name the sums of the powers of the roots: R1 = r + s + t + u =
1, R2 = r2 + s2 + t2 + u2 = 2, R3 = r3 + s3 + t3 + u3 = 3, and
R4 = r4 + s4 + t4 + u4 = 4. The coefficient c = −(rst + rsu +
rtu + stu). To find c, look at combinations of a, b, and R# s that
produce terms with three factors of the roots r, s, t, and u; that
is, look at things like R3 or R1 b or R2 a. Keep in mind that the
first two useful equalities were − a = R1 and −2b = aR1 + R2 .
Observe that in fact −3c = bR1 + aR2 + R3 . Put in the values we
know for the right side and −3c = (− 12 )(1) + (−1)(2) + 3 = 21 .
Thus c = − 61 . Similarly, −4d = cR1 + bR2 + aR3 + R4 = − 61 +
1
(− 12 )(2) + (−1)(3) + 4 = − 16 and d = 24
. We have determined
the last coefficient that we need.
The identities relating the Rs and the a, b, c and d are very old
identities called the Newton-Girard formulae. The relationships
between the roots of a polynomial and the coefficients are also
very old and are known as Vieta’s formulae. The R# s are sometimes called power functions and the expressions rs + rt + ru +
st + su + tu, rst + rsu + rtu + stu and so on are known as symmetric functions in the roots r, s, t and u. Students were not
expected to have seen these before! The challenge was to work
out the various relationships.
The contest writing team this year included Oscar Levin, Rich Morrow, Richard Grassl, Katie Diaz, contest director Ricardo Diaz, and
Rocke Verser, who submitted Problems 3, 10, and 11.
UNC MATH CONTEST SOLUTIONS
7