UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST

Transcription

1 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: 2 3 x 5 4 x 7 2 = 250 x N 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? 4. The ones digit in the number 2 4 = 16 is 6. a. What is the ones digit in the number 2 6? b. What is the ones digit in the number 2 8? c. What is the ones digit in the number ? A C D B 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? A F B C D E 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. B Over A P Q

2 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 < n 2! 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 100 th meal? Give your answer as a fraction c /d, expressed in lowest terms. 10. Treasure Chest You have a long row of boxes. The 1 st 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 100 th 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 20 th 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

3

4 Solutions for First Round Fall 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 = N = 980. If you factor 250, you get Dividing both sides of the equation by gives = N. Multiplying these out gives N = = = 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 x 2 +y 2 =7 2 (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 =5 2 (BC is the hypotenuse and is given to have length 5). We can rewrite this as y 2 =5 2 (x 6) 2 and then substitute in to the other equation and solve: x 2 + (5 2 (x 6) 2 )=7 2 x (x 2 12x + 36) = 49 12x 11 = 49 x =5 4. Check the last digit of small powers of 2: 2 1 = 2, 2 2 = 4, 2 3 = 8, 2 4 = 16, 2 5 = 32, 2 6 = 64, 2 7 = 128, 2 8 = 256,.... It appears the pattern among last digits is 2, 4, 8, 6, 2, 4, 8, 6,.... This pattern repeats every 4 numbers. When 2 n has an n divisible by 4, the last digit is a 6. Now 2008 is divisible by 4, so we expect to end in a 6, so will end in a 2, will end in a 4, and most importantly 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 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 = 720, and if you had 4 angles greater than 180, that would already be over 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).

5 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 c = 51 and d = 101. After day one, 3 4 of your original lump remains. After day two, 8 9 of what you had at the end of day one is left, so of your original lump is left. Continuing, after the 100th day, the fraction of what you started with remaining is ( )( )( ) ( n 2 ) ( ) n We stop at because on day n we eat of what we currently have. To simplify this (n+1) 2 long product, note that n 2 1=(n 1)(n + 1). So we can rewrite: (1 3) (2 4) (3 5) (2 2) (3 3) (4 4) (n 1)(n + 1) n n ( ) ( ) The product collapses - there is massive cancellation. All that remains is = Here is a table of which boxes have how many coins: Box numbers: Coins per box: (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 The number of segments in the 20th figure is 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.

6 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 = 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 = 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 = 3540.

7 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, then what is the largest perfect square that is a factor of? 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 is divisible by? 5. What is the remainder when is divided by seven? over

8 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. Treasure Chest. You have a long row of boxes. The 1 st 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 for which the total number of coins exceeds 20120? (Remember to count the first box.) 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 4 th 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 4 th powers of the roots is 4.

9 (1.A.) 36 UNC MATH CONTEST SOLUTIONS FINAL ROUND JANUARY = 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 = 14 2 Factor into prime factors, so that the equation becomes = N. Cancel common factors and deduce = N. The largest perfect square factor of N is constructed by taking each prime factor to the largest available even power. You get = 14 2 = 196. (2.) 1/324 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 Each of these has probability 1/6 4, so the total probability is 4/6 4 = 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 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: = 126. FIGURE 1. Path-counting, step by step Solution (ii). ( 3 3 ) + (4 3 ) + (5 3 ) + (6 3 ) + (7 3 ) + (8 3 ) = 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 ( 7 3 ). In general, to reach the intersection of Avenue D and N th 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 ( N+2 3 ). Compute this for N = First, Second,..., Sixth Streets, and add.

10 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 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 ( 9 4 )=126 paths from which to choose. 56%&5 56%&7 56%&8 56%&9.:;&!; +(<&!; -=<&!; A;?&!; !;BC FIGURE 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 FIGURE 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 o o #1 #2 #3 #4 #5 #6 FIGURE 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 ( 9 4 )=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 ( 9 4 ). 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} is ( n+3 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.

11 UNC MATH CONTEST SOLUTIONS 3 (4.A) 289 Use long division to get (n ) (n + 11) =(n 2 11n + 121)+ 300 (n + 11). The polynomial n 2 11n is an integer for all integer values of n. The largest n for which 300 is integer is the n that (n+11) makes the denominator 300: n = = 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 = Each factor of 300 can be written as 2 r 3 s 5 t with r = 0, 1, 2; s = 0, 1; and t = 0, 1, 2. There are = 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, hence mod 7; and similarly mod 7. Now tabulate the powers working mod 7, and look for cyclic patterns: k = k k 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 mod 6, and deduce mod 7. Similarly, reduce mod 3, and deduce mod 7. Therefore = 7 0 mod 7. This problem is an extension of Problem 4 on the First Round Factoring into primes, 600 = 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 π 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 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

12 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: A 1 E 3 D 2 G 1 F 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. (9.a) (N 1)2 N FIGURE 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 π 3. Angle GDA is 60 degrees and the sector in the larger circle is one sixth of that circle, hence has area 3π 2. Desired area= 4 3 π 3 3π 2 =4 3 11π On any single roll, 5/6 of the time you win some points. On average this win W adds ( )/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 WL WWL 6 ( 5 6 )2 1 6 WWWL 9 ( 5 6 ) Taking the sum of these values, weighted by the probability of each case, gives S = (3 1 6 ) k=1 kpk where p = 5 6. This is FIGURE 6. Coins in boxes Let T N = 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 = 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 T N. 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 = N 2 S N = 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 T N : investigate the effect of doubling the sum that defines T n. 2 T N = N 2 N+1 T N = N 2 N

13 UNC MATH CONTEST SOLUTIONS 5 Subtract to get 2T N T N = (0 1)2 +(1 2)4 +(2 3) (N 1 N)2 N + N2 N+1 = N2 N+1 ( N ) = N2 N+1 (S N 1) = N2 N+1 (2 N+1 2) = (N 1)2 N (9.b) 2201 Note that if (N 1)2 N , 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 = = 18434, which is So far we have taken all boxes up to all the boxes with 10 coins. We still need to get = 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 2 11 boxes with 11 coins each, so there are plenty available.) Thus we need S = = 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. (10.b) ( n+3 4 ) c B a c c a a A FIGURE 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. C b b b

14 6 FINAL ROUND JANUARY 2012 B c a d d d A C FIGURE 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 number of such lists is ( n+3 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) x 4 x x2 1 6 x The polynomial P(x) =(x r)(x s)(x t)(x u) has roots r, s, t, and u. Multiply out: P(x) =x 4 (r + s + t + u)x 3 +(rs+ rt+ ru + st + su + tu)x 2 b (rst + rsu + rtu + stu)x + rstu. The coefficient a of x 3 is a = (r + s + t + u), the coefficient b of x 2 is b =(rs + rt + ru + st + su + tu), the coefficient c of x is c = (rst + rsu + rtu + stu), and the constant term d is d = rstu. We are told that r + s + t + u = 1 and r 2 + s 2 + t 2 + u 2 = 2 and r 3 + s 3 + t 3 + u 3 = 3 and r 4 + s 4 + t 4 + u 4 = 4. Therefore a = 1. Next observe that (r + s + t + u) 2 (r 2 + s 2 + t 2 + u 2 )= 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 = 1 2. That is, the coefficient of x2 is b = 1 2. We use similar observations to find c and d. It will be convenient to name the sums of the powers of the roots: R 1 = r + s + t + u = 1, R 2 = r 2 + s 2 + t 2 + u 2 = 2, R 3 = r 3 + s 3 + t 3 + u 3 = 3, and R 4 = r 4 + s 4 + t 4 + u 4 = 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 R 3 or R 1 b or R 2 a. Keep in mind that the first two useful equalities were a = R 1 and 2b = ar 1 + R 2. Observe that in fact 3c = br 1 + ar 2 + R 3. Put in the values we know for the right side and 3c =( 1 2 )(1) + ( 1)(2)+3 = 1 2. Thus c = 1 6. Similarly, 4d = cr 1 + br 2 + ar 3 + R 4 = ( 1 2 )(2) + ( 1)(3)+4 = 1 6 and d = 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.

15 UNC MATH CONTEST SOLUTIONS 7