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

Transcription

1 Twenty-third nnual UNC Math Contest First Round November, 0 Rules: 90 minutes; no electronic devices The positive integers are,, 3,, rectangle 0 feet by 00 feet has a fence around its perimeter There are posts every feet along each of the four sides, arranged so that there is one post at each corner How many posts are there in all around the perimeter of the rectangle The bottom edge of the region at the left is a half circle whose diameter has length six The top edge is made of three smaller, congruent half circles whose diameters lie on the diameter of the bigger half circle What is the area inside the region C B 3 The square in the figure has area 36 Points and B are midpoints of sides What is the area of the triangle BC How many pairs of consecutive positive integers have product less than 300 For example, 3x= is less than 300 and 0x=0 is less than 300 How many such pairs are there (Count 0x and x0 as the same pair) The large circle at left is split into two congruent regions by two half circles that meet each other at the center of the large circle and meet the large circle at points directly above and below the center There is a straight line that simultaneously bisects (ie cuts in half) the area of both the regions What is the slope of the line TURN PGE OVER

2 6 (a) If two different integers are taken at random from among the integers,,, 000, what is the probability that the sum of the two integers is odd (b) If three different integers are taken at random from among the integers,,, 000, what is the probability that the sum of the three integers is odd 7 Earl E Bird drives the same route to work each day and leaves at 8 am He finds that if he goes 0 miles per hour the whole way he is 3 minutes late If he goes 60 miles per hour the whole way, he is 3 minutes early What speed must he go in order to arrive exactly on time 8 Suppose p(x) =x + 3x 7 and q(x) =x x 99 If a value of x is chosen at random from the interval 00 apple x apple 00, then what is the probability that q(p(x)) is negative 9 The Fibonacci numbers are,,, 3,, 8, (fter the first two Fibonacci numbers, each Fibonacci number is equal to the sum of the two before it) If F is the 0 th Fibonacci number, find the remainder when F F is divided by 7 D G B E H C F I 0 code is typed by always jumping from a position on the keypad to another adjacent position, diagonally, vertically, or horizontally The moving finger, having writ, must move on: it cannot stay frozen in position to pick a letter twice in a row The first letter can be any letter on the keypad Examples The code EIE can be created, but B, CF, and EII cannot lso, BC and CB are different codes (a) How many different three-letter codes can be created (b) How many different four-letter codes can be created END OF CONTEST

3 University of Northern Colorado Mathematics Contest University of Northern Colorado Mathematics Contest 0-0 Problems of First Round with Solutions rectangle 0 feet by 00 feet has a fence around its perimeter There are posts every feet along each of the four sides, arranged so that there is one post at each corner How many posts are there in all around the perimeter of the rectangle nswer: 8 Solution: The perimeter is The number of posts is The bottom edge of the region below is a half circle whose diameter has length six The top edge is made of three smaller, congruent half circles whose diameters lie on the diameter of the bigger half circle What is the area inside the region nswer: Solution: The area is 3 The square in the figure has area 36 Points and B are midpoints of sides What is the area of the triangle BC C B nswer: Problems are duplicated and solved by Ming Song (msongmath@yahoocom)

4 University of Northern Colorado Mathematics Contest Solution: The red-shaded part, the green-shaded part and the blue-shaded part have areas equal to,, and of the area of the square respectively C B So triangle BC s area is of the area of the square The answer is How many pairs of consecutive positive integers have product less than 300 For example, is less than 300 and is less than 300 How many such pairs are there (Count and as the same pair) nswer: 6 Solution: Note that and So there are 6 pairs,, 3, 3,, 6 7 The large circle at left is split into two congruent regions by two half circles that meet each other at the center of the large circle and meet the large circle at points directly above and below the center There is a straight line that simultaneously bisects (ie cuts in half) the area of both the regions What is the slope of the line nswer: Problems are duplicated and solved by Ming Song (msongmath@yahoocom)

5 University of Northern Colorado Mathematics Contest Solution: If we draw the other two small semicircles, we see the four regions of the same area: So each of two original regions (green and yellow parts) is equal to two small circles in area The half of one region is equal to one small circle in area Let O be the center Let be the lowest point of the large circle O Draw O Draw radius OB cutting the right original region into equal parts in area Then the sector OB of the large circle must equal half the area of one small circle Since the large circle B has an area equal to times the area of one small circle Therefore, sector OB must be of the circle That is, Draw line along OB obviously dividing both original regions into two parts of equal area O The slope of OB is º B Problems are duplicated and solved by Ming Song (msongmath@yahoocom) 3

6 University of Northern Colorado Mathematics Contest 6 (a) If two different integers are taken at random from among the integers,,, 000, what is the probability that the sum of the two integers is odd (b) If three different integers are taken at random from among the integers,,, 000, what is the probability that the sum of the three integers is odd nswer: (a), (b) Solution to (a): Take the first number If it is odd, we need the next to be even for the sum to be odd If it is even, we need the next to be odd for the sum to be odd For whatever it is there are 00 numbers to match it such that the sum is odd There are 999 numbers remaining So the desired probability is Solution to (a): There are ways to choose two numbers For the sum to be odd, one must be odd and the other must be even There are 00 ways to choose an odd number and 00 ways to choose an even number So there are ways to choose an odd number and an even number The desired probability is Solution to (b): There are 00 odd numbers and 00 even numbers n odd number + is en even number, and an even number is an odd number There is a one-to-one correspondence between the odd sums and the even sums The probability is exactly a half for each The answer is Or we can understand it as follows For the sum to be odd, we have two cases Case : all numbers are odd Case : one is odd, and the other two are even For the sum to be even, we have two cases as well Case : all numbers are even Case : one is even, and the other two are odd Since there are 00 odd numbers and 00 even numbers, the chances for the sum to be odd and to be even are equal Problems are duplicated and solved by Ming Song (msongmath@yahoocom)

7 University of Northern Colorado Mathematics Contest 7 Earl E Bird drives the same route to work each day and leaves at 8 am He finds that if he goes 0 miles per hour the whole way he is 3 minutes late If he goes 60 miles per hour the whole way, he is 3 minutes early What speed must he go in order to arrive exactly on time nswer: 8 miles per hour Solution : Let x in miles be the distance one way The time difference for the two trips is We obtain minutes So Driving at 60 miles per hour he needs minutes To be on time, he should drive for minutes So driving speed should be / (/60) = 8 mph Solution : The time for coming on time is the average of the times needed for the two trips at 0 miles per hour and 60 miles per hour respectively So the speed for coming on time is a kind of average speed of these two trips The answer to this kind of question is not the usual arithmetic mean average In this problem the answer is not the arithmetic mean, 0 It is instead the reciprocal of the average of the reciprocals of the two speeds This is called the harmonic mean of the speeds Let v be the average speed in miles per hour We have We get This is the answer 8 Suppose and If a value of x is chosen at random from the interval, then what is the probability that is negative nswer: Solution: Note that For, we must have Let Then That is, We have Let Then That is, We have or Problems are duplicated and solved by Ming Song (msongmath@yahoocom)

8 University of Northern Colorado Mathematics Contest Therefore, for, we must have or This can be seen in the figure: x The total interval length for is The length of interval is Therefore, the desired probability is = 0 9 The Fibonacci numbers are,,, 3,, 8, (fter the first two Fibonacci numbers, each Fibonacci number is equal to the sum of the two before it) If F is the 0 th Fibonacci nswer: Solution: number, find the remainder when is divided by 7 Let be the i th Fibonacci number for Let us write the Fibonacci sequence in mod 7:,,, 3,,,, 0,,,,,, 0,,, We see the period of length 6 in mod 7 Note that mod 6 So mod 7 Then is either or mod 7 depending on being odd or even Look at the Fibonacci numbers again:,,, 3,, 8, Every third number is even That is, mod if and only if 0 mod 3 Note that mod 3 So is odd Then mod 7 The answer is 6 Problems are duplicated and solved by Ming Song (msongmath@yahoocom) 6

9 University of Northern Colorado Mathematics Contest 0 code is typed by always jumping from a position on the keypad to another adjacent position, diagonally, vertically, or horizontally The moving finger, having writ, must move on: it cannot stay frozen in position to pick a letter twice in a row The first letter can be any letter on the keypad Examples: the code EIE can be created, but B, CF, and EII cannot lso, BC and CB are different codes (a) (b) nswer: (a) 00, (b) 9 Solution: Let B C D E F G H I How many different three-letter codes can be created How many different four-letter codes can be created be the number of codes of length n starting from a corner letter (one of, C, I, and G) Let be the number of codes of length n starting from an edge letter (one of B, F, H, and D) Let be the number of codes of length n starting from the center letter E Let Then is the total number of codes of length n We need to find and From a corner letter, we may go to one of two edge letters or to the center letter So we have the following recursion: Similarly, we have Obviously,,, and Calculate the recursions with the following table: We obtain and n () () (3) Problems are duplicated and solved by Ming Song (msongmath@yahoocom) 7

10 University of Northern Colorado Mathematics Contest lthough it was not asked for, we will find the general formula for If you have questions about the technique below, please us or look some things up These techniques are beyond what it is expected participants would know for this contest Plug () into (): Plug () into (3): () That is, Plug () and () into () () That is, (6) The characteristic equation of (6) is (7) By testing, is a root So we can factor out from the left side of (7): We have or From the second equation we have That is, So, or We obtain three characteristic values of (7):,, and So can be expressed as (8) With three initial values,, and, we can determine, B, and C However, I would like to introduce Obviously, Plugging,, into (8), we have respectively,, Solving for, B, and C we obtain Problems are duplicated and solved by Ming Song (msongmath@yahoocom) 8

11 University of Northern Colorado Mathematics Contest,, and Plugging these into (8) we have That is,, or From () and () We get the formula for in the closed form: Do you believe it If not, plug,, 3, and and see what you get The Moving Finger writes and, having writ, Moves on: nor all your Piety nor Wit Shall lure it back to cancel half a Line, Nor all your Tears wash out a Word of it - Omar Khayyam Problems are duplicated and solved by Ming Song (msongmath@yahoocom) 9

12 Twenty-third nnual UNC Math Contest Final Round January 3, 0 Three hours; no electronic devices Show your work nswers must be justified to receive full credit We hope you enjoy thinking about these problems, but you are not expected to do them all The positive integers are,, 3,, polynomial is quadratic if its highest power term has power two The sum of three consecutive integers is What is the smallest of the three integers Find the area of the shaded region The outer circle has radius 3 The shaded region is outlined by half circles whose radii are and and whose centers lie on the dashed diameter of the big circle 3 If P is a polynomial that satisfies P(x + ) =x + 7x + 9, then what is P(x) (Hint: P is quadratic) Tarantulas, B, and C start together at the same time and race straight along a 00 foot path, each running at a constant speed the whole distance When reaches the end, B still has 0 feet more to run When B reaches the end, C has 0 feet more to run How many more feet does Tarantula C have to run when Tarantula reaches the end termite nest has the shape of an irregular polyhedron The bottom face is a quadrilateral The top face is another polygon The sides comprise 9 triangles, 6 quadrilaterals, and pentagon The nest has 0 vertices on its sides and bottom, not counting the several around the top face How many edges does the top face have You may use Euler s polyhedral identity, which says that on a convex polyhedron the number of faces plus the number of vertices is two more than the number of edges ( vertex is a corner point and an edge is a line segment along which two faces meet) 6 How many ordered pairs (n, m) of positive integers satisfying m < n apple 0 have the property that their product mn is less than 0 TURN PGE OVER

13 7 (a) Give an example of a polyhedron whose faces can be colored in such a way that each face is either blue or gold, no two gold faces meet along an edge, and the total area of all the blue faces is half the total area of all the gold faces blue face may meet another blue face along an edge, and any colors may meet at vertices Describe your polyhedron and also describe how to assign colors to the faces (b) Show that if the faces of a polyhedron are colored in such a way that each face is either blue or gold and no two gold faces meet along an edge, and if the polyhedron contains a sphere inside it that is tangent to each face, then the total area of all the blue faces is at least as large as the total area of all the gold faces 8 garden urn contains 8 colored beetles: 6 red beetles, numbered from to 6, and yellow beetles, numbered from to Beetles wander out of the urn in random order, one at a time, without any going back in What is the probability that the sequence of numbers on the first four beetles to wander out is steadily increasing, that is, that the number on each beetle to wander out is larger than the number on the beetle before and that no number is repeated Give your answer as a fraction in lowest terms You may leave the numerator and denominator in a factored form 9 Starting at the node in the center of the diagram, an orb spider moves along its web It is permissible for the spider to backtrack as often as it likes, in either direction, on segments it has previously travelled On each move, the spider moves along one of the segments (curved or straight) to some adjacent node that is different from the node that it currently occupies (a) How many different five-move paths start at the center node and end at the center node (b) How many different seven-move paths start at the center node and end at the center node 0 (a) You want to arrange 8 biologists of 8 different heights in two rows for a photograph Each row must have biologists Height must increase from left to right in each row Each person in back must be taller than the person directly in front of him How many different arrangements are possible (b) You arrange biologists of different heights in two rows of 6, with the same conditions on height as in part (a) How many different arrangements are possible Remember to justify your answers (c) You arrange n biologists of n different heights in two rows of n, with the same conditions on height as in part (a) Give a formula in terms of n for the number of possible arrangements BONUS: You arrange biologists of different heights in three rows of, with the same conditions on height as in part 0(a) for all three rows How many different arrangements are possible END OF CONTEST

14 Twenty-third nnual UNC Math Contest Final Round January 3, 0 SOLUTIONS The sum of three consecutive integers is What is the smallest of the three integers NSWER: 7 SOLUTION: The sum of three consecutive integers is three times the average, or middle, integer, so the middle integer is /3 = 8 and the smallest is 7 nother approach is to write the three integers as n, n+, and n+ and compute =n+(n+)+(n+)=3n+3 so =3n and 7=n Find the area of the shaded region The outer circle has radius 3 The shaded region is outlined by half circles whose radii are and and whose centers lie on the dashed diameter of the big circle NSWER: 3p SOLUTION: The area of the shaded region is the area of a circle of diameter radius minus the area of a circle of radius = p p =3p One elegant way to see this is to reflect the top half of the diagram left to right 3 If P is a polynomial that satisfies P(x + ) =x + 7x + 9, then what is P(x) NSWER: P(x) =x 3x + 7 SOLUTION: P is clearly quadratic Let P(x) =x + Bx + C Then P(x + ) =(x + ) + B(x + )+C = x + x + + Bx + B + C = x +( + B)x +( + B + C) Therefore, =, + B = 7, and + B + C = 9 Solve for, B, and C Tarantulas, B, and C start together at the same time and race straight along a 00 foot path, each running at a constant speed the whole distance When reaches the end, B still has 0 feet more to run When B reaches the end, C has 0 feet more to run How many more feet does Tarantula C have to run when Tarantula reaches the end NSWER: 8 SOLUTION: In the time travels 00 feet, B travels 90 feet Thus B travels at a speed 9/0 the speed of Similarly, C travels at a speed 8/0 the speed of B and (8/0)(9/0)=7/00 the speed of When has gone 00 feet, C has gone 7 feet and has 00-7=8 feet to go

15 termite nest has the shape of an irregular polyhedron The bottom face is a quadrilateral The top face is another polygon The sides comprise 9 triangles, 6 quadrilaterals, and pentagon The nest has 0 vertices on its sides and bottom, not counting the several around the top face How many edges does the top face have You may use Euler s polyhedral identity, which says that on a convex polyhedron the number of faces plus the number of vertices is two more than the number of edges ( vertex is a corner point and an edge is a line segment along which two faces meet) NSWER: 8 SOLUTION: F = bottom + top + 9 triangles + 6 quadrilaterals + pentagon= 8 faces Let n be the number of edges around the open top Then there are also n vertices on the rim and the total number of vertices is V = 0 + n Count 9 3 edges on triangles, 7 edges on quadrilaterals (including the bottom), and so on Observe that this will count each edge two times So E = n = n = 60 + n Using the Euler identity, we obtain n =(/)(60 + n)+ or n = 8 6 How many ordered pairs (n, m) of positive integers satisfying m < n apple 0 have the property that their product mn is less than 0 NSWER: 97 SOLUTION: Note that 0 0 = 000 so for n = 0, m =,, 3,, 0 There are 0 such pairs For n = 9 clearly m can be 0 Try 9 = 009 For n = 9, m =,, 3,, For n = 8 clearly m can be Try 8 = 06 No! For n = 8, m =,, 3,, For n = 7 m =,, 3,, For n = 6 m =,, 3,, 3 For n = m =,, 3,, From now on, all m apple n will work dding up, we get plus The first sum is / = = 990 and the second is = 07 The total is 97 lternative way of looking at this one: There are 0 9 = ordered pairs (n, m) that satisfy m < n apple 0 Because most of them will satisfy the inequality mn < 0, it is easier to concentrate on counting the exceptions (shown in the shaded region ) that satisfy mn > 0 but n apple 0, hence m > = The bounding curve nm = 0 is a hyperbola, mirrorsymmetric across the line m = n By mirror symmetry, the tangent line through ( p 0), p 0) has slope The shaded region is convex, so it lies above this tangent line, inside the triangular region defined by m + n > p = 897 Since we seek only integer solutions, this narrows the field of candidates to m + n 90 There are 30 lattice points in the triangular

16 region that lie on or above this border-line It seems probable that most of these candidates will lie in the shaded region, but now we must check carefully if any of the points on or above the border-line are false solutions (not actually on the shaded region) Points (m, n) on this line can be written in parametric form as n = + t, m = t Since ( + t)( t) =0 t it is easy to create this table of values for (n, m, n m) for the small values of t of interest: n = m = 3 0 n m = Note that the last two points (9, ) and (0, 0) lie outside the shaded region Thus there are 30 = 8 candidates that remain One can check easily (numerically) that the lattice points immediately above these two discarded false solutions are true solutions; that is, all candidates have been classified, and all lattice point solutions in the shaded region are now accounted for Take -8= 97 7 (a) Give an example of a polyhedron whose faces can be colored in such a way that each face is either blue or gold, no two gold faces meet along an edge, and the total area of all the blue faces is half the total area of all the gold faces blue face may meet another blue face along an edge, and any colors may meet at vertices Describe your polyhedron and also describe how to assign colors to the faces (b) Show that if the faces of a polyhedron are colored in such a way that each face is either blue or gold and no two gold faces meet along an edge, and if the polyhedron contains a sphere inside it that is tangent to each face, then the total area of all the blue faces is at least as large as the total area of all the gold faces NSWER: (a) One possible example is a square box with large top and bottom and small height, with the top and bottom colored gold and the four sides blue If the top and bottom are squares of side s, then the total area colored gold is s If the height of the box is h, then the total area colored blue is h s It is easy to select s and h to make h s equal to half of s Set h s = s or h = s Choose, say, h = and s = (b) Triangulate each face by connecting the point of tangency on the face to each vertex of the face Look at a gold face and one of the triangles into which it has been cut The gold triangle meets a blue triangle along the edge where the two faces meet Consider the two triangles and the radii that go from the tangent points on the two faces (=the apexes of the triangles) to the center of the sphere By symmetry, the two triangles are congruent This means that for each triangle colored gold, there is a congruent triangle colored blue Therefore, the total area colored blue is at least as large as the total area colored gold G More detail: We are told that each face has a point of F tangency to the sphere Each face is perpendicular to the radius that connects the center of the sphere to the point of tangency Consider a pair of faces that meet P along an edge, as in the diagram G Using the radii, which are perpendicular to the respective faces, we can find some congruent triangles Call F C the far end of the edge in the diagram P and the center of the sphere C Call the point of tangency of one face F and the point of tangency of the other face G The segments FC and GC are radii and so have the same length The segments FP and GP lie in the faces so they are perpendicular to the radii to those two faces This makes triangles CFP and CGP both right with right angles at F and G They have the common hypotenuse PC and

17 they each have legs the length of the radius of the sphere This makes them congruent (by the Hypotenuse-Leg rule for right triangles) and tells us that FP and GP have the same length P G There is another pair of congruent right triangles: Just F Q replace P with the other end of the edge, Q: Triangles C CFQ and CGQ are also congruent This will give us the result about color Consider a face colored gold- the pentagon in the example, say Triangulate it by connecting the point of tangency to each vertex Consider one of the gold triangles: we know it meets a blue face and that the blue face contains a triangle that is congruent to the gold one Therefore, the amount of surface area that is blue is at least as large as the amount that is colored gold 8 garden urn contains 8 colored beetles: 6 red beetles, numbered from to 6, and yellow beetles, numbered from to In random order, beetles wander out of the urn, one at a time, without any going back in What is the probability that the sequence of numbers on the first four beetles to wander out is steadily increasing, that is, that the number on each beetle to wander out is larger than the number on the beetle before and that no number is repeated Give your answer as a fraction in lowest terms You may leave the numerator and denominator in a factored form NSWER: 7/(8 7 6) or 7/896 SOLUTION: There are 8 numbered beetles that can wander out of the urn and therefore ways for a sequence of four beetles to emerge There are two copies of each number apple 6 and a single copy of each of the numbers 7 Categorize the possibilities in which the numbers increase into disjoint cases Case : ll beetles have numbers apple 6 There are C(6, ) combinations of distinct numbers apple 6 Once we have selected the numbers, there is only one way to list them in increasing order Each number apple 6 occurs on different beetles in the urn There are C(6, ) ways to create strictly increasing sequences of distinct numbers apple 6 Case The first 3 beetles have numbers apple 6 and the last one out has a number 7 There are C(6, 3) combinations of 3 distinct numbers apple 6 Each of the 3 numbers occurs on beetles There are 6 possibilities for the fourth number and the fourth number chosen is on only one beetle: 3 C(6, 3)C(6, ) Case 3 The first beetles have numbers apple 6 and the last have numbers 7: C(6, )C(6, ) Case The first beetle has a number apple 6 and the last three have numbers 7: C(6, )C(6, 3) Case ll four beetles have numbers 7: C(6, ) Total number of possibilities that give increasing sequences: = ( 6 ) + 3 ( 6 3 )(6 ) + ( 6 )(6 ) + (6 )(6 3 ) + (6 ) =( ) ( 3 )( 6 3 )(6)+( ) (6)( 6 3 ) = 3 78 Total number of random sequences: Probability of the sequence being increasing is the quotient: (3 78)/(8 7 6 ) =78/(8 7 6 ) =7/(8 7 6) =7/896 Note on generating functions: Those students familiar with the binomial theorem may recognize the sum ( 6 ) + 3 ( 6 3 )(6 ) + ( 6 )(6 ) + (6 )(6 3 ) + (6 )

18 as the coefficient of x in the product ( + x) 6 ( + x) 6 In fact, the problem can be worked by the method of generating functions, as follows Think of picking an increasing sequence of numbers on the emerging beetles as selecting either 0 or beetle with the number, then selecting either 0 or beetle with the number, and so on Since there are two beetles with each of the numbers apple 6, the generating function for each of those numbers is + x There is just one beetle with each of the numbers 7, so the generating function for each of those numbers is + x The combined generating function for increasing sequences is thus ( + x) 6 ( + x) 6 The number of increasing sequences of length is the coefficient of x Finish the problem by finding that coefficient and taking the quotient with (Finding the coefficient is a bit of work) 9 Starting at the node in the center of the diagram, an orb spider moves along its web It is permissible for the spider to backtrack as often as it likes, in either direction, on segments it has previously travelled On each move, the spider moves along one of the segments (curved or straight) to some adjacent node that is different from the node that it currently occupies (a) How many different five-move paths start at the center node and end at the center node (b) How many different seven-move paths start at the center node and end at the center node NSWER: (a) 6 SOLUTION: (a) The various types of paths can be listed and counted Let O stand for a move to any outside point from the inside, I stand for a move to the inside (center) point, and X stand for a move from any outside point to any other outside point along any path There are three cases: OXXXI, OXIOI, OIOXI with the following numbers of different paths, respectively, 3 = 9, 3 3 = 36, 3 3 = 36 Total=6 NSWER: (b) 700 SOLUTION: (b) The various types of paths can be listed and counted, as before This time the possibilities are OIOIOXI, OIOXIOI, OIOXXXI, OXIOIOI, OXIOXXI, OXXIOXI, OXXXXOI, OXXXXXI Count as before lternatively, work as in the keypad question on the First Round Label the center node x and the peripheral nodes as a, b, c Let n, B n, C n, and X n represent the number of paths of length n that start at the center x and end at the nodes a, b, c, x respectively Now extend the paths one extra step, and use the diagram to deduce this highly symmetrical system of recurrence equations: n+ = B n + C n + X n B n+ = n + C n + X n C n+ = n + B n + X n X n+ = n + B n + C n ; hence X n+ = n+ + B n+ + C n+ dd the top three equations to deduce that the X values satisfy X n+ = n+ + B n+ + C n+ = X n+ + 3X n Next use this recurrence equation to eventually solve for X 7, as follows Note that by inspection of the network, X = 0, X = 3 Insert these into the recurrence formula to deduce X 3 = (3)+3(0) = Likewise X = ()+3(3) =7, X = (7)+3() = = 6, X 6 = 7 and finally X 7 = 700

19 0 (a) You want to arrange 8 biologists of 8 different heights in two rows for a photograph Each row must have biologists Height must increase from left to right in each row Each person in back must be taller than the person directly in front of him How many different arrangements are possible (b) You arrange biologists of different heights in two rows of 6, with the same conditions on height as in part (a) How many different arrangements are possible Remember to justify your answers (c) You arrange n biologists of n different heights in two rows of n, with the same conditions on height as in part (a) Give a formula in terms of n for the number of possible arrangements NSWER: (a) (b) 3 n n n n n (c) = = = n + = (n)! n + n n n n n + n + n (n + )!n! These are the Catalan numbers and there are many acceptable forms for them ny of the forms listed here, for instance, are acceptable answers SOLUTION: (a) Either list the arrangements or refer to the reasoning shown for either of parts (b) or (c) Here is a systematic list in a lexicographic order: FFFFBBBB; FFFBFBBB; FFFBBFBB; FFFBBBFB; FFBFFBBB; FFBFBFBB; FFBFBBFB; FFBBFFBB; FFBBFBFB; FBFFFBBB; FBFFBFBB; FBFFBBFB; FBFBFFBB; FBFBFBFB Keep in mind as you list that acceptable strings have the Fs and Bs and the Bs never outnumber the Fs as you move left to right SOLUTION:(b) We can use a better system than listing Line up the biologists by height Start filling in seats Each new person must go in the next empty seat in either the front row or the back row Look at the patterns and keep track of how many of the seating arrangements begin with j Fs and k Bs We will construct a table (See below) When j=k we have two full rows: this is what (a) and (b) ask about, j=k= and j=k=6 There will be no arrangements with k bigger than j (more B s than F s so far) so all entries above the diagonal are 0 F & 0B or F & 0B, etc are easy There is one arrangement with each of these, so the first column is all s Next column, the column with in the back row: F & B one way F & B two ways Notice that to get F & B then just before you added the most recent biologist, you had either F & B and added an F or you had F & 0B and added a B This means each entry is the sum of entry to left + entry above Fill in the table 0B B B 3B B B 6B F F F F F 8 0 6F The F & B entry is the from part (a) For part (b) we want the 6F & 6B entry, 3

20 SOLUTION: (c) First select n of the n biologists to go in the front row There are ( n n ) ways to make this selection Given a selection, the front row people and the back row people can be put in order so that height increases left to right Some of these seating arrangements will satisfy the condition that every person in the back row is taller than the person in front and some of the arrangements will not We will count how many violate this condition and subtract that from ( n n ) Number the biologists in order of height and then write down FFBBFB to record whether each person is seated in the front row or the back row ll ( n n ) of our arrangements have equal numbers of F and B, because we selected n for the front row The strings that satisfy the last requirement are exactly the strings that have the property that as you count from left to right, the total number of F s never exceeds the total number of B s We want to count the strings that violate this We will associate to each of the "bad" strings one unique string of n + F s and n B s in such a way that each "bad" string has such an unbalanced string and that each such n + /n string has exactly one "bad" string associated with it Then all we have to do is count how many of these unbalanced strings there are, the ones with n + F s and n B s That is, of course, ( n n ) Here is the association: given a "bad" string, move from left to right until you first have more F s than B s Leave that first excess F in place To the left of that F you will have equal numbers of F s and B s To the right of that F you will have one more B than F In every place to the right of the F, change every F to a B and every B to an F The new string will have n + F s and n B s To check that the assignment is reversible, consider a string with n + F s and n B s s you move from left to right, there is a first place the number of F s exceeds the number of B s To the left of that F, the string is balanced Switch all the F s and B s to the right of this F The resulting string will be the "bad" balanced string that went with the unbalanced string Therefore, the number of good strings is ( n n ) ) This expression is an acceptable answer It is also easily simplified to ( n n n+ (n n )

21 BONUS: You arrange biologists of different heights in three rows of, with the same conditions on height as in part 0(a) for all three rows How many different arrangements are possible NSWER 6 SOLUTION: Work as in part (b), but with a third index for the third row We will call the rows, B, and C The recursive rule for filling in the table is similar to the rule in part (b), but for a three-dimensional array, displayed below in a compressed format as follows: the upper left number in each box is for 0C; the next one is for C, and so on The last entry in the table represents (, B, 3C) =6 However, this will also equal the next recursive entry: (, B, C) =6 0B B B 3B B c c 3c ! c c 3c

22 Solution to 0 UNC Final by Ming Song nswer: 7 Solution: The number at the middle is 8 3 The smallest number is 7 nswer: 3 Solution: The shaded area is 3 3 nswer: x x 3x 7 Solution: p Let y x Then x y So y y 7 y 9y 3y7 p Therefore, p x x 3x 7 The speed ratio of to B is 00 : 90 The speed ratio of B to C is 00 : 80, 0 : 8, or 90 : 7 Then the speed ratio of to C is 00 : 7 When runs 00 feet, C runs 7 feet C has to run 8 feet more The answer is 8 nswer: 8 Solution: Let x to the number of vertices on the top face Then the top face has x edges Let V be the number of faces of the polyhedron, F be the number of faces, and E be the number of edges Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission

23 Solution to 0 UNC Final by Ming Song ccording to the Euler Theorem we have V F E In this problem, We have the equation: Solving for x we obtain x 8 The answer is 8 V 0 x, F 968, x936 x E 30 x 0 x nswer: 97 Solution: Note that 0 and If m n, we always have mn 0 There are 990 ways to assign values for m and n in this range If n 6, we must have m 3 by noticing If n 7, we must have m by noticing If n 8, we must have m by noticing If n 9, we must have m by noticing If n 0, we must have m 0 by noticing Therefore, the answer is Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission

24 Solution to 0 UNC Final by Ming Song 7 Solution to (a): Color a cube with two opposite faces being golden and the other faces to be blue Solution to (): In D-geometry we have P PB is P is a point outside a circle and P and PB are tangent to the circle at and B respectively P B We will prove a lemma, a similar conclusion, in 3D-geometry Lemma: Plane m and n intersect at line l sphere is tangent to both planes at and B respectively C l D B n m Let C and D are two points on line l Then Proof: CDBCD Draw the plane through C,, and B The plane cuts the sphere with a circle C and CB are tangent to the circle at and B respectively ccording to the conclusion in D-Geometry, we have C CB C l D B n m Similarly, we have D DB Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 3

25 Solution to 0 UNC Final by Ming Song With CD is common, we have CDBCD Particularly, triangle CD and BCD have the same area Now the solution to (b) follows Consider all golden-blue edges The blue area is at least as much as the golden area 8 7 nswer: 896 Solution : Put beetles in a line with numbers from small to large With the same number the red beetle is to the left of the yellow beetle There are 8 76 ways for 8 beetles to wander out 8 There are ways to choose four beetles There are number 6 6 ways to choose four beetles with a pair which have the same 6 There are ways to choose four beetles with two pairs each of which have the same number 8 So there are ways to choose four beetles all of which have different numbers Put this four beetles in with right order with the numbers increasing Therefore, the answer to the problem is Solution : The denominator is 8 76, obviously The numerator is the coefficient of x in x 6 x 6 That is, Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission

26 Solution to 0 UNC Final by Ming Song Therefore, the desired probability is Solution (Recursion): B C D Let a n be the number of ways of paths of length n such that the spider comes back to when the spider is at right now Let b n be the number of ways of paths of length n such that the spider comes back to when the spider is at B right now Define c n and d n likewise By the symmetry, we have b n c d We can establish the following recursion: So we have Plugging () into () we obtain With () we have the same recursion for a n: n n a b c d, n n n n b a n n c n dn a () n 3b n b a () n n bn b (3) a n n bn 3b n an 3a n With a 0 and a 3, we calculate with the recursion: a 330, 3 Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission

27 Solution to 0 UNC Final by Ming Song a 33 7, a 73 6, a 637 7, a The answer to (a) is 6, and the answer to (b) is 700 Solution : B C D Let x be the number of moves that the spider moves along a radius, and let y be the number of moves that the spider moves among B, C, and D Problem (a): We have x y Case : x and y 3 In the first step, the spider must go out along a radius There are 3 choices In the last step, the spider must go in along the radius where it is There is one choice In any of three other moves, there are choices The number of ways in this case is Case : x and y In the first step, the spider must go out along a radius There are 3 choices In the last step, the spider must go in along the radius where it is There is one choice In between there are two chances for the spider go in and then immediately go out with 3 choices The spider has one move among points B, C, and D There are choices in this move The number of ways in this case is The total number of ways is Problem (b): We have x y 7 Case : x and y In the first step, the spider must go out along a radius There are 3 choices Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 6

28 Solution to 0 UNC Final by Ming Song In the last step, the spider must go in along the radius where it is There is one choice In any of five other moves, there are choices The number of ways in this case is Case : x and y In the first step, the spider must go out along a radius There are 3 choices In the last step, the spider must go in along the radius where it is There is one choice In between there are four chances for the spider go in and then immediately go out with 3 choices The spider has three moves among points B, C, and D There are choices in each move 3 The number of ways in this case is Case 3: x 6 and y In the first step, the spider must go out along a radius There are 3 choices In the last step, the spider must go in along the radius where it is There is one choice In between there are three ways for the spider go in and then immediately go out with 3 choices twice The spider has one move among points B, C, and D There are choices in this move The number of ways in this case is The total number of ways is Solution 3 (Maxtrix): 3 We define the matrix: The entry a ij ( i, j ) is the number of ways of the spider to move from i to j Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 7

29 Solution to 0 UNC Final by Ming Song Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 8 Then the entry ij b (, j i ) in matrix n is the number of ways of the spider to reach j from i in m moves Let us calculate: , , , The answer to (a) is 6, and the answer to (b) is 700

30 Solution to 0 UNC Final by Ming Song 0 Let us consider to arrange to 6 Look at the one-to-one correspondence for the five arrangements: The number of arrangements for numbers to n satisfying the given conditions is equal to the number of shortest routes in the n by n gird from the left-bottom corner to the right-top corner without passing the diagonal as shown Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 9

31 Solution to 0 UNC Final by Ming Song So we can easily get the answer by marking numbers: The answer to Problem (a) is, and the answer to problem (b) is 3 If you know the Catalan numbers, the general formula is For n, n n n This is the answer to problem (a) For n 6, This is the answer to problem (b) Copyright 0 Ming Song , msongmath@yahoocom Don t use or copy without permission 0