2012 IUPUI High School Mathematics Contest

Transcription

1 2012 IUPUI High School Mathematics Contest Mathematics & Information Technology Presented by The IUPUI Department of Mathematical Sciences STUDENT PRIZES One 1 st place prize $500 Five 2 nd place prizes $200 each Ten 3 rd place prizes $100 each A full four-year academic tuition scholarship will be awarded to the first place prizewinner. The winner must be directly admitted to the Purdue School of Science at IUPUI to major in any discipline and attend full-time. Scholarships in the amount of $2,000 per year will be awarded to the remaining cash prizewinners who are directly admitted to the Purdue School of Science at IUPUI and attend full-time. These scholarships are renewable for four years, given satisfactory academic performance. Honorable mentions will receive a gift. All entrants will receive certificates honoring their participation. QUESTIONS 1) Your goal is to color the gray roads below, and to give a set of instructions to reach the house in the center. There should be one red and one blue road leading out of each house. One set of instructions must work for everybody and it should be at most six words long, with each word either red or blue. A person starting at any of the five houses must finish at the central house after completing the full set of instructions, and can visit a house more than once along the way. You are not allowed to change the color of the existing blue line or red line. MATHEMATICS DEPARTMENT AWARDS Schools awarded the 1st place trophy in previous years: 2011 Brownsburg High School 2010 Carmel High School 2009 Park Tudor High School 2008 Carmel High School 2007 Hamilton Southeastern H.S Carmel High School 2005 Carmel High School 2004 Carmel High School 2003 Hamilton Southeastern H.S Hamilton Southeastern H.S Ben Davis High School 2000 Carmel High School 1999 Roncalli High School 1998 Brebeuf Jesuit Preparatory CEREMONY Prizewinners will be invited to an awards ceremony at IUPUI on Friday, May 11, 2012 from 4:00-8:00 p.m. Parents and teachers will also be invited. The program will feature refreshments, a tour of IUPUI s Advanced Visualization Lab, a special presentation by Albert William, Research Associate, School of Informatics and Advanced Visualization Lab, and will end with an awards presentation. ELIGIBILITY This contest is open to students attending high school (grades 9-12) in the 15-county area of central Indiana: Bartholomew, Boone, Brown, Clinton, Hamilton, Hancock, Hendricks, Howard, Johnson, Madison, Marion, Morgan, Putnam, Shelby and Tipton. 2) Start with the number N 1 = You can produce a new number N 2 by first choosing any k = 1, 2, 3, 4, 5, 6, 7, 8, 9, then taking the product k N 1, and finally removing any 1's appearing in the resulting decimal expansion. (If you end up with leading zeros in front of your number, remove them as well for example would become 370.) Now repeat this process (choosing whatever k you like at each step) to form a sequence of numbers N 1, N 2, N 3 What is the smallest number that you can reach? How? 3) Show that there are infinitely many triples of positive integers (a,b,n) with 0 a, b < n so that (aa n ) 2 = bbbb n. Here, the repeated a's and b's are digits of a number, and the subscript n indicates that the number is written in base n. 4) Ariadne wants to design a gigantic puzzle for the 2012 Pan-Galactic Expo. She has an empty cube 2012 in. on each side, and plans to add k partitions to form a maze. Each partition is a 1 1 in. wall or floor/ceiling separating two contiguous in. cells within the big cube. The final maze is designed so that every cell is reachable from any other cell in an unique way. Find k and justify your answer. 5) Write an essay of 500 to 700 words (complete with bibliography) on how Mathematics is related to Information Technology. Co Chairs: Jeffrey Watt, Roland Roeder Questions written by William Cross and Rodrigo Perez of the IUPUI Department of Mathematical Sciences Contact Information: IUPUI High School Mathematics Contest Department of Mathematical Sciences 402 North Blackford Street, LD 270 Indianapolis, IN (317) 274-MATH or contest@math.iupui.edu ENTRIES: Mail your entry by Monday, April 9, 2012 to the address listed under Contact Information. You may obtain a copy of the questions, instructions for entering, and the cover page from your math teacher or the contest website: Solve the questions, giving your reasoning, not just the answers. Entries will be judged by professors in the IUPUI Department of Mathematical Sciences. Judging will be based on elegance of solution as well as correctness.

2 2012 IUPUI High School Math Contest Solutions Problem 1. Your goal is to color the grey roads below, and to give a set of instructions to reach the house in the center. There should be one red and one blue road leading out of each house. There can be at most six words in the instructions, and each word must be either red or blue. A person starting at any of the five houses must finish at the central house after completing the full set of instructions, and can visit a house more than once along the way. Solutions to Problem 1. There are many solutions, which can be found either by trial and error, or by systematically keeping track of your choices by means of a tree diagram. One solution is shown below. You can check that the instructions blue, red, red, red, blue, blue will lead someone from any of the houses to the house in the center. 1

3 Problem 2. Start with the number N 1 = You can produce a new number N 2 by first choosing any k = 1, 2, 3, 4, 5, 6, 7, 8, 9, then taking the product k N 1, and finally removing any 1 s appearing in the resulting decimal expansion. (If you end up with leading zeros in front of your number, remove them as well for example would become 370.) Now repeat this process (choosing whatever k you like at each step) to form a sequence of numbers N 1, N 2, N 3,.... What is the smallest number that you can reach? How? Solution to Problem 2. We claim that for any N 1 0 we can reach 0 by choosing some suitable sequence of k s. It is sufficient to show that for any integer N > 0 there is some k {1, 2, 3,..., 9} so that taking the product k N, and finally removing any 1 s results in a number that is strictly smaller than N. (Repeating the process N times, we would reach 0.) If there are any 1 s in the decimal expansion of N, then we take k = 1. After removing any 1 s from k N = N, the resulting number will have fewer digits than N, hence it will be strictly smaller than N. If there are no 1 s in the decimal expansion of N, then there exists some m 0 so that 2 10 m N < 10 m+1. We let k be the smallest element of {2, 3,..., 9} so that 10 m+1 k N. Then, since k is the smallest element with this property, we have k N < 10 m+1 + N. Since N < 10 m+1, k N < 2 10 m+1, implying that the first digit of k N is a 1. Therefore, after removing all 1 s from the decimal expansion of k N we get a number N satisfying N k N 10 m+1 < N. Problem 3. Show that there are infinitely many triples of positive integers (a, b, n) with 0 a, b < n so that (aa n ) 2 = bbbb n. Here, the repeated a s and b s are digits of a number, and the subscript n indicates that the number is written in base n. Solution to Problem 3. Note: for any n 1, a = b = 0 is a solution. However, we consider these trivial solutions, so below we consider a, b > 0. 2

4 We ll first find necessary conditions that the a, b, and n must satisfy, in order to narrow down the possibilities. Notice that since a and b are digits of a number written in base n, we must have a n 1 and b n 1. We have We also have aa n = a n + a = a(n + 1), so (aa n ) 2 = a 2 (n + 1) 2. bbbb n = b n 3 + b n 2 + b n + 1 = b(n + 1)(n 2 + 1). So, the condition that (aa n ) 2 = bbbb n is equivalent to a 2 (n + 1) 2 = b(n + 1)(n 2 + 1), which can be written more simply as a 2 = b(n 2 + 1)/(n + 1). (1) Now n = (n + 1)(n 1) + 2, so the greatest common factor between n and n + 1 is either 1 or 2. If they are relatively prime, then b must be a multiple of n + 1, since the right hand side of Equation 1 is an integer. This is impossible, since b n 1. So, the greatest common factor between n 2 +1 and n+1 must be two and b must be a multiple of (n + 1)/2. Since 0 b < n, we conclude that b = (n + 1)/2. Therefore, a 2 = (n 2 + 1)/2, which implies 2a 2 n 2 = 1. (2) Our derivation shows that Equation 2 is a necessary conditions for a, b, and n to satisfy (aa n ) 2 = bbbb n. However, you can directly check that if a and n are any solution to Equation 2 then a, b = (n + 1)/2, and n satisfy (aa n ) 2 = bbbb n. So, we must show that there are infinitely many distinct solutions to Equation 2. Some experimentation yields the following sequence of solutions for (a, n): (1, 1), (5, 7), (29, 41), (169, 239), (985, 1393). Using this sequence of solutions, you may guess the following linear recursion satisfied by the solutions (a i+1, n i+1 ) = (3a i + 2n i, 4a i + 3n i ), (3) 3

5 which should be used with the initial conditions (a 1, n 1 ) = (1, 1). We will prove by induction that (a i, n i ) is a solutions to Equation (2) for each i 1. For i = 1, we clearly have = 1. Suppose for some i that (a i, n i ) is a solution to Equation 2. Then, 2a 2 i+1 n2 i+1 = 2(3a i + 2n i ) 2 (4a i + 3n i ) 2 = 18a 2 i + 24a in i + 8n 2 i 16a2 i 24a ib i 9n 2 i = 2a2 i n2 i = 1. Therefore, by the Principle of Mathematical Induction, we conclude that for every i 1 that (a i, n i ) is a solution to Equation 2. Finally, note that Equation 3 implies for each i that a i, n i 1 and therefore that a i+1 > 3a i, forcing the sequence of a i to increase, thus giving us infinitely many solutions. (Similarly, the n i are also forced to increase.) Remark: there are many other recursive relationships possible (for example, several students discovered a k = 6a k 1 a k 2 and n k = 6n k 1 n k 2 ) and the inductive proof would be similar for each. The recursion given in Equation 3 was found by rewriting Equation 2 as (n + a 2)(n a 2) = 1. Our smallest solution, (a, n) = (1, 1), corresponds to So, for any i 1, we have that Letting (1 + 2)(1 2) = 1. (1 + 2) 2i 1 (1 2) 2i 1 = ( 1) 2i 1 = 1. n i + a i (2) = (1 + 2) 2i 1, (4) we find our sequence of solutions (a i, n i ) given above. The recursion can then be derived from Equation 4, since ( n i+1 + a i+1 2 = n i + a i 2 )(1 + ) 2 )( 2 = (n i + a i ) 2 = (4a i + 3n i ) + (3a i + 2n i ) 2. Problem 4. Ariadne wants to design a gigantic puzzle for the 2012 Pan-Galactic Expo. She has an empty cube 2012 in. on each side, and plans to add k partitions 4

6 to form a maze. Each partition is a 1 1 in. wall or floor/ceiling separating two contiguous in. cells within the big cube. The final maze is designed so that every cell is reachable from any other cell in an unique way. Find k and justify your answer. Solution to Problem 4. We interpret the statement every cell is reachable from any other cell in an unique way to mean that for any two cells there is at most one path between them that goes through each cell at most once. Thus, our maze is allowed to have dead ends. Let us suppose that all partitions on the outer faces of the big cube are already given to us. We will only count how many interior partitions need to be added to make the maze. Instead of counting how many interior partitions to add, let s suppose that all of the interior partitions are already present and count how many of them we need to remove. The total number of interior partitions is To see that thus number is correct, notice that each interior partition is contained in some plane that intersects the cube in a total of interior partitions. We ll call this a sheet of partitions. There are 3 possible orientations of each sheet and for each orientation there are 2011 choices of sets of cells to separate. Let s describe the desired maze by a graph G having vertices, with each vertex corresponding to a cell of our maze. Two vertices will be connected by an edge if you can walk from between their corresponding cells in one step. (I.e. there is an edge between two vertices if the corresponding cells share a face that is not blocked by a partition.) The condition that every cell is reachable from any other cell in an unique way means that G is connected and has no cycles (i.e., there are no closed paths that meet every vertex other than the initial/terminal vertex exactly once). Such a graph G is called a tree. A proof by induction can be used to show that a tree with n vertices has exactly n 1 edges. (Try it yourself.) Therefore, our tree G has edges. Since we are starting with all of the interior partitions, each edge corresponds to a partition that we had to remove; i.e., we must remove exactly partitions. 5

7 However, we were supposed to start with an empty cube and then to add partitions. The total number of interior partitions that one needs to add is therefore equal to: ( ) = Remark: some students expressed the answer as 1 2 ( ) ( ) = It is obtained by counting the total number of interior partitions in a different way. 6

8 2012 IUPUI HIGH SCHOOL MATH CONTEST First Prize Winner Rebecca Chen, 12 th Grade, Park Tudor School. Teacher: Sarah Webster Second Prize Winners Yushi Homma, 10 th Grade, Carmel High School. Teacher: Janice Mitchener Lyndon Ji, 12 th Grade, Carmel High School. Teacher: Catherine Mytelka, 11 th Grade, Park Tudor School. Teacher: Sarah Webster Melinda Song, 11 th Grade, Carmel High School. Teacher: Janice Mitchener Marc WuDunn, 9 th Grade, International School of Indiana. Teacher: Catherine Zvinevich Third Prize Winners Gabriela Borges, 12 th Grade, Avon High School. Teacher: Courtney Guth Brian Ertl, 12 th Grade, Hamilton Southeastern High School. Teacher: Letitia McCallister Simon Jones, 10 th Grade, Hamilton Southeastern High School. Teacher: Lisa Boyl Usama Kamran, 11 th Grade, Fishers High School. Teacher: John Drozd Akshay Kumar, 11 th Grade, Carmel High School. Teacher: Peter Beck Colin Mothersead, 11 th Grade, Avon High School. Teacher: Anthony Record Samual Patterson, 11 th Grade, Carmel High School. Teacher: Janice Mitchener Natasha Rollings, 12 th Grade, Avon High School. Teacher: Anthony Record Clayton Thomas, 10 th Grade, Fishers High School. Teacher: Kathleen Robeson Chase Thompson, 11 th Grade, Hamilton Southeastern High School. Teacher: Letitia McCallister Phillip Witcher, 11 th Grade, Hamilton Southeastern High School. Teacher: Letitia McCallister Weston Wright, 12 th Grade, Avon High School. Teacher: Anthony Record Honorable Mention Winners Olivia Cane, 11 th Grade, Fishers High School. Teacher: John Drozd Steven Cochran, 12 th Grade, Avon High School. Teacher: Courtney Guth Ian Conklin, 12 th Grade, Avon High School. Teacher: Anthony Record Chase Costin, 10 th Grade, North Central High School. Teacher: Jan Wendt Bridget Curtin, 11 th Grade, Avon High School. Teacher: Anthony Record Zita Erbowor-Becksen, 12 th Grade, Avon High School. Teacher: Anthony Record

9 Honorable Mention Winners (cont d) Kelsey Hay, 10 th Grade, Hamilton Southeastern High School. Teacher: Lisa Boyl Courtney McDermott, 12 th Grade, Avon High School. Teacher: Courtney Guth Abdulaziz Mohamed, 12 th Grade, MTI School of Knowledge. Teacher: Heba El Shakmak Elizabeth Neibert, 12 th Grade, Avon High School. Teacher: Courtney Guth Collin Nguyen, 11 th Grade, Avon High School. Teacher: Anthony Record Dhara Patel, 11 th Grade, Avon High School. Teacher: Anthony Record Bradley Patz, 11 th Grade, Avon High School. Teacher: Anthony Record Sharmila Paul, 10 th Grade, Carmel High School. Teacher: Laura Diamente Haley Priest, 12 th Grade, Avon High School. Teacher: Courtney Guth Mason Swofford, 8 th Grade, Hamilton Southeastern High School. Teacher: Lisa Boyl Daniel Tucek, 12 th Grade, Fishers High School. Teacher: John Drozd Madeline Weber, 12 th Grade, Avon High School. Teacher: Anthony Record Rui Xiao, 12 th Grade, Avon High School. Teacher: Anthony Record Amanda Zolcak, 12 th Grade, Avon High School. Teacher: Anthony Record Additional Competitors Qasim Alam, 10 th Grade, MTI School of Knowledge. Teacher: Heba El Shakmak Brandon Alcom, 12 th Grade, Avon High School. Teacher: Courtney Guth Alexander Deal, 12 th Grade, Avon High School. Teacher: Courtney Guth Jasmine Dhami, 12 th Grade, Avon High School. Teacher: Courtney Guth Nicole Enyart, 12 th Grade, Union Bible Academy. Teacher: Rick Herring Logan Hausman, 12 th Grade, Avon High School. Teacher: Courtney Guth Lindsey Hill, 12 th Grade, Avon High School. Teacher: Courtney Guth Jiaming Lin, 12 th Grade, Avon High School. Teacher: John Drozd Christopher Long, 12 th Grade, Avon High School. Teacher: Courtney Guth Mohamed Mohamed, 12 th Grade, MTI School of Knowledge. Teacher: Heba El Shakmak Rebecca Pyle, 10 th Grade, Union Bible Academy. Teacher: Rick Herring Daniel Seach, 10 th Grade, Hamilton Southeastern High School. Teacher: Lisa Boyl Nicholas Sparzo, 11 th Grade, Avon High School. Teacher: Anthony Record Hunter Stephenson, 12 th Grade, Avon High School. Teacher: Courtney Guth