Please see pages 2 and 3 for the list of problems and contest details

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1 Student Prizes One 1st place prize: $300 and full 4-year tuition scholarship* Five 2nd place prizes: $150 each and $2000 scholarship** Ten 3rd place prizes: $100 each and $2000 scholarship** Honorable mentions will receive a gift. All entrants will receive certificates honoring participation. *In order to receive the 4-year scholarship, the winner must be directly admitted to the IUPUI School of Science, major in any discipline in the School of Science, and attend full-time. **Scholarships in the amount of $2000 per year awarded to remaining cash prize winners who are directly admitted to the IUPUI School of Science and attend full-time. Renewable for four years given satisfactory academic performance Avon 2009 Park Tudor 2004 Carmel 1999 Roncalli 2014 MTI School of Knowledge 2013 Avon 2008 Carmel 2003 Hamilton SE 1998 Brebeuf Previous First Place Winners 2012 Carmel 2007 Hamilton SE 2002 Hamilton SE Previous Spirit Award Winners 2011 Brownsburg 2006 Carmel 2001 Ben Davis 2013 Fishers 2012 Avon 2010 Carmel 2005 Carmel 2000 Carmel All participants, parents, and teachers will be invited to an awards ceremony at IUPUI on the afternoon of Friday, April 24th, The program will feature refreshments, a special presentation by Dr. Robert Worth, Professor of Neurosurgery, Indiana University School of Medicine and Adjunct Professor, IUPUI Department of Mathematical Sciences. The contest is open to all Indiana High School Students (grades 9-12). Co Chairs: Jeffrey Watt and Roland Roeder Please see pages 2 and 3 for the list of problems and contest details

2 IUPUI Department of Mathematical Sciences, 402 North Blackford Street, LD 270 Indianapolis, IN (317) or Submissions must be received by Friday, March 6th, 2015 in order to be considered. Details are listed on the required cover sheet, which is posted on the website. You may choose to solve one problem, several, or all five of the problems. Give your reasoning, not just the answers, and cite your sources and references appropriately. Entries will be judged by faculty in the IUPUI Department of Mathematical Sciences based on elegance of solutions as well as correctness. 1. The Greek architect Iupuius is designing the main entrance for his masterpiece: The Speedway Neuromath Library. He wants to have 14 columns aligned in a row to hold 7 arches. Each arch will be supported by 2 columns, but the columns do not need to be contiguous. Instead, the arches may extend on top of each other as in the examples below. Iupuius wants to draw all possible arch configurations to decide which looks nicer. His only constraint is to avoid an arch that covers all others; thus, the example on the bottom would be ruled out. How many arch configurations are there? How many if he decides to use 20 columns to hold 10 arches? 2. It is impossible to find an equilateral triangle all of whose vertices have integer coordinates, but one can get really close. For instance, the points (0, 0), (3, 11), and (11, 3) form an isosceles triangle with sides 130, 130, and 128. a) Show that there are infinitely many positive integers n for which there are pairs of positive integers r and s such that the points (0, 0), (r, s), and (s, r) form a triangle with sides n, n, and n + 1. b) Show that there are no positive integer pairs r, s such that the points (0, 0), (r, s), and (s, r) form a triangle with sides n, n, and n Find all functions f that take integer input and give integer output, and which satisfy the formula: f(a + b + f(b)) = f(a) + 2b 4. You are given an arbitrary triangle ABC inscribed in a circle. Show how to construct, using only a compass and straight edge, a triangle DEF that is similar to ABC, with corresponding sides parallel, and such that D lies on the segment BC, while E, F lie on the circular arc from B to C that does not contain A. 5. Write an essay of 500 to 700 words (complete with references) on an application of mathematics to study the brain. Please see Page 3 for the TEAM problem!

3 New This Year: Team Problem! There are many configurations of six squares that can be folded to form a cube; let us call them patterns. One such pattern is shown in the diagram below. Patterns are assumed to be connected, so, for example, 6 copies of a 1x1 square is not a pattern. You are given a sheet of paper, and tasked with cutting patterns from it to make as many cubes as possible. You are not allowed to do any re-gluing s after cutting. How many can you make? Teams of two or more students from the same high school may compete. No student can be listed on more than one team (in this case, each team s submission will be disqualified). Students are allowed to collaborate, but teachers are not allowed to help. Please include with your submission the team-specific cover sheet that is available at

4 2015 IUPUI HIGH SCHOOL MATH CONTEST SOLUTIONS 1. PROBLEM 1 The Greek architect Iupuius is designing the main entrance for his masterpiece: The Speedway Neuromath Library. He wants to have 14 columns aligned in a row to hold 7 arches. Each arch will be supported by 2 columns, but the columns do not need to be contiguous. Instead, the arches may extend on top of each other as in the examples below. Iupuius wants to draw all possible arch configurations to decide which looks nicer. His only constraint is to avoid an arch that covers all others; thus, the example on the bottom would be ruled out. How many arch configurations are there? How many if he decides to use 20 columns to hold 10 arches? Let s use the symbol C n to denote the number of configurations with exactly n arches (i.e., with 2n columns), but WITHOUT the last restriction about one arch covering all others. The first few values of C n are easy to find by trial and error: C 1 = 1 : C 2 = 2 :, C 3 = 5 :,,,, These are the Catalan numbers ; a famous sequence that shows up in many applications. To find a recursive formula for C n notice that the left-most arch L in any n-arch configuration splits the picture into two shorter configurations: one (possibly empty) covered by L, and another (possibly empty) outside and to the right of L. For example, in, the left-most arch covers, and leaves on the right. 1

5 IUPUI HIGH SCHOOL MATH CONTEST The arch L can cover any number of smaller arches, from 0 to n 1, leaving to the right a complementary number of arches, from n 1 down to 0. It follows that the total number of different arch configurations will be C n = C 0 C n 1 +C 1 C n 2 + +C n 1 C 0 (here, C 0 denotes a configuration with 0 arches, and we count it as equal to 1). For instance, C 3 = C 0 C 2 + C 1 C 1 + C 2 C 0 = = 5 (compare the examples above). Now we can use our formula to find C 7 : C 4 = C 0 C 3 +C 1 C 2 +C 2 C 1 +C 3 C 0 = = 14. C 5 = C 0 C 4 +C 1 C 3 +C 2 C 2 +C 3 C 1 +C 4 C 0 = = 42. C 6 = C 0 C 5 +C 1 C 4 +C 2 C 3 +C 3 C 2 +C 4 C 1 +C 5 C 0 = = 132. C 7 = C 0 C 6 +C 1 C 5 +C 2 C 4 +C 3 C 3 +C 4 C 2 +C 5 C 1 +C 6 C 0 = = 429. But remember that Iupuius wants to avoid configurations where one arch covers all others. Then we have to remove C 6 invalid configurations to get a grand total of = 297. For the case of 10 arches we can compute C 10 C 9 = =

6 2015 IUPUI HIGH SCHOOL MATH CONTEST 3 2. PROBLEM 2 It is impossible to find an equilateral triangle all of whose vertices have integer coordinates, but one can get really close. For instance, the points (0,0), (3,11), and (11,3) form an isosceles triangle with sides 130, 130, and 128. (a) Show that there are infinitely many positive integers n for which there are pairs of positive integers r and s such that the points (0,0), (s,r), and (r,s) form a triangle with sides n, n, and n + 1. (b) Show that there are no positive integer pairs r,s such that the points (0,0), (s,r), and (r,s) form a triangle with sides n, n, and n 1. Let s look at the second part first. We want to prove that there are no integer solutions to 2(r s) 2 (r 2 + s 2 ) = 1. This is easy. Simplifying and taking both sides mod 4, we get r 2 +s 2 = 3. However, integer squares can only be 0 or 1 mod 4, so the only possible sums are 0, 1, and 2. The first part is harder but there are several approaches. First, one might discover by trial and error that the following pairs work: (1,0), (4,1), (15,4), (56,15). At this point, it s not too hard to discover that the sequence 0,1,4,15,56,... satisfies the recurrence relationship f (n) = 4 f (n 1) f (n 2). Using mathematical induction: Base case: 2(1 0) 2 = ( )+1 or, if you prefer, 2(4 1) 2 = ( ) + 1. Inductive step: Does assuming that Or, more simply, does assuming that 2( f (n) f (n 1)) 2 = ( f (n) 2 + f (n 1) 2 ) + 1, 2( f (n 1) f (n 2)) 2 = ( f (n 1) 2 + f (n 2) 2 ) + 1? f (n) 2 4 f (n) f (n 1) + f (n 1) 2 = 1 f (n 1) 2 4 f (n 1) f (n 2) + f (n 2) 2 = 1 Using the recursion, we restate f (n) as 4 f (n 1) f (n 2) to get: Does 16 f (n 1) 2 8 f (n 1) f (n 2)+ f (n 2) 2 4(4 f (n 1) 2 f (n 1) f (n 2))+ f (n 1) 2 = 1 assuming that f (n 1) 2 4 f (n 1) f (n 2) + f (n 2) 2 = 1?

7 IUPUI HIGH SCHOOL MATH CONTEST Simplifying the expression, we find that the terms are exactly what we want - f (n 1) 2 4 f (n 1) f (n 2) + f (n 2) 2, which was assumed to equal 1. This recursion can also be discovered/generated by a trigonometric argument: If (0, 0), (s,r), and (r,s) are the vertices of an equilateral triangle, then r/s = tan15 (or tan75); tan15 and tan75 are 2 3, respectively, and these are the roots of the quadratic equation x 2 4x + 1 = 0. So if we assume that we are looking for a recursion of some sort which produces consecutive ratios near 2+ 3, we might well start by looking for a base case and expecting that use of the recursive equation f (n) 4 f (n 1) + f (n 2) = 0.

8 2015 IUPUI HIGH SCHOOL MATH CONTEST 5 3. PROBLEM 3 Find all functions f that take integer input and give integer output, and which satisfy the formula f (a + b + f (b)) = f (a) + 2b. Claim 1: f (0) = 0. By taking a = b = 0, we get that f (0) = f ( f (0)). By taking b = 0, we see that f (a + f (0)) = f (a) for any a, which implies that f ( f (0)) = f (2 f (0)). The first equality then yields that f (2 f (0)) = f (0 + f (0) + f ( f (0))), which must be equal to f (0) + 2 f (0) = 3 f (0). That is, f (0) = 3 f (0) and the claim follows. Claim 2: if f (a) = a, then f (na) = na for all integer n. According to Claim 1, it holds that 2a = f (0 + a + f (a)) = f (2a). This establishes the base cases for n = 1, 2. Assume now that the claim is true for all i = 1, 2,..., k. Then f ((k + 1)a) = f ((k 1)a + a + f (a)) = f ((k 1)a) + 2a = (k + 1)a. Thus, the claim holds for all positive integers n by the Principle of Mathematical Induction. Since a = f (a) = f ( a + a + f (a)) = f ( a) + 2a, it holds that f ( a) = a. Therefore, the previous reasoning shows that f ( na) = na for all positive integers n, which finishes the proof of the claim. Claim 3: f (n) = n f (1) for all non-negative integers n. It holds that f (1 + f (1)) = f ( f (1)) = 2. Then f (n) + 2 = f (n f (1)) = f (n f (1) + f (1 + f (1))) = f (n 2) + 2(1 + f (1)), which gives is f (n) = f (n 2) + 2 f (1). Since f (1) = 1 f (1) and f (0) = 0 f (1) by Claim 1, Claim 3 follows from the Principle of Mathematical Induction. Claim 4: either f (n) = n for all integers n or f (n) = 2n for all integers n. Let A = 2 + f (1). Then f ( A ) = A. Indeed, f (A) = f ( f (1)) = f (1) + 2 = A and f ( A) = A where the second equality follows from Claim 2. By Claim 3, f ( A ) = A f (1). Hence, if A 0, then f (1) = 1 and respectively f (n) = n for all integers n by Claim 3. Otherwise, A = 0, which means that f (1) = 2. Consequently, f (n) = 2n for all non-negative integers n by Claim 3. Finally, notice that f ( n) = f (n + f (n)) = f (0 + n + f (n)) = 2n for all positive integers n, which finishes the proof of Claim 4.

9 IUPUI HIGH SCHOOL MATH CONTEST 4. PROBLEM 4 You are given an arbitrary triangle ABC inscribed in a circle. Show how to construct, using only a compass and straight edge, a triangle DEF that is similar to ABC, with corresponding sides parallel, and such that D lies on the segment BC, while E and F lie on the circular arc from B to C that does not contain A. This problem would be a lot easier if it specified that ABC was isosceles, because then we d have a specific point on BC to work from and known angles to use to determine the other two points. Strangely thinking about the problem we don t have makes it easier to solve the problem we do have, because we can use exactly this technique to construct a rectangle with BC as a base, A on the opposite side, and then construct a similar RECTANGLE meeting the constraints we want. The reason this is helpful is that a similar triangle to ABC can be inscribed in a similar rectangle to the one in which ABC will shortly be inscribed. For the purposes of this solution, I will take basic constructions as given. First, construct a line parallel to BC through A. Next, construct perpendiculars to BC at B and C so that we have the rectangle in question. Call the rectangle C B BC (with A between C and B ). Then construct the midpoints of BC and B C. Call them G and G, respectively, see Figure 1. B 0 B G 0 G A C 0 C FIGURE 1. Now we create a similar rectangle to C B BC such that two points lie on the circular arc between B and C and the other two are between B and C. Construct line segments G B and G C, and then construct lines through G parallel to these segments. The intersection with these two lines emanating from point G (E and F) are the base of our similar rectangle (and, hence, triangle), see Figure 2.

10 2015 IUPUI HIGH SCHOOL MATH CONTEST 7 B 0 B G 0 E 0 G F 0 E F A C 0 C FIGURE 2. Getting the point corresponding to the vertex A can be done in many ways but by far the simplest is to find the intersection of BE and CF. Call this point A. Then the point at which AA intersects BC is D, see Figure 3. B 0 B E 0 E D A0 F 0 F A C 0 C FIGURE 3.

11 2015 Team Problem Solution Problem: There are many configurations of six squares that can be folded to form a cube; let us call them patterns. You are given a sheet of paper, and tasked with cutting patterns from it to make as many cubes as possible. Solution: I was easily able to work out how to get 14 in my head, using just two cube- forms which could each tile rows very densely: It was only late in the game when I discovered I could fit in a 15th cube- form as follows, by fitting in a capital T in the largest gap (one of two choices for that gap, actually). There are, as it happens, 11 different hexomino cube- forms (this is easy to Google): And of the two team solutions which presented 15 cube- forms, each used a much less limited palette, the first using five different cube- forms in its rendering and the second using seven. (Their solutions are presented below.) Well done!

12 As an aside, because each of the 35 hexominoes (including the 11 cube- forms) can be used to tile the plane, as n gets large the number of cube- forms of ANY SINGLE variety that can be cut from an nxn sheet is asymptotically equal to n2/6.

13 2015 IUPUI HIGH SCHOOL MATH CONTEST First Place Winners Gabriel Paree-Huff, 11 th Grade, Fishers High School. Teacher: John Drozd Michael Wang, 10 th Grade, Carmel High School. Teacher: Janice Mitchener Second Place Winners Gabriel Caldwell, 11 th Grade, Avon High School. Teacher: Anthony Record Anthony Ji, 10 th Grade, Carmel High School. Teacher: Janice Mitchener Vishaal Mali, 11 th Grade, Avon High School. Teacher: Anthony Record Dominique WuDunn, 10 th Grade, International School of Indiana. Teacher: Olivier Bernadac Third Place Winners Samuel Canner, 11 th Grade, Center Grove High School. Teacher: Karen Fruits Emma Caress, 10 th Grade, Brebeuf Jesuit Preparatory School. Teacher: Iris Manton Nichole Cochran, 9 th Grade, Logos Christian Academy Homeschool. Teacher: Kathy Cochran Zachary Cochran, 11 th Grade, Logos Christian Academy Homeschool. Teacher: Kathy Cochran Katharina Fransen, 11 th Grade, Fishers High School. Teacher: John Drozd Kireina Gray, 12 th Grade, Warren Central High School. Teacher: Alissa Horn Bharat Gummalla, 10 th Grade, Carmel High School. Teacher: Janice Mitchener Selena Qian, 10 th Grade, Carmel High School. Teacher: Linda Jones Halie Szilagyi, 11 th Grade, Fishers High School. Teacher: John Drozd Emily Tong, 11 th Grade, Fishers High School. Teacher: John Drozd Essay Winners Kireina Gray, 12 th Grade, Warren Central High School. Teacher: Alissa Horn Matthew Rennekamp, 9 th Grade, Seymour High School. Teacher: Laura Handloser Award for Best Solution to Problem 1 Phoebus Yang, 10 th Grade, Hamilton Southeastern High School. Teacher: Cynthia Cooper

14 Honorable Mention Winners Haley Drabek, 12 th Grade, Avon High School. Teacher: Anthony Record Nathan Gossmann, 11 th Grade, Avon High School. Teacher: Anthony Record Audrie Hillis, 12 th Grade, Avon High School. Teacher: Courtney Guth Daniel Kobold, 12 th Grade, Avon High School. Teacher: Anthony Record Olivia Korte, 10 th Grade, Fishers High School. Teacher: Erin Ingwersen Celena Langlois, 11 th Grade, Avon High School. Teacher: Anthony Record Matthew Rennekamp, 9 th Grade, Seymour High School. Teacher: Laura Handloser Joshua Roth, 10 th Grade, Fishers High School. Teacher: John Drozd Angelina Shi, 11 th Grade, Avon High School. Teacher: Anthony Record Elizabeth Shi, 12 th Grade, Avon High School. Teacher: Anthony Record Phoebus Yang, 10 th Grade, Hamilton Southeastern High School. Teacher: Cynthia Cooper Additional Competitors James Cerone, 11 th Grade, Fishers High School. Teacher: John Drozd Aysha Chaudhary, 11 th Grade, Fishers High School. Teacher: Erin Ingwersen Christopher Costelle, 12 th Grade, Avon High School. Teacher: Anthony Record Nolan Crist, 11 th Grade, Avon High School. Teacher: Anthony Record Suzanne Gatons, 12 th Grade, Avon High School. Teacher: Anthony Record Camille Goodwyn, 11th Grade, Avon High School. Teacher: Anthony Record Debra Hughes, 11 th Grade, Avon High School. Teacher: Anthony Record Issei Kobayashi, 12th Grade, Avon High School. Teacher: Anthony Record Michael Kuzma, 11 th Grade, Avon High School. Teacher: Anthony Record Andrew Langford, 9 th Grade, Avon High School. Teacher: Jeffrey Osterman Olivia Lazaro, 10 th Grade, Avon High School. Teacher: Kyle Meunier Evan Long, 12 th Grade, Fishers High School. Teacher: John Drozd Katherine Long, 10 th Grade, Carmel High School. Teacher: Laura Diamente Jacob Miller, 12 th Grade, West Washington High School. Abhinav Ramkumar, 10 th Grade, Carmel High School. Teacher: Laura Diamente Jeremy Rasor, 11 th Grade, Avon High School. Teacher: Anthony Record Abdul Saltagi, 12 th Grade, Fishers High School. Teacher: John Drozd

15 Additional Competitors (con t) Christa Serratos, 10 th Grade, Hammond Academy of Science & Technology. Teacher: Adam Erler Tyler Shanahan, 11 th Grade, Avon High School. Teacher: Anthony Record Haowei Shi, 9 th Grade, Avon High School. Teacher: Jeffrey Osterman Alexandra Short, 11 th Grade, Avon High School. Teacher: Anthony Record Adam Shumaker, 10 th Grade, Hamilton Southeastern High School. Teacher: Cynthia Cooper Damanveer Singh, 11 th Grade, Avon High School. Teacher: Anthony Record Morgan Soultz, 12 th Grade, Avon High School. Teacher: Courtney Guth Grant Sprout, 11 th Grade, Avon High School. Teacher: Anthony Record Tristan Strobel, 11 th Grade, Fishers High School. Teacher: Erin Ingwersen Alex Todd, 11 th Grade, Fishers High School. Teacher: Erin Ingwersen Sreya Vemuri, 11 th Grade, Carmel High School. Teacher: Janice Mitchener Maria Witcher, 10 th Grade, Hamilton Southeastern High School. Teacher: Cynthia Cooper First Place Team Awards Fishers Team 1, Gabriel Paree-Huff and Daniel Vance, Fishers High School C-Squared Z&N Cochran, Zachary Cochran and Nichole Cochran, Logos Christian Academy Homeschool School Award Carmel High School Sprit Award Avon High School