Caculus Môn toán

The Calculus Manifesto Delong Meng Sixth Hour March 2008 Contents There Are Always Options i Intermediate Value Theorem 1.1 Background 1.2 Examples 1.3 Homework Exercises 1 Formal Definition of the 2.1 Background 2.2 Examples 2.3 Homework Exercises 5 Derivative Ugly Girls Are Very Friendly 10 3.1 Background 10 3.2 Examples 12 3.3 Homework Exercises 13 Good Looking Girls Are Dumb 4.1 Background 4.1.1 September 18, 1967 4.1.2 I Love Mr.Barnes 4.1.3 Emmi, Hannah, and Leyla 4.1.4 Jason with Kristin and Erin 4.1.5 Convexity and Jensen 4.1.6 Car Burning Inequality 4.2 Examples 4.3 Homework Exercises 16 16 16 16 17 18 19 19 20 22 How to Cheat on Optimization Test 5.1 Joyful Joyful, We Adore Spams 5.1.1 The Ideal Girls Law 5.1.2 The System 5.1.3 You Can’t Beat the System 5.1.4 I Have Cremated the System 25 25 25 25 25 26 CONTENTS 5.2 5.3 5.4 The 6.1 6.2 6.3 How to Beat the System 5.2.1 The Enemy of My Enemy 5.2.2 The Mean Method Examples Homework Exercises 26 26 26 27 28 Dream Team 30 Background 30 Examples 31 Homework Exercises 32 Two Journals 34 7.1 My Best Area 34 7.2 Worst Grade 34 Extra Tricky Math Magics 35 8.1 Points on a Circle 35 8.2 Number Guessing 36 A Moment In MATh 37 CONTENTS There Are Always Options Before working on this book, Delong would facebook all night This project has been more addicting than writing lengthy facebook notes On average, Delong spent at least an hour on a page, and of course there are only forty pages Therefore, Delong hopes that the reader would enjoy reading and appreciate the problems in this book A calculus textbook can be technical and boring, but there are always options This book shows that calculus exists in every aspect of our life especially FACEBOOK! Even though many people have not taken calculus yet, but this book gave them the enthusiasm to take Ms.O’Dell’s class next year (To Delong) I can’t wait to write about you in my calculus project! And I’m going to be talk about how mean you are! —Lisa Chapter is submitted to Mathematical Reflection, and the editor wrote: hey, your article is interesting in terms of the math content i’ll see what ivan thinks about it (if it gets published (which is likely) i am pretty sure that all the girls in mr barnes’ class stuff is going to get edited out ) somebody who helps edit for MR (=mathematical reflections) is named ivan borsenco and yes i looked at that chapter or whatever you submitted because i help edit articles its pretty funny but i think that all the inside jokes or whatever are out i like the idea of writing a book it is a good way to ’master’ material i try and write problems :D haha i love how like among all of the math there are like random pictures and aim conversations :D —Alex Anderson Calculus is by the people, for the people, and of the people Therefore, this book presents real life examples and applications and calculus Facebook is a great and convenient way for people to post their comments Let’s take the following as an example of the people’s voice: i ii THERE ARE ALWAYS OPTIONS Ok, suggestions for your new baby’s name: For Asian ”traditional purposes”, each name will sound like a common house appliance or some other loud noise ——————————Ding Dong - doorbell Ching Ching - bicycle Bang - person with a gun Ding Dang Ding - coin dropped in a bowl Gong - um, a gong Ding Ding Ding - small bell Ding Clang Dong Clash Ding - shattered glass Ding Dong Ding Dong Ding Dong Ding Dong Ding - very impatient person at the doorbell Delong Delong Delong - a cowbell (how coincidental) Shwiiiing! - Samurai sword Dooooooo-lyyy - Delong calling for his wife Clash! Clang! Clash! - sword fight Smack!!! - Antoine and his wife Cling Dling Clang - broken vase ——————————– I hope they’re helpful These are some of the notes that Delong wrote on Facebook: • Preface to My Book • A Moment in MATh • publishing my calculus project • Pwned by Neal again • A problem about Dooly in my chapter • How to Cheat on Optimization Quiz • I love Mr.Barnes The first two chapters are more conventional, and the rest are more jocular Since there are always options, let’s move on to Chapter 1! Chapter Intermediate Value Theorem 1.1 Background Limits and continuity is the first step for first-year calculus students After grasping these concepts, the following thereom would be interesting and useful tools in solving mathematical problems Theorem 1: Intermediate Value Theorem If f is continuous on the closed interval [a, b] and k is any number between f (a) and f(b), then there is at least one number c in [a, b] such that f (c) = k Figure 1.1: Intermediate Value Theorem A first glance at this formal statement may seem confusing, however, all the theorem said is that a continuous graph on an interval would hit all the values CHAPTER INTERMEDIATE VALUE THEOREM in its range (In a fancy way speaking, the function is surjective or onto in that interval.) A particular case of this theorem is very useful: Theorem 2: Existence of Zeros of A Polynomial If f is a polynomial with f (a) > and f (b) < 0, then there exist a root (or zero) in the interval [a, b] Looking at the theorems alone may be boring, but the application of the Intermediate Value Theorem has intriguing effects in problem solving With that, on to the problems! 1.2 Examples Example 1: Prove that the function f (x) = sin x has a value c ∈ [ π6 , π2 ] such that f (c) = 12 Solution: Since f (x) = sin x is continuous on the interval [ π6 , π2 ], and π π < = f( ) f( ) = < 12 by the Intermediate Value Thereom, there exist a c ∈ [ π6 , π2 ] such that f (c) = 12 Example 2: Show that there exit a c ∈ [1, 4] such that f (c) = 6, where x2 + x f (x) = x−1 First Solution: Similar to our previous attempts, we try to show that f (1) < < f (4) However, this time only the second half is true since f (6) = 36+6 6−1 = 42 > Now the question is: how can we get around the problem that f (1) is undefined? The idea is to try a slightly stronger result than the problem proposes For instance, notice that f( ) = 25 + 52 = where a, k, l are real numbers and < l < k < Solution: This inequality seems intimidating, but if we fix l and consider the LHS as a cubic polynomial in k, namely f (k) Let’s first examine the coefficient of k : −2l−2l3 +4l2 −a2 l2 +2l(1−l) (1 − l)2 + a2 l = −2l[(1−l)2 + a2 l − (1 − l)4 + (1 − l)2 a2 l] = −2l[ (1 − l)4 + (1 − l)2 a2 l + a 4l − (1 − l)4 + (1 − l)2 a2 l] < Therefore, f (k) is a cubic of k with a negative coefficient of k , so limk→∞ f (k) = −∞ Furthermore, f (0) = f (l) = 0, so and l are two roots We wish to prove that the third root lies in [1, ∞] by showing that f (1) > Notice that f (1) = (l − 1)4 + (l − 1)2 [(l − 1)2 + a2 l] − 2(1 − l)3 (1 − l)2 + a2 l = [(l − 1)2 − (1 − l) (1 − l)2 + a2 l]2 > The Intermediate Value Theorem tells us that there is a zero of the polynomial between and infinity, namely, z ∈ (1, ∞)! Now we can make a table of the signs of f (k) k (−∞, 0) (0,l) (l, z) (z, ∞) f (k) + + As we can see, f (k) > for all k ∈ (l, 1) Even though this problem seems unsolvable at first, the intermediate value theorem kills this monstrous inequality! Comment: This origin of this problem comes from Math Olympiad Training Program Project Z MOCK Test Problem The original formulation is actually a geometry problem which can be solved with the knowledge of excircles or trig bash However, the Intermediate Value Theorem conquered this problem algebraically 1.3 Homework Exercises Exercise 1: Use the Intermediate Value Theorem to show that the polynomial function f (x) = x3 + 2x − has a zero in the interval [0,1] Solution: Note that f is continuous because it’s a polynomial Since f (0) = −1 < < = f (1) it follows that there must be some c ∈ [0, 1] such that f (c) = Exercise 2: Given integers A, B, C and equation Ax2 + Bx + C = 0, where C > and A + C < B Suppose √ that D and E are positive √ integers, and E is not a perfect square, x = D + E > 0, and let x ¯ = D − E Show that CHAPTER INTERMEDIATE VALUE THEOREM −1 < x ¯ < Solution: Let f (t) = At2 + Bt + C Since x is an irrational root of f (t), the conjugate of x must also be a root, i.e x ¯ Notice that f (0) = C > and f (−1) = A − B + C < The problem now invokes the applicaton of Intermediate Value Theorem: there exist a root of f in the interval (−1, 0) and since x > 0, the root must be x ¯ One interesting fact about this problem is that if x = a1 + a2 + a3 + a1 + 1 a2 +··· is the periodic continued fraction expansion of x, then −1 < x ¯ < Comment: This problem appeared in the 2007 American Regional Math League Power Round as well as Math Olympiad Program Blue Test Problem Less than five students solved this problem during the MOP test, and we have just given a simple solution with the Intermediate Value Theorem 26 CHAPTER HOW TO CHEAT ON OPTIMIZATION TEST in this chapter so that the optimization test would be like a breeze Obviously you can’t cheat on a homework quiz, but it’s better to cheat than to repeat as Mr.Barnes’ high school coach would say 5.1.4 I Have Cremated the System Delong had totally burned the optimization test! He was 120 percent confident that his answers were correct, but on the homework quiz he was only BLANK percent sure about his answers, where BLANK denotes a number less than 100 Delong had only shared his (actually well known) optimization method with the smartest kid in Baton Rouge, Neal, whom we are going to discuss more in the next chapter 5.2 How to Beat the System 5.2.1 The Enemy of My Enemy First let’s look at this exceedingly hard math concept: x2 ≥ for all real numbers x Since we can’t beat the system, let’s prove this using the system The derivative of x2 is 2x, so the function is increasing for positive x and decreasing for negative x We deduce that the minimum value occur at x = which is Thus we have proven that x2 ≥ Now let’s try to beat the system The friend of my friend is my friend, and the enemy of my enemy is also my friend Therefore, x2 ≥ See how simple it is to work without calculus? Simply replacing x by a − b: (a − b)2 ≥ ⇔ a2 + b2 ≥ 2ab would solve a lot of optimization problems 5.2.2 The Mean Method The following inequality is mean because it trivializes Ms.O’Dell’s optimization quiz Theorem 8: Mean Inequality Given nonnegative real numbers a1 , a2 , · · · , an , then √ a1 + a2 + · · · + an ≥ n a1 a2 · · · an n This inequality allows us to find the upper bound of the product with a given sum Furthurmore, we not need to calculate the zero of the derivative anymore We can instead simply assume all the values are equal 5.3 EXAMPLES 5.3 27 Examples Example 1: [Log Contest 2008] Leyla wants to know the maximum possible area of a rectangular field surrounded by a 200m wall and a 1000m electric fence However, due to time constraint during the contest, she did not have time to read this chapter (Actually this chapter is written after she took the test.) Moreover, Delong didn’t get the correct answer What answer did Neal get? Follow the System: Since the total perimeter of the rectangle is 1200m, we can assume the sides are x and 600 − x To maximize the function f (x) = x(600 − x), we need to find the zero of the derivative f (x) = 600 − 2x = ⇒ x = 300 So the extreme value occur at x = 300 Since f (x) = −2, the function is concave down, and therefore f (300) = 90000 is indeed the minimum Neal’s answer is therefore 90000 Beat the System: By the Mean Inequality, x(600 − x) ≤ x + (600 − x) = 300 ⇒ x(600 − x) ≤ 90000 Example 2: [Mandelbrot 2007] Given that xy(x−y) = 3, what is the minimum of x2 + y ? Follow the System: Both Delong and Neal used derivative method in the contest and neither solved this problem Let’s try to take the derivative of xy(x − y) = This is like grits hit the fan We need to first express y in terms of x, and then find the derivative y x − yx2 + = y = y = x2 ± √ x4 − 12x 2x√ √ d (x ± x4 − 12x) · 2x − 2(x2 ± x4 − 12x dx ) 4x2 The details are left as an exercise for the reader (and the grader.) After this step, both Neal and Delong basically gave up because they knew that this is definitely not the intention of the problem proposer Beat the System: Using the Mean Inequality, there is a one line solution: √ x2 + y = (x − y)2 + xy + xy ≥ 3 (x − y)2 xy · xy = 28 5.4 CHAPTER HOW TO CHEAT ON OPTIMIZATION TEST Homework Exercises Exercise 1: Brian wants Dooly to be the main character even though Dooly said she would kill Delong if he dares to mention her in this chapter For the sake of pretension, let’s pretend Brian and Dooly are building a four dimensional rectangle with 800m of wire What’s the maximum possible volume? Note: The inspiration of this problem is 2005 National Mathcounts Masters Round Follow the System: We first need to image a four dimensional rectangle Consider a one dimensional rectangle (i.e a segment) Sliding this segment on a plane would create a two dimensional rectangle Sliding this two dimensional rectangle upward would create a three dimensional rectangle (i.e a box.) Now sliding this box would create the four dimensional rectangle Alternately, we can consider a cloned box connecting corresponding vertices with the original box Since a three dimensional rectangle has vertices and 12 edges, the four dimensional rectangle would have + 12 × = 32 edges Assume the four dimensions are a, b, c, d Then each edge appears times and 8(a+b+c+d) = 800, and we need to find the maximum of abcd As we have tried before, let’s take the derivative of a function But wait, what function should we use? We have four variables, and only Ashton is reading ahead about multivariable calculus The best thing we can here is to fix two variables, namely, c and d Now we have f (a) = abcd = a × (100 − c − d − a)cd Again, this is a concave down parabola, but we still need to find the zero of f (a) = cd(100 − c − d − 2a) We get a = b = 50 − c+d would maximize f (a) Thus, if we fix two variables, then the other two must be equal in order to maximize the volume Now we fix a and b, and we deduce that c = d What now? Can we fix a and c and claim that b = d and therefore all four variables are equal? That idea deserves a round of applause because it’s a hundred and ten percent wrong We need to use another function to finish this problem Since now we know that a = b, c = d, and a + b + c + d = 2a + 2b = 100, we can set ga = abcd = a2 (50 − a)2 = a2 (2500 − 100a + a2 ) Now back to the system: g (a) = 5000a − 300a2 + 4a3 = 4a(a − 25)(a − 50) = The maximum is either g(0) = g(50) = or g(25) = 254 Obviously 254 > Therefore, the greatest possible volume is 254 Beat the System: Since a + b + c + d = 100, √ abcd ≤ a+b+c+d = 25 Therefore, the greatest possible volume is 254 5.4 HOMEWORK EXERCISES 29 Exercise 2: Lucy said that she’s not going to talk to Delong for the rest of the year if he dares to mention her cheek Let’s assume that her cheek is shrinking along with her ugliness What is the minimum possible sum of her ugliness and the ratio of her meanness over her dumbness? Figure 5.1: Raymond and Lucy Follow the System: By the Ideal Girls Law, we want to find the minimum of π The widehat symbol was introduced in Chapter Suppose that Lucy + Lucy π f (Lucy) = Lucy + Lucy then f (Lucy) = − π Lucy implies Lucy = √ =0 π and f (Lucy) = π Lucy >0 √ √ for positive Lucy Therefore, the minimum value of f (Lucy) is f ( π) = π Beat the System: Again, we’ll show a one line solution: Lucy + π Lucy ≥2 Lucy · π Lucy √ =2 π Chapter The Dream Team 6.1 Background The title of this chapter has been changed Calculus Triumphs Over Trig to The Hot War to Math Genius, and finally, Delong decided to use the Dream Team of two math genius at Baton Rouge High This chapter is on integration Right now, consider integration as an average value of infinitely many numbers So far, Neal is pwning Delong in every contest in the year 2008 (LSU, Mandelbrot, APMO, Log 1, and AIME) Instead of containment, Delong and Neal are escalating the race for the 2008 IMO team We are going to explore two MOP test problems that Neal probably doesn’t know because he’s too lazy to read over MOP solution manuel Figure 6.1: Math Genius Integration is basically the opposite of differentiation The derivative of the integral is the function itself The integral over an interval of a function is like the average value of the function divided by the length of the interval The difference of the values of the integral function on the two ends of the interval is equal the the definite integral 30 6.2 EXAMPLES 6.2 31 Examples b Example 0: Show that b is an integer if and only if sin 2πxdx = Solution: Let’s first show the necessary part Since the derivative of cos x is cos 2πx − sin x, the derivative of cos 2πxis −2π sin 2πx So the derivative of − is 2π sin 2πx Now looking backward, we deduce that b sin 2πxdx = − b cos 2πx 2π =− cos 2πb − cos 2π Therefore cos b = cos = 1, and b must be an integer Obviously, if b is an integer, then the integral is 0, which proves the sufficiency Example 1: What is the average value of a rectangular tile with vertices (0, 0)(a, 0)(0, b)(a, b) where at least one of a, b is an integer, given that each point (x, y) in the coordinate plane is assigned a value of sin(2πx) sin(2πy)? Solution: Without Loss of Generality, let’s assume b is an integer, then by b the previous problem, sin 2πxdx = This means that the average value of b each vertical line of the rectangle is Notice that the values of points on one vertical line is always a multiple of another vertical line So the average value of the whole rectangle is a a sin 2πxdx × b b sin 2πxdx = 0 Therefore, by the assignment of the points, if a rectangular tile has an integer side, then the average values it covers is Example 2: Show that if a rectangular board can be tiled by smaller rectangular tiles each of which has at least one integer sides, then the tiled rectangular board has at least one integer side Solution: Similar to the previous example, each rectangular tile covers an average number of Therefore, the average value of the points on the entire rectangular board must be Now we can let the board have vertices (0, 0)(a, 0)(0, b)(a, b) We know that a a sin 2πxdx × b b sin 2πxdx = 0 So we want to show that at least one of a, b is an integer This is a consequence of example because either a a sin 2πxdx = 0 32 CHAPTER THE DREAM TEAM or b b sin 2πxdx = 0 Which means either a or b is an integer Comment: This problem comes from Yufei Zhao’s Blue MOP lecture note A paper in the American Math Monthly showed 14 proofs of this result! 6.3 Homework Exercises Problem to pwn: Evaluate xn dx Solution: Basically, we want to find a function whose derivative is xn We xn+1 is xn know that the derivative of xn+1 is (n + 1)xn , so the derivative of n+1 Now we can calculate the exact value by subtracting two end point values xn dx = xn+1 n+1 = n+1 More pwning problem: Prove that f ·g =f ·g− f ·g (6.1) Solution: Recall that the integral is basically the opposite of the derivative Let’s take the derivative of both sides f · g = (f · g) − f · g This is the product rule of differentiation! Indeed, the product rule of differentiation states that d (f (x)g(x)) = f (x)g(x) + g(x)g (x) dx Rearranging the terms would get the desired result Comment: This is also called integration by parts which is analogous to summation by parts or Abel’s Summation Formula: n n Sk (bk − bk+1 ) ak bk = Sn bn + k=1 (6.2) k=1 where Sk = a1 + a2 + · + ak In Zach Abel’s Blue MOP handout, he claimed that the equivalence of 6.1 and 6.2 can be established directly: one way with Riemann sums, and the other with appropriate choice of functions However, Delong still couldn’t understand 6.3 HOMEWORK EXERCISES 33 this claim He even posted this on AoPS forum, and no one has given an answer yet Problem to be pwned by: Prove that for all nonnegative integers k ≤ n (1 − x)k xn−k dx = (n + 1) n k Solution: [Due to Oleg Golberg] We proceed by induction on k The base case k = is solved in problem Now suppose this is true up to k, then by the result of problem 1 (1 − x)k+1 xn−k−1 dx = (1 − x)k+1 · xn−k n−k − −(k + 1)(1 − x)k · xn−k n−k = = = d+1 (1 − x)k xn−k dx n−k d+1 n − k (n + 1) nk n (n + 1) k+1 Comment: In a 2005 Mathcamp lecture, Paul Zeitz showed a specially case of this problem (n = 2001, k = 1), and he claimed that it can be solved without calculus He said that a 2001 Bay Area Math Olympiad that used the above result can be solved combinatorially Indeed, we frist generate a random number ≤ x ≤ 1, and then make a coin such that the probability of getting a head is x The probability of getting k heads out of n flips is (1 − x)k xn−k dx This suggests that the probablity of getting any number of heads are equal However, Delong could not figure out the combinatorial interpretation, so he asked Neal and posted it on AoPS Obviously, no answers yet Chapter Two Journals 7.1 My Best Area My favorite area so far is optimization As presented in chapter 5, I could solve optimization problems without calculus On the optimization quiz, I first used the usual calculus method to find the answer Then I used the Arithmetic Geometric Mean Inequality to check my answer In this way, I can guarantee that my answers are 120% correct Even though my chapter on optimization may not be appealing to everyone, it is indeed very useful Once you understand the concept of the Mean Inequality, optimization problems are all pwned by non-calculus solution, especially the problems about fences or rectangles When given the sum of two numbers, I immediately know that their product is the greatest when the two numbers are equal because (a + b)2 ≥ 2ab ⇔ (a − b)2 ≥ Of course, this can be solved with calculus, but that’s like using a big cannon ball to shoot a little insect 7.2 Worst Grade My worst grade was actually a negative zero, but let’s just say my weakest area is related rate On the related rate quiz, I forgot the formula for the volume of a sphere! I knew that it’s either R3 π or R3 π I also wasn’t sure about whether the surface area of a sphere is 3R2 π or 4R2 π On the quiz, I suddenly realized that I haven’t used these geometry formulae for more than x years, where x is a positive integer I was trying to derive the formula Since I didnt’ know integration then, I had to guess at the end I randomly guessed the wrong formula, and obviously I didn’t well on the quiz 34 Chapter Extra Tricky Math Magics Both magic tricks in this chapter come from the 2007 Russian Math Olympiad 8.1 Points on a Circle Ms.O’Dell and Ashley together are performing the following trick to sixth hour class A circle is drawn on the board, and the class can mark as many points on the circle as they want while Ms.O’Dell is outside the room Then Ashley is going to erase a point that the class had just marked Ms.O’Dell then comes in and immediately points out a semi-circle that contains the erased point How did she pick the semi-circle? Solution: This trick is actually not hard After the points are marked on the circle, Ashley is going to look at the distances between two consecutive points She then choose two consecutive points A and B clockwise such that their distance is the greatest Now she’s going to erase B Now Ms.O’Dell comes in, and she looks for the two consecutive points that are furthest to each other, namely, A and C where C is the point after B Now Ms.O’Dell simply start from C and go counter clockwise until she has a semi-circle Figure 8.1: Gargi, Delong, DOOOOOOOOOOLY, Ashley 35 36 8.2 CHAPTER EXTRA TRICKY MATH MAGICS Number Guessing Ms.O’Dell and Gargi are performing this magic to fifth hour The class is going to write a 2008 digit number on the board while Ms.O’Dell is outside Gargi is going to cover two consecutive digits When Ms.O’Dell comes in, she immediately know what the two digits are How did she and Mary perform the trick? Solution: Gargi first index the digits through 2008 Suppose the sum of all numbers with an odd index is s mod 10, and the sum of all numbers with an even index is t mod 10 Assume that Gargi can add, she is going to cover the two numbers with indices p and p + 1, where p = 10s + t When Ms.O’Dell comes in, she immediately know what s and t because she knows p Now with the knowledge of s and the rest of the odd index numbers she can figure out the covered odd index number Similarly, she can calculate the covered even index number Chapter A Moment In MATh This is a story about the 2008-2009 MATh (Mu Alpha Theta) President Figure 9.1: The next Mu Alpha Theta President In 2006 Catholic High School tournament, both Delong and Shela competed in Algebra II The award ceremony was a little different because they call up everyone who won a trophy in alphabetical order first Then they distribute the trophies from last to first place Everyone in the audience, especially those from Glasgow Middle School could remember that the result was: Third place from Baton Rouge High, Adam Mendrela Second place from Baton Rouge High, Delong Meng The whole audience was shocked and in first place from Baton Rouge High This is Shela’s most commemorable moment in MATh history which she will remember even in her eighties 37 List of Figures 1.1 Intermediate Value Theorem 2.1 Tangent to a curve 3.1 My ”you-look-stupid-but-I’m-not-going-to-be-a-bother-and-tell-youabout-it” expression-Dina 10 4.1 Cute girls: Emmi, Hannah, Leyla, and Dooly 17 5.1 Raymond and Lucy 29 6.1 Math Genius 30 8.1 Gargi, Delong, DOOOOOOOOOOLY, Ashley 35 9.1 The next Mu Alpha Theta President 37 38 Index 102 Combinatorial Problems, 21 Formal Definition of the Derivative, definitons Degree of Ugliness, 11 Degree of Ugliness, 11 derivative by rationalization, Abel Summation Formula, 32 Adam, 37 Annie, 18 Antoine’s Thereom, 10 Arf Arf, 14 Ashton, 18, 28 Erin, 18 Ethan, 22 Existence of Zeros of A Polynomial, Balkan Math Olympiad 1998, 12 Brandon, 18 Formal Definition of the Derivative, Car Burning Inequality, 19 Catholic High Tournament 2006, 37 Chinese Math Olympiad 1988, Coach Antoine, 14, 16 comments Adam, 15 Amol, 15 Andrew, 15 Dooly, 15 George, 15 Hudson, 15 Ivan, 15 Johnny, 15 Leyla, 15 Myeisha, 10 Rhonna, 15 Sheila, 15 Shela, 15 continued fraction expansion, convexity, 19 girls Aabha, 17 Aastha, 17 Ashley, 35 Chloe, 14 Devynn, 25 Dina, 10 Dooly, 12 Emmi, 14, 17 Erin, 14 Gargi, 36 Hannah, 17 Jerica, 14 Kristin, 18 Leyla, 12 Megan, 19 Shela, 10, 12 Shelby, 13 Ideal Girls Law, 25 Integration by Parts, 32 Intermediate Value Theorem, Ivan, 18 definitions convexity, 19 convexity and concavity, 19 Degree of Ugliness, 11 Jason, 18 39 40 Jenson’s Inequality, 19 Justin, 20 Mean Inequality, 26 microwave generation, 16 MOCK Test 2007, monochromatic triangles, 21 MOP/ARML 2007, Neal, 30 Raymond, 12 Russian Math Olympiad 2007, 35 September 18, 1967, 16 Sweden Math Olympiad 2005, 14 theorem Ideal Girls Law, 25 theorems Antoine’s Thereom, 10 Car Burning Inequality, 19 Existence of Zeros of A Polynomial, Ideal Girls Law, 25 Intermediate Value Theorem, Jenson’s Inequality, 19 Mean Inequality, 26 Ugly Girls Thereom, 11 Titu Andreescu, 22 ugly girls are very friendly, 10 Ugly Girls Thereom, 11 Zuming Feng, 22 INDEX