AMC 10:12 Cheat Sheet

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Duke University**We aren't endorsed by this school
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MATH 281S
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Industrial Engineering
Date
Dec 17, 2024
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10
Uploaded by KidBravery2213
AMC 10/12 Formulas & StrategiesMichel LiaoOctober 2021Contents1Algebra31.1Mean, Median, and Mode. . . . . . . . . . . . . . . . . . . . . .31.2Arithmetic Sequences. . . . . . . . . . . . . . . . . . . . . . . .31.3Geometric Sequences. . . . . . . . . . . . . . . . . . . . . . . . .31.4Special Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . .41.5Algebraic Manipulations. . . . . . . . . . . . . . . . . . . . . . .41.5.1Quadratic Factorizations. . . . . . . . . . . . . . . . . . .41.5.2Cubic Factorizations. . . . . . . . . . . . . . . . . . . . .41.5.3Sophie Germain’s Identity. . . . . . . . . . . . . . . . . .42Counting & Probability52.1Definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52.1.1Factorials and Arrangements. . . . . . . . . . . . . . . .52.2Combinatoric Strategies. . . . . . . . . . . . . . . . . . . . . . .52.2.1Complementary Counting. . . . . . . . . . . . . . . . . .52.2.2Overcounting. . . . . . . . . . . . . . . . . . . . . . . . .52.2.3Casework. . . . . . . . . . . . . . . . . . . . . . . . . . .52.3Advanced Concepts. . . . . . . . . . . . . . . . . . . . . . . . . .52.3.1Rearrangements and Counting. . . . . . . . . . . . . . .52.3.2Stars and Bars. . . . . . . . . . . . . . . . . . . . . . . .62.3.3Binomial Theorem. . . . . . . . . . . . . . . . . . . . . .62.3.4Combinatorial Identities. . . . . . . . . . . . . . . . . . .62.3.5Pigeonhole Principle. . . . . . . . . . . . . . . . . . . . .73Probability and Expected Value73.0.1Geometric Probability. . . . . . . . . . . . . . . . . . . .73.0.2Principle of Inclusion-Exclusion (PIE). . . . . . . . . . .73.0.3Bijections, Recursion, and States. . . . . . . . . . . . . .84Number Theory84.1Primes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .84.2GCD & LCM. . . . . . . . . . . . . . . . . . . . . . . . . . . . .94.3Modular Arithmetic. . . . . . . . . . . . . . . . . . . . . . . . .91
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5Geometry106Trigonometry107Logarithms108Complex Numbers109General Tips10PrefaceAuthor & MotivationThroughout my junior high/high school career, I’ve learned that the AMC’s arehard. There’s no way around it. “Genius” solvers tend to trivialize problems,and I always wondered how: How did you think of the solution? What promptedyou to think of the solution? AmIeven able to achieve things like this? Theanswer: experience.It really does all boil down to experience. The more problems you do, themore insight you gain. You see enough problems that are similar to each otherso that you begin to form connections.Often, “genius” solutions only take achange in perspective or mindset. That, however, is not easy to achieve–that’swhy we have competition math.So, I wrote this summary as notes for myself and for you.This is heav-ily inspired by Sohil Rathi’s The Book of Math Formulas and Strategies, but Idecided to remove some rudimentary (and hard) ideas.1If these ideas aren’t‘rudimentary’ for you, I implore you to stay with competition math and workuntil even the formulas and strategies outlined here are trivial.Math Beyond CompetitionCompetition math really is amazing, but you are limited to many concepts thatstray away from ‘higher-level’ mathematics, such as college/grad-level courses.I encourage you to not only explore through competition math, but also dabblein higher-level math courses–they really are a pleasure. Math is not random for-mulas pumped from a genius’s brain, but a set of concepts that are logically andproved. Often, these proofs are more satisfying than their rote memorization,formula counterparts.1I’ve decided that the ‘hard ideas’ aren’t necessary to do well on either of these exams, butas AMC’s problems become harder and harder, they may be necessary.2
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1Algebra1.1Mean, Median, and ModeMean =a1+a2+. . .+ann.Mode = Most Frequent Number.Median = Middle Term.If number of terms is even, Median = Average of Middle terms.To find which term is the median, take the total terms, add one, anddivide by two.21.2Arithmetic SequencesGiven the arithmetic sequencea, a+d, a+ 2d, . . .:an=a1+ (n-1)d.n=an-a1d+ 1.Average =a1+an2=a1+a2+. . .+ann.S=a1+an2·n=2a1+ (n-1)d2·n.1.3Geometric SequencesGiven the geometric sequencea, ar, ar2, . . .:a+ar+ar2+. . .=a1-ra+ar+ar2+. . .+arn=a1-rn+11-r.2If the number terms in the sequence is even, then you take the floor and ceiling of thedecimal you get.3
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1.4Special Series1 + 2 + 3 +. . .+n=n(n+ 1)2.1 + 3 + 5 +. . .+ (2n-1) =n2.2 + 4 + 6 +. . .+ 2n=n(n+ 1).12+ 22+ 32+. . .+n2=n(n+ 1)(2n+ 1)6.13+ 23+. . .+n3=n(n+ 1)22.1.5Algebraic Manipulations1.5.1Quadratic Factorizationsx2-y2= (x-y)(x+y)(x+y)2=x2+ 2xy+y2= (x-y)2+ 4xy(x-y)2=x2-2xy+y2= (x+y)2-4xy(x+y)2+ (x-y)2= 2(x2+y)2(x+y)2-(x-y)2= 4xy(x+y+z)2=x2+y2+z2+ 2(xy+xz+yz)1.5.2Cubic Factorizationsx3+y3= (x+y)(x2-xy+y2)x3-y3= (x-y)(x2+xy+y2)(x+y)3=x3+ 3xy(x+y) +y3(x-y)3=x3-3xy(x-y) +y3x3+y3+z3-3xyz= (x+y+z)(x2+y2+z2-xy-xz-yz)1.5.3Sophie Germain’s Identityx4+ 4y4= (x2-2xy+ 2y2)(x2+ 2xy+ 2y2).4
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2Counting & Probability2.1Definitions2.1.1Factorials and ArrangementsThe number of ways to arrangenobjects in a line is given byn!.The number of ways to arrangenobjects in a circle is (n-1)!, since youmust divide bynrotations.2.2Combinatoric Strategies2.2.1Complementary CountingComplementary counting is counting what we don’t want and subtracting thatfrom the total possible cases. “At least” in problem statements hints at usingcomplementary counting.2.2.2OvercountingOvercounting is when we count more than what we need and subtract the casesthat we don’t need to arrive at our answer.Note that we usually combineovercounting with PIE.2.2.3CaseworkMany C&P problems can be solved using casework, which is solving a problemthrough dividing the general problem into cases and summing them at the end.Often, you should combine casework with other counting strategies.Whenlooking for your cases, try and make cases easy to compute.2.3Advanced Concepts2.3.1Rearrangements and CountingThe number of rectangles of all sizes in a rectangular grid of sizemnism+ 12◆✓n+ 12,since two horizontal lines and two vertical lines create a unique rectangle.There aren4ways to find an intersection point of two diagonals in ann-gon because wechoose four points, of which there is only one way to form an intersectionby connecting pairs of those four.5
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We can get from (0,0) to (x, y) inx+yx=x+yyways.Think of arranging u’s and d’s for ups and downs.2.3.2Stars and BarsWe can distributekindistinguishable objects intondistinguishable binswith the formulan+k-1k.When a problem has indistinguishable objects, we think of stars and bars.We can use stars and bars to solve linear equations likea+b+c= 4,assuming thata, b, c0.You can manipulate a counting problem into an equation problem.2.3.3Binomial TheoremFor non-negativen, we have(x+y)n=n0xny0+n1xn-1y1+n2xn-2y2+. . .+nnx0yn.Corollary:n0+n1+. . .+nn= 2n.2.3.4Combinatorial IdentitiesVandermonde’s Identity:n0◆✓mm+n1◆✓mm-1+. . .+nm◆✓m0=m+nn.Pascal’s Identity:nk+nk+ 1=n+ 1k+ 1.Hockey Stick Identity:kk+k+ 1k+. . .+nk=n+ 1k+ 1.6
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2.3.5Pigeonhole Principle3Probability and Expected ValueProbability =Successful OutcomesTotal Outcomes3.The expected value of eventXisXxi·P(xi),wherexidenote the possible values ofXandP(xi) denote the probabilitythey occur.Linearity of Expectation (regardless ifXis an independent or dependentevent):E[x1+x2+. . .+xn] =E[x1] +E[x2] +. . .+E[xn].3.0.1Geometric ProbabilityWe use geometric probability when there are an infinite or non-discrete numberof cases. We denote total possible outcomes using lengths, areas, or volumes.Try to solve a geometric probability problem by1. Try a couple examples for various cases, including testing boundary cases.2. Generalize from these examples to try to find a region of successful out-comes.3. Use geometry to find the length/area/volume of this region.3.0.2Principle of Inclusion-Exclusion (PIE)PIE is a method we use to systematically overcount and correct for overcounting.|A1[A2|=|A1|+|A2|-|A1\A2|.|A1[A2[A3|=|A1|+|A2|+|A3|-|A1\A2|-|A1\A3|-|A2\A3|+|A1\A2\A3|.3Although this seems trivial, it’s very important in more advanced applications of proba-bility.7
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3.0.3Bijections, Recursion, and StatesStatesWe use states to try to find a probability of a “win” from dierent positionsor turns:1. Assign variables to the probabilities of winning from the dierent positions2. Write your equations for the probability of winning from each of thesepositions in terms of variables and constants3. Solve the system of equations4Number Theory4.1PrimesTo check ifnis prime, we check all the primes that are less than or equaltopn.Every integer has a unique prime factorization of the formp311·pe22·. . .·p3kk,which has(e1+ 1)(e2+ 1). . .(ek+ 1)factors.2Last digit is even3Sum of digits is divisible by 34Last 2 digits divisible by 45Last digit is 0 or 56Divisible by 2 and 37Divide by 7 repeatedly8Last 3 digits are divisible by 89Sum of digits is divisible by 910Last digit is 011Calculate the sum of odd digits (O) and even digits (E). If|O-E|is divisible by 11, then the number is also divisibleby 1112Divisible by 3 and 415Divisible by 3 and 5We count the number of factors of primepinn! with the following formula:np1+np2+np3+. . . .8
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4.2GCD & LCMGCD is found by taking the lowest exponents of the prime factorizationsofmandn.LCM is found by taking the highest exponents of the prime factorizationsofmandn.GCD of two numbers must be a factor of the LCM.gcd(m, n)·lcm(m, n) =mn.gcd(ac, bc) =c·gcd(a, b).When you draw a diagonal of a rectangular grid, the number of unitsquares it will pass through isa+b-gcd(a, b), whereaandbare thegrid’s side lengths.Euclidean Algorithm:gcd(x, y) = gcd(x, y-kx).Euler’s Totient Function: If a numbernhas the prime factorizationpe11·pe22. . . penn,thenφ(n) =n1-1p1◆ ✓1-1p2. . .1-1pn,whereφ(n) denotes the number of positive integers less than or equal tonthat are relatively prime ton.Euler’s Totient Theorem:aφ(n)1(modn)igcd(a, n) = 1.4.3Modular ArithmeticIfax(modn) andby(modn), thenabxy(modn).Ifax(modn), thenamxm(modn).9
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5Geometry6Trigonometry7Logarithms8Complex Numbers9General TipsUse tables and Venn Diagrams.10
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