AIME

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Every rule and concept in AIME, with its definition and a link to the lesson that teaches it. Search by term, by idea, or by lesson title.

Glossary (64)

Addition and double-angle
sin⁡(a±b)=sin⁡acos⁡b±cos⁡asin⁡b\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b; cos⁡(a±b)=cos⁡acos⁡b∓sin⁡asin⁡b\cos(a\pm b)=\cos a\cos b\mp\sin a\sin b; tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan(a+b)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}; sin⁡2a=2sin⁡acos⁡a\sin 2a = 2\sin a\cos a, cos⁡2a=1−2sin⁡2a\cos 2a = 1-2\sin^2 a.
Trig Formulas That Win › Sum, Double, Triple
Algebra and polynomials
Newton's sums; roots of unity filter; ∏k=1n−1(1−ωk)=n\prod_{k=1}^{n-1}(1 - \omega^k) = n and ∏sin⁡kπn=n2n−1\prod\sin\tfrac{k\pi}{n} = \tfrac{n}{2^{n-1}}; Lagrange interpolation; finite differences (degree-dd polynomial has constant ddth differences); floor: $∑ k/m…
Formulas & Methods › Number Theory & Algebra
Area ratios: same height, same base
Triangles sharing an apex have areas in the ratio of their bases; triangles on the same base have areas in the ratio of their heights.
Cevians & Mass Points › The Ratio-Chasing Recipes
Assign masses so the ratios balance
On a segment split m:nm:n, put mass nn at the near end and mm at the far end; masses add at intersection points.
Cevians & Mass Points › Mass Points in Practice
Binomial theorem
(x+y)n=∑k=0n(nk)xn−kyk(x+y)^n = \sum_{k=0}^n\dbinom nk x^{n-k}y^k.
Advanced Counting › Building Recurrences
Ceva's theorem
Cevians ADAD, BEBE, CFCF (with D,E,FD, E, F on sides BCBC, CACA, ABAB) are concurrent if and only if BDDC⋅CEEA⋅AFFB=1\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1.
Geometry: Coordinates & Complex › Cevians: Ceva & Menelaus
Chinese Remainder Theorem
If the moduli are pairwise coprime, a system x≡ai(modmi)x\equiv a_i\pmod{m_i} has a unique solution modulo ∏mi\prod m_i.
Advanced Number Theory › Systems of Congruences
Choosing between them — and mass points
Concurrency of three cevians ⇒\Rightarrow Ceva.
Geometry: Coordinates & Complex › Cevians: Ceva & Menelaus
Circle power tools together
Ptolemy's theorem (AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC in a cyclic quadrilateral) and the Power of a Point (AP⋅PB=CP⋅PDAP \cdot PB = CP \cdot PD for chords through PP) often chain together in one problem.
Telescoping & Recursion › Making Long Computations Short
Condition on the last step
Count by splitting on the final move: a tiling ending in a square leaves an−1a_{n-1} ways, one ending in a domino leaves an−2a_{n-2}, so an=an−1+an−2a_n = a_{n-1}+a_{n-2}.
Advanced Counting › Building Recurrences
Counting
Bijections (Dyck paths, compositions, stars and bars with bounds by PIE); recursion by first choice; generating functions in easy cases; Burnside for symmetric colorings 1∣G∣∑∣Fix(g)∣\tfrac{1}{|G|}\sum|\mathrm{Fix}(g)|; Vandermonde and hockey stick; distributing with …
Formulas & Methods › Counting, Probability & Expectation
Distinct roots
For an+2=p an+1+q ana_{n+2}=p\,a_{n+1}+q\,a_n, solve x2=px+qx^2=px+q.
Recurrences & Characteristic Equations › The Characteristic Equation Method
Divisor sums are multiplicative
d(n)=∏(ai+1)d(n) = \prod(a_i+1), σ(n)=∏piai+1−1pi−1\sigma(n) = \prod\dfrac{p_i^{a_i+1}-1}{p_i-1}, and both are multiplicative across coprime factors.
Advanced Number Theory › Valuations, Divisor Sums and Digits
Euler's theorem and totient
If gcd⁡(a,n)=1\gcd(a,n)=1, then aφ(n)≡1(modn)a^{\varphi(n)}\equiv 1\pmod n, where φ(n)=n∏p∣n(1−1p)\varphi(n) = n\prod_{p\mid n}(1-\tfrac1p).
Advanced Number Theory › Orders and Totients
Evaluation tricks
Set x=1x = 1 to sum the coefficients; x=−1x = -1 for the alternating sum; p(1)+p(−1)2\tfrac{p(1) + p(-1)}2 for the even-index coefficients.
Polynomials › Roots, Factors and Clever Evaluations
Extracting every nth coefficient
To sum the coefficients of x0,xn,x2n,…x^0, x^n, x^{2n},\dots in PP, average PP over the nnth roots of unity: 1n∑jP(ωj)\tfrac1n\sum_{j} P(\omega^j).
Polynomials › The Roots of Unity Filter
Fibonacci and Binet
Fibonacci is the recurrence x2=x+1x^2 = x+1 with roots φ=1+52\varphi = \tfrac{1+\sqrt5}{2} and φ^=1−52\hat\varphi = \tfrac{1-\sqrt5}{2}: Fn=φn−φ^n5F_n = \dfrac{\varphi^n - \hat\varphi^n}{\sqrt5}.
Recurrences & Characteristic Equations › The Characteristic Equation Method
Find FF with tk=F(k+1)−F(k)t_k = F(k + 1) - F(k)
Useful differences: $k · k!
Telescoping & Recursion › Telescoping Beyond Fractions
Floor functions and base representations
⌊x⌋\lfloor x\rfloor problems are block counts: ∑k=12026⌊k7⌋\sum_{k=1}^{2026}\lfloor\tfrac k7\rfloor groups kk by quotient; "nn divisible by ⌊n⌋\lfloor\sqrt n\rfloor" gives exactly three per square block.
Complete Syllabus & Gap Fillers › Gap Fillers
Generating functions
Set F(x)=∑anxnF(x) = \sum a_nx^n, multiply the recurrence by xnx^n, sum, and solve for FF; partial fractions recovers ana_n.
Recurrences & Characteristic Equations › Solving Recurrences Two Ways
Generating functions, the easy cases
The coefficient of xnx^n in (1+x)a(1+x)b(1 + x)^a(1 + x)^b is Vandermonde (∑(ak)(bn−k)=(a+bn)\sum\binom ak\binom{b}{n-k} = \binom{a+b}{n}; ∑(10k)2=(2010)\sum\binom{10}{k}^2 = \binom{20}{10}).
Complete Syllabus & Gap Fillers › Gap Fillers
Geometry
Power of a point; radical axis; Ptolemy; Brahmagupta; Stewart; mass points and Ceva/Menelaus; Heron; rs=abc4Rrs = \tfrac{abc}{4R}; Euler OI2=R(R−2r)OI^2 = R(R - 2r); area ratios via shared angles [ADE][ABC]=AD⋅AEAB⋅AC\tfrac{[ADE]}{[ABC]} = \tfrac{AD\cdot AE}{AB\cdot AC}; coordinates with a …
Formulas & Methods › Geometry & Trigonometry
Iterated functions and periodicity
Digit sums preserve residues mod 99 and shrink fast — f(f(f(72026)))f(f(f(7^{2026}))) is a single digit found by size bounds plus mod 99.
Complete Syllabus & Gap Fillers › Gap Fillers
Legendre's formula and trailing zeros
vp(n!)=⌊np⌋+⌊np2⌋+⋯v_p(n!) = \left\lfloor\tfrac np\right\rfloor + \left\lfloor\tfrac n{p^2}\right\rfloor + \cdots; trailing zeros of n!n! count v5(n!)v_5(n!). Why it works.
Advanced Number Theory › Valuations, Divisor Sums and Digits
Linear recursions
When each state comes from the previous ones — like an=an−1+an−2a_n = a_{n-1} + a_{n-2} — compute forward from the base cases rather than hunting for a formula.
Telescoping & Recursion › Making Long Computations Short
Linearity of expectation
The expected value of a sum equals the sum of expected values, even when the parts depend on each other.
Advanced Counting › Expected Value
Mass points for ratios ALONG a cevian
Assign masses inversely to the side ratios so each cevian foot balances; then APPD=mass(D)mass(A)\dfrac{AP}{PD}=\dfrac{\text{mass}(D)}{\text{mass}(A)} at the intersection PP.
Cevians & Mass Points › The Ratio-Chasing Recipes
Menelaus, for a line crossing the triangle
For a transversal meeting BCBC, CACA, ABAB at DD, EE, FF: BDDC⋅CEEA⋅AFFB=1\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1 with signed ratios (the unsigned product is 11 with an odd number of external points).
Cevians & Mass Points › Mass Points in Practice
Menelaus's theorem
A transversal line crossing lines BCBC, CACA, ABAB of triangle ABCABC at DD, EE, FF satisfies BDDC⋅CEEA⋅AFFB=1\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1 (unsigned; with signed ratios the product is −1-1).
Geometry: Coordinates & Complex › Cevians: Ceva & Menelaus
Newton's identities
Power sums pk=∑rikp_k = \sum r_i^k come from the Vieta sums eje_j: p1=e1p_1 = e_1, p2=e1p1−2e2p_2 = e_1p_1 - 2e_2, p3=e1p2−e2p1+3e3p_3 = e_1p_2 - e_2p_1 + 3e_3.
Polynomials › Symmetric Functions of Roots
Number theory
CRT; orders and primitive roots; φ\varphi, τ\tau, σ\sigma multiplicative; Legendre and Kummer; LTE vp(an−bn)=vp(a−b)+vp(n)v_p(a^n - b^n) = v_p(a - b) + v_p(n); Wilson; quadratic residues (−1-1 QR iff p≡1 mod 4p \equiv 1 \bmod 4); base-bb digit sums ≡n(modb−1)\equiv n \pmod{b-1}; $(a^m…
Formulas & Methods › Number Theory & Algebra
Order of an element
The order of aa mod nn is the least d>0d>0 with ad≡1a^d\equiv 1, and it always divides φ(n)\varphi(n).
Advanced Number Theory › Orders and Totients
Pairing tricks for long sums
Pair to a constant: sin⁡2k∘+sin⁡2(90−k)∘=1\sin^2 k^\circ + \sin^2(90-k)^\circ = 1; tan⁡k∘tan⁡(90−k)∘=1\tan k^\circ\tan(90-k)^\circ = 1; (1+tan⁡k∘)(1+tan⁡(45−k)∘)=2(1+\tan k^\circ)(1+\tan(45-k)^\circ) = 2.
Trig Formulas That Win › Products & Sums That Collapse
Partial fractions first
A rational term with factored denominator splits into pieces whose difference telescopes: 1k(k+2)=12(1k−1k+2)\dfrac{1}{k(k+2)} = \tfrac12\left(\dfrac1k - \dfrac1{k+2}\right).
Telescoping & Recursion › Finding the Difference That Telescopes
Pick the tool by what is given
Given many lengths and a right angle: coordinates.
Geometry: Coordinates & Complex › Lengths by Coordinates, Trig and Power of a Point
Probability and expectation
Linearity with indicators; expected waiting times by first-step equations; Markov-chain states for "until two in a row"; geometric probability (unit square regions, broken stick); conditional probability by restricting the sample space; symmetry arguments f…
Formulas & Methods › Counting, Probability & Expectation
Product-to-sum and sum-to-product
2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B); sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\tfrac{A+B}2\cos\tfrac{A-B}2.
Trig Formulas That Win › Sums That Collapse
Proof of Ptolemy's theorem
In cyclic ABCDABCD, mark EE on diagonal BDBD with ∠BCE=∠ACD\angle BCE = \angle ACD.
Heavy Machinery, Proved › Ptolemy's Theorem, Proved
Proof of Stewart's theorem
Let cevian AD=dAD = d hit BCBC with BD=mBD = m, DC=nDC = n.
Heavy Machinery, Proved › Stewart's Theorem and the Bisector, Proved
Proof: the arithmetic mean–geometric mean inequality for two terms
a+b2≥ab\dfrac{a+b}{2} \ge \sqrt{ab} for a,b≥0a, b \ge 0, with equality iff a=ba = b.
Heavy Machinery, Proved › Two More Proofs Worth Carrying
Proof: the distances from one vertex multiply to nn
Divide xn−1x^n - 1 by x−1x - 1: ∏k=1n−1(x−ωk)=1+x+⋯+xn−1\prod_{k=1}^{n-1}(x - \omega_k) = 1 + x + \cdots + x^{n-1}.
Heavy Machinery, Proved › Roots of Unity, Proved
Proof: the law of cosines
c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C.
Heavy Machinery, Proved › Two More Proofs Worth Carrying
Reading equality: when Ptolemy becomes an inequality
For a NON-cyclic quadrilateral the same construction gives AC⋅BD<AB⋅CD+AD⋅BCAC \cdot BD < AB \cdot CD + AD \cdot BC — so equality is a TEST for being cyclic.
Heavy Machinery, Proved › Ptolemy's Theorem, Proved
Repeated roots
If the characteristic equation has a double root rr, the general solution gains a factor of nn: an=(A+Bn) rna_n = (A + Bn)\,r^n.
Recurrences & Characteristic Equations › The Characteristic Equation Method
Rotations by multiplication
Rotating a point zz about the origin by θ\theta is multiplication by eiθe^{i\theta}; about a center cc, use c+(z−c)eiθc + (z-c)e^{i\theta}.
Geometry: Coordinates & Complex › Complex Numbers as Points
Set up states, then solve the linear system
Name pAp_A, pBp_B, … for the probability of the goal from each state; write one equation per state from its transitions; solve.
Advanced Counting › Probability With States and Symmetry
Shift, flip, or reduce
For an=c an−1+da_n = c\,a_{n-1} + d with c≠1c \ne 1, subtract the fixed point L=d1−cL = \tfrac{d}{1 - c}: then an−L=c(an−1−L)a_n - L = c(a_{n-1} - L), a geometric sequence.
Recurrences & Characteristic Equations › Constant Terms, Substitutions and Periodicity
Shoelace formula
A polygon with vertices (xi,yi)(x_i,y_i) in order has area 12∣∑i(xiyi+1−xi+1yi)∣\tfrac12\left|\sum_i (x_iy_{i+1} - x_{i+1}y_i)\right|.
Geometry: Coordinates & Complex › Analytic Geometry
Symmetry collapses the work
If the labels can be permuted without changing the process, the corresponding probabilities are equal — solve one and multiply.
Advanced Counting › Probability With States and Symmetry
Telescoping products
A product telescopes when each factor is a ratio of consecutive values: ∏k=2nk−1k=1n\prod_{k=2}^{n}\dfrac{k-1}{k} = \dfrac1n.
Telescoping & Recursion › Finding the Difference That Telescopes
Telescoping sums
Partial fractions turn 1k(k+1)\dfrac{1}{k(k+1)} into 1k−1k+1\dfrac1k - \dfrac{1}{k+1}, so the sum telescopes: ∑k=1n1k(k+1)=1−1n+1\sum_{k=1}^{n} \dfrac{1}{k(k+1)} = 1 - \dfrac{1}{n+1}.
Telescoping & Recursion › Making Long Computations Short
Telescoping with cotangents and tangents
arctan⁡a−arctan⁡b=arctan⁡a−b1+ab\arctan a - \arctan b = \arctan\dfrac{a-b}{1+ab}, and cot⁡θ−cot⁡2θ=csc⁡2θ\cot\theta - \cot2\theta = \csc2\theta.
Trig Formulas That Win › Sums That Collapse
The 60-degree product identity
sin⁡θ sin⁡(60∘−θ) sin⁡(60∘+θ)=sin⁡3θ4\sin\theta\,\sin(60^\circ-\theta)\,\sin(60^\circ+\theta)=\dfrac{\sin 3\theta}{4}, and the same shape holds for cosine and tangent (tan⁡\tan version multiplies to tan⁡3θ\tan 3\theta).
Trig Formulas That Win › Products & Sums That Collapse
The angle bisector length, derived from Stewart
A bisector from AA splits BCBC in ratio c:bc : b (bisector ratio theorem), so m=acb+cm = \tfrac{ac}{b+c}, n=abb+cn = \tfrac{ab}{b+c}.
Heavy Machinery, Proved › Stewart's Theorem and the Bisector, Proved
The characteristic equation
For an=pan−1+qan−2a_n = pa_{n-1} + qa_{n-2}, solve r2=pr+qr^2 = pr + q: distinct roots give an=Ar1n+Br2na_n = Ar_1^n + Br_2^n, a repeated root gives an=(A+Bn)rna_n = (A + Bn)r^n.
Recurrences & Characteristic Equations › Solving Recurrences Two Ways
The checklist
Integer answers 000000–999999, fifteen problems, three hours.
Complete Syllabus & Gap Fillers › What the AIME Tests
The factor and remainder theorems, used backwards
p(a)=0  ⟺  (x−a)∣pp(a) = 0 \iff (x-a) \mid p; the remainder of pp on division by x−ax - a is p(a)p(a).
Polynomials › Roots, Factors and Clever Evaluations
The law of cosines as a length machine
c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C; with cos⁡C\cos C known from another triangle, one unknown length falls out.
Geometry: Coordinates & Complex › Lengths by Coordinates, Trig and Power of a Point
The universal problem-solving loop
Read twice.
Formulas & Methods › Geometry & Trigonometry
Trigonometry
Product-to-sum and sum-to-product; tan⁡\tan addition; ∑k=1n−1cos⁡2kπn=−1\sum_{k=1}^{n-1}\cos\tfrac{2k\pi}{n} = -1; cos⁡π7cos⁡2π7cos⁡3π7=18\cos\tfrac\pi7\cos\tfrac{2\pi}7\cos\tfrac{3\pi}7 = \tfrac18 (doubling trick); telescoping via tan⁡(a)−tan⁡(b)=sin⁡(a−b)cos⁡acos⁡b\tan(a) - \tan(b) = \tfrac{\sin(a - b)}{\cos a\cos b}; $ x…
Formulas & Methods › Geometry & Trigonometry
Triple angle in disguise
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta and sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta.
Trig Formulas That Win › Sum, Double, Triple
Turn every length ratio into an area ratio
Two triangles with the same apex and bases on one line have areas in the ratio of their bases.
Cevians & Mass Points › Areas from Ratios
Which theorem for which picture
Three cevians through ONE point: Ceva, BDDC⋅CEEA⋅AFFB=1\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1.
Cevians & Mass Points › The Ratio-Chasing Recipes
Why xn−1x^n - 1 factors over the nnth roots of unity
By De Moivre (proved by induction on the angle-addition formula), the nn numbers ωk=cis2πkn\omega_k = \text{cis}\tfrac{2\pi k}{n} all satisfy xn=1x^n = 1; a degree-nn polynomial has at most nn roots, so these are ALL of them and xn−1=∏k(x−ωk)x^n - 1 = \prod_k (x - \omega_k).
Heavy Machinery, Proved › Roots of Unity, Proved