AMC 10

Search & glossary

Every rule and concept in AMC 10, with its definition and a link to the lesson that teaches it. Search by term, by idea, or by lesson title.

Glossary (64)

3D geometry
Cube diagonals (s3s\sqrt3), cone and pyramid volumes (13\tfrac13 base ×\times height), sphere (43πr3\tfrac43\pi r^3, surface 4πr24\pi r^2), and "unfold the surface" for shortest paths on boxes.
Complete Syllabus & Gap Fillers › Gap Fillers
AM–GM and when equality holds
For nonnegative aia_i: a1+⋯+ann≥a1⋯ann\dfrac{a_1+\cdots+a_n}{n} \ge \sqrt[n]{a_1\cdots a_n}, with equality exactly when all aia_i are equal.
Algebraic Weapons › Inequalities That Decide Problems
Angle bisectors as an equidistance locus
The bisectors of the angles between A1x+B1y+C1=0A_1x + B_1y + C_1 = 0 and A2x+B2y+C2=0A_2x + B_2y + C_2 = 0 satisfy A1x+B1y+C1A12+B12=±A2x+B2y+C2A22+B22\dfrac{A_1x + B_1y + C_1}{\sqrt{A_1^2+B_1^2}} = \pm\dfrac{A_2x + B_2y + C_2}{\sqrt{A_2^2+B_2^2}} — two perpendicular lines, one per sign.
Circles & Lines › The Bisector Between Two Lines
Arithmetic series: proof of Sn=n2(a1+an)S_n = \tfrac{n}{2}(a_1 + a_n)
Write the series forwards and backwards and add: each column gives a1+ana_1 + a_n (what one end gains, the other loses), so 2Sn=n(a1+an)2S_n = n(a_1 + a_n).
Where the Formulas Come From › The Series Formulas, Derived
Arrangements with repeated letters
n!n1! n2!⋯\dfrac{n!}{n_1!\,n_2!\cdots} for a word with repeated letters; divide by the arrangements of the identical items.
Counting: Inclusion–Exclusion & Stars and Bars › Distributions, Arrangements and Overcounting
Cauchy–Schwarz in the useful form
(a12+⋯+an2)(b12+⋯+bn2)≥(a1b1+⋯+anbn)2(a_1^2+\cdots+a_n^2)(b_1^2+\cdots+b_n^2) \ge (a_1b_1+\cdots+a_nb_n)^2.
Algebraic Weapons › Inequalities That Decide Problems
Circles and polygons
Inscribed angle =12= \tfrac12 arc; tangent ⊥\perp radius; power of a point PA⋅PB=PC⋅PD=PT2PA\cdot PB = PC\cdot PD = PT^2; chord length 2r2−d22\sqrt{r^2 - d^2}.
Formulas & Methods › Geometry
Circles: the four angle facts
Inscribed angle is half its arc; angles from the same arc are equal; a tangent makes a right angle with the radius; the angle between a tangent and a chord equals the inscribed angle on the other side.
Complete Syllabus & Gap Fillers › Gap Fillers
Congruence arithmetic
a≡b(modn)a\equiv b\pmod n means n∣(a−b)n\mid(a-b).
Number Theory: Modular Arithmetic › Working Modulo n
Coordinates and 3D
Distance, midpoint, slope; perpendicular slopes multiply to −1-1; point-to-line ∣ax0+by0+c∣a2+b2\tfrac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}; shoelace for polygon area.
Formulas & Methods › Geometry
Counting
Permutations, combinations, (nk)=(nn−k)\binom nk = \binom{n}{n-k}, Pascal (nk)=(n−1k−1)+(n−1k)\binom nk = \binom{n-1}{k-1} + \binom{n-1}{k}.
Formulas & Methods › Counting, Probability & Number Theory
Denest m+2n\sqrt{m + 2\sqrt n} and clear radicals with conjugates
If x+y=mx + y = m and xy=nxy = n with x≥y≥0x \ge y \ge 0, then m+2n=x+y\sqrt{m + 2\sqrt n} = \sqrt x + \sqrt y and m−2n=x−y\sqrt{m - 2\sqrt n} = \sqrt x - \sqrt y.
Algebraic Weapons › Radicals: Conjugates and Denesting
Derivation by completing the square
From ax2+bx+c=0ax^2 + bx + c = 0: divide by aa, move the constant, and complete the square: (x+b2a)2=b2−4ac4a2\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}.
Where the Formulas Come From › The Quadratic Formula, Derived
Derivation: the number of subsets is 2n2^n
A set with nn elements has 2n2^n subsets, and ∑k(nk)=2n\sum_{k}\dbinom nk = 2^n.
Where the Formulas Come From › Two More Derivations Worth Knowing
Derivation: the sum of the first nn squares
12+22+⋯+n2=n(n+1)(2n+1)61^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}.
Where the Formulas Come From › Two More Derivations Worth Knowing
Difference of squares, everywhere
a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) — the most-used identity in competition algebra.
Algebraic Weapons › Identities That Do the Work
Distributing identical objects
Nonnegative solutions of x1+⋯+xk=nx_1+\cdots+x_k=n number (n+k−1k−1)\dbinom{n+k-1}{k-1}; positive solutions number (n−1k−1)\dbinom{n-1}{k-1}.
Counting: Inclusion–Exclusion & Stars and Bars › Stars and Bars
Expected value, the fast way
Linearity: E[X+Y]=E[X]+E[Y]E[X + Y] = E[X] + E[Y] with NO independence needed.
Complete Syllabus & Gap Fillers › Gap Fillers
Exponents, radicals, logs (light)
aman=am+na^ma^n = a^{m+n}, (am)n=amn(a^m)^n = a^{mn}, a−n=1ana^{-n} = \tfrac{1}{a^n}, a1/n=ana^{1/n} = \sqrt[n]a.
Formulas & Methods › Algebra
Fermat and Euler for exponents
ap−1≡1(modp)a^{p-1} \equiv 1 \pmod p for a prime p∤ap \nmid a; more generally aφ(n)≡1(modn)a^{\varphi(n)} \equiv 1 \pmod n when gcd⁡(a,n)=1\gcd(a,n) = 1, with φ(100)=40\varphi(100) = 40.
Number Theory: Modular Arithmetic › Chinese Remainder and Last-Two-Digit Problems
Fermat's little theorem
If pp is prime and p∤ap\nmid a, then ap−1≡1(modp)a^{p-1}\equiv 1\pmod p.
Number Theory: Modular Arithmetic › Working Modulo n
Focus and directrix
The parabola x2=4pyx^2 = 4py has focus (0,p)(0, p) and directrix y=−py = -p; every point is equidistant from the two.
Circles & Lines › Circles Meet Parabolas
Geometric series: proof of Sn=a 1−rn1−rS_n = a\,\dfrac{1 - r^n}{1 - r}
Multiply the sum by rr and subtract: Sn−rSn=a−arnS_n - rS_n = a - ar^n — every middle term appears once in each copy and cancels.
Where the Formulas Come From › The Series Formulas, Derived
Intersecting a circle with a conic
Substitute the parabola into the circle and solve the resulting equation in u=x2u = x^2 (or yy): each POSITIVE root uu contributes two points (x=±ux = \pm\sqrt u), u=0u = 0 one point, and negative roots none.
Circles & Lines › Circles Meet Parabolas
Manipulate without solving
Symmetric expressions reduce to S=r+sS = r+s and P=rsP = rs: r2+s2=S2−2Pr^2+s^2 = S^2 - 2P and 1r+1s=SP\tfrac1r+\tfrac1s = \tfrac SP.
Algebra & Polynomials › Polynomial Roots
Mass points
Hang weights at the vertices so each side balances at its division point (weight ×\times length equal on both sides); the weights reveal all segment ratios.
Geometry: Power of a Point & Mass Points › Mass Points and Coordinates
Nearest point of a circle to a line
If the center is at distance d>rd > r from a line, the circle's nearest point to the line is at distance d−rd - r and the farthest at d+rd + r — both along the perpendicular from the center.
Circles & Lines › Circles: Chords, Tangents, Power
Number theory
Modular arithmetic; Fermat ap−1≡1a^{p-1} \equiv 1; Euler aφ(n)≡1a^{\varphi(n)} \equiv 1; φ(n)=n∏(1−1p)\varphi(n) = n\prod(1 - \tfrac1p).
Formulas & Methods › Counting, Probability & Number Theory
Probability and expected value
Conditional P(A∣B)=P(A∩B)P(B)P(A\mid B) = \tfrac{P(A\cap B)}{P(B)}; independent events multiply.
Formulas & Methods › Counting, Probability & Number Theory
Proof of Power of a Point (two chords)
Chords ABAB and CDCD meet at PP.
Where the Formulas Come From › Circle Power, Proved
Proof of the inscribed angle theorem
First the case where one chord is a diameter: the triangle from the center is isosceles (two radii), so the central angle, an exterior angle of it, equals the sum of two equal base angles — twice the inscribed angle.
Where the Formulas Come From › Circle Power, Proved
Proof of Vieta's formulas
If rr and ss are the roots, the quadratic factors as a(x−r)(x−s)=a(x2−(r+s)x+rs)a(x - r)(x - s) = a\left(x^2 - (r+s)x + rs\right).
Where the Formulas Come From › The Quadratic Formula, Derived
Proof: a+b2≥ab\dfrac{a + b}{2} \ge \sqrt{ab} for a,b≥0a, b \ge 0
Squares are never negative: (a−b)2≥0\left(\sqrt a - \sqrt b\right)^2 \ge 0.
Where the Formulas Come From › AM–GM, Proved
Quadratics and Vieta
x=−b±b2−4ac2ax = \tfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}; roots sum to −ba-\tfrac ba, multiply to ca\tfrac ca; vertex at x=−b2ax = -\tfrac{b}{2a}.
Formulas & Methods › Algebra
Ratios: lengths, areas, volumes
Similar figures with ratio kk have areas in ratio k2k^2 and volumes in ratio k3k^3.
Geometry: Power of a Point & Mass Points › Similar Triangles, Areas and the Shoelace Formula
Remainder and factor theorems
The remainder of P(x)P(x) divided by x−ax-a is P(a)P(a); so x−ax-a is a factor exactly when P(a)=0P(a) = 0.
Algebra & Polynomials › Polynomial Roots
Sequences and telescoping
Arithmetic: Sn=n(a1+an)2S_n = \tfrac{n(a_1 + a_n)}{2}.
Formulas & Methods › Algebra
Simon's Favourite Factoring Trick
xy+ax+by=c  ⟺  (x+b)(y+a)=c+abxy + ax + by = c \iff (x+b)(y+a) = c + ab: add the missing constant and factor.
Algebra & Polynomials › Vieta, Symmetric Sums and Clever Factoring
Squares have few remainders
Every square is 00 or 1(mod4)1 \pmod 4, and 00, 11 or 4(mod8)4 \pmod 8; mod 33 it is 00 or 11.
Number Theory: Modular Arithmetic › Squares, Remainders and Impossibility
Symmetric sums via Vieta
By Vieta's formulas, everything symmetric in two roots can be written using x+yx + y and xyxy: for instance x2+y2=(x+y)2−2xyx^2 + y^2 = (x+y)^2 - 2xy.
Algebraic Weapons › Identities That Do the Work
Symmetry shortcut
If swapping x↔yx \leftrightarrow y (or another visible reflection) exchanges the two lines, then the mirror line of that reflection is automatically one of the bisectors.
Circles & Lines › The Bisector Between Two Lines
Tangency is a distance statement
A line is tangent to a circle exactly when the center's distance to the line EQUALS the radius; it misses when the distance exceeds the radius, and cuts a chord when it is smaller.
Circles & Lines › Distance: Point to Line
The checklist
Everything on the AMC 8, plus: quadratics and Vieta; polynomial remainder/factor theorems; systems of equations; exponents and radicals; basic logarithms (rarely) and no trigonometry; modular arithmetic, bases, divisor counting; counting with inclusion–excl…
Complete Syllabus & Gap Fillers › What the AMC 10 Tests
The constant product
From a point PP, along any line through PP meeting a circle at A,BA,B, the product PA⋅PBPA\cdot PB is constant; for a tangent of length tt, t2=PA⋅PBt^2 = PA\cdot PB.
Geometry: Power of a Point & Mass Points › Power of a Point
The identity, and what it looks like in Pascal's triangle
Claim.
Identities and Techniques, Proved › The Hockey Stick Identity
The one move behind every identity in this topic
Almost every binomial identity a contest asks for is proved the same way: find a set of objects, count it one way, count it another way, and set the two answers equal.
Identities and Techniques, Proved › Counting the Same Thing Twice
The point-to-line distance formula
The distance from (x0,y0)(x_0, y_0) to the line Ax+By+C=0Ax + By + C = 0 is d=∣Ax0+By0+C∣A2+B2d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}; parallel lines Ax+By=C1Ax+By=C_1, Ax+By=C2Ax+By=C_2 sit ∣C1−C2∣A2+B2\dfrac{|C_1 - C_2|}{\sqrt{A^2+B^2}} apart.
Circles & Lines › Distance: Point to Line
The principle
∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|.
Counting: Inclusion–Exclusion & Stars and Bars › Inclusion–Exclusion
The shoelace formula
For vertices (x1,y1),…,(xn,yn)(x_1,y_1), \dots, (x_n,y_n) in order, the area is 12∣∑(xiyi+1−xi+1yi)∣\tfrac12\left|\sum (x_iy_{i+1} - x_{i+1}y_i)\right|.
Geometry: Power of a Point & Mass Points › Similar Triangles, Areas and the Shoelace Formula
The substitution u=x+1xu = x + \tfrac{1}{x}
From u=x+1xu = x + \tfrac1x: x2+1x2=u2−2x^2 + \tfrac{1}{x^2} = u^2 - 2 and x3+1x3=u3−3ux^3 + \tfrac{1}{x^3} = u^3 - 3u.
Algebraic Weapons › Identities That Do the Work
The three coordinate workhorses
Distance (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}; midpoint (x1+x22,y1+y22)\left(\tfrac{x_1+x_2}{2}, \tfrac{y_1+y_2}{2}\right); a circle (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.
Circles & Lines › Coordinates: Distance, Midpoint, and the Circle Through Three Points
The three right-triangle facts
Chord at distance dd from the center: length 2r2−d22\sqrt{r^2 - d^2}.
Circles & Lines › Circles: Chords, Tangents, Power
The universal problem-solving loop
Read twice.
Formulas & Methods › Geometry
Trade xx and yy for s=x+ys = x + y and p=xyp = xy
If a system is unchanged by swapping xx and yy, rewrite it in s=x+ys = x + y and p=xyp = xy using x2+y2=s2−2px^2 + y^2 = s^2 - 2p and x3+y3=s3−3psx^3 + y^3 = s^3 - 3ps.
Algebra & Polynomials › Symmetric Systems: Solve for the Sum and Product
Triangles
Heron: s(s−a)(s−b)(s−c)\sqrt{s(s-a)(s-b)(s-c)}.
Formulas & Methods › Geometry
Two congruences at once
If gcd⁡(m,n)=1\gcd(m,n) = 1, the pair x≡a(modm)x \equiv a \pmod m, x≡b(modn)x \equiv b \pmod n has exactly one solution modulo mnmn.
Number Theory: Modular Arithmetic › Chinese Remainder and Last-Two-Digit Problems
Using the equality case on purpose
AM–GM problems are usually about WHERE equality happens: to optimize, arrange the terms so they can all be equal, then read off the extreme value.
Where the Formulas Come From › AM–GM, Proved
Vieta for any degree
For a monic polynomial with roots r1,…,rnr_1,\dots,r_n: ∑ri=−an−1\sum r_i = -a_{n-1}, ∑i<jrirj=an−2\sum_{i<j} r_ir_j = a_{n-2}, and ∏ri=(−1)na0\prod r_i = (-1)^n a_0.
Algebra & Polynomials › Polynomial Roots
Vieta's relations
For ax2+bx+cax^2+bx+c: r1+r2=−bar_1+r_2 = -\tfrac ba, r1r2=car_1r_2 = \tfrac ca.
Algebra & Polynomials › Vieta, Symmetric Sums and Clever Factoring
When in doubt, use coordinates
Place the figure on axes; use the distance formula, perpendicular slopes, and the shoelace area formula.
Geometry: Power of a Point & Mass Points › Mass Points and Coordinates
When to place the axes yourself
A configuration with a right angle, a midpoint, or a line of symmetry can be placed on the axes for free: put the right angle at the origin and the legs on the axes.
Circles & Lines › Coordinates: Distance, Midpoint, and the Circle Through Three Points
Why the discriminant decides everything
In the derivation, the only square root taken is b2−4ac\sqrt{b^2 - 4ac}.
Where the Formulas Come From › The Quadratic Formula, Derived
Why the squares are the interesting case
The hockey stick adds binomial coefficients.
Identities and Techniques, Proved › Sums of Squares and Cubes, from Binomial Coefficients