Olympiad Toolkit

A growing collection of essential formulas, identities, and techniques for mathematical olympiad problem solving.

Showing 156 toolkits
  1. 1.
    1+2+⋯+n=n(n+1)21+2+\cdots+n=\frac{n(n+1)}{2}
  2. 2.
    12+22+⋯+n2=n(n+1)(2n+1)61^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}
  3. 3.
    13+23+⋯+n3=(n(n+1)2)21^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2
  4. 4.
    an=a1rn−1a_n=a_1r^{n-1}
    an2=an−1an+1a_n^2=a_{n-1}a_{n+1}
    Sn=a1+a1r+a1r2+⋯+a1rn−1=a1(rn−1)r−1S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}=\frac{a_1(r^n-1)}{r-1}
    ∣r∣<1:S∞=a1+a1r+a1r2+⋯=a11−r|r|<1:\quad S_\infty=a_1+a_1r+a_1r^2+\cdots=\frac{a_1}{1-r}
  5. 5.
    11⋅2+12⋅3+⋯+1n(n+1)=(1−12)+⋯+(1n−1n+1)=1−1n+1\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)}=\left(1-\frac{1}{2}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}
  6. 6.
    (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2
  7. 7.
    (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2
  8. 8.
    (a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3
  9. 9.
    (a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3
  10. 10.
    a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b)
  11. 11.
    a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)
  12. 12.
    a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)
  13. 13.
    an−bn=(a−b)(an−1+an−2b+⋯+bn−1)a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)
  14. 14.
    n odd:an+bn=(a+b)(an−1−an−2b+⋯+bn−1)n\text{ odd}:\quad a^n+b^n=(a+b)\left(a^{n-1}-a^{n-2}b+\cdots+b^{n-1}\right)
  15. 15.
    n even:an−bn=(a+b)(an−1−an−2b+⋯−bn−1)n\text{ even}:\quad a^n-b^n=(a+b)\left(a^{n-1}-a^{n-2}b+\cdots-b^{n-1}\right)
  16. 16.
    (kk)+(k+1k)+(k+2k)+⋯+(nk)=(n+1k+1)\binom{k}{k}+\binom{k+1}{k}+\binom{k+2}{k}+\cdots+\binom{n}{k}=\binom{n+1}{k+1}
  17. 17.
    (m0)(nk)+(m1)(nk−1)+(m2)(nk−2)+⋯+(mk)(n0)=(m+nk)\binom{m}{0}\binom{n}{k}+\binom{m}{1}\binom{n}{k-1}+\binom{m}{2}\binom{n}{k-2}+\cdots+\binom{m}{k}\binom{n}{0}=\binom{m+n}{k}
  18. 18.
    a3+b3+c3=(a+b+c)(a2+b2+c2−ab−ac−bc)+3abca^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc
  19. 19.
    x2+(a+b)x+ab=(x+a)(x+b)x^2+(a+b)x+ab=(x+a)(x+b)
  20. 20.
    (a+b)n=(n0)an+(n1)an−1b+(n2)an−2b2+⋯+(nn)bn(a+b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n
  21. 21.
    ax2+bx+c=0  ⇒  x=−b±b2−4ac2aax^2+bx+c=0\;\Rightarrow\;x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
  22. 22.
    Addition and Subtraction.\text{Addition and Subtraction.}
  23. 23.
    xy+ax+by=c  ⇒  (x+b)(y+a)=c+abxy+ax+by=c\;\Rightarrow\;(x+b)(y+a)=c+ab
    GSFFT (General Form)\text{GSFFT (General Form)}
    axy+bx+cy=daxy+bx+cy=d
    a2xy+abx+acy=ada^2xy+abx+acy=ad
    (ax+c)(ay+b)=ad+bc(ax+c)(ay+b)=ad+bc
  24. 24.
    (x1+x2+⋯+xn)2=x12+⋯+xn2+2x1x2+⋯+2xn−1xn(x_1+x_2+\cdots+x_n)^2=x_1^2+\cdots+x_n^2+2x_1x_2+\cdots+2x_{n-1}x_n
  25. 25.
    an=a1+(n−1)da_n=a_1+(n-1)d
    Sn=a1+a2+⋯+an=n2(a1+an)S_n=a_1+a_2+\cdots+a_n=\frac{n}{2}(a_1+a_n)
    an=an−1+an+12a_n=\frac{a_{n-1}+a_{n+1}}{2}
    Number of elements=last−firstd+1\text{Number of elements}=\frac{\text{last}-\text{first}}{d}+1
  26. 26.
    abc−ab−ac−bc+a+b+c−1=(a−1)(b−1)(c−1)abc-ab-ac-bc+a+b+c-1=(a-1)(b-1)(c-1)
  27. 27.
    a2+ab+ac+bc=(a+b)(a+c)a^2+ab+ac+bc=(a+b)(a+c)
  28. 28.
    x1,…,xn∈R+x_1,\dots,x_n\in\mathbb{R}^+
    max⁡{xi}≥x12+x22+⋯+xn2n≥x1+x2+⋯+xnn≥x1x2⋯xnn≥n1x1+1x2+⋯+1xn≥min⁡{xi}\max\{x_i\}\ge\sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}\ge\frac{x_1+x_2+\cdots+x_n}{n}\ge\sqrt[n]{x_1x_2\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}\ge\min\{x_i\}
    Equality holds if and only if all xi are equal.\text{Equality holds if and only if all }x_i\text{ are equal.}
  29. 29.
    [x]: Integer part{x}: Fractional part[x]:\text{ Integer part}\qquad\{x\}:\text{ Fractional part}
    We can write x as:\text{We can write }x\text{ as:}
    x=k+r,k∈Z,  r∈R,  0≤r<1x=k+r,\qquad k\in\mathbb{Z},\;r\in\mathbb{R},\;0\le r<1
  30. 30.
    ax2+bx+c=0ax^2+bx+c=0
    r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1r_2=\frac{c}{a}
    ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0
    r1+r2+r3=−bar_1+r_2+r_3=-\frac{b}{a}
    r1r2+r1r3+r2r3=car_1r_2+r_1r_3+r_2r_3=\frac{c}{a}
    r1r2r3=−dar_1r_2r_3=-\frac{d}{a}
    anxn+an−1xn−1+⋯+a1x+a0=0a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0
    r1+r2+⋯+rn=−an−1anr_1+r_2+\cdots+r_n=-\frac{a_{n-1}}{a_n}
    r1r2+r1r3+⋯+rn−1rn=an−2anr_1r_2+r_1r_3+\cdots+r_{n-1}r_n=\frac{a_{n-2}}{a_n}
    ⋯\cdots
    r1r2⋯rn−1+⋯+r2r3⋯rn=(−1)n−1a1anr_1r_2\cdots r_{n-1}+\cdots+r_2r_3\cdots r_n=(-1)^{n-1}\frac{a_1}{a_n}
    r1r2⋯rn=(−1)na0anr_1r_2\cdots r_n=(-1)^n\frac{a_0}{a_n}
  31. 31.
    f(x)=ax2+bx+c,Δ=b2−4acf(x)=ax^2+bx+c,\qquad \Delta=b^2-4ac
    ConditionNumber of Real RootsSign of f(x)f(x)
    Δ>0\Delta>02
    sign(a)0−sign(a)0sign(a)r₁r₂
    Δ=0\Delta=01
    sign(a)0sign(a)r
    Δ<0\Delta<00
    sign(a)
  32. 32.
    A(x)÷B(x):  A(x)=B(x)Q(x)+R(x),deg⁡(R(x))<deg⁡(B(x))A(x)\div B(x):\;A(x)=B(x)Q(x)+R(x),\quad \deg(R(x))<\deg(B(x))
    ⇒ If B(x)=0, then A(x)=R(x)\Rightarrow\text{ If }B(x)=0,\text{ then }A(x)=R(x)
  33. 33.
    n=p1α1p2α2⋯pkαkn=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}
    Number of positive divisors of n:  (α1+1)(α2+1)⋯(αk+1)\text{Number of positive divisors of }n:\;(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_k+1)
  34. 34.
    n=p1α1p2α2⋯pkαkn=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}
    (1+p1+p12+⋯+p1α1)(1+p2+p22+⋯+p2α2)⋯(1+pk+pk2+⋯+pkαk)(1+p_1+p_1^2+\cdots+p_1^{\alpha_1})(1+p_2+p_2^2+\cdots+p_2^{\alpha_2})\cdots(1+p_k+p_k^2+\cdots+p_k^{\alpha_k})
    =p1α1+1−1p1−1×p2α2+1−1p2−1×⋯×pkαk+1−1pk−1=\frac{p_1^{\alpha_1+1}-1}{p_1-1}\times\frac{p_2^{\alpha_2+1}-1}{p_2-1}\times\cdots\times\frac{p_k^{\alpha_k+1}-1}{p_k-1}
  35. 35.
    n=p1α1p2α2⋯pkαkn=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}
    t=(α1+1)(α2+1)⋯(αk+1)t=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_k+1)
    Product of positive divisors of n:  nt/2\text{Product of positive divisors of }n:\;n^{t/2}
  36. 36.
    sin⁡α=sin⁡β  ⇒  α=β+360k,  180−β+360k,  k∈Z\sin\alpha=\sin\beta\;\Rightarrow\;\alpha=\beta+360k,\;180-\beta+360k,\;k\in\mathbb{Z}
    cos⁡α=cos⁡β  ⇒  α=±β+360k,  k∈Z\cos\alpha=\cos\beta\;\Rightarrow\;\alpha=\pm\beta+360k,\;k\in\mathbb{Z}
    tan⁡α=tan⁡β  ⇒  α=β+180k,  k∈Z\tan\alpha=\tan\beta\;\Rightarrow\;\alpha=\beta+180k,\;k\in\mathbb{Z}
    cot⁡α=cot⁡β  ⇒  α=β+180k,  k∈Z\cot\alpha=\cot\beta\;\Rightarrow\;\alpha=\beta+180k,\;k\in\mathbb{Z}
  37. 37.
    sin⁡(90∘−α)=cos⁡α\sin(90^\circ-\alpha)=\cos\alpha
    cos⁡(90∘−α)=sin⁡α\cos(90^\circ-\alpha)=\sin\alpha
    sin⁡(180∘−α)=sin⁡α\sin(180^\circ-\alpha)=\sin\alpha
    cos⁡(180∘−α)=−cos⁡α\cos(180^\circ-\alpha)=-\cos\alpha
    sin⁡(−α)=−sin⁡α\sin(-\alpha)=-\sin\alpha
    cos⁡(−α)=cos⁡α\cos(-\alpha)=\cos\alpha
    ⋯\cdots
  38. 38.
    sin⁡2α+cos⁡2α=1\sin^2\alpha+\cos^2\alpha=1
    sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
    cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
    sin⁡2α=2sin⁡αcos⁡α\sin2\alpha=2\sin\alpha\cos\alpha
    cos⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha
    ⋯\cdots
  39. 39.
    sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
    sin⁡α−sin⁡β=2cos⁡(α+β2)sin⁡(α−β2)\sin\alpha-\sin\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)
    cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
    cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)\cos\alpha-\cos\beta=-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)
  40. 40.
    sin⁡pcos⁡q=12(sin⁡(p+q)+sin⁡(p−q))\sin p\cos q=\frac{1}{2}(\sin(p+q)+\sin(p-q))
    cos⁡pcos⁡q=12(cos⁡(p+q)+cos⁡(p−q))\cos p\cos q=\frac{1}{2}(\cos(p+q)+\cos(p-q))
    sin⁡psin⁡q=12(cos⁡(p−q)−cos⁡(p+q))\sin p\sin q=\frac{1}{2}(\cos(p-q)-\cos(p+q))
  41. 41.
    Dual Pythagorean Theorem figure
    CB2−CA2=EB2−EA2=FB2−FA2=DB2−DA2CB^2-CA^2=EB^2-EA^2=FB^2-FA^2=DB^2-DA^2
  42. 42.
    Angle Bisector Theorem figure
    BDDC=cb\frac{BD}{DC}=\frac{c}{b}
  43. 43.
    5x+1(x−1)(x+2)=Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}
    Method 1: Comparing coefficients\text{Method 1: Comparing coefficients}
    Method 2: Heaviside Cover-Up Method\text{Method 2: Heaviside Cover-Up Method}
  44. 44.
  45. 45.
    Treat adjacent objects as one block\text{Treat adjacent objects as one block}
    Example: logarithm\text{Example: }\boxed{\mathrm{log}}\mathrm{arithm}
    7!×3!7!\times3!
  46. 46.

    Definition

    a∣b  ⟺  there exists an integer k s.t. b=aka\mid b\iff \text{there exists an integer }k\text{ s.t. }b=ak

    Properties

    1.1.
    a∣b,  b∣c⟹a∣ca\mid b,\; b\mid c \Longrightarrow a\mid c
    2.2.
    a∣b,  a∣c⟹a∣mb+nca\mid b,\; a\mid c \Longrightarrow a\mid mb+nc
    3.3.
    gcd⁡(a,b)=1,  a∣bc⟹a∣c\gcd(a,b)=1,\; a\mid bc \Longrightarrow a\mid c
    4.4.
    p∣ab⟹p∣a or p∣b,p prime.p\mid ab \Longrightarrow p\mid a \text{ or } p\mid b,\qquad p\text{ prime.}
  47. 47.

    Definition

    a≡b(modm)  ⟺  m∣(a−b)a\equiv b\pmod m\iff m\mid(a-b)

    Properties

    1.1.
    a≡b(modm),  b≡c(modm)⟹a≡c(modm)a\equiv b\pmod m,\; b\equiv c\pmod m\Longrightarrow a\equiv c\pmod m
    2.2.
    a≡b(modm)⟹an≡bn(modm),n∈Z+a\equiv b\pmod m\Longrightarrow a^n\equiv b^n\pmod m,\quad n\in\mathbb Z^+
    3.3.
    ac≡bc(modm)  ⟺  a≡b(modmgcd⁡(m,c))ac\equiv bc\pmod m\iff a\equiv b\pmod{\frac{m}{\gcd(m,c)}}
    4.4.
    a≡b(modm),  d∣m⟹a≡b(modd)a\equiv b\pmod m,\; d\mid m\Longrightarrow a\equiv b\pmod d
  48. 48.

    Two Sets

    ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|

    Three Sets

    ∣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|

    General Formula

    ∣A1∪⋯∪An∣=∑i∣Ai∣−∑i<j∣Ai∩Aj∣+∑i<j<k∣Ai∩Aj∩Ak∣−⋯|A_1\cup\cdots\cup A_n|=\sum_i|A_i|-\sum_{i<j}|A_i\cap A_j|+\sum_{i<j<k}|A_i\cap A_j\cap A_k|-\cdots

    Exactly One Set

    PA1,A2=∣A1∣+∣A2∣−2∣A1∩A2∣P_{A_1,A_2}=|A_1|+|A_2|-2|A_1\cap A_2|
    PA1,A2,A3=∣A1∣+∣A2∣+∣A3∣−2∣A1∩A2∣−2∣A1∩A3∣−2∣A2∩A3∣+3∣A1∩A2∩A3∣P_{A_1,A_2,A_3}=|A_1|+|A_2|+|A_3|-2|A_1\cap A_2|-2|A_1\cap A_3|-2|A_2\cap A_3|+3|A_1\cap A_2\cap A_3|
  49. 49.
    φ(n)=∣{1≤i≤n∣gcd⁡(i,n)=1}∣\varphi(n)=|\{1\le i\le n\mid\gcd(i,n)=1\}|

    φ(n)\varphi(n) is the number of integers from 11 to nn that are relatively prime to nn.

    If n=p1α1p2α2⋯pkαkn=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k},

    φ(n)=n(1−1p1)(1−1p2)⋯(1−1pk)\varphi(n)=n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\left(1-\frac{1}{p_k}\right)
  50. 50.

    Let nn be a positive integer. If gcd⁡(n,a)=1\gcd(n,a)=1,

    aφ(n)≡1(modn)a^{\varphi(n)}\equiv1\pmod n
  51. 51.
    TransformationExample
    Horizontal Shift Right 3y=f(x−3)y=f(x-3)
    Horizontal Shift Left 3y=f(x+3)y=f(x+3)
    Vertical Shift Up 3y=f(x)+3y=f(x)+3
    Vertical Shift Down 3y=f(x)−3y=f(x)-3
    Horizontal Stretch by Factor 2y=f(x/2)y=f(x/2)
    Horizontal Compression by Factor 2y=f(2x)y=f(2x)
    Vertical Stretch by Factor 2y=2f(x)y=2f(x)
    Vertical Compression by Factor 2y=12f(x)y=\frac{1}{2}f(x)
    Reflection in x-axisy=−f(x)y=-f(x)
    Reflection in y-axisy=f(−x)y=f(-x)
  52. 52.

    Triangle Area Formulas

    1.1.
    [ABC]=12Base⋅Height[ABC]=\frac12\text{Base}\cdot\text{Height}
    2.2.
    [ABC]=12bcsin⁡A[ABC]=\frac12 bc\sin A
    3.3.
    s=a+b+c2,[ABC]=s(s−a)(s−b)(s−c)s=\frac{a+b+c}{2},\qquad [ABC]=\sqrt{s(s-a)(s-b)(s-c)}
    4.4.
    [ABC]=a234(equilateral triangle)[ABC]=\frac{a^2\sqrt3}{4}\qquad\text{(equilateral triangle)}

    Quadrilateral Area Formulas

    1.1.
    [ABCD]=12AC⋅BDsin⁡α[ABCD]=\frac12 AC\cdot BD\sin\alpha
    2.2.
    [ABCD]=h(AB+CD)2(trapezoid)[ABCD]=\frac{h(AB+CD)}{2}\qquad\text{(trapezoid)}
    3.3.
    [ABCD]=bh(parallelogram)[ABCD]=bh\qquad\text{(parallelogram)}

    Circle Formulas

    1.1.
    A=πr2,C=2πrA=\pi r^2,\qquad C=2\pi r
    2.2.
    Asector=α360πr2A_{\text{sector}}=\frac{\alpha}{360}\pi r^2
    3.3.
    ℓarc=α360⋅2πr\ell_{\text{arc}}=\frac{\alpha}{360}\cdot2\pi r
  53. 53.
    Polygon with vertices A_1 through A_n
    Area=12∣x1y2+x2y3+⋯+xn−1yn+xny1−y1x2−y2x3−⋯−yn−1xn−ynx1∣\text{Area}=\tfrac{1}{2}\bigl|x_1y_2+x_2y_3+\cdots+x_{n-1}y_n+x_ny_1-y_1x_2-y_2x_3-\cdots-y_{n-1}x_n-y_nx_1\bigr|
  54. 54.

    Pair up the elements:

    Gauss pairing 1+100, 2+99, ... equals 50 x 101

    or

    S=1+2+3+⋯+98+99+100S=1+2+3+\cdots+98+99+100
    S=100+99+98+⋯+3+2+1S=100+99+98+\cdots+3+2+1
    2S=101+101+⋯+1012S=101+101+\cdots+101
    2S=100×1012S=100\times101

    Therefore:

    S=100×1012S=\frac{100\times101}{2}
  55. 55.
    θ\thetasin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\thetacot⁡θ\cot\theta
    0∘0^\circ001100Undefined
    30∘30^\circ12\tfrac{1}{2}32\tfrac{\sqrt3}{2}33\tfrac{\sqrt3}{3}3\sqrt3
    45∘45^\circ22\tfrac{\sqrt2}{2}22\tfrac{\sqrt2}{2}1111
    60∘60^\circ32\tfrac{\sqrt3}{2}12\tfrac{1}{2}3\sqrt333\tfrac{\sqrt3}{3}
    90∘90^\circ1100Undefined00
    180∘180^\circ00−1-100Undefined
    270∘270^\circ−1-100Undefined00

    15° and 75°

    Anglesin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\thetacot⁡θ\cot\theta
    15∘15^\circ6−24\dfrac{\sqrt{6}-\sqrt{2}}{4}6+24\dfrac{\sqrt{6}+\sqrt{2}}{4}2−32-\sqrt{3}2+32+\sqrt{3}
    75∘75^\circ6+24\dfrac{\sqrt{6}+\sqrt{2}}{4}6−24\dfrac{\sqrt{6}-\sqrt{2}}{4}2+32+\sqrt{3}2−32-\sqrt{3}

    18°, 36°, 54°, and 72°

    Anglesin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\thetacot⁡θ\cot\theta
    18∘18^\circ5−14\dfrac{\sqrt{5}-1}{4}10+254\dfrac{\sqrt{10+2\sqrt{5}}}{4}15+25\dfrac{1}{\sqrt{5+2\sqrt{5}}}5+25\sqrt{5+2\sqrt{5}}
    36∘36^\circ10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4}1+54\dfrac{1+\sqrt{5}}{4}5−25\sqrt{5-2\sqrt{5}}15−25\dfrac{1}{\sqrt{5-2\sqrt{5}}}
    54∘54^\circ1+54\dfrac{1+\sqrt{5}}{4}10−254\dfrac{\sqrt{10-2\sqrt{5}}}{4}15−25\dfrac{1}{\sqrt{5-2\sqrt{5}}}5−25\sqrt{5-2\sqrt{5}}
    72∘72^\circ10+254\dfrac{\sqrt{10+2\sqrt{5}}}{4}5−14\dfrac{\sqrt{5}-1}{4}5+25\sqrt{5+2\sqrt{5}}15+25\dfrac{1}{\sqrt{5+2\sqrt{5}}}
  56. 56.
    Triangle ABC with sides a, b, c
    a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A
  57. 57.
    Cyclic quadrilateral ABCD with diagonals

    A quadrilateral ABCDABCD is cyclic if and only if:

    AB⋅CD+AD⋅BC=AC⋅BDAB\cdot CD+AD\cdot BC=AC\cdot BD
  58. 58.
    Central angle
    Inscribed angle
    Tangent-chord angle
    Interior angle
    Exterior angle
  59. 59.
    Opposite angles
    Opposite angles
    1.∠A+∠C=180∘1.\quad \angle A+\angle C=180^\circ
    Equal angles
    2.∠A1=∠C12.\quad \angle A_1=\angle C_1
    Intersecting chords
    3.PA⋅PC=PB⋅PD3.\quad PA\cdot PC=PB\cdot PD
    Power of a point
    4.PA⋅PD=PB⋅PC4.\quad PA\cdot PD=PB\cdot PC
  60. 60.
    Right triangle ABC with right angle at A and angle alpha at C
    sin⁡α=Opposite SideHypotenuse=ABBC\sin\alpha=\frac{\text{Opposite Side}}{\text{Hypotenuse}}=\frac{AB}{BC}
    cos⁡α=Adjacent SideHypotenuse=ACBC\cos\alpha=\frac{\text{Adjacent Side}}{\text{Hypotenuse}}=\frac{AC}{BC}
    tan⁡α=Opposite SideAdjacent Side=ABAC\tan\alpha=\frac{\text{Opposite Side}}{\text{Adjacent Side}}=\frac{AB}{AC}
    cot⁡α=Adjacent SideOpposite Side=ACAB\cot\alpha=\frac{\text{Adjacent Side}}{\text{Opposite Side}}=\frac{AC}{AB}
    Unit circle showing sin alpha and cos alpha
    Unit circle showing tan alpha and cot alpha
    Unit circle showing tan alpha and cot alpha
  61. 61.
    3,9:n≡(sum of digits)3,9:\quad n\equiv\text{(sum of digits)}
    11:n≡(alternating sum of digits)11:\quad n\equiv\text{(alternating sum of digits)}
    2,5,10:last digit2,5,10:\quad \text{last digit}
    4,25,100:last 2 digits4,25,100:\quad \text{last 2 digits}
    8,125,1000:last 3 digits8,125,1000:\quad \text{last 3 digits}
  62. 62.

    Constraints can be based on the number of elements, on the size of the elements, or on other conditions.

    But usually, if you base your answer on the most limiting case, it will make the problem easier to solve.

  63. 63.
    ∣ ∣ ∣a∣−∣b∣+∣c∣−∣d∣ ∣ ∣≡a+b+c+d(mod2)\bigl|\,|\,|a|-|b|+|c|-|d|\,|\,\bigr|\equiv a+b+c+d\pmod{2}
  64. 64.
    Trapezoid ABCD with altitudes drawn from A and B to base DC

    When working with a trapezoid, drawing the altitudes creates right triangles and often simplifies the problem.

    Isosceles trapezoid with altitudes and equal side lengths
    x=CD−AB2x=\frac{CD-AB}{2}
  65. 65.
    Tangent and radius
    1.OP⊥tangent1.\quad OP\perp\text{tangent}
    Two tangents
    2.AB=AC2.\quad AB=AC
    Externally tangent circles
    3.O1O2=R1+R23.\quad O_1O_2=R_1+R_2
    Internally tangent circles
    4.O1O2=R2−R14.\quad O_1O_2=R_2-R_1
  66. 66.

    1) Square it.

    2) Use a changing variable.

  67. 67.

    1 dimension

    0≤x≤10\le x\le 1
    P ⁣(13<x<12)=12−13=16P\!\left(\tfrac{1}{3}<x<\tfrac{1}{2}\right)=\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{1}{6}
    Number line from 0 to 1 with segment between 1/3 and 1/2 highlighted

    2 dimensions

    0≤x,y≤10\le x,y\le 1
    P(∣x−y∣≥12)=2×12×122=14P(|x-y|\ge \tfrac{1}{2})=2\times\frac{\tfrac{1}{2}\times\tfrac{1}{2}}{2}=\tfrac{1}{4}
    Unit square with two shaded triangles representing |x-y| >= 1/2

    3 dimensions

    0≤x,y,z≤10\le x,y,z\le 1
    P(x+y+z≤1)=12×12×123=16P(x+y+z\le 1)=\frac{\tfrac{1}{2}\times\tfrac{1}{2}\times\tfrac{1}{2}}{3}=\tfrac{1}{6}
    Unit cube with tetrahedron representing x+y+z <= 1 shaded
  68. 68.
    Stars and Bars
    x1+x2+⋯+xk=n,xi≥1x_1+x_2+\cdots+x_k=n,\qquad x_i\ge1
    (n−1k−1)\binom{n-1}{k-1}
    x1+x2+⋯+xk=n,xi≥aix_1+x_2+\cdots+x_k=n,\qquad x_i\ge a_i
    xi′=xi−ai+1x_i'=x_i-a_i+1
    (n−∑ai+k−1k−1)\binom{n-\sum a_i+k-1}{k-1}
    x1+x2+⋯+xk≤n,xi≥1x_1+x_2+\cdots+x_k\le n,\qquad x_i\ge1
    (nk)\binom{n}{k}
    xi≤M  ⇒  Total−Badx_i\le M\;\Rightarrow\;\text{Total}-\text{Bad}
  69. 69.
    A,B⊆{1,2,…,n}A,B\subseteq\{1,2,\ldots,n\}
    A∩B=∅A\cap B=\varnothing

    Instead of considering the subsets, we should consider the elements.

    Each element has 3 possible choices for AA and BB:

    AB✓✓✓××✓\begin{array}{c|c} A & B\\ \hline \checkmark & \checkmark\\ \checkmark & \times\\ \times & \checkmark \end{array}

    Therefore,

    3⋅3⋅3⋯3⏟n times=3n.\underbrace{3\cdot 3\cdot 3\cdots 3}_{n\text{ times}}=3^n.
  70. 70.
    1.an=an−1+f(n)1.\quad a_n=a_{n-1}+f(n)
    Write an−an−1=f(n) and telescope.\text{Write }a_n-a_{n-1}=f(n)\text{ and telescope.}
    2.an=can−1+f(n)2.\quad a_n=ca_{n-1}+f(n)
    bn=ancnb_n=\frac{a_n}{c^n}
    Then telescope bn.\text{Then telescope }b_n.
    3.an=nan−1+f(n)3.\quad a_n=na_{n-1}+f(n)
    bn=ann!b_n=\frac{a_n}{n!}
    Then telescope bn.\text{Then telescope }b_n.
  71. 71.

    Start with:

    1⋅21+2⋅22+3⋅23+⋯+(n−1)2n−1+n 2n.1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots+(n-1)2^{n-1}+n\,2^n.

    Write the following equations in a triangular arrangement:

    21+22+23+⋯+2n−1+2n ⟶ 2n+1−21  22+23+⋯+2n−1+2n ⟶ 2n+1−22  23+⋯+2n−1+2n ⟶ 2n+1−23  ⋮  2n−1+2n ⟶ 2n+1−2n−1 2n ⟶ 2n+1−2n\begin{array}{l} 2^1+2^2+2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^1\\[2pt] \qquad\ \ 2^2+2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^2\\[2pt] \qquad\qquad\ \ 2^3+\cdots+2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^3\\[2pt] \qquad\qquad\qquad\qquad\ \ \vdots\\[2pt] \qquad\qquad\qquad\qquad\qquad\ \ 2^{n-1}+2^n\ \longrightarrow\ 2^{n+1}-2^{n-1}\\[2pt] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\ 2^n\ \longrightarrow\ 2^{n+1}-2^n \end{array}

    Then add them:

    =n 2n+1−(21+22+23+⋯+2n−1+2n).=n\,2^{n+1}-\left(2^1+2^2+2^3+\cdots+2^{n-1}+2^n\right).

    Since

    21+22+⋯+2n−1+2n=2n+1−21,2^1+2^2+\cdots+2^{n-1}+2^n=2^{n+1}-2^1,

    we get

    =n 2n+1−(2n+1−21)=n\,2^{n+1}-(2^{n+1}-2^1)
    =(n−1)2n+1+21.=\boxed{(n-1)2^{n+1}+2^1}.
  72. 72.
  73. 73.
    Placing numbers
    1.Each point =sum of its predecessor(s).1.\quad \text{Each point }=\text{sum of its predecessor(s).}
    Permutation
    2.#paths=(R+U)!R! U!2.\quad \#\text{paths}=\frac{(R+U)!}{R!\,U!}
    Levels
    3.Choose the transition point at each level.3.\quad \text{Choose the transition point at each level.}
  74. 74.

    30-60-90 Triangle

    30-60-90 right triangle
    Sides are a, a32, a2\text{Sides are } a,\ \frac{a\sqrt3}{2},\ \frac{a}{2}

    45-45-90 Triangle

    45-45-90 right isosceles triangle
    Sides are a22, a22, a\text{Sides are }\frac{a\sqrt2}{2},\ \frac{a\sqrt2}{2},\ a
  75. 75.

    aa and bb are positive integers.

    78.1

    If ab=x2ab=x^2 and gcd⁡(a,b)=1\gcd(a,b)=1, then a=y2, b=z2a=y^2,\ b=z^2.

    78.2

    If ab=x2ab=x^2 and gcd⁡(a,b)=d\gcd(a,b)=d, then a=dy2, b=dz2a=dy^2,\ b=dz^2.


    pp is a prime number.

    78.3

    If ab=px2ab=px^2 and gcd⁡(a,b)=1\gcd(a,b)=1, then a=py2, b=z2a=py^2,\ b=z^2 or a=y2, b=pz2a=y^2,\ b=pz^2.

    78.4

    If ab=px2ab=px^2 and gcd⁡(a,b)=d\gcd(a,b)=d, then a=dpy2, b=dz2a=dpy^2,\ b=dz^2 or a=dy2, b=dpz2a=dy^2,\ b=dpz^2.

  76. 76.
    gcd⁡(a+kb,b)=gcd⁡(a,b)\gcd(a+kb,b)=\gcd(a,b)
    gcd⁡(a,b)=gcd⁡(a mod b,b)\gcd(a,b)=\gcd(a\bmod b,b)
  77. 77.

    Equation of a Circle

    Circle with center and radius
    A circle with center (x0,y0) and radius r:\text{A circle with center }(x_0,y_0)\text{ and radius }r:
    (x−x0)2+(y−y0)2=r2(x-x_0)^2+(y-y_0)^2=r^2

    Finding the Equation of a Tangent Line

    Tangent line through an external point to a circle

    Let the tangent line through the point P(x₀, y₀) have slope m.

    ℓ: y−y0=m(x−x0)\ell:\ y-y_0=m(x-x_0)
    (x−x1)2+(y−y1)2=r2(x-x_1)^2+(y-y_1)^2=r^2

    Substitute the line equation into the circle equation. Since a tangent line intersects the circle at exactly one point, set the discriminant equal to zero and solve for m.

    Δ=0\Delta=0
  78. 78.

    Example

    3ab+2=6b,3bc+2=5c,3ca+2=4a3ab+2=6b,\qquad 3bc+2=5c,\qquad 3ca+2=4a

    Express two variables in terms of the third:

    c=4a−23a,b=26−3ac=\frac{4a-2}{3a},\qquad b=\frac{2}{6-3a}

    Substitute into the remaining equation:

    (7a−8)(a−1)=0(7a-8)(a-1)=0

    Substitute the solutions back:

    a=1⇒abc=49a=1\Rightarrow abc=\frac49
    a=87⇒abc=23a=\frac87\Rightarrow abc=\frac23
  79. 79.

    Cube

    Cube
    TSA=6a2\mathrm{TSA}=6a^2
    V=a3V=a^3

    Rectangular Prism

    Cuboid
    TSA=2(ab+ac+bc)\mathrm{TSA}=2(ab+ac+bc)
    V=abcV=abc

    Prism

    Prism
    V=h⋅AreabaseV=h\cdot\mathrm{Area}_{\mathrm{base}}

    Sphere

    Sphere
    TSA=4πr2\mathrm{TSA}=4\pi r^2
    V=43πr3V=\frac43\pi r^3

    Cylinder

    Cylinder
    TSA=2πr(r+h)\mathrm{TSA}=2\pi r(r+h)
    V=πr2hV=\pi r^2h

    Cone

    Cone
    TSA=πr(r+l)\mathrm{TSA}=\pi r(r+l)
    l=r2+h2l=\sqrt{r^2+h^2}
    V=13πr2hV=\frac13\pi r^2h

    Pyramid

    Pyramid
    V=13Areabase⋅hV=\frac13\mathrm{Area}_{\mathrm{base}}\cdot h

    Regular Tetrahedron

    Regular tetrahedron
    TSA=a23\mathrm{TSA}=a^2\sqrt3
    V=a3212V=\frac{a^3\sqrt2}{12}

    Triangular Prism

    Triangular prism with triangular base and height
    V=12Areabase×LV=\frac12\text{Area}_{\text{base}}\times L
  80. 80.
    Diagonal of an m by n grid crossing lattice lines

    If you draw the diagonal of an m×nm\times n grid, then the diagonal is split into m+n−gcd⁡(m,n)m+n-\gcd(m,n) parts.

  81. 81.
    Triangle ABC with cevian AP
    [APB][APC]=BPPC\frac{[APB]}{[APC]}=\frac{BP}{PC}
    Triangle ABC with point L on cevian AP
    [ALB][ALC]=BPPC\frac{[ALB]}{[ALC]}=\frac{BP}{PC}
  82. 82.
    Menelaus configuration with transversal cutting triangle ABC
    Menelaus configuration with transversal cutting extensions
    DBDC⋅ECEA⋅FAFB=1\frac{DB}{DC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1

    if and only if D,E,FD,E,F are collinear.

  83. 83.

    For an nn-sided polygon,

    Sum of interior angles=180∘(n−2)\text{Sum of interior angles}=180^\circ(n-2)

    Regular Hexagon

    For a regular hexagon with side length aa:

    a)

    ∠A=180∘(6−2)6=120∘\angle A=\frac{180^\circ(6-2)}{6}=120^\circ

    b)

    Regular hexagon divided into six equilateral triangles
    [ABCDEF]=6(a234)=3a232[ABCDEF]=6\left(\frac{a^2\sqrt{3}}{4}\right)=\frac{3a^2\sqrt{3}}{2}

    c)

    Regular hexagon with diagonal AC of length a√3
    AC=a3AC=a\sqrt{3}

    d)

    Regular hexagon with long diagonal AD of length 2a
    AD=2aAD=2a

    Regular Octagon

    For a regular octagon with side length aa:

    a)

    ∠A=180∘(8−2)8=135∘\angle A=\frac{180^\circ(8-2)}{8}=135^\circ

    b)

    Regular octagon inscribed in a square of side a+a√2
    [ABCDEFGH]=(a+a2)2−4(a2/22)[ABCDEFGH]=(a+a\sqrt{2})^2-4\left(\frac{a^2/2}{2}\right)
    =a2+2a2+2a22−a2=2a2(1+2)=a^2+2a^2+2a^2\sqrt{2}-a^2=2a^2(1+\sqrt{2})

    c)

    Regular octagon with diagonal AC and 135° angle at B
    AC=a2+2AC=a\sqrt{2+\sqrt{2}}
    AC2=a2+a2−2a2cos⁡135∘=2a2+2a222AC^2=a^2+a^2-2a^2\cos 135^\circ=2a^2+2a^2\frac{\sqrt{2}}{2}
    =2a2+a22=a2(2+2)=2a^2+a^2\sqrt{2}=a^2(2+\sqrt{2})

    d)

    Regular octagon with diagonal AD of length a(√2+1)
    AD=a2+a=a(2+1)AD=a\sqrt{2}+a=a(\sqrt{2}+1)

    e)

    Regular octagon with diagonal AE and right triangle ADE
    AE=a4+22AE=a\sqrt{4+2\sqrt{2}}
    AE2=a2+AD2=a2+(a(2+1))2AE^2=a^2+AD^2=a^2+\left(a(\sqrt{2}+1)\right)^2
    =a2+a2(3+22)=a2(4+22)=a^2+a^2(3+2\sqrt{2})=a^2(4+2\sqrt{2})
  84. 84.
    v=dtv=\frac{d}{t}
    v: speedd: distancet: timev:\text{ speed}\qquad d:\text{ distance}\qquad t:\text{ time}
  85. 85.
    Triangle ABC with side lengths a, b, and c
    a<b+ca<b+c
    b<a+cb<a+c
    c<a+bc<a+b
  86. 86.

    Mean

    Mean=Sum of ValuesNumber of Values\text{Mean}=\frac{\text{Sum of Values}}{\text{Number of Values}}
    Sum=Mean×Number of Values\text{Sum}=\text{Mean}\times\text{Number of Values}

    Median, Mode, and Range

    Median: the middle value after sorting

    Mode: the value that occurs most frequently

    Range=Maximum−Minimum\text{Range}=\text{Maximum}-\text{Minimum}

    Effect of Transforming Every Value

    OperationMeanMedianModeRange
    +k+k+k+k+k+k+k+kSame
    −k-k−k-k−k-k−k-kSame
    ×k\times k×k\times k×k\times k×k\times k×∣k∣\times |k|
    ÷k, k≠0\div k,\ k\ne0÷k\div k÷k\div k÷k\div k÷∣k∣\div |k|
  87. 88.
    If f is continuous on [a,b] and f(a)f(b)<0, then ∃ c∈(a,b) such that f(c)=0.\text{If }f\text{ is continuous on }[a,b]\text{ and }f(a)f(b)<0,\text{ then }\exists\,c\in(a,b)\text{ such that }f(c)=0.
  88. 89.
    A polynomial of degree n with real coefficients has at most n real roots.\text{A polynomial of degree }n\text{ with real coefficients has at most }n\text{ real roots.}
  89. 90.
    Similar triangles
    △ABC∼△A′B′C′\triangle ABC\sim\triangle A'B'C'
    AA\text{AA}
    SAS\text{SAS}
    SSS\text{SSS}
  90. 91.
    E(X)=∑xx Pr⁡(X=x)E(X)=\sum_x x\,\Pr(X=x)
  91. 92.

    Slope, Line Equation, and Distance

    Two points on a line with slope, line equation, and distance formulas
    mAB=y2−y1x2−x1m_{AB}=\frac{y_2-y_1}{x_2-x_1}
    y−y1=m(x−x1)y-y_1=m(x-x_1)
    AB=(x2−x1)2+(y2−y1)2AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
    Perpendicular lines with slopes m1 and m2
    Perpendicular lines:m1m2=−1\text{Perpendicular lines:}\qquad m_1m_2=-1

    Distance from a Point to a Line

    Perpendicular distance from a point to a line
    For P(x0,y0) and ℓ:ax+by+c=0,\text{For }P(x_0,y_0)\text{ and }\ell:ax+by+c=0,
    d(P,ℓ)=∣ax0+by0+c∣a2+b2d(P,\ell)=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}

    Distance Between Two Parallel Lines

    Distance between two parallel lines
    ℓ1:ax+by+c1=0,ℓ2:ax+by+c2=0\ell_1:ax+by+c_1=0,\qquad \ell_2:ax+by+c_2=0
    d(ℓ1,ℓ2)=∣c1−c2∣a2+b2d(\ell_1,\ell_2)=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}

    Distance from a Point to a Plane

    Perpendicular distance from a point to a plane
    For P(x0,y0,z0) and α:ax+by+cz+d=0,\text{For }P(x_0,y_0,z_0)\text{ and }\alpha:ax+by+cz+d=0,
    d(P,α)=∣ax0+by0+cz0+d∣a2+b2+c2d(P,\alpha)=\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}
  92. 93.
    ab=cd\frac{a}{b}=\frac{c}{d}
    a+bb=c+dd\frac{a+b}{b}=\frac{c+d}{d}
    aa+b=cc+d\frac{a}{a+b}=\frac{c}{c+d}
    ab−a=cd−c\frac{a}{b-a}=\frac{c}{d-c}
    a−ba+b=c−dc+d\frac{a-b}{a+b}=\frac{c-d}{c+d}
    a1b1=a2b2=⋯=anbn=k⟹λ1a1+λ2a2+⋯+λnanλ1b1+λ2b2+⋯+λnbn=k\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}=k\quad\Longrightarrow\quad\frac{\lambda_1a_1+\lambda_2a_2+\cdots+\lambda_na_n}{\lambda_1b_1+\lambda_2b_2+\cdots+\lambda_nb_n}=k
  93. 94.
    L≤Answer≤UL\leq \text{Answer}\leq U
    L=U⟹Answer=LL=U\Longrightarrow\text{Answer}=L
  94. 95.
    (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1
    c2=a2−b2c^2=a^2-b^2
    e=cae=\frac ca
    Area=πab\text{Area}=\pi ab
  95. 96.
    P(r)=0⟹(x−r)∣P(x)P(r)=0\Longrightarrow (x-r)\mid P(x)
  96. 97.
  97. 98.
    (anan−1⋯a1a0)b=a0+a1b+a2b2+⋯+anbn(a_na_{n-1}\cdots a_1a_0)_b=a_0+a_1b+a_2b^2+\cdots+a_nb^n
  98. 99.
    a<b⟹a+c<b+ca<b\Longrightarrow a+c<b+c
  99. 100.
    Dn=n!(10!−11!+12!−13!+⋯+(−1)nn!)D_n=n!\left(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^n}{n!}\right)
  100. 101.
    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
  101. 102.
    sin⁡A1sin⁡A2=BPPC⋅bc=BPPC⋅sin⁡Bsin⁡C\frac{\sin A_1}{\sin A_2}=\frac{BP}{PC}\cdot\frac{b}{c}=\frac{BP}{PC}\cdot\frac{\sin B}{\sin C}
  102. 103.
    DE∥BC⟹ADAB=AEAC=DEBC,ADAB=AEAC⟹DE∥BCDE\parallel BC\Longrightarrow\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC},\qquad \frac{AD}{AB}=\frac{AE}{AC}\Longrightarrow DE\parallel BC
  103. 104.
    MN∥AB∥CD,MN=AB+CD2MN\parallel AB\parallel CD,\qquad MN=\frac{AB+CD}{2}
  104. 105.
    ax+by=nax+by=n
    gcd⁡(a,b)∣n\gcd(a,b)\mid n
    x=x1+b′k,y=y1−a′kx=x_1+b'k,\qquad y=y_1-a'k
  105. 106.

    Find min{PA+PB}.

    Reflection Method

    CQ=CQ′CQ=CQ'
    BL=BL′BL=BL'
    Reflection Method in square
  106. 107.
    a+b5=0⟹a=b=0a+b\sqrt5=0\Longrightarrow a=b=0
    a+b5=c+d5⟹a=c, b=da+b\sqrt5=c+d\sqrt5\Longrightarrow a=c,\ b=d
  107. 108.
    ax≡b(modm)ax\equiv b\pmod m

    Reduce the coefficient of x to solve.

  108. 109.
    a≡r(modm),a≡r(modn)a\equiv r\pmod m,\qquad a\equiv r\pmod n
    ⟺a≡r(modlcm⁡(m,n))\Longleftrightarrow a\equiv r\pmod{\operatorname{lcm}(m,n)}

    Combine congruences by listing values or substituting one congruence into the other.

    a≡2(mod5),a≡3(mod11)⟹a≡47(mod55)a\equiv2\pmod5,\qquad a\equiv3\pmod{11}\Longrightarrow a\equiv47\pmod{55}
    a≡5(mod19),a≡11(mod47)⟹a≡575(mod893)a\equiv5\pmod{19},\qquad a\equiv11\pmod{47}\Longrightarrow a\equiv575\pmod{893}
  109. 110.

    Find the last 3 digits of 23578 × 16327

    Finding last digits of multiplication
  110. 111.
    2341≡  ?(mod125)2^{341}\equiv\; ?\pmod{125}
    2341=2×(5−1)1702^{341} = 2\times(5-1)^{170}
  111. 112.
    Minimum and Maximum Distances with Circles
    min⁡{PA}=PB,max⁡{PA}=PC\min\{PA\}=PB,\qquad \max\{PA\}=PC
    Distance between two circles
    min⁡{PQ}=BC,max⁡{PQ}=AD\min\{PQ\}=BC,\qquad \max\{PQ\}=AD
  112. 113.
    Graph of sine function
    Graph of cosine function
    Graph of tangent function
    Graph of cotangent function
  113. 114.

    Graph of ∣x∣+∣y∣=5|x|+|y|=5:

    Graph of |x|+|y|=5
  114. 115.

    Graph of y=12xy=\dfrac{1}{2x}:

    Graph of y = 1/(2x)
  115. 116.
    Radical axis
  116. 117.

    A functional graph represents a function

    f:X→Xf:X\to X

    by drawing a directed edge

    x→f(x)x\to f(x)

    For example, assume X={1,2,3,…,14}X=\{1,2,3,\ldots,14\} and

    Example of a functional graph for a function on the set X = {1, 2, 3, ..., 14}
  117. 118.

    A set partition divides a set of distinct elements into nonempty, disjoint groups whose union is the entire set.

    For example, the number of ways to divide 1515 people as shown is

    Partition of 15 people into three groups of 2, one group of 4, and one group of 5
    (152)(132)(112)(94)(55)3!=15!2!3 4! 5! 3!\frac{\binom{15}{2}\binom{13}{2}\binom{11}{2}\binom{9}{4}\binom{5}{5}}{3!}=\frac{15!}{2!^3\,4!\,5!\,3!}
  118. 119.

    Definition

    Probability of AA given BB is

    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}
    Venn diagram of events A and B with their intersection highlighted

    Bayes' Theorem

    P(A∣B)=P(A)P(B∣A)P(B)P(A\mid B)=\frac{P(A)P(B\mid A)}{P(B)}
  119. 120.

    Let BD=xBD=x, then

    x=a2+c2−b22ax=\frac{a^2+c^2-b^2}{2a}
    Triangle ABC with altitude AD to BC, where BD = x, BC = a, AC = b, and AB = c
  120. 121.
    (x′y′)=(hk)+(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(x−hy−k) \begin{pmatrix} x'\\ y' \end{pmatrix} = \begin{pmatrix} h\\ k \end{pmatrix} + \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x-h\\ y-k \end{pmatrix}
    OP=OP′,∠POP′=θ OP=OP', \qquad \angle POP'=\theta
    Rotation of point P(x,y) about O(h,k) through angle theta to P'(x',y')
  121. 122.

    Coloring can reveal patterns or invariants in problems involving boards, tilings, and moves.

    Checkerboard Coloring

    Color the squares alternately black and white, so squares sharing a side have different colors.

    An 8 by 8 checkerboard with alternating black and white squares and rows and columns numbered from 1 to 8

    A square in row rr and column cc can be colored according to the parity of

    r+cr+c

    • r+cr+c even → white
    • r+cr+c odd → black

  122. 123.

    When we want to prove that some selected objects must have a certain relationship, it can be useful to split all possible objects into carefully chosen groups.

    Key Idea

    1. Determine the relationship you want to force.
    2. Split the objects into groups based on that relationship.

  123. 124.

    For example, this applies to:

    • a subset and its complement
    • a graph and its complement
    • selected and unselected objects
    • an event and its complementary event

  124. 125.

    The centroid of a triangle is the point where its three medians intersect.

    If G is the centroid of △ABC and M is the midpoint of BC, then

    Triangle ABC with medians AM and BN intersecting at G
    AG:GM=2:1AG:GM=2:1
    AG=23AMGM=13AMAG=\frac23 AM \qquad GM=\frac13 AM

    The three medians divide the triangle into six equal-area triangles.

    G=(x1+x2+x33,y1+y2+y33)G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)
  125. 126.
    Section Formula
    ACCB=mn\frac{AC}{CB}=\frac{m}{n}
    C=mB+nAm+nC=\frac{mB+nA}{m+n}

    A=(x1,y1),B=(x2,y2)A=(x_1,y_1),\qquad B=(x_2,y_2)
    C=(nx1+mx2m+n,ny1+my2m+n)C=\left(\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n}\right)

    Midpoint Formula

    C=A+B2C=\frac{A+B}{2}
    C=(x1+x22,y1+y22)C=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
  126. 127.
    Projection and reflection of a point across a line

    Projection onto a Line

    H=(x0−a(ax0+by0+c)a2+b2,y0−b(ax0+by0+c)a2+b2)H=\left(x_0-\frac{a(ax_0+by_0+c)}{a^2+b^2},y_0-\frac{b(ax_0+by_0+c)}{a^2+b^2}\right)

    Reflection Across a Line

    P′=(x0−2a(ax0+by0+c)a2+b2,y0−2b(ax0+by0+c)a2+b2)P'=\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right)

    Special Cases

    (x,y)⟶(x,−y)across the x-axis(x,y)\longrightarrow(x,-y)\qquad \text{across the }x\text{-axis}
    (x,y)⟶(−x,y)across the y-axis(x,y)\longrightarrow(-x,y)\qquad \text{across the }y\text{-axis}
    (x,y)⟶(y,x)across y=x(x,y)\longrightarrow(y,x)\qquad \text{across }y=x
    (x,y)⟶(−y,−x)across y=−x(x,y)\longrightarrow(-y,-x)\qquad \text{across }y=-x
    (x,y)⟶(2h−x,y)across x=h(x,y)\longrightarrow(2h-x,y)\qquad \text{across }x=h
    (x,y)⟶(x,2k−y)across y=k(x,y)\longrightarrow(x,2k-y)\qquad \text{across }y=k
  127. 128.

    Equation of a Plane

    ax+by+cz=dax+by+cz=d
    n=(a,b,c)\mathbf{n}=(a,b,c)
    A plane with a normal vector

    Point-Normal Form

    a(x−x0)+b(y−y0)+c(z−z0)=0a(x-x_0)+b(y-y_0)+c(z-z_0)=0

    Plane Through Three Points

    AB→=B−A,AC→=C−A\overrightarrow{AB}=B-A,\qquad \overrightarrow{AC}=C-A
    n=AB→×AC→\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}

    Parallel and Perpendicular Planes

    (a1,b1,c1)=k(a2,b2,c2)(a_1,b_1,c_1)=k(a_2,b_2,c_2)
    a1a2+b1b2+c1c2=0a_1a_2+b_1b_2+c_1c_2=0

    Angle Between Two Planes

    cos⁡θ=∣n1⋅n2∣∣n1∣∣n2∣\cos\theta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}

    Intercept Form

    xp+yq+zr=1\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1
  128. 129.

    Angle Bisector Plane

    Angle bisector planes of two intersecting planes

    ∣a1x+b1y+c1z+d1∣a12+b12+c12=∣a2x+b2y+c2z+d2∣a22+b22+c22\frac{|a_1x+b_1y+c_1z+d_1|}{\sqrt{a_1^2+b_1^2+c_1^2}}=\frac{|a_2x+b_2y+c_2z+d_2|}{\sqrt{a_2^2+b_2^2+c_2^2}}
    a1x+b1y+c1z+d1a12+b12+c12=±a2x+b2y+c2z+d2a22+b22+c22\frac{a_1x+b_1y+c_1z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm\frac{a_2x+b_2y+c_2z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}
  129. 130.

    Dot Product

    Geometric interpretation of dot product

    u→⋅v→=u1v1+u2v2+⋯+unvn\overrightarrow{u}\cdot\overrightarrow{v}=u_1v_1+u_2v_2+\cdots+u_nv_n
    u→⋅v→=∣u→∣∣v→∣cos⁡θ\overrightarrow{u}\cdot\overrightarrow{v}=|\overrightarrow{u}||\overrightarrow{v}|\cos\theta
    u→⊥v→  ⟺  u→⋅v→=0\overrightarrow{u}\perp\overrightarrow{v}\iff\overrightarrow{u}\cdot\overrightarrow{v}=0
    proj⁡v→u→=u→⋅v→∣v→∣2v→\operatorname{proj}_{\overrightarrow{v}}\overrightarrow{u}=\frac{\overrightarrow{u}\cdot\overrightarrow{v}}{|\overrightarrow{v}|^2}\overrightarrow{v}
  130. 131.

    Rational Root Theorem

    P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0
    r=pq,gcd⁡(p,q)=1r=\frac pq,\qquad \gcd(p,q)=1
    p∣a0,q∣anp\mid a_0,\qquad q\mid a_n

    Key Idea

    Use divisibility conditions to reduce the possible rational roots of a polynomial

  131. 132.

    Homogeneous Expressions

    An expression is homogeneous if every term has the same total degree

    2ax2+x2b=b2a2ax^2+x^2b=b^2a

    When all terms have the same degree, divide by a suitable power of one variable and introduce ratios

    r=ax,t=bxr=\frac{a}{x},\qquad t=\frac{b}{x}

    Key Idea

    When all terms have the same total degree:

    1. Divide by a suitable power of one nonzero variable
    2. Replace ratios of variables by new variables
    3. Solve the simpler problem involving fewer variables

  132. 133.

    Symmetry and Fixed Cases Principle

    When a set of cases has a symmetry, we can often pair each case with another case that behaves in the opposite or corresponding way.

    The cases that are unchanged by the symmetry are called fixed cases. These must be considered separately.


    Key Idea

    Pair cases that are transformed into each other by a symmetry. Handle fixed cases separately.


    Applications

    Counting problems with opposite pairs, sign choices, and symmetry arguments

  133. 134.

    Sum of Binomial Coefficients

    (n0)+(n1)+(n2)+⋯+(nn)=2n\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n

    Sums of Even and Odd Binomial Coefficients

    (n0)+(n2)+(n4)+⋯=(n1)+(n3)+(n5)+⋯=2n−1\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots=2^{n-1}

    Applications

    Simplifying binomial sums and solving combinatorial counting problems

  134. 135.

    Definition

    u→×v→=(u2v3−u3v2, u3v1−u1v3, u1v2−u2v1)\overrightarrow{u}\times\overrightarrow{v}=(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1)

    Geometric Meaning

    ∣u→×v→∣=∣u→∣∣v→∣sin⁡θ|\overrightarrow{u}\times\overrightarrow{v}|=|\overrightarrow{u}||\overrightarrow{v}|\sin\theta

    The cross product produces a vector perpendicular to both vectors

    Right-Hand Rule

    To find the direction of A→×B→\overrightarrow{A}\times\overrightarrow{B}, point the fingers of your right hand in the direction of A→\overrightarrow{A} and curl them toward B→\overrightarrow{B}. Your thumb points in the direction of A→×B→\overrightarrow{A}\times\overrightarrow{B}.

    Right-hand rule for the cross product showing vectors A and B and the perpendicular vector C equals A cross B

    Applications

    Finding areas, volumes, normal vectors, and checking whether vectors are parallel

  135. 136.

    Definition

    A function assigns exactly one output to each input.

    f:A→Bf:A\to B
    Function mapping from domain A to codomain B

    Domain and Range

    Domain⁡(f)={x:f(x) is defined}\operatorname{Domain}(f)=\{x:f(x)\text{ is defined}\}
    Range⁡(f)={f(x):x∈Domain⁡(f)}\operatorname{Range}(f)=\{f(x):x\in\operatorname{Domain}(f)\}

    One-to-One Functions

    f(x1)=f(x2)⟹x1=x2f(x_1)=f(x_2)\Longrightarrow x_1=x_2

    Inverse Functions

    f(a)=b⟺f−1(b)=af(a)=b\Longleftrightarrow f^{-1}(b)=a
  136. 137.

    A rate compares how much of something is completed with the amount of time required.

    If aa units are completed in bb units of time, the rate is

    ab.\frac{a}{b}.

    Combined Rates

    When several people or machines work at the same time, their rates can be added:

    rcombined=r1+r2+⋯+rn.r_{\text{combined}}=r_1+r_2+\cdots+r_n.

    Amount Completed

    amount=rate×time.\text{amount}=\text{rate}\times\text{time}.
  137. 138.

    Geometric transformations map points in the plane to new points.

    Translation

    (x,y)⟼(x+a,y+b)(x,y)\longmapsto(x+a,y+b)

    Rotation

    (x,y)⟼(xcos⁡θ−ysin⁡θ, xsin⁡θ+ycos⁡θ)(x,y)\longmapsto(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)
    Counterclockwise rotation of a point about the origin

    Reflection

    (x,y)⟼(y,x)across y=x(x,y)\longmapsto(y,x)\qquad\text{across }y=x

    Dilation

    (x,y)⟼(kx,ky)(x,y)\longmapsto(kx,ky)
  138. 139.

    A vector is a quantity that has both magnitude and direction.

    If

    A=(x1,y1),B=(x2,y2),A=(x_1,y_1),\qquad B=(x_2,y_2),

    then

    AB→=(x2−x1,y2−y1).\overrightarrow{AB}=(x_2-x_1,y_2-y_1).

    Magnitude

    v=(a,b)⟹∣v∣=a2+b2.\mathbf{v}=(a,b)\qquad\Longrightarrow\qquad |\mathbf{v}|=\sqrt{a^2+b^2}.

    Vector Addition

    (a,b)+(c,d)=(a+c,b+d).(a,b)+(c,d)=(a+c,b+d).
    Geometric interpretation of vector addition

    Scalar Multiplication

    k(a,b)=(ka,kb),∣kv∣=∣k∣ ∣v∣.k(a,b)=(ka,kb),\qquad |k\mathbf{v}|=|k|\,|\mathbf{v}|.

    Points on a Segment

    p=(1−t)a+tb,0≤t≤1.\mathbf{p}=(1-t)\mathbf{a}+t\mathbf{b},\qquad 0\le t\le1.
  139. 140.

    Floor Function

    The floor of x is the greatest integer less than or equal to x.

    ⌊x⌋≤x<⌊x⌋+1\boxed{\lfloor x\rfloor\le x<\lfloor x\rfloor+1}
    Graph of the floor function

    Ceiling Function

    The ceiling of x is the smallest integer greater than or equal to x.

    ⌈x⌉−1<x≤⌈x⌉\boxed{\lceil x\rceil-1<x\le\lceil x\rceil}
    Graph of the ceiling function

    x∉Z⟹⌈x⌉=⌊x⌋+1x\notin\mathbb Z\Longrightarrow\lceil x\rceil=\lfloor x\rfloor+1
  140. 141.

    Addition Principle

    For mutually exclusive cases, add the numbers of possibilities.

    n1+n2+⋯+nkn_1+n_2+\cdots+n_k

    Multiplication Principle

    For successive choices, multiply the numbers of possibilities.

    n1n2⋯nkn_1n_2\cdots n_k
    Counting choices using the Multiplication Principle

    Complement Principle

    Desired=Total−Undesired\text{Desired}=\text{Total}-\text{Undesired}
  141. 142.

    A permutation is an arrangement of objects in which order matters.

    Arranging All Objects

    n!n!

    Arranging Some of the Objects

    P(n,r)=n!(n−r)!P(n,r)=\frac{n!}{(n-r)!}

    Identical Objects

    n!n1!n2!⋯nk!\frac{n!}{n_1!n_2!\cdots n_k!}
  142. 143.

    A combination is a selection of objects in which order does not matter.

    (nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}

    The notation (nr)\binom{n}{r} is read as “n choose r”.


    Symmetry Identity

    (nr)=(nn−r)\binom{n}{r}=\binom{n}{n-r}

    Choosing the objects to include is equivalent to choosing the objects to exclude.


    Pascal's Identity

    (nr)=(n−1r)+(n−1r−1)\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}

    Faster Calculation

    (107)=(103)=10⋅9⋅83⋅2⋅1=120\binom{10}{7}=\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120
  143. 144.

    Probability measures how likely an event is to occur.

    0≤P(A)≤10\le P(A)\le1

    Sample Space and Events

    The sample space SS is the set of all possible outcomes, and an event AA is a collection of outcomes from the sample space.


    Equally Likely Outcomes

    P(A)=∣A∣∣S∣P(A)=\frac{|A|}{|S|}

    Complement

    P(Ac)=1−P(A)P(A^c)=1-P(A)

    Probability of A or B

    P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)

    Independent Events

    P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)
  144. 145.

    A step-by-step algorithm for finding square roots by grouping digits into pairs, subtracting squares, and determining each successive digit.

    Square-root long-division algorithm illustrated with 4047 and 954.
  145. 146.

    For two nonnegative numbers with a fixed sum, their product increases and the sum of their squares decreases as the numbers become closer together.

    Moving two numbers closer together while keeping their sum fixed.
    a+b=S⟹ab≤S24\boxed{a+b=S\Longrightarrow ab\le\frac{S^2}{4}}
  146. 147.

    The maximum size of an antichain of subsets of an n-element set is

    (n⌊n/2⌋)\boxed{\binom{n}{\lfloor n/2\rfloor}}
  147. 148.

    If x and y are nonnegative integers, d is squarefree, and

    x+y=kd,\sqrt{x}+\sqrt{y}=k\sqrt{d},

    then for some nonnegative integers a and b,

    x=da2,y=db2,a+b=k\boxed{x=da^2,\quad y=db^2,\quad a+b=k}
  148. 149.

    An invariant is a quantity or property that remains unchanged throughout a process.

    I(S0)=I(S1)=I(S2)=⋯I(S_0)=I(S_1)=I(S_2)=\cdots

    If the initial and target configurations have different invariant values, the target is impossible to reach.

    I(initial)≠I(target)⟹impossible\boxed{I(\text{initial})\ne I(\text{target})\Longrightarrow\text{impossible}}
  149. 150.

    A rational number can be written as

    ab,a,b∈Z,b≠0.\frac ab,\qquad a,b\in\mathbb Z,\quad b\ne0.

    A real number that cannot be written in this form is irrational.

    37, 0, 0.125⏟Rational2, π⏟Irrational\underbrace{\frac37,\ 0,\ 0.125}_{\text{Rational}}\qquad\underbrace{\sqrt2,\ \pi}_{\text{Irrational}}
    xxyyx+yx+yxyxy
    RationalRationalRationalRational
    RationalIrrationalIrrationalMay be either
    IrrationalIrrationalMay be eitherMay be either
  150. 151.

    Mathematical induction proves statements for every integer in a specified range using a base case and an inductive step.

    Base Case

    P(n0)P(n_0)

    Inductive Step

    P(k)⟹P(k+1)P(k)\Longrightarrow P(k+1)

    Extensions include strong induction, step-size induction, binary induction, and Cauchy induction.

    Four quadrants with the original missing cell and a central L-shaped tromino (16.3.png)
  151. 152.

    The Fibonacci sequence starts with 0 and 1, and each subsequent term is the sum of the previous two.

    F0=0,F1=1,Fn+2=Fn+1+FnF_0=0,\quad F_1=1,\quad F_{n+2}=F_{n+1}+F_n
    0, 1, 1, 2, 3, 5, 8, 13, 21,…0,\ 1,\ 1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ldots
    nn0011223344556677
    FnF_n001111223355881313
  152. 153.

    Every real square is nonnegative. This helps prove inequalities and find minimum values.

    x2≥0(x∈R)x^2\ge0\qquad(x\in\mathbb{R})
    a2+b2−2ab=(a−b)2≥0a^2+b^2-2ab=(a-b)^2\ge0
    x2−6x+13=(x−3)2+4≥4x^2-6x+13=(x-3)^2+4\ge4
  153. 154.
    Stewart's Theorem in triangle ABC, where AD = d, BD = m, DC = n, AB = c, and AC = b.
    a(d2+mn)=mb2+nc2\boxed{a(d^2+mn)=mb^2+nc^2}
  154. 155.

    The incenter is the intersection of the three internal angle bisectors of a triangle.

    Triangle ABC with incenter I and its incircle of radius r.
    r=[ABC]s\boxed{r=\frac{[ABC]}{s}}
  155. 156.
    π=3.14159…\pi=3.14159\ldots
    2=1.41421…\sqrt{2}=1.41421\ldots
    3=1.73205…\sqrt{3}=1.73205\ldots
    e=2.71828…e=2.71828\ldots