I'm looking for useful lower-level mathematical formulas that are useful for a variety of problems and help in high-school math competitions.
Some formulas that I've found very useful so far are
- Vieta's formulas (Quadratics: $r_1+r_2=\frac{-b}a$, $r_1r_2=\frac ca$) (Cubics: $r_1+r_2+r_3=\frac{-b}a, r_1r_2+r_2r_3+r_3r_1=\frac ca, r_1r_2r_3=\frac{-d}a$
- Euclidian Algorithim $\gcd(2322, 654)=\gcd(654,360), \;\gcd(654,360)=\gcd(360,294)...$
- Shoelace Theorem $P=\frac 12((a_1b_2,\;a_2b_3\;...\;a_nb_1)-(b_1a_2,\;b_2a_3\;...\;b_na_1))$ where $(a_1b_1),\;(a_2b_2)\;...\;(a_nb_n)$ are the points of a polygon listed in a clockwise order and $P$ is the area of the polygon
- Heron's Formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ where $A$ is a triangle with side lengths $a,\;b,$ and $c$, and $s$ is the semi-perimeter, or $\dfrac{a+b+c}2$
- Ptolemy's Theorem $ab+cd=ef$, when $a,\;b,\;c,$ and $d$ are the side lengths of a cyclic quadrilateral and $e$ and $f$ are its diagonals.
- Fermat's Little Theorem $a^{p-1}\equiv 1 \pmod p$ where $a$ is an integer, $p$ is a prime number, and $a$ is not divisible by $p$
- Formula for a median of a triangle: In $\Delta ABC$, the length of the median to side $BC$ is $\frac 12 \sqrt{2AB^2+2AC^2-BC^2}$
- Distance from a point to a line: $ ax + by + c = 0, \left(x_0, y_0\right) \implies D_{\text{min}} = \dfrac{\left|ax_0 + by_0 + c\right|}{\sqrt{a^2 + b^2}} $
- Stewart's Theorem: Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices $A$, $B$, $C$, respectively. If cevian $AD$ is drawn so that $BD = m$, $DC = n$ and $AD = d$, we have that $b^2m + c^2n = amn + d^2a$.
- Volume of a tetrahedron with perpendicular sides $x,y,z$: $\dfrac{xyz}6$
The towers of Hanoi problem.
A puzzle with $n$ discs of different sizes arranged so that smaller discs always go on top of bigger discs.
We have to move the discs, one at a time, so that the tower is reassembled at peg B or C. When moving a disc we are only allowed to put a small disc on a larger one.
We want to know the minimum number of moves to achieve this.
Let $H(n)$ be the number of moves with $n$ discs.
Hence $H(1)=1$.
For a general case, we move $n-1$ discs to form a tower at peg B. So we need $H(n-1)$ moves.Then move remaining discs at A to C, then move discs from B onto larger discs at C. This again requires $H(n-1)$ moves.
$\therefore H(n) = 2H(n+1) +1$, for $n \ge 2$.