Geometry and trigonometry

Notation:

- the side lengths of the triangle $ABC$, $AB=c$, $BC=a$, $CA=b$;

$AD={{h}_{a}}\,\,\left( AD\bot BC,\,{{h}_{a}} \right.$ the length of the altitude from $\left. A,D\in BC \right)$;

$AD={{m}_{a}}\left( BD=DC,{{m}_{a}} \right.$ the length of the median from $\left. A,D\in \left( BC \right) \right)$;

$AD={{b}_{a}}\left( \measuredangle BAD=\measuredangle CAD,\,\,{{b}_{a}} \right.$ the length of the bisector from $\left. A,D\in \left( BC \right) \right)$;

- $\frac{a+b+c}{2}=p$ (p – the semiperimeter of the triangle ABC);

- R – the radius of the circle circumscribed about a polygon;

- r – the radius of the circle inscribed in a polygon;

- ${{l}_{n}}$ – the side of the regular polygon with n sides;

- ${{a}_{n}}$ – the apothem of the regular polygon with n sides;

- P – the perimeter of the polygon;

- ${{A}_{lat}}$ – the lateral area;

- ${{A}_{tot}}$ – the total area, also written A;

- V – the volume.

The triangle

Geometric inequalities:

1. $m\left( \measuredangle MBA \right)>M\left( \measuredangle A \right),m\left( \measuredangle MBA \right)>m\left( \measuredangle C \right),\measuredangle MBA$ is an exterior angle;

$m\left( \measuredangle MBA \right)=m\left( \measuredangle A \right)+m\left( \measuredangle C \right);$

2. $a+b>c,\,\,b+c>a,\,\,a+c>b;$

3. $a>\left| b-c \right|,\,\,b>\left| c-a \right|,\,\,c>\left| a-b \right|;$

4. ${{m}_{a}}<\frac{b+c}{2};$

5. $p<{{m}_{a}}+{{m}_{b}}+{{m}_{c}}<P$.

The bisector theorem $\left( \measuredangle BAD=\measuredangle DAC \right);$

$\frac{BD}{DC}=\frac{AB}{AC};\,\,BD=\frac{a\cdot c}{b+c};\,\,DC=\frac{a\cdot b}{b+c};$

Notes:

1. The centre of the circle circumscribed about a triangle is the point where the perpendicular bisectors meet;

2. The orthocentre of the triangle is the point where the altitudes meet.

3. The centroid of the triangle is the point where the medians meet;

2. The centre of the circle inscribed in a triangle is the point where the angle bisectors meet;

Convex polygons

The sum ${{S}_{n}}$ of the measures of the angles of a convex polygon with n sides: ${{S}_{n}}=\left( n-2 \right)\cdot {{180}^{0}};$

A regular polygon can be inscribed in a circle and circumscribed about another circle.

Concave polygon — Convex polygon

Metric relations in a triangle

The right-angled triangle$ABC\left( m\left( \measuredangle A \right)={{90}^{0}},AB\bot BC \right)$

1. Pythagoras' theorem: ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$;

2. The leg theorem: ${{b}^{2}}=a\cdot CD,\,\,{{c}^{2}}=a\cdot BD;$

3. The altitude theorem: $h_{a}^{2}=BD\cdot DC;$

4.${{h}_{a}}=\frac{b\cdot c}{a},\, \,{{h}_{b}}=b,\, \,{{h}_{c}}=c;$

5.${{m}_{a}}=\frac{a}{2},\, \,m_{b}^{2}={{a}^{2}}-\frac{3}{4}{{b}^{2}}, \,\,m_{c}^{2}={{a}^{2}}-\frac{3}{4}{{c}^{2}};$

6.${{b}_{a}}=\frac{b\cdot c}{b+c}\cdot \sqrt{2},\, \,{{b}_{b}}=c\cdot \sqrt{\frac{2a}{a+c}},\,\, {{b}_{c}}=b\cdot \sqrt{\frac{2a}{a+c}};$

7.${{A}_{ABC}}=\frac{b\cdot c}{2};$ 8.$R=\frac{a}{2};$ 9.$r=\frac{b\cdot c}{a+b+c};$

10. Relations expressed through trigonometric functions:

$b=a\cdot \sin B, \,\,b=a\cdot \cos C,\, \,b=c\cdot tgB,\, \,b=c\cdot ctgC$.

$\sin C=\frac{cateta opus\breve{a}}{ipotenuz\breve{a}}=\frac{c}{a}=\cos B$	$\sin B=\frac{cateta opus\breve{a}}{ipotenuz\breve{a}}=\frac{b}{a}=\cos C$
$\cos C=\frac{cateta al\breve{a}turat\breve{a}}{ipotenuz\breve{a}}=\frac{b}{a}$	$\cos B=\frac{cateta al\breve{a}turat\breve{a}}{ipotenuz\breve{a}}=\frac{c}{a}$
$tgC=\frac{cateta opus\breve{a}}{cateta al\breve{a}turat\breve{a}}=\frac{c}{b}$	$tgB=\frac{cateta opus\breve{a}}{cateta al\breve{a}turat\breve{a}}=\frac{b}{c}$
$ctgC=\frac{cateta al\breve{a}turat\breve{a}}{cateta opus\breve{a}}=\frac{b}{c}$	$ctgB=\frac{cateta al\breve{a}turat\breve{a}}{cateta opus\breve{a}}=\frac{c}{b}$

The equilateral triangle $ABC\,\left( a=b=c \right);$

1.${{h}_{a}}={{m}_{a}}={{b}_{a}}=\frac{a\sqrt{3}}{2}$; 2.${{A}_{\Delta ABC}}=\frac{{{a}^{2}}\sqrt{3}}{4};$

3.$R=\frac{a\sqrt{3}}{3};$ 4.$r=\frac{a\sqrt{3}}{6}$.

The general triangle $ABC\left( AD\bot BC \right)$;

1. The generalised Pythagoras theorem:

a)${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2a\cdot BD$, if $m\left( {\hat{B}} \right)<{{90}^{0}};$

b) ${{b}^{2}}={{a}^{2}}+{{c}^{2}}+2a\cdot BD$, if $m\left( {\hat{B}} \right)>{{90}^{0}};$

2. Stewart's relation $O\in \left( BC \right)$:

${{b}^{2}}\cdot BO+{{c}^{2}}-{{a}^{2}}\cdot AO=a\cdot BO\cdot CO$;

3.$m_{a}^{2}=\frac{2\left( {{b}^{2}}+{{c}^{2}} \right)-{{a}^{2}}}{4}$;

4.${{h}_{a}}=\frac{2}{a}\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}$;

5.${{b}_{a}}=\frac{2}{b+c}\sqrt{p\left( p-a \right)bc}$;

6.${{A}_{ABC}}=\frac{a\cdot {{h}_{a}}}{2}=S$;

7.$S=\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}$;

8.$R=\frac{abc}{4S}$; 9.$r=\frac{S}{p}$.

Relations expressed through trigonometric functions

1. The sine rule: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$;

2. The cosine rule (the generalised Pythagoras theorem):

${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\,\,\cos A$ $\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$;

3. The tangent rule: $tg\frac{A-B}{2}\cdot tg\frac{C}{2}=\frac{a-b}{a+b}$;

4.$S=\frac{ab\,\,\sin C}{2},\,\,S=\frac{{{a}^{2}}\sin B\sin C}{2\sin A}$,

$S=p\left( p-a \right)tg\frac{A}{2},\,\text{ }\,S=2{{R}^{2}}\sin A\sin B\sin C$;

5.$p=4R\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$;

6.${{h}_{a}}=2R\sin B\sin C$;

7.$m_{a}^{2}={{R}^{2}}\left( {{\sin }^{2}}A+4\cos A\sin B\sin C \right)$;

8.${{b}_{a}}=\frac{2bc}{b+c}\cos \frac{A}{2}$;

9.$\cos \frac{A}{2}=\sqrt{\frac{p\left( p-a \right)}{bc}}$;

10.$\sin \frac{A}{2}=\sqrt{\frac{\left( p-b \right)\left( p-c \right)}{bc}}$;

11.$tg\frac{A}{2}=\sqrt{\frac{\left( p-b \right)\left( p-c \right)}{p\left( p-a \right)}}$.

Elements of analytic geometry

By a Cartesian coordinate system in the plane we mean a rectangular system $\left( O,\vec{i},\vec{j} \right)$, in which O is the origin and $\vec{i}\text{ i }\vec{j}$ are the unit vectors of the two axes, of the abscissa and of the ordinate respectively; a plane carrying a Cartesian system is called a Cartesian plane. The set $\{\vec{i},\vec{j}\}$ is a basis of the set of vectors in the plane. In such a system, a vector $\overrightarrow{OM}$ is expressed uniquely in the form

$\overrightarrow{OM}=x\vec{i}+y\vec{j}, x, y\in \mathbb{R}$

Segments

1. The distance between two points $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$:

$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$;

The distance between the points ${{M}_{1}}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and ${{M}_{2}}\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in space is ${{M}_{1}}{{M}_{2}}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$.

2. The gradient of the line $AB: {{m}_{AB}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$;

3. The coordinates of the point M$\left( x,y \right)$, the midpoint of the segment AB;

;

$x=\frac{{{x}_{1}}+{{x}_{2}}}{2},\,y=\frac{{{y}_{1}}+{{y}_{2}}}{2}$

4. The coordinates of the point M dividing the segment $\left( AB \right)$ in the ratio k:

$x=\frac{{{x}_{1}}+k{{x}_{2}}}{1+k},\,\text{ }\,y=\frac{{{y}_{1}}+k{{y}_{2}}}{1+k}$

The equation of a line.

1. Lines parallel to the coordinate axes:

$\left( d \right):x=a\,\,\left( d||Oy \right),\,\,\,\,\left( d \right):y=a\,\,\,\,\left( d||Ox \right)$;

2. The line determined by the point ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}} \right)$ and the non-zero vector $\bar{a}\left( u,v \right):\left( d \right):\bar{r}={{\bar{r}}_{0}}+t\bar{a},t\in R$, ${{\bar{r}}_{0}}$ – the position vector M of the line d;

$\left( d \right):\left\{ \begin{aligned} & x={{x}_{0}}+ut \\ & y={{y}_{0}}+vt \\ \end{aligned} \right.,t\in R$, the parametric equations;

3. The explicit equation: $y=mx+n$, $m\in {{R}^{*}},n\in R,$ $m$ – the gradient, $n$ – the y-intercept;

4. The intercept equation: $\frac{x}{a}+\frac{y}{b}-1=0,\left( a,b\in {{R}^{*}} \right)$, where the points A(a,0) and B(0,b) are the intercepts;

5. The equation of the line of gradient m through the point $M\left( {{x}_{0}},{{y}_{0}} \right)$:

$y-{{y}_{0}}=m\left( x-{{x}_{0}} \right),\left( m\ne 0 \right)$;

6. The equation of the line determined by the points

$A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right):y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right),$

$\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}},\left( {{x}_{1}}\ne {{x}_{2}},{{y}_{1}}\ne {{y}_{2}} \right)$, or $\left| \begin{matrix} x & y & 1 \\ {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ \end{matrix} \right|=0$

7. The general equation: $ax+by+c=0;$

8. The area of the triangle $ABC\left( A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right),C\left( {{x}_{3}},{{y}_{3}} \right) \right):$

${{A}_{ABC}}=\frac{1}{2}\left| \Delta \right|$, where $\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$;

if $\Delta =0$ then $A,B,C$ are collinear;

9. The relative position of the lines $\left( {{d}_{1}} \right)$ and $\left( {{d}_{2}} \right)$:

$\left( {{d}_{1}} \right):{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and $\left( {{d}_{2}} \right):{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$

${{d}_{1}}={{d}_{2}}$, if $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$;

${{d}_{1}}||{{d}_{2}}$, if $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$;

${{d}_{1}}\ne {{d}_{2}}$ and ${{d}_{1}}\cap {{d}_{2}}\ne \emptyset$, if $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$;

10. The distance from the point $M\left( {{x}_{0}},{{y}_{0}} \right)$ to the line $\left( h \right):ax+by+c=0$:

$d\left( M,h \right)=\frac{\left| a{{x}_{0}}+b{{y}_{0}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$

11. The angle $\alpha$ determined by the lines

$\left( {{d}_{1}} \right):y={{m}_{1}}x+{{n}_{1}}$ and $\left( {{d}_{2}} \right):y={{m}_{2}}x+{{n}_{2}}$:

$tg\alpha =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|,\left( {{m}_{1}}{{m}_{2}}\ne -1 \right)$;

${{d}_{1}}\bot {{d}_{2}}$, if ${{m}_{1}}{{m}_{2}}\ne -1$;

Equations of the plane

The general equation of a plane in three-dimensional space is $Ax+By+Cz+D=0$, where A,B,C are not all zero. The position vector with coordinates (A,B,C) is perpendicular to the plane $Ax+By+Cz+D=0.$

The equation of the plane through the point $\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ is:

$A\left( x-{{x}_{0}} \right)+B\left( y-{{y}_{0}} \right)+C\left( z-{{z}_{0}} \right)=0$.

The equation of the plane through 3 non-collinear points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$, $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is $\left| \begin{matrix} x \\ {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix}\begin{matrix} y \\ {{y}_{1}} \\ {{y}_{2}} \\ {{y}_{3}} \\ \end{matrix}\begin{matrix} z \\ {{z}_{1}} \\ {{z}_{2}} \\ {{z}_{3}} \\ \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{matrix} \right|=0$. The condition for three points with coordinates $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$, $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ to be non-collinear is .

Two planes $p:Ax+By+Cz+D=0$ and $p':A'x+B'y+C'z+D'=0$ with $\frac{A}{A'}\ne \frac{B}{B'}$ or $\frac{B}{B'}\ne \frac{C}{C'}$ or $\frac{A}{A'}\ne \frac{C}{C'}$ intersect in a line.

Equations of a line in space

The parametric equations of the line determined by the point ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ and the direction vector $\vec{v}\left( l,m,n \right)$ are $d:\left\{ \begin{aligned} & x={{x}_{0}}+\lambda \cdot l \\ & y={{y}_{0}}+\lambda \cdot m \\ & z={{z}_{0}}+\lambda \cdot n \\ \end{aligned} \right.$, where $\lambda \in \mathbb{R}$.

The line determined by the point ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ and the direction vector $\vec{v}\left( l,m,n \right)$ can be described by the canonical equations: $\frac{x-{{x}_{0}}}{l}=\frac{y-{{y}_{0}}}{m}=\frac{z-{{z}_{0}}}{n}$.

Let ${{M}_{1}}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, ${{M}_{2}}\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ be points of the line d. Then the canonical equations of the line $d:\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$, ${{x}_{1}}\ne {{x}_{2}}$, ${{y}_{1}}\ne {{y}_{2}}$, ${{z}_{1}}\ne {{z}_{2}}$.

Let the lines ${{d}_{1}}$ and ${{d}_{2}}$ be given by the canonical equations $\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}$ and $\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}$ respectively. The angle $\varphi$ formed by the lines ${{d}_{1}}$ and ${{d}_{2}}$ is given by the formula: $\cos \varphi =\frac{{{l}_{1}}\cdot {{l}_{2}}+{{m}_{1}}\cdot {{m}_{2}}+{{n}_{1}}\cdot {{n}_{2}}}{\pm \sqrt{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}\cdot \sqrt{l_{2}^{2}+m_{2}^{2}+n_{2}^{2}}}$.

The relative position of a line and a plane

Let $d:\frac{x-{{x}_{0}}}{l}=\frac{y-{{y}_{0}}}{m}=\frac{z-{{z}_{0}}}{n}$ and the plane: $Ax+By+Cz+D=0$.

1. If $Al+Bm+Cn\ne 0$, d meets the plane in a point.

2. If $Al+Bm+Cn=0$ and $A{{x}_{0}}+B{{y}_{0}}+C{{z}_{0}}+d\ne 0$, then $d||??$

3. If $Al+Bm+Cn=0$ and $A{{x}_{0}}+B{{y}_{0}}+C{{z}_{0}}+d=0$, then $d\subset ??$.

The angle between a line and a plane

Let the line d be given by the equations $\frac{x-{{x}_{0}}}{l}=\frac{y-{{y}_{0}}}{m}=\frac{z-{{z}_{0}}}{n}$ and the plane by the equation $Ax+By+Cz+D=0$. Let $\varphi$ be the angle between the line d and the plane.

We have: $\sin \varphi =\frac{\left| A\cdot l+B\cdot m+C\cdot n \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}\cdot \sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$.

The distance from a point $M\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ to a plane is:

$d=\frac{\left| A{{x}_{0}}+B{{y}_{0}}+C{{z}_{0}}+D \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$, if the equation of the plane is $Ax+By+Cz+D=0$.

The planes with equations ${{A}_{1}}x+{{B}_{1}}y+{{C}_{1}}z+D=0$ and ${{A}_{2}}x+{{B}_{2}}y+{{C}_{2}}z+D=0$ have the cosine of the angle between them given by the formula:

$\cos \varphi =\frac{\left| {{A}_{1}}{{A}_{2}}+{{B}_{1}}{{B}_{2}}+{{C}_{1}}{{C}_{2}} \right|}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}\cdot \sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}}$.

The planes $:{{A}_{\,1}}x+{{B}_{1}}y+{{C}_{1}}z+{{D}_{1}}=0$ and $:{{A}_{\,2}}x+{{B}_{2}}y+{{C}_{2}}z+{{D}_{2}}=0$ are parallel if $\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}\ne \frac{{{D}_{1}}}{{{D}_{2}}}$ (${{A}_{\,1}}$ and ${{A}_{\,2}}$, ${{B}_{\,1}}$ and ${{B}_{\,2}}$, or ${{C}_{\,1}}$ and ${{C}_{\,2}}$ respectively may be simultaneously zero).

The area of the triangle with vertices ${{M}_{1}}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, ${{M}_{2}}\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$, ${{M}_{3}}\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is:

${{A}_{\,\,\left[ {{M}_{1}}{{M}_{2}}{{M}_{3}} \right]}}=\frac{1}{2}\sqrt{\Delta _{1}^{2}+\Delta _{2}^{2}+\Delta _{3}^{2}}$, with ${{\Delta }_{1}}=\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|$, ${{\Delta }_{2}}=\left| \begin{matrix} {{x}_{1}} & {{z}_{1}} & 1 \\ {{x}_{2}} & {{z}_{2}} & 1 \\ {{x}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|$,

${{\Delta }_{3}}=\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$

* The volume of the tetrahedron with vertices ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$, ${{M}_{1}}\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, ${{M}_{2}}\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$, ${{M}_{3}}\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is

$V=\frac{1}{6}|\left| \begin{matrix} {{x}_{0}} \\ {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ \end{matrix}\begin{matrix} {{y}_{0}} \\ {{y}_{1}} \\ {{y}_{2}} \\ {{y}_{3}} \\ \end{matrix}\begin{matrix} {{z}_{0}} \\ {{z}_{1}} \\ {{z}_{2}} \\ {{z}_{3}} \\ \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{matrix} \right||$ ($\frac{1}{6}$ of the absolute value of the determinant).

Conics

In a double cone of revolution whose height is $h\to \infty$, the cross-sections are called conics, and they may be: a point, two intersecting lines, a circle, an ellipse, a hyperbola or a parabola, as follows:

1. first, the double cone uncut

2. the cutting plane passes through the apex and meets neither nappe of the cone

3. the cutting plane passes through the apex and meets the nappes of the cone

4. the cutting plane meets only one nappe of the cone and cuts every generator

5. the cutting plane meets both nappes of the cone

6. the cutting plane meets only one nappe of the cone and is parallel to a generator

1. The circle

Equation of the circle: ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$;

if $M\left( a,b \right)=O\left( 0,0 \right):{{x}^{2}}+{{y}^{2}}={{r}^{2}}$;

Normal equation: ${{x}^{2}}+{{y}^{2}}+mx+ny+p=0$, where $a=-\frac{m}{2}$, $b=-\frac{n}{2}$ and ${{r}^{2}}=\frac{1}{4}\left( {{m}^{2}}+{{n}^{2}} \right)-p$.

The parametric equations are: $\begin{aligned} & x(\theta )=a+r\cos (\theta ) \\ & y(\theta )=b+r\sin (\theta );\theta \in [0,2\pi ]. \\ \end{aligned}$

$\begin{aligned} & x(t)=\frac{1-{{t}^{2}}}{1+{{t}^{2}}};\text{ }y(t)=\frac{2t}{1+{{t}^{2}}} \\ & \\ \end{aligned}$

Equation of the tangent at the point $M\left( {{x}_{0}},{{y}_{0}} \right),M\in cercului$; $x{{x}_{0}}+y{{y}_{0}}-{{r}^{2}}=0$

Conics referred to their axes of symmetry

1. The ellipse E: $F\left( c,0 \right),F'\left( -c,0 \right),A\left( a,0 \right),A'\left( -a,0 \right),$ $B\left( 0,b \right),B'\left( 0,-b \right),$

Equation of the ellipse: $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-1=0,{{b}^{2}}+{{c}^{2}}={{a}^{2}}$

Equation of the tangent at the point $M\left( {{x}_{0}},{{y}_{0}} \right),M\in E$; $\frac{x{{x}_{0}}}{{{a}^{2}}}+\frac{y{{y}_{0}}}{{{b}^{2}}}-1=0$

The parametric equations are: $\begin{aligned} & x(\theta )=a\cos (\theta ) \\ & y(\theta )=b\sin (\theta );\theta \in [0,2\pi ]. \\ \end{aligned}$

The ellipse is the orthogonal projection of the circle centred at the origin of radius $r=a$, lying in a plane making with the plane of the ellipse the angle $\alpha$ given by the relation $\cos (\alpha )=\frac{b}{a}$.

The ellipse is the locus of the points for which the sum of the distances to two fixed points, called foci, is constant. $MF+MF'=2a,M\in E;$

The eccentricity of the ellipse is: $e=\frac{c}{a}<1$.

The area of the ellipse is: $S=\pi ab$.

2. The hyperbola H: $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1=0,{{c}^{2}}-{{b}^{2}}={{a}^{2}}$

$\left| TF-TF' \right|=2a,M\in H$;

Equation of the hyperbola: $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1=0,{{c}^{2}}={{a}^{2}}+{{b}^{2}}$;

Equation of the tangent at ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}} \right),{{M}_{0}}\in H$: $\frac{x{{x}_{0}}}{{{a}^{2}}}-\frac{y{{y}_{0}}}{{{b}^{2}}}-1=0$

The hyperbola with centre A(p,q), referred to the system xOy, has the equation:

$\frac{{{(x-p)}^{2}}}{{{a}^{2}}}-\frac{{{(y-q)}^{2}}}{{{b}^{2}}}-1=0.$

The parametric equations are: $\begin{aligned} & x(t)=\frac{a}{2}(t+\frac{1}{t}), \\ & y(t)=\frac{b}{2}(t-\frac{1}{t});t\in \mathbb{R} \\ \end{aligned}$

Equation of the conjugate hyperbola: $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}+1=0,{{c}^{2}}={{a}^{2}}+{{b}^{2}}$;

3. The parabola P: $F\left( \frac{P}{2},0 \right),h:x=-\frac{P}{2}$ (h – the directrix): $d\left( M,h \right)=MF,M\in P$;

Equation of the parabola: ${{y}^{2}}=2px;$

Equation of the tangent at ${{M}_{0}}\left( {{x}_{0}},{{y}_{0}} \right),{{M}_{0}}\in P$: $y{{y}_{0}}=p\left( x+{{x}_{0}} \right)$.

The parametric equations are: $\begin{aligned} & x(t)=\frac{{{t}^{2}}}{2p}, \\ & y(t)=t;t\in \mathbb{R} \\ \end{aligned}$

Tangent to a conic from a point P outside the conic

The equation of a line through $P({{x}_{p}},{{y}_{p}})$ with gradient m is $y-{{y}_{p}}=m(x-{{x}_{p}})$. Intersect this line with the conic, that is, write the system formed by their equations:

$\left\{ \begin{aligned} & y-{{y}_{p}}=m(x-{{x}_{p}}) \\ & ''ecuaia conicei'' \\ \end{aligned} \right.$

Solve the system and impose the condition that the tangent meets the conic in a single point ($\Delta =0$).

Intersection of two conics, or of a conic and a line

The intersections are found by solving the system formed from the equations of the two conics. The resulting solution gives the points of intersection.