Handbook · Chapter 4

Geometry

Triangles, quadrilaterals, circles and power of a point, polygons, cyclic quadrilaterals, 3-D solids, the advanced triangle formulas and coordinate geometry.

Book pages 60–112 · 120 entries

4.1 Triangles

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Theorem 4.1.1 · Using base and height

A triangle with base $b$ and height $h$ has an area of

\frac{1}{2} \cdot b \cdot h.

Learn it:Theorem 30.1.1 · Using base and height Section 30.1 Area of a Triangle

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Theorem 4.1.2 · Heron’s Formula

A triangle with sides $a$ , $b$ , $c$ and semiperimeter $s = \frac{a+b+c}{2}$ has an area of

\sqrt{s(s-a)(s-b)(s-c)}.

Learn it:Theorem 30.1.2 · Heron's Formula

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Definition 4.1.3 · In-radius

The inradius of a triangle is the radius of the circle inscribed in the triangle.

Learn it:Definition 30.1.4 · In-radius

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Theorem 4.1.4 · Using inradius

A triangle with inradius $r$ (the radius of the circle that can be inscribed in the triangle) and semiperimeter $s$ has an area of

\text{Area} = \text{inradius} \cdot \text{semiperimeter} = rs.

Learn it:Theorem 30.1.6 · Using inradius

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Remark 4.1.5

If we know the area of the triangle and its semiperimeter, we can apply the inradius formula to find the inradius: $r = \dfrac{\text{Area}}{s}$ .

Learn it:Remark 30.1.7

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Definition 4.1.6 · Circumradius

The circumradius of a triangle is the radius of the circle that the triangle is inscribed in.

Learn it:Definition 30.1.9 · Circumradius

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Theorem 4.1.7 · Using circumradius

A triangle with circumradius $R$ (the radius of the circle that the triangle can be inscribed in) and sides $a$ , $b$ , $c$ has an area of

\frac{abc}{4R}.

Learn it:Theorem 30.1.10 · Using circumradius

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Remark 4.1.8

Similar to the inradius: if we know all 3 sides of a triangle, we can apply Heron’s formula to get the area and then easily calculate the circumradius, $R = \dfrac{abc}{4 \cdot \text{Area}}$ .

Learn it:Remark 30.1.11

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Theorem 4.1.9 · Using Trigonometry

A triangle with 2 sides $a$ and $b$ and angle $C$ between those sides has an area of

\frac{1}{2} \cdot a \cdot b \cdot \sin C.

Learn it:Theorem 43.1.1 · Area of a Triangle Using Trigonometry Section 43.1 Area of a Triangle Using Trigonometry

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Theorem 4.1.10 · Pick’s Theorem

If a polygon has vertices with integer coordinates (lattice points), then the area of the polygon is

i + \frac{b}{2} - 1,

where $i$ is the number of lattice points inside the polygon and $b$ is the number of lattice points on the boundary of the polygon.

Basically, the area is the number of interior lattice points, plus half the number of boundary lattice points, minus $1$ .

Learn it:Theorem 30.1.14 · Pick's Theorem Theorem 38.4.1 · Pick's Theorem

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Theorem 4.1.11 · Shoelace Theorem using coordinates

Suppose the polygon $P$ has vertices $(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$ , listed in clockwise order. Then the area $A$ of $P$ is

A = \frac{1}{2}\left|S_1 - S_2\right|,

where

S_1 = a_1 b_2 + a_2 b_3 + \cdots + a_n b_1,

S_2 = b_1 a_2 + b_2 a_3 + \cdots + b_n a_1.

You can also go in counterclockwise order, as long as you take the absolute value of the answer.

The Shoelace Theorem gets its name from the picture you get when you list the coordinates in a column, repeat the first pair at the bottom, and mark the pairs of coordinates to be multiplied:

\begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \\ \vdots & \vdots \\ a_n & b_n \\ a_1 & b_1 \end{array}

The criss-cross lines between the two columns look like the laces of a shoe.

Learn it:Theorem 30.1.12 · Shoelace Theorem using coordinates Theorem 38.3.1 · Shoelace Theorem using coordinates

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Remark 4.1.12 · Intuitive Way of Thinking about Shoelace Theorem

Steps to the Shoelace Theorem:

Line up all of your polygon’s coordinates in a vertical column.
Repeat your first coordinate at the bottom of the column.
Let $A$ be the sum of the products of all the rightward (down-right) diagonal pairs.
Let $B$ be the sum of the products of all the leftward (down-left) diagonal pairs.
Find

\frac{1}{2}|A - B|.

Learn it:Remark 30.1.13 · Intuitive Way of Thinking about Shoelace Theorem Remark 38.3.2 · Intuitive Way of Thinking about Shoelace Theorem

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Theorem 4.1.13

If the side length of an equilateral triangle is $a$ , then

\text{Height of the triangle} = \frac{\sqrt{3}}{2}a.

This follows directly from the 30-60-90 triangle.

\text{Area of the triangle} = \frac{\sqrt{3}}{4}a^2.

Learn it:Theorem 31.1.1 · Equilateral Triangle Section 31.1 Equilateral Triangle

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Theorem 4.1.14

If the side length (each leg) of a 45-45-90 triangle is $a$ , then

\text{hypotenuse} = \sqrt{2} \times \text{side length} = \sqrt{2}\,a

\text{Area of the triangle} = \frac{1}{2} \times \text{side length}^2 = \frac{1}{2}a^2.

Learn it:Theorem 31.2.3 · 45-45-90 Triangle

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Theorem 4.1.15

If the short leg of a 30-60-90 triangle is $a$ , then

\text{Long leg of the triangle} = \sqrt{3} \times \text{short leg} = \sqrt{3}\,a

\text{hypotenuse of the triangle} = 2 \times \text{short leg} = 2a

\text{Area} = \frac{\sqrt{3}}{2} \times \text{short leg}^2 = \frac{\sqrt{3}}{2}a^2.

Learn it:Theorem 31.2.4 · 30-60-90 Triangle

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Theorem 4.1.16

If the three sides of a triangle are $13$ , $14$ and $15$ , it can be divided into two right triangles with side lengths

5, 12, 13 \quad\text{and}\quad 9, 12, 15.

\text{Area of this triangle} = \frac{1}{2} \times 14 \times 12 = 84.

Learn it:Theorem 31.2.5 · 13-14-15 Triangle

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Theorem 4.1.17 · Pythagorean Theorem

A right triangle with legs $a$ and $b$ and hypotenuse $c$ satisfies

c^2 = a^2 + b^2.

Learn it:Theorem 31.2.1 · Pythagorean Theorem Section 31.2 Right Triangles

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Fact 4.1.18 · Important Pythagorean Triples

Important Pythagorean triples:

3, 4, 5 \qquad 5, 12, 13 \qquad 7, 24, 25

8, 15, 17 \qquad 9, 40, 41 \qquad 20, 21, 29

If all numbers in a Pythagorean triple are multiplied by a constant, the resulting numbers still form a Pythagorean triple. For example, these are all Pythagorean triples: $3, 4, 5$ ; $6, 8, 10$ ; $9, 12, 15$ ; $12, 16, 20$ ; $15, 20, 25$ .

Learn it:Fact 31.2.2 · Important Pythagorean Triples

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Theorem 4.1.19 · Special Properties of right triangles

In a right triangle $ABC$ where $B$ is the right angle and $BD$ is the altitude to the hypotenuse, the following triangles are similar:

\triangle ABC \sim \triangle ADB \sim \triangle BDC.

The length of the perpendicular to the hypotenuse is

BD = \frac{AB \cdot BC}{AC}.

Also note that:

AD \cdot CD = BD^2

AD \cdot AC = AB^2

CD \cdot CA = CB^2

In the figure, $AB = 4$ , $BC = 3$ , $AC = 5$ , so $BD = \frac{12}{5}$ , $AD = \frac{16}{5}$ and $CD = \frac{9}{5}$ .

Correction

The book prints $BD = \sqrt{\dfrac{AB \cdot BC}{AC}}$ ; the square root is wrong. Twice the area is $AB \cdot BC = AC \cdot BD$ , so $BD = \dfrac{AB \cdot BC}{AC}$ (the book's own figure has $BD = \frac{12}{5} = \frac{4 \cdot 3}{5}$ ).

Learn it:Theorem 32.0.3 · Special Properties of Right Triangles

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Theorem 4.1.20 · Median in a Right Triangle

In a right triangle $ABC$ with the right angle at $B$ , let the median from $B$ meet $AC$ at the point $P$ . Then

AP = BP = CP.

Basically, in a right triangle $AC$ is a diameter of the circumcircle, and $PA$ , $PB$ and $PC$ are radii of the circumcircle.

Learn it:Theorem 30.2.2 · Median in a Right Triangle

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Concept 4.1.21 · Similarity Test

Two triangles are similar if the three angles in the triangles are the same. In other words, the triangles are the same shape multiplied by a scale factor.

In general, triangles are similar if:

AA similarity: two angles of the triangles are the same (which means the third angle is equal too).
SAS similarity (Side Angle Side): two sides are proportional and the angle between the sides is equal.
SSS similarity (Side Side Side): all three sides are proportional.
HL similarity (Hypotenuse Leg): in a right triangle, the hypotenuse and a leg are proportional.
LL similarity (Leg Leg): in a right triangle, the two legs are proportional.

Warning: SSA does not mean the triangles are similar.

An easy way to detect similar triangles: the bases of the triangles are parallel and the sides of the triangles are collinear (see the figure).

Learn it:Concept 32.0.2 · Similarity Test Section 32.0 Similar Triangles

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Theorem 4.1.22

For similar triangles:

All the angles of the triangles are the same.
All corresponding sides have the same ratio.
The ratio of the areas is the square of the side length ratio.

Learn it:Theorem 32.0.1 · Similar Triangles

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Theorem 4.1.23 · Angle Bisector Theorem

If the line $AD$ bisects angle $A$ of triangle $ABC$ (with $D$ on $BC$ ), then

\frac{AB}{BD} = \frac{AC}{CD}.

Learn it:Theorem 32.1.1 · Angle Bisector Theorem Section 32.1 Angle Bisector Theorem

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Theorem 4.1.24

Viviani’s Theorem. If $P$ is a point inside an equilateral triangle $ABC$ , then the sum of the distances from $P$ to the sides of the triangle is equal to the length of its altitude.

Basically, if the height of the triangle is $h$ , and $PQ$ , $PR$ and $PS$ are the perpendiculars to $AB$ , $BC$ and $AC$ respectively, then

PQ + PR + PS = h.

Learn it:Section 31.1 Equilateral Triangle

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Theorem 4.1.25 · van Schooten’s theorem

Let $P$ be a point on the minor arc $BC$ of the circumcircle of equilateral triangle $ABC$ . Then

PA = PB + PC.

Learn it:Theorem 34.4.3 · Ptolemy's Theorem

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Theorem 4.1.26

For a point $P$ inside an equilateral triangle $ABC$ with side length $s$ ,

\begin{aligned} &3\left(PA^4 + PB^4 + PC^4 + s^4\right) \\ &\quad = \left(PA^2 + PB^2 + PC^2 + s^2\right)^2. \end{aligned}

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4.2 Quadrilaterals

Theorem 4.2.1 · Area of a Square

Any square with side length $s$ has an area of

s^2

and a perimeter of

4s.

Learn it:Theorem 33.1.1 · Area of a Square

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Theorem 4.2.2 · Area of a Rectangle

Any rectangle with base $b$ and height $h$ has an area of

bh

and a perimeter of

2b + 2h.

Learn it:Theorem 33.2.1 · Area of a Rectangle

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Theorem 4.2.3 · British Flag Theorem

If a point $P$ is chosen inside rectangle $ABCD$ , then

PA^2 + PC^2 = PB^2 + PD^2.

( $A$ and $C$ are opposite corners, and so are $B$ and $D$ .)

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Theorem 4.2.4 · Area of a Rhombus

A rhombus with diagonals $d_1$ and $d_2$ has an area of

\frac{1}{2}d_1 d_2

and a perimeter of

2\sqrt{d_1^2 + d_2^2}.

Learn it:Theorem 33.3.1 · Area of a Rhombus Concept 33.3.2 · Rhombus

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Theorem 4.2.5 · Area of a Parallelogram

A parallelogram with base $b$ and height $h$ has an area of

bh.

Learn it:Theorem 33.4.1 · Area of a Parallelogram Concept 33.4.2 · Parallelogram Properties

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Theorem 4.2.6 · Area of a Trapezoid

A trapezoid with bases $b_1$ and $b_2$ and height $h$ has an area of

\frac{b_1 + b_2}{2} \cdot h.

Learn it:Theorem 33.5.1 · Area of a Trapezoid

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4.3 Circles

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Theorem 4.3.1 · Area and Circumference

A circle with radius $r$ has

\text{Area} = \pi r^2

\text{Circumference} = 2\pi r

Learn it:Theorem 34.1.1 · Area and Circumference Section 34.1 Circle Area

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Theorem 4.3.2 · Arcs of a circle

For an arc of a circle with radius $r$ and angle $a^\circ$ :

\text{Area of a sector} = \pi r^2 \times \frac{a^\circ}{360^\circ}

that is, $\pi \times \text{radius}^2 \times$ the fraction of the circle in the sector.

\text{Arc length} = 2\pi r \times \frac{a^\circ}{360^\circ}

that is, $2\pi \times \text{radius} \times$ the fraction of the circle in the sector.

Learn it:Theorem 34.1.2 · Arcs of a circle Section 34.2 Circle Length

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Definition 4.3.3 · Angle of an arc

The angle of an arc is the angle that the arc makes at the center of the circle.

Learn it:Definition 34.1.3 · Angle of an arc

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Theorem 4.3.4 · Inscribed Arc Theorem

The angle formed by an arc at the center (the arc angle) is double the angle it forms at any point on the edge of the circle:

\text{central angle} = 2 \times \text{inscribed angle}.

Learn it:Theorem 29.1.7 · Inscribed Arc Theorem Section 29.1c Angles in Circles

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Corollary 4.3.5 · Inscribed Right Triangle

An inscribed triangle with a diameter as one side is always a right triangle.

Learn it:Corollary 29.1.8 · Inscribed Right Triangle

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Definition 4.3.6 · Chord

A chord is a line segment between any two distinct points on the circle. The diameter of the circle is the longest chord in the circle.

Learn it:Definition 29.1.9 · Chord

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Theorem 4.3.7

The perpendicular bisector of any chord passes through the center. In the figure, the perpendicular bisectors of $AB$ and $CD$ intersect at the center $O$ .

Learn it:Theorem 29.1.10

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Corollary 4.3.8

Congruent chords are equidistant from the center of a circle.
If two chords in a circle are congruent, then their intercepted arcs are congruent.
If two chords in a circle are congruent, then they determine two central angles that are congruent.

Learn it:Corollary 29.1.11

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Theorem 4.3.9

Two secants from an outside point $P$ cut the circle at $A, B$ and $C, D$ ( $A$ and $C$ nearer to $P$ ). The angle at $P$ is half the difference of the two arcs it cuts off:

\angle APC = \frac{\overset{\frown}{BD} - \overset{\frown}{AC}}{2}.

Learn it:Theorem 29.1.12

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Theorem 4.3.10

If two chords $AB$ and $CD$ intersect at $P$ , then $\angle BPC$ and $\angle APD$ are equal to the average of the two arcs:

\angle BPC = \angle APD = \frac{\overset{\frown}{BC} + \overset{\frown}{AD}}{2}.

Learn it:Theorem 29.1.13

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Theorem 4.3.11

If a tangent line through $R$ touches the circle at $Q$ , and a chord $QP$ is drawn, then $\angle RQP$ is equal to half the arc $QP$ inside the angle.

Learn it:Theorem 29.1.14

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Remark 4.3.12

Circles are really useful for angle chasing, so keep an eye out for the inscribed arc theorem: it can be used in many angle chasing problems.

Learn it:Remark 29.1.15

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Remark 4.3.13

A useful trick for solving angle chasing problems with regular polygons is to draw a circle around the polygon and use the inscribed arc theorem.

Learn it:Remark 29.1.16

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Theorem 4.3.14

Equal chords mark out equal arcs.

This basically means that if you have 2 chords of the same length, the arcs (and sectors) of the circle they mark out are equal.

Learn it:Theorem 29.1.17

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Definition 4.3.15 · Tangent

A tangent is any line from a point external to the circle that just touches the circle.

Learn it:Definition 29.1.18 · Tangent

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Theorem 4.3.16 · Right Angle Tangency Point

If you connect the center of a circle to the point where the circle and a line are tangent, they will form a right angle.

Learn it:Theorem 29.1.19 · Right Angle Tangency Point

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Remark 4.3.17

This property is very useful in circle problems, as it allows us to work with right angles. In addition, another helpful technique is drawing useful radii to various points in your diagram, as that opens up new information to work with.

Learn it:Remark 29.1.20 Concept 34.2.1

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4.4 Power of a Point

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Theorem 4.4.1 · Power of Point For 2 Tangents

From a given point $P$ external to a circle, the two tangents to the circle are equal:

PS = PT.

Learn it:Theorem 34.3.1 · Power of a Point for 2 tangents Section 34.3 Power of a Point

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Theorem 4.4.2 · Power of Point For 2 Secants

If $AB$ and $CD$ are two chords (or secants) of a circle that intersect at a point $P$ , then the line segments satisfy

PA \cdot PB = PC \cdot PD.

Learn it:Theorem 34.3.2 · Power of a Point inside the circle Theorem 34.3.3 · Power of a Point outside the circle

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Theorem 4.4.3 · Power of a Point For Secants and Tangents

If $P$ is a point external to the circle, $PT$ is a tangent to the circle, and the secants $AB$ and $CD$ pass through $P$ , then the line segments satisfy

PA \cdot PB = PC \cdot PD = PT^2.

Learn it:Theorem 34.3.3 · Power of a Point outside the circle Section 34.3 Power of a Point

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Remark 4.4.4

It’s a common mistake to get confused and think $PA \cdot AB = PC \cdot CD$ . Just remember that all the line segments must have the point $P$ as an endpoint (maybe that’s why it is called the “Power of a Point”).

Learn it:Section 34.3 Power of a Point

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Theorem 4.4.5

If $AB$ and $CD$ are two chords of a circle with center $O$ that intersect at a point $P$ inside the circle, then

PA \cdot PB = PC \cdot PD = r^2 - OP^2,

where $r$ is the radius of the circle.

Learn it:Theorem 34.3.2 · Power of a Point inside the circle

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Theorem 4.4.6

If $AB$ and $CD$ are two secants of a circle with center $O$ that intersect at a point $P$ outside the circle, then

PA \cdot PB = PC \cdot PD = OP^2 - r^2,

where $r$ is the radius of the circle.

Learn it:Theorem 34.3.3 · Power of a Point outside the circle

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Remark 4.4.7

Power of a point is useful when dealing with circles and chord lengths.

Learn it:Remark 34.3.4

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Definition 4.4.8 · Curvature

The curvature of a circle is $1$ divided by its radius: $\dfrac{1}{r}$ .

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Theorem 4.4.9

In a tangential quadrilateral (one in which a circle can be inscribed), the two sums of the lengths of opposite sides are the same:

AB + CD = AD + BC.

The two tangents from each corner are equal (Theorem 4.4.1), so with tangent lengths $a$ , $b$ , $c$ , $d$ from $A$ , $B$ , $C$ , $D$ , both sums equal $a + b + c + d$ .

Learn it:Theorem 34.3.1 · Power of a Point for 2 tangents

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Theorem 4.4.10 · Descartes Theorem

Let the radii of 3 mutually externally tangent circles be $r_1$ , $r_2$ and $r_3$ . Let the radius of the big circle around them that is tangent to all three (green in the figure) be $s_1$ . Then

\begin{aligned} &\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} - \frac{1}{s_1}\right)^2 \\ &\quad = 2\left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{s_1^2}\right). \end{aligned}

The term $\frac{1}{s_1}$ is negative because the green circle is internally tangent to the other circles, which means its curvature is negative.

Also, let the radius of the small circle in the gap between them (red) be $s_2$ . Then

\begin{aligned} &\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{s_2}\right)^2 \\ &\quad = 2\left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{s_2^2}\right). \end{aligned}

The term $\frac{1}{s_2}$ is positive because the red circle is externally tangent to the other circles, which means its curvature is positive.

In the figure, $r_1 = r_2 = 3$ , $r_3 = 2$ , $s_1 = 6$ and $s_2 = \frac{2}{5}$ .

Correction

The book prints each squared curvature as $\frac{1}{r_1}^{2}$ , which reads as $\frac{1^2}{r_1} = \frac{1}{r_1}$ ; it means $\left(\frac{1}{r_1}\right)^2 = \frac{1}{r_1^2}$ (and likewise for $r_2$ , $r_3$ , $s_1$ , $s_2$ ).

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Remark 4.4.11

This formula also works if one of the circles is a line instead of a circle. In that case, we can just say $\frac{1}{r} = 0$ .

For example, two circles of radius $r$ that touch each other and a line leave room for a circle of curvature $k$ with $\left(\frac{2}{r} + k\right)^2 = 2\left(\frac{2}{r^2} + k^2\right)$ , so $k = \frac{4}{r}$ : its radius is $\frac{r}{4}$ .

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Theorem 4.4.12 · Carnot’s Theorem

In a triangle $ABC$ with circumcenter $O$ , circumradius $R$ and inradius $r$ , let $P$ , $Q$ and $R$ be the feet of the perpendiculars from $O$ to the sides $AB$ , $BC$ and $CA$ (here $OR$ is a distance to the point $R$ ; $R$ alone is the circumradius). The signed sum of these distances is

OP + OQ + OR = R + r.

Sign: a distance counts as negative when the segment from $O$ to the side lies outside the triangle. For example, if angle $A$ is obtuse, $O$ lies across $BC$ from $A$ , the segment $OQ$ lies outside the triangle, and

OP + OR - OQ = R + r.

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4.5 Area of Complex Shapes

Concept 4.5.1

Tricks to finding the area of complex shapes:

Divide the shape into "nicer" areas which are easier to calculate.
Extend lines. You generally want to extend lines when they form nicer shapes or areas to work with, such as triangles.
Break up areas. A common way to do so is to drop altitudes, as doing so generally allows you to form right triangles.

Learn it:Concept 37.1.1 · Tricks to finding the area of complex shapes Section 37.1 Area of Complex Shapes

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Remark 4.5.2

A common technique is to find the area of the shape one way, then find the area again in terms of a variable (like an altitude, the inradius, the circumradius, etc.) and solve for that variable.

Learn it:Remark 37.1.2

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4.6 Length of complex shapes

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Concept 4.6.1

Finding length of complex shapes:

Equal angles mean equal lengths, and vice versa.
Be on the lookout for $90^\circ$ angles, as you can use the Pythagorean theorem.
Split the length into multiple components by using some of these techniques:
- drawing extra lines;
- dropping altitudes;
- extending lines to create similar triangles, special triangles, etc., and then subtracting the extra length.

Learn it:Concept 37.2.1 · Finding length of complex shapes Section 37.2 Length of Complex Shapes

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4.7 Angle Chasing

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Concept 4.7.1 · Angle Chasing Tricks

The sum of the angles in a triangle is $180^\circ$ .
A triangle with 2 equal angles has the opposite sides equal, and a triangle with 2 equal sides has the opposite angles equal (isosceles triangle).
Opposite angles at intersecting lines are equal.
Corresponding angles at parallel lines are equal.
The angle an arc makes at the center of a circle is double the angle it makes at the boundary of the circle.

Learn it:Section 29.1a Lines, Triangles and Isosceles Chains Theorem 29.1.7 · Inscribed Arc Theorem

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Concept 4.7.2 · Complementary Angle

Complementary angles are a pair of angles with a sum of $90^\circ$ : $a + b = 90^\circ$ .

Learn it:Concept 29.1.1 · Complementary Angles

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Concept 4.7.3 · Supplementary Angle

Supplementary angles are a pair of angles with a sum of $180^\circ$ : $a + b = 180^\circ$ .

Learn it:Concept 29.1.2 · Supplementary Angles

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Concept 4.7.4 · Intersecting lines

When two lines intersect, the vertical angles are equal. Vertical angles are each of the pairs of opposite angles made by two intersecting lines. "Vertical" in this case means they share the same vertex (corner point), not the usual meaning of up-down.

Learn it:Concept 29.1.3 · Intersecting Lines

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Concept 4.7.5

Parallel lines: when a line crosses two parallel lines, corresponding angles are equal. So every angle at the two crossings is either $a$ or $b$ , with $a + b = 180^\circ$ .

Learn it:Concept 29.1.4 · Parallel Lines

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4.8 Polygons

Theorem 4.8.1

For a polygon with $n$ sides:

sum of the interior angles: $(n-2)\cdot 180^\circ$
each interior angle of a regular polygon: $\dfrac{n-2}{n}\cdot 180^\circ$
each exterior angle of a regular polygon: $\dfrac{360^\circ}{n}$

Learn it:Theorem 35.0.1 Theorem 29.1.5

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Fact 4.8.2 · Important Interior Angles

Important interior angles of regular polygons:

Number of sides	Interior angle
3	$60^\circ$
4	$90^\circ$
5	$108^\circ$
6	$120^\circ$
8	$135^\circ$
9	$140^\circ$
10	$144^\circ$

Learn it:Fact 35.0.2 · Important Interior Angles Fact 29.1.6 · Important Interior Angles

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Theorem 4.8.3

For a regular hexagon with side $s$ :

sum of the interior angles: $(6-2)\cdot 180^\circ = 720^\circ$
each interior angle: $\dfrac{6-2}{6}\cdot 180^\circ = 120^\circ$
each exterior angle: $\dfrac{360^\circ}{6} = 60^\circ$
area: $6\cdot\dfrac{\sqrt{3}}{4}s^2$
long diagonal: $2s$

Learn it:Theorem 35.1.1 · Regular Hexagon

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Theorem 4.8.4

For a regular octagon with side $s$ :

sum of the interior angles: $(8-2)\cdot 180^\circ = 1080^\circ$
each interior angle: $\dfrac{8-2}{8}\cdot 180^\circ = 135^\circ$
each exterior angle: $\dfrac{360^\circ}{8} = 45^\circ$
area: $2(1+\sqrt{2})s^2$

Correction

The book prints the interior angle as $\frac{(8-2)}{6}\cdot 180 = 135$ ; that fraction equals $180$ . The denominator is the number of sides, $8$ : $\frac{(8-2)}{8}\cdot 180^\circ = 135^\circ$ .

Learn it:Theorem 35.2.1 · Regular Octagon

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Remark 4.8.5

A regular hexagon can be divided into $6$ congruent equilateral triangles (join the center to the six corners).

Learn it:Remark 35.2.2

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4.9 Cyclic Quadrilaterals

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Theorem 4.9.1 · Properties of a Cyclic quadrilateral

In a cyclic quadrilateral (all four corners on one circle), the sum of opposite angles is $180^\circ$ :

p + q = 180^\circ

r + s = 180^\circ

Learn it:Theorem 34.4.1 · Properties of a cyclic quadrilateral Theorem 29.1.21 · Properties of a Cyclic Quadrilateral

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Theorem 4.9.2 · Ptolemy’s Theorem

In a cyclic quadrilateral $ABCD$ , the product of the diagonals equals the sum of the products of both pairs of opposite sides:

AC \cdot BD = AB \cdot CD + AD \cdot BC

Learn it:Theorem 34.4.3 · Ptolemy's Theorem

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Theorem 4.9.3 · Brahmagupta’s Formula

In a cyclic quadrilateral with side lengths $a, b, c, d$ , the area is

A = \sqrt{(s-a)(s-b)(s-c)(s-d)},

where $s = \dfrac{a+b+c+d}{2}$ is the semiperimeter.

Basically, to find the area of a cyclic quadrilateral:

Find the perimeter and divide by 2.
Subtract each of the side lengths from it to get 4 values.
Multiply your 4 values.
Take the square root of your product.

Learn it:Theorem 34.4.4 · Brahmagupta's Formula

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4.10 3D Geometry

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Theorem 4.10.1 · Volume and Surface Area of a cube

For a cube with side length $a$ :

V = a^3

S = 6a^2

\text{space diagonal} = \sqrt{3}\,a

In words: the volume is the side length cubed, the surface area is $6$ times the side length squared, and the space diagonal is $\sqrt{3}$ times the side length.

Learn it:Theorem 36.1.1 · Volume and Surface Area of a Cube Section 36.1 Cube

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Theorem 4.10.2 · Volume and Surface Area of a rectangular prism

For a rectangular prism (box) with length $l$ , width $w$ and height $h$ :

V = l \times w \times h

S = 2(lw + wh + lh)

\text{diagonal} = \sqrt{l^2 + w^2 + h^2}

The volume is the product of all 3 dimensions; the space diagonal runs from a corner to the opposite corner.

Learn it:Theorem 36.2.1 · Volume and Surface Area of a Rectangular Prism Section 36.2 Rectangular Prism

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Theorem 4.10.3 · Volume and Surface Area of a cylinder

For a cylinder with radius $r$ and height $h$ :

V = \pi r^2 h

\begin{aligned} S &= 2\pi r^2 + 2\pi r h \\ &= 2\pi r(r + h) \end{aligned}

Learn it:Theorem 36.3.1 · Volume and Surface Area of a Cylinder Section 36.3 Cylinder

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Theorem 4.10.4 · Volume and Surface Area of a cone

For a cone with radius $r$ , height $h$ and slant height $s$ :

V = \frac{1}{3}\pi r^2 h

\begin{aligned} S &= \pi r^2 + \pi r s \\ &= \pi r(r + s) \end{aligned}

In words: the volume is $\frac13 \cdot \pi \cdot \text{radius}^2 \cdot \text{height}$ , and the surface area is $\pi \cdot \text{radius}^2 + \pi \cdot \text{radius} \times \text{slant height}$ , where the slant (lateral) height $s$ is measured along the side.

Learn it:Theorem 36.4.1 · Volume and Surface Area of a Cone Section 36.4 Cone

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Remark 4.10.5

The slant height $s$ can be calculated by the formula

s = \sqrt{r^2 + h^2},

that is, $\text{slant height} = \sqrt{\text{radius}^2 + \text{height}^2}$ .

Learn it:Remark 36.4.2 · Slant Height

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Theorem 4.10.6 · Volume and Surface Area of a sphere

For a sphere with radius $r$ :

V = \frac{4}{3}\pi r^3

S = 4\pi r^2

Learn it:Theorem 36.5.1 · Volume and Surface Area of a Sphere Section 36.5 Sphere

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Theorem 4.10.7 · Volume of a tetrahedron

For any tetrahedron:

V = \frac{1}{3} \cdot \text{base area} \cdot \text{height}

Learn it:Theorem 36.6.1 · Volume of a Tetrahedron Section 36.6 Tetrahedron

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Theorem 4.10.8 · Volume of a regular tetrahedron (all sides equal)

A regular tetrahedron (all six edges equal to $s$ ) has volume

V = \frac{\sqrt{2}}{12}s^3

Learn it:Theorem 36.6.2 · Volume of a Regular Tetrahedron

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Theorem 4.10.9 · Volume of a pyramid

For any pyramid:

V = \frac{1}{3} \cdot \text{base area} \cdot \text{height}

Learn it:Theorem 36.7.1 · Volume of a Pyramid Section 36.7 Pyramid

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Theorem 4.10.10 · Volume of a regular pyramid (all sides equal)

When the pyramid has a square base and all the edges (all eight) are equal to $s$ , its volume is

V = \frac{\sqrt{2}}{6}s^3

Learn it:Theorem 36.7.2 · Volume of a Regular Square Pyramid

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Concept 4.10.11 · Cavalieri’s principle

If two solids have equal altitudes, and the sections made by planes parallel to their bases at the same distance from their respective bases are always equal in area, then the volumes of the two solids are equal.

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Theorem 4.10.12 · Euler’s Polyhedron Formula

E = F + V - 2

That is, $\text{Edges} = \text{Faces} + \text{Vertices} - 2$ .

Learn it:Theorem 7.5.2 · Euler's Polyhedron Formula Section 36.8 Volume of Complex Polyhedrons

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Remark 4.10.13

If you ever forget this theorem, just think of a cube and remember that it has $6$ faces, $8$ vertices, and $12$ edges: $12 = 6 + 8 - 2$ .

Learn it:Remark 7.5.3

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Remark 4.10.14

In a polyhedron, if we know (or can work out) how many faces or edges come out of each vertex, and how many vertices each face or edge has, we can easily calculate the number of faces or edges from the number of vertices, and vice versa (each edge is counted twice when you add up around every vertex or every face).

Learn it:Concept 7.5.1 · Counting Edges Twice Remark 7.5.4

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4.11 Advanced Formulas

Definition 4.11.1 · Median

A median is a segment connecting a vertex of a triangle to the midpoint of the opposite side.

Learn it:Definition 30.2.1 · Median

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Definition 4.11.2 · Centroid

In a triangle, the intersection of all 3 medians is the centroid.

Learn it:Definition 30.2.3 · Centroid

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Theorem 4.11.3

The centroid of a triangle lies on each median, $\frac{2}{3}$ of the way from the vertex to the midpoint of the opposite side.

Learn it:Theorem 30.2.4 · Centroid splits each median 2 : 1

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Definition 4.11.4 · Circumcenter

The circumcenter of a triangle is the intersection of all 3 perpendicular bisectors (a perpendicular bisector is the line that bisects a segment and is perpendicular to it). This point is also the center of the circumcircle, and it is equidistant from all three vertices.

Learn it:Definition 30.1.8 · Circumcenter

Book p. 105

Definition 4.11.5 · Incenter

The incenter of a triangle is the intersection of all the angle bisectors. This point is also the center of the incircle, and it is equidistant from all three sides.

Learn it:Definition 30.1.3 · Incenter

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Theorem 4.11.6

The inradius $r$ of a right triangle is

r = \frac{1}{2}(a + b - c),

where $a$ and $b$ are the legs of the triangle and $c$ is the hypotenuse.

Learn it:Theorem 30.1.5 · Inradius of a right triangle

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Definition 4.11.7 · Cevian

A cevian is any segment from a vertex of a triangle to the opposite side. Medians and angle bisectors are special cases of cevians.

Learn it:Definition 30.2.5 · Cevian

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Theorem 4.11.8 · Ceva’s Theorem

In a triangle $ABC$ , let $D$ , $E$ , $F$ be points on lines $BC$ , $CA$ , $AB$ respectively. Lines $AD$ , $BE$ , $CF$ are concurrent if and only if

\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1.

The cevians need not lie within the triangle: the point where they meet may be outside it. Then read each ratio as signed (negative when the point lies outside its side), so exactly two of $D, E, F$ are outside. (With signed ratios, a product of $1$ can also mean the three lines are parallel.)

Correction

The book allows $D, E, F$ outside the sides but states the converse with plain lengths. Then a product of $1$ does not guarantee concurrency: with $D$ , $E$ inside their sides and $F$ moved to the outside point with the same length ratio $AF/FB$ , the product is still $1$ but the lines do not meet. The app reads the ratios as signed.

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Remark 4.11.9

A way to remember this is that you are going around the triangle multiplying ratios: $B \to D \to C \to E \to A \to F \to B$ .

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Theorem 4.11.10 · Menelaus’s Theorem

If a line crosses side $BC$ of $\triangle ABC$ at $P$ , side $AB$ at $R$ , and the extension of $AC$ at $Q$ , then

\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.

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Theorem 4.11.11 · Stewart’s Theorem

Given a triangle $ABC$ with sides of length $a$ , $b$ , $c$ opposite vertices $A$ , $B$ , $C$ respectively. If the cevian $AD$ is drawn so that $BD = m$ , $DC = n$ and $AD = d$ , then

man + dad = bmb + cnc,

that is, $a(mn + d^2) = b^2 m + c^2 n$ .

Learn it:Theorem 30.2.6 · Stewart's Theorem Section 30.2 Triangle Properties

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Remark 4.11.12

A way to remember this is the saying "A man and his dad put a bomb in the sink": $man + dad = bmb + cnc$ .

Learn it:Remark 30.2.7

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Corollary 4.11.13 · Stewart’s For Angle Bisector

If $AD$ is an angle bisector, then

d^2 + mn = bc.

This follows from Stewart’s Theorem and the Angle Bisector Theorem.

Learn it:Corollary 30.2.8 · Stewart's for Angle Bisector

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Corollary 4.11.14 · Stewart’s Theorem For Medians

If $AD$ is a median, then

d^2 = \frac{1}{2}(b^2 + c^2) - \frac{1}{4}a^2.

Learn it:Corollary 30.2.9 · Stewart's Theorem for Medians

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Theorem 4.11.15 · Euler’s Geometry Theorem

The distance $d$ between the circumcenter and the incenter of a triangle is given by

d^2 = R(R - 2r)

or equivalently

\frac{1}{R-d} + \frac{1}{R+d} = \frac{1}{r},

where $R$ and $r$ denote the circumradius and inradius respectively.

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4.12 Coordinate Geometry

Definition 4.12.1 · Equation of a Line

The equation of a line is

ax + by + c = 0.

Learn it:Definition 38.1.1 · Equation of a Line

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Definition 4.12.2 · Slope-Intercept Form

An equation of a line in slope-intercept form is

y = mx + b,

where $m$ is the slope and $b$ is the $y$ -intercept.

Learn it:Definition 38.1.2 · Slope-Intercept Form

Book p. 110

Theorem 4.12.3 · Slope of a line through 2 points

The slope of a line passing through 2 points $(x_1, y_1)$ and $(x_2, y_2)$ is

\frac{y_2 - y_1}{x_2 - x_1}.

Learn it:Theorem 38.1.3 · Slope of a line through 2 points

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Theorem 4.12.4 · Slope of a line through angle

The slope of a line with an angle of $\theta$ above the $x$ -axis is

\tan\theta.

Learn it:Theorem 38.1.4 · Slope of a line through angle

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Theorem 4.12.5 · Distance between 2 points

The distance between 2 points $(x_1, y_1)$ and $(x_2, y_2)$ is

\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.

Learn it:Theorem 38.1.5 · Distance between 2 points

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Theorem 4.12.6 · Point to line formula

The distance between a point $(x_0, y_0)$ and a line $ax + by + c = 0$ is

\frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}.

Learn it:Theorem 38.1.6 · Point to line formula

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Remark 4.12.7

Be careful not to get the equation of the line confused with $ax + by = c$ .

Learn it:Remark 38.1.7

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Remark 4.12.8

Note that this distance is the shortest possible distance from the point to the line, which is the length of the perpendicular.

Learn it:Remark 38.1.8

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Remark 4.12.9

This formula is a bit confusing, so an easy way to remember the numerator is that it’s just the equation of the line with the coordinates of the point plugged in as the $x$ and $y$ values.

Learn it:Remark 38.1.9

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Theorem 4.12.10 · Equation of a Circle

A circle with center $(a, b)$ and radius $r$ has equation

(x - a)^2 + (y - b)^2 = r^2.

Learn it:Theorem 38.2.1 · Equation of a Circle Section 38.2 Circle in Coordinate Plane

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Theorem 4.12.11 · Equation of a Parabola

A parabola with vertex $(a, b)$ has an equation of

y = k(x - a)^2 + b

for any real non-zero constant $k$ .

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Theorem 4.12.12 · Equation of an Ellipse

An ellipse has an equation of

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,

where the lengths of the 2 radii (semi-axes) of the ellipse are $a$ and $b$ .

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Remark 4.12.13

Note that when $a = b$ , it becomes a circle.

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Theorem 4.12.14 · Equation of a Hyperbola

A hyperbola has an equation of

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.

Its vertices are $(\pm a, 0)$ , and its asymptotes $y = \pm\frac{b}{a}x$ are the diagonals of the $2a \times 2b$ rectangle centered at the origin; the curve stays outside that rectangle. A hyperbola consists of 2 symmetric parts that are reflections of each other across the $y$ -axis.

Correction

The book says the hyperbola "can fit inside" a rectangle of width $a$ and length $b$ . A hyperbola is unbounded and lies outside the guide rectangle, which measures $2a \times 2b$ ; the rectangle's diagonals are the asymptotes $y = \pm\frac{b}{a}x$ .

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