Math Core

Synthetic tricks are elegant, but they aren't always available at 2 a.m. on problem 11. Coordinates and trigonometry are the reliable backup: with a smart setup, almost any triangle problem becomes a computation. This module collects the setups and formulas that make those computations short.

Choosing coordinates

A good coordinate system does most of the work. Some habits:

Put a vertex at the origin and a side along the $x$ -axis: $B = (0, 0)$ , $C = (a, 0)$ , $A = (p, q)$ .
Use symmetry: an isosceles triangle or a rectangle should be centered on an axis.
Put a circle's center at the origin.
Prefer integer coordinates. The $13$ - $14$ - $15$ triangle fits perfectly as $B = (0, 0)$ , $C = (14, 0)$ , $A = (5, 12)$ .

To place $A$ when you know all three sides, solve $p^2 + q^2 = c^2$ and $(p - a)^2 + q^2 = b^2$ . Subtracting gives $p = \dfrac{a^2 + c^2 - b^2}{2a}$ right away.

Area and distance formulas

Shoelace formula

A polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ in order around it has area

$\frac{1}{2}\left|\sum_{i=1}^{n} (x_i y_{i+1} - x_{i+1} y_i)\right|, \qquad (x_{n+1}, y_{n+1}) = (x_1, y_1).$

Why it works. The term $\tfrac12(x_i y_{i+1} - x_{i+1} y_i)$ is the signed area of the triangle with vertices $O$ , $P_i$ , $P_{i+1}$ : positive when you turn counterclockwise from $P_i$ to $P_{i+1}$ and negative otherwise. Walking around a polygon, the triangles fanned out from $O$ add up to the polygon's area, and any part outside the polygon is counted once positively and once negatively, so it cancels.

Distance from a point to a line. The distance from $(x_0, y_0)$ to the line $ax + by + c = 0$ is

$\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$

The vector $(a, b)$ is perpendicular to the line. Moving from $(x_0, y_0)$ a distance $t$ along the unit normal changes $ax + by + c$ by $t\sqrt{a^2 + b^2}$ , so you reach the line (where it's $0$ ) when $|t| = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ .

The trigonometry toolkit

For a triangle with sides $a, b, c$ opposite angles $A, B, C$ , circumradius $R$ , inradius $r$ and semiperimeter $s$ :

Formula	Name
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$	extended law of sines
$c^2 = a^2 + b^2 - 2ab\cos C$	law of cosines
$K = \tfrac12 ab\sin C = \dfrac{abc}{4R} = rs = \sqrt{s(s-a)(s-b)(s-c)}$	area

Proof of the extended law of sines. Draw the diameter $BB'$ of the circumcircle. Inscribed angles on arc $BC$ are equal, so $\angle BB'C = A$ (or $180^\circ - A$ if $A$ is obtuse, with the same sine). Triangle $BB'C$ has a right angle at $C$ because $BB'$ is a diameter. So $a = BC = BB' \sin A = 2R\sin A$ .

The others follow: $K = \tfrac12 ab\sin C = \tfrac12 ab \cdot \dfrac{c}{2R} = \dfrac{abc}{4R}$ , and splitting the triangle into three triangles from the incenter, each with height $r$ , gives $K = \tfrac12 r(a + b + c) = rs$ .

Stewart's theorem

If $D$ is on side $BC$ with $BD = m$ , $DC = n$ (so $a = m + n$ ) and $AD = d$ , then

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

Proof. Let $\theta = \angle ADB$ , so $\angle ADC = 180^\circ - \theta$ and the cosines are opposite. The law of cosines in triangles $ABD$ and $ACD$ :

$c^2 = m^2 + d^2 - 2md\cos\theta, \qquad b^2 = n^2 + d^2 + 2nd\cos\theta.$

Multiply the first by $n$ and the second by $m$ and add; the cosine terms cancel:

$c^2 n + b^2 m = mn(m + n) + d^2(m + n) = a(d^2 + mn). \quad \blacksquare$

Special cases worth memorizing: the median is $m_a^2 = \dfrac{2b^2 + 2c^2 - a^2}{4}$ , and the angle bisector (with $m : n = c : b$ ) is $t_a^2 = bc\left(1 - \dfrac{a^2}{(b + c)^2}\right)$ .

The 13-14-15 triangle with B = (0, 0), C = (14, 0), A = (5, 12). The incenter is (6, 4) and the circumcenter is (7, 33/8).Open in grapher →

Worked example: The 13-14-15 triangle

Find the area, inradius and circumradius of the triangle with sides $13$ , $14$ , $15$ .

With $A = (5, 12)$ above base $BC = 14$ , the area is $\tfrac12 \cdot 14 \cdot 12 = 84$ . Then $s = 21$ , so $r = \dfrac{84}{21} = 4$ , and $R = \dfrac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \dfrac{2730}{336} = \dfrac{65}{8}$ .

In coordinates: the circumcenter is on $x = 7$ (the perpendicular bisector of $BC$ ), at $(7, y)$ with $49 + y^2 = 4 + (12 - y)^2$ , so $y = \dfrac{33}{8}$ and $R^2 = 49 + \dfrac{1089}{64} = \dfrac{4225}{64}$ . ✓

Worked example: A median with Stewart

In a triangle with $AB = 7$ , $AC = 9$ , $BC = 8$ , find the length of the median from $A$ .

$m_a^2 = \dfrac{2 \cdot 81 + 2 \cdot 49 - 64}{4} = \dfrac{196}{4} = 49$ , so the median has length $7$ .

Worked example: Trig with an altitude

In triangle $ABC$ the altitude $AD$ has $BD = 4$ and $DC = 6$ , and $\angle BAC = 45^\circ$ . Find $AD$ .

Let $h = AD$ , and split $\angle A$ into $\alpha = \angle BAD$ and $\beta = \angle DAC$ , with $\tan\alpha = \dfrac{4}{h}$ and $\tan\beta = \dfrac{6}{h}$ . Then

$1 = \tan(\alpha + \beta) = \frac{4/h + 6/h}{1 - 24/h^2} = \frac{10h}{h^2 - 24}.$

So $h^2 - 10h - 24 = 0$ , and $h = 12$ .

Common mistake

When you place a triangle in coordinates, check which side is opposite which vertex. In the incenter formula $I = \dfrac{aA + bB + cC}{a + b + c}$ , each vertex is weighted by the length of the opposite side. Mixing these up is the most common coordinate bug.

Tip

Before a long coordinate bash, estimate the answer with a rough sketch. If your exact answer is $\dfrac{65}{8} \approx 8.1$ and your picture says "about $8$ ", you can trust the algebra.

Practice

Practice 1

Find the area of the pentagon with vertices $(1, 2)$ , $(9, 1)$ , $(12, 8)$ , $(6, 13)$ and $(0, 7)$ , in that order.