Math Core

On the AMC 10 and 12, "find the area" is rarely just base times height. The winning move is usually picking the right formula for the information you have, or using area as a bridge to a length you actually want: an altitude, an inradius, a circumradius. This module collects the formulas every serious AMC student knows cold, and the area-ratio tricks that make them powerful.

The formula toolkit

Label the triangle with sides $a, b, c$ opposite angles $A, B, C$ , and semiperimeter $s = \dfrac{a + b + c}{2}$ . Write $[ABC]$ for the area of triangle $ABC$ .

You know	Use	Formula
a side and its altitude	base–height	$[ABC] = \tfrac12 a h_a$
two sides and the included angle	sine formula	$[ABC] = \tfrac12 ab \sin C$
all three sides	Heron	$[ABC] = \sqrt{s(s-a)(s-b)(s-c)}$
the inradius $r$	inradius formula	$[ABC] = rs$
the circumradius $R$	circumradius formula	$[ABC] = \dfrac{abc}{4R}$

Each one comes from the first. The sine formula is base–height with $h = b \sin C$ . The inradius formula comes from cutting the triangle into three pieces from the incenter: each piece has a side as its base and $r$ as its height, so the area is $\tfrac12 ar + \tfrac12 br + \tfrac12 cr = rs$ .

Area is a bridge

Compute the area once, in whatever way is easiest. Then read off any length you need:

$h_a = \frac{2[ABC]}{a}, \qquad r = \frac{[ABC]}{s}, \qquad R = \frac{abc}{4[ABC]}.$

Heron's formula is at its best on the "nice" triangles that the AMC loves because their areas are integers: $13$ - $14$ - $15$ (area $84$ ), $5$ - $5$ - $6$ (area $12$ ), $5$ - $5$ - $8$ (area $12$ ), $10$ - $17$ - $21$ (area $84$ ), $9$ - $10$ - $17$ (area $36$ ). The $13$ - $14$ - $15$ triangle is two right triangles glued along the altitude to the side of length $14$ : a $5$ - $12$ - $13$ and a $9$ - $12$ - $15$ .

The 13-14-15 triangle: the altitude of length 12 splits the base of 14 into 5 and 9.

Worked example: One triangle, every length

A triangle has sides $13$ , $14$ and $15$ . Find its area, its inradius, its circumradius and its shortest altitude.

$s = 21$ , so Heron gives $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$ .

Inradius: $r = \dfrac{84}{21} = 4$ .
Circumradius: $R = \dfrac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \dfrac{2730}{336} = \dfrac{65}{8}$ .
The shortest altitude goes to the longest side: $h = \dfrac{2 \cdot 84}{15} = \dfrac{56}{5}$ .

Worked example: The sine formula

Two sides of a triangle have lengths $8$ and $10$ , and the angle between them is $150^\circ$ . What is the area?

$[ABC] = \tfrac12 \cdot 8 \cdot 10 \cdot \sin 150^\circ = 40 \cdot \tfrac12 = 20$ .

Notice that an angle of $30^\circ$ would give the same area, because $\sin 150^\circ = \sin 30^\circ$ . Supplementary angles always give the same area; keep that in mind when a problem has two possible triangles.

Area ratios

Many AMC area problems never ask for a single area. They ask for a ratio, and the tool is this:

Same height, areas proportional to bases

Two triangles with the same height have areas in the ratio of their bases. In particular, if $D$ is on side $BC$ of triangle $ABC$ , then

$\frac{[ABD]}{[ADC]} = \frac{BD}{DC}.$

More generally, if $D$ is on ray $AB$ and $E$ is on ray $AC$ , then $\dfrac{[ADE]}{[ABC]} = \dfrac{AD}{AB} \cdot \dfrac{AE}{AC}$ (use the sine formula with the shared angle $A$ ).

Worked example: Chaining ratios

Triangle $ABC$ has area $60$ . Point $D$ is on $BC$ with $BD : DC = 2 : 3$ , and $E$ is the midpoint of $AD$ . Find $[ABE]$ .

Triangles $ABD$ and $ADC$ share the height from $A$ , so $[ABD] = \tfrac25 \cdot 60 = 24$ . Triangles $ABE$ and $EBD$ share the height from $B$ and have equal bases $AE = ED$ , so $[ABE] = \tfrac12 \cdot 24 = 12$ .

Worked example: Area gives a hidden constant

Point $P$ lies inside an equilateral triangle with side $6$ . What is the sum of the distances from $P$ to the three sides?

Connect $P$ to the three vertices, cutting the triangle into three triangles with base $6$ and heights $d_1, d_2, d_3$ . Then

$\tfrac12 \cdot 6 (d_1 + d_2 + d_3) = \tfrac{\sqrt3}{4} \cdot 36 = 9\sqrt3,$

so $d_1 + d_2 + d_3 = 3\sqrt3$ , the height of the triangle, no matter where $P$ is.

Common mistake

In $[ABC] = \tfrac12 ab \sin C$ , the angle must be the one between the two sides you use. If you know two sides and a non-included angle, find the included angle first (or use a different formula).

Tip

Before running Heron on big numbers, look for an altitude that splits the triangle into two Pythagorean triangles. For $10$ - $17$ - $21$ , the altitude to the side $21$ is $8$ , splitting it into $6 + 15$ (a $6$ - $8$ - $10$ and an $8$ - $15$ - $17$ ).

Practice

Practice 1

What is the inradius of a triangle with side lengths $5$ , $5$ and $6$ ?

Practice 2

Two sides of a triangle have lengths $7$ and $12$ , and the angle between them measures $30^\circ$ . What is the area of the triangle?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A triangle has side lengths $10$ , $17$ and $21$ . What is the length of its shortest altitude?

Practice 4

Triangle $ABC$ has area $90$ . Point $D$ lies on $\overline{AB}$ with $AD : DB = 1 : 2$ , and point $E$ lies on $\overline{AC}$ with $AE : EC = 3 : 2$ . What is the area of triangle $ADE$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

What is the circumradius of a triangle with side lengths $5$ , $5$ and $8$ ?

Practice 6

A right triangle has perimeter $60$ and inradius $5$ . What is the length of its hypotenuse?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Triangle $ABC$ has $AB = 10$ , $AC = 16$ and area $40\sqrt3$ . What is the sum of all possible values of $BC^2$ ?

Practice 8

In triangle $ABC$ , point $D$ is on $\overline{BC}$ with $BD : DC = 1 : 3$ , point $E$ is on $\overline{CA}$ with $CE : EA = 1 : 3$ , and point $F$ is on $\overline{AB}$ with $AF : FB = 1 : 3$ . Segments $\overline{AD}$ , $\overline{BE}$ and $\overline{CF}$ bound a small triangle in the middle. What fraction of the area of $ABC$ is this small triangle?

Real contest practice

2018 AMC 10A, Problem 24: area of a quadrilateral cut out by midpoints and an angle bisector, using area ratios and the sine formula.
2017 AMC 10B, Problem 21: inradii of two triangles, found from $[ABC] = rs$ .