Math Core

The Pythagorean theorem is the most-used tool in contest geometry. Any time a problem has a right angle, or you can create one by drawing an altitude, a missing length is usually one $a^2 + b^2 = c^2$ away. Strong competitors also recognize the common right triangles on sight, which turns many problems into mental math.

The Pythagorean theorem

In a right triangle with legs $a$ and $b$ and hypotenuse $c$ (the side across from the right angle),

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

It also works backward: if the sides of a triangle satisfy $a^2 + b^2 = c^2$ , the triangle has a right angle opposite $c$ .

Pythagorean triples

Whole-number solutions show up constantly. Memorize these, along with their multiples:

Triple	Common multiples
$3, 4, 5$	$6, 8, 10$ ; $9, 12, 15$ ; $15, 20, 25$
$5, 12, 13$	$10, 24, 26$
$8, 15, 17$	$16, 30, 34$
$7, 24, 25$	$14, 48, 50$
$20, 21, 29$

If a right triangle has legs $12$ and $16$ , notice that it is $4$ times a $3, 4, 5$ triangle, so the hypotenuse is $20$ . No squaring needed.

Special right triangles

Two right triangles have fixed shapes, so their side ratios never change.

Special right triangles

$45^\circ$ - $45^\circ$ - $90^\circ$ : sides in the ratio $1 : 1 : \sqrt{2}$ . The hypotenuse is $\sqrt{2}$ times a leg. This is half of a square cut along its diagonal.
$30^\circ$ - $60^\circ$ - $90^\circ$ : sides in the ratio $1 : \sqrt{3} : 2$ . The hypotenuse is twice the shortest leg, and the longer leg is $\sqrt{3}$ times the shortest leg. This is half of an equilateral triangle.

An equilateral triangle with side $s$ has height $\dfrac{\sqrt{3}}{2}s$ and area $\dfrac{\sqrt{3}}{4}s^2$ .

Both come straight from the Pythagorean theorem. For example, cut an equilateral triangle with side $2$ in half: the halves have hypotenuse $2$ and short leg $1$ , so the other leg is $\sqrt{4 - 1} = \sqrt{3}$ .

The triangle inequality

Three lengths form a triangle only if each one is less than the sum of the other two. In practice, check that the longest side is less than the sum of the other two. So if two sides are $a$ and $b$ , the third side $x$ must satisfy $|a - b| < x < a + b$ .

Worked example: The sliding ladder

A $25$ -foot ladder leans against a vertical wall with its foot $7$ feet from the wall. The top slides down $4$ feet. How far does the foot slide out?

The ladder before (top at 24, foot at 7) and after sliding.

At first the top is $\sqrt{25^2 - 7^2} = \sqrt{576} = 24$ feet high (a $7, 24, 25$ triangle). After sliding, it is $20$ feet high. The foot is now $\sqrt{25^2 - 20^2} = 15$ feet from the wall (a $15, 20, 25$ triangle). The foot slid $15 - 7 = 8$ feet.

Worked example: Area of an isosceles triangle

Find the area of a triangle with sides $13$ , $13$ and $10$ .

The altitude to the base splits the triangle into two right triangles.

The altitude to the side of length $10$ cuts it into two halves of $5$ , since the triangle is isosceles. Each half is a right triangle with hypotenuse $13$ and leg $5$ , so the altitude is $12$ . The area is $\dfrac{1}{2} \cdot 10 \cdot 12 = 60$ .

Worked example: An equilateral triangle

An equilateral triangle has side length $6$ . Find its height and its area.

The height splits an equilateral triangle into two 30-60-90 triangles.

The height splits it into two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles with hypotenuse $6$ and short leg $3$ . The height is the long leg, $3\sqrt{3}$ . The area is $\dfrac{1}{2} \cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3}$ , which matches $\dfrac{\sqrt{3}}{4} \cdot 6^2 = 9\sqrt{3}$ .

Worked example: Counting possible third sides

A triangle has sides of length $5$ and $9$ . How many whole-number lengths are possible for the third side?

The third side $x$ must satisfy $9 - 5 < x < 9 + 5$ , so $4 < x < 14$ . The possible values are $5, 6, \dots, 13$ : that's $9$ lengths.

Common mistake

The hypotenuse is always the longest side, across from the right angle. If a problem gives two sides of a right triangle, check which one is the hypotenuse before you add or subtract squares. Legs $5$ and $12$ give hypotenuse $13$ , but leg $5$ and hypotenuse $12$ give other leg $\sqrt{119}$ .

Tip

Before squaring big numbers, divide out a common factor. For legs $36$ and $48$ , divide by $12$ to get $3$ and $4$ . The hypotenuse is $12 \cdot 5 = 60$ .