Lesson · Metric toolkit
The metric toolkit
The selection rounds rarely hand you a length; they hide it in a figure. The way out is almost always the same three moves: name an unknown, find a right triangle that contains it, and let Pythagoras write the equation. This lesson collects the standing equipment - where the right triangles come from, the two triangles whose sides you should know by heart, and how to leave a result exact.
Pythagoras writes equations
Principle
To compute a length, do not hunt for a formula that names it. Call it \(x\), find a right triangle whose three sides can be written with \(x\) and known lengths, and write \(a^2 + b^2 = c^2\). At that moment the geometry is over; what remains is algebra. If one triangle does not determine \(x\), a second right triangle sharing a side with the first supplies a second equation.
The right triangles are rarely printed in the figure - you drop them in. The standard sources: a perpendicular from a point to a line, an altitude of a triangle, half of an isosceles triangle cut along its axis of symmetry, and, in circles, the radius drawn to the end of a chord: the perpendicular from the centre bisects the chord, so half the chord, the distance to the centre and the radius always close up a right triangle.
Example
A chord of a circle of radius \(5\) has length \(8\). How far is it from the centre? Drop the perpendicular from the centre \(O\) to the chord; its foot \(N\) is the chord's midpoint, so \(|NA| = 4\) for an endpoint \(A\). In the right triangle \(ONA\) the hypotenuse is the radius: \(|ON|^2 = 5^2 - 4^2 = 9\), and the distance is \(3\).
The altitude to the hypotenuse
Theorem
In a right triangle with legs \(a\), \(b\) and hypotenuse \(c\), let the altitude \(h\) from the right angle split \(c\) into the pieces \(p\) (adjacent to \(a\)) and \(q\) (adjacent to \(b\)). Then
The altitude cuts the triangle into two smaller ones, each right-angled and each sharing an acute angle with the whole - so both are similar to the whole triangle and to each other, and the three relations are the resulting side ratios.
Example
Legs \(3\) and \(4\): the hypotenuse is \(5\), and the area computed two ways gives \(3 \cdot 4 = 5h\), so \(h = \frac{12}{5}\). From \(a^2 = pc\) the pieces are \(p = \frac{9}{5}\) and \(q = \frac{16}{5}\), and indeed \(pq = \frac{144}{25} = h^2\).
In practice the fastest route to \(h\) is usually the one just used: the area written two ways gives \(ab = ch\) in one line. But the configuration itself - one altitude, three similar triangles - is worth recognising on sight, because it converts a single right triangle into a small system of solvable equations.
Lookup triangles and clean surds
Two right triangles occur so often that solving them each time wastes competition minutes. Both come from cutting a familiar figure in half.
| Triangle | Sides | How it announces itself |
|---|---|---|
| half a square | \(1 : 1 : \sqrt{2}\) | isosceles and right; any diagonal of a square, any \(45^\circ\) angle |
| half an equilateral triangle | \(1 : \sqrt{3} : 2\) | a \(30^\circ\) or \(60^\circ\) angle - or a hypotenuse exactly twice a leg |
The last column works in both directions. If a computation returns a right triangle whose hypotenuse is twice one leg, you have just been handed the angles \(30^\circ\) and \(60^\circ\) for free - and often an equilateral triangle hiding nearby.
Principle
Results stay exact. Pull squares out of roots, \(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\); clear a root from a denominator with the conjugate, \(\frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} = \sqrt{2}-1\); and compare two positive surd expressions by squaring both, never by approximating either.
Warning
No calculator sits on the competition desk, so \(\sqrt{3} \approx 1.73\) is not a step of a solution - and a rounded value can point at the wrong answer when two quantities are close. Keep \(\sqrt{3}\) and \(\pi\) as letters to the very end and settle comparisons with algebra: \(2\sqrt{3} > 3\) because \(12 > 9\).
Warm-up
Example
The legs of a right triangle measure \(1\) and \(2\). Compute the altitude to the hypotenuse, and give the result with a rationalised denominator.
Begin with the side you were not given
Pythagoras first: the hypotenuse is \(c = \sqrt{1^2 + 2^2} = \sqrt{5}\). Do not touch a decimal - \(\sqrt{5}\) is a perfectly good name for it.
The area does not care which base you choose
Computed from the legs the area is \(\frac{1 \cdot 2}{2} = 1\); computed from the hypotenuse it is \(\frac{\sqrt{5}\, h}{2}\). Equating the two gives \(h = \frac{2}{\sqrt{5}}\).
The answer is not yet in final form
Multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\): \(h = \frac{2\sqrt{5}}{5}\). As a check, the pieces of the hypotenuse are \(p = \frac{1}{\sqrt{5}}\) and \(q = \frac{4}{\sqrt{5}}\) from \(a^2 = pc\), and \(pq = \frac{4}{5} = h^2\), as the altitude theorem demands.