If you follow me on Twitter, you might have noticed that I’m a fan of legal blogger David Allen Green – not because I always agree with him, but because even when I don’t, he sets out a clear and compelling case for why I should.

He’s famed for writing in one-sentence paragraphs, because there’s “no place to hide” – it forces him to distil his argument down to its absolute key points, avoiding Johnsonesque word salad.

So, I wondered, what can the world of legal writing tell us about mathematical communication?

### IRAC

I stumbled on an interesting cheat-sheet by Izzy (theheartbeating), about a “problem-solving technique for legal issues((I would say it’s as much a presentation technique, but tomatoes, edible berries of the plant Solanum lycopersicum.))” called IRAC.

So what is it ((For those of you who don’t want to read the sheet, for some reason))?

It stands for:

• Issue
• Rule
• Analysis
• Conclusion

That is:

• State the problem you’re faced with
• State the rule or rules you rely on in your answer
• Write down how the rule applies to the problem

### An example

Let’s try it, based on a Madas IYGB A-level question (it’s question 1 on this paper). I’ve already solved the question, and am now trying to communicate my solution clearly.

#### Issue

Here, I’m just going to paraphrase the question. A diagram wouldn’t go amiss.

Square ABCD has sides of length 2. Point M is the midpoint of CD. Points A, B and M lie on a circle. What is the radius of the circle?

#### Rule

I’m going to state the main rule I rely on in my solution.

The intersecting chord theorem states that if two chords of a circle, PQ and RS, meet at a point X inside the circle, then $|PX||XQ| = |RX||XQ|$.

### Analysis

A run down of my reasoning, in one-word paragraphs.

• Let MN be the diameter of the circle perpendicular to CD.
• Let X be the point where MN meets AB.
• We know that:
• $|AX| = |XB| = 1$, since the diameter bisects the chord
• $|MX| = 2$, since MX is the shortest distance between AB and CD
• Applying the rule as $|AX||XB| = |MX||XN|$, we get $(1)(1) = (2)|XN|$
• Thus $|XN| = \frac{1}{2}$
• The length of the diameter is therefore $2 + \frac{1}{2} = \frac{5}{2}$.

### Conclusion

And finally…

The radius of the circle is half of the diameter, or $\frac{5}{4}$

I think that’s a clear and tidy way to present the solution to a question! In an exam, you may not have the time or inclination to go through the rigmarole, but exams aren’t proper maths. If you’re trying to communicate your thinking – to someone else, or to future-you – then taking the time to refine and express your thoughts clearly will pay off.