STEP I/2001/1 – The points A, B, and C lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle ABC must be less than or equal to cm.
1. The working from the point where we have narrowed it down to equilateral triangles with a vertex in a corner of a square can also be done by trig, by saying:
side of triangle
That’s fine. Doing it by Pythagoras is a bit simpler and draws on less knowledge (you don’t have to remember your cos(A-B) formula).
2. The main part of any good solution, however, is the reasoning which explains why you can narrow it down to equilateral triangles with a vertex in a corner of a square. The working from that point on is more or less like A level. The argument up to that point not technical or difficult, but it’s the sort of thing you do a lot in STEP and scarcely at all in A level.
3. How do you know to go that way? You can make a good guess that the triangles of maximum shortest side will be equilateral, because the triangle of maximum shortest side for a given area is equilateral. Then sketching a few triangles-in-squares on rough paper will convince you that you can always enlarge a triangle by moving one vertex into a corner.