# STEP I 2001 Q.1 (triangle in square)

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 $\sqrt{6}-\sqrt{2}$ cm.

Notes

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 $= \frac{1}{\cos {15{^\circ}}}$

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.