151.--THE JOINER'S PROBLEM.

[Illustration]

Nothing could be easier than the solution of this puzzle--when you know

how to do it. And yet it is apt to perplex the novice a good deal if he

wants to do it in the fewest possible pieces--three. All you have to do

is to find the point A, midway between B and C, and then cut from A to D

and from A to E. The three pieces then form a square in the manner

shown. Of course, the proportions of the original figure must be

correct; thus the triangle BEF is just a quarter of the square BCDF.

Draw lines from B to D and from C to F and this will be clear.

Amusements in Mathematics · 第 659 / 970 题