1.  The easiest way is to arrange the eighteen matches as in Diagrams 1 and 2, making the length of the perpendicular AB equal to a match and a half.  Then, if the matches are an inch in length, Fig. 1 contains two square inches and Fig. 2 contains six square inches—­4 x 11/2.  The second case (2) is a little more difficult to solve.  The solution is given in Figs. 3 and 4.  For the purpose of construction, place matches temporarily on the dotted lines.  Then it will be seen that as 3 contains five equal equilateral triangles and 4 contains fifteen similar triangles, one figure is three times as large as the other, and exactly eighteen matches are used.

[Illustration:  Figures 1, 2, 3, 4.]

205.—­THE SIX SHEEP-PENS.

[Illustration] Place the twelve matches in the manner shown in the illustration, and you will have six pens of equal size.

206.—­THE KING AND THE CASTLES.

There are various ways of building the ten castles so that they shall form five rows with four castles in every row, but the arrangement in the next column is the only one that also provides that two castles (the greatest number possible) shall not be approachable from the outside.  It will be seen that you must cross the walls to reach these two.

[Illustration:  The King and the Castles]

207.—­CHERRIES AND PLUMS.

There are several ways in which this problem might be solved were it not for the condition that as few cherries and plums as possible shall be planted on the north and east sides of the orchard.  The best possible arrangement is that shown in the diagram, where the cherries, plums, and apples are indicated respectively by the letters C, P, and A. The dotted lines connect the cherries, and the other lines the plums.  It will be seen that the ten cherry trees and the ten plum trees are so planted that each fruit forms five lines with four trees of its kind in line.  This is the only arrangement that allows of so few as two cherries or plums being planted on the north and east outside rows.

[Illustration]

208.—­A PLANTATION PUZZLE.

The illustration shows the ten trees that must be left to form five rows with four trees in every row.  The dots represent the positions of the trees that have been cut down.

[Illustration]

209.—­THE TWENTY-ONE TREES.

I give two pleasing arrangements of the trees.  In each case there are twelve straight rows with five trees in every row.

[Illustration:  Figure 1, Figure 2.]

210.—­THE TEN COINS.

The answer is that there are just 2,400 different ways.  Any three coins may be taken from one side to combine with one coin taken from the other side.  I give four examples on this and the next page.  We may thus select three from the top in ten ways and one from the bottom in five ways, making fifty.  But we may also select three from the bottom and one from the top in fifty ways.  We may thus select the four coins in one hundred ways, and the four removed may be arranged by permutation in twenty-four ways.  Thus there are 24 x 100 = 2,400 different solutions.