Page:Amusements in mathematics.djvu/201

Rh length, Fig. 1 contains two square inches and Fig. 2 contains six square inches—4×1½. 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.

205.—THE SIX SHEEP-PENS.

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

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.

207.—CHERRIES AND PLUMS.

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.

208.— A PLANTATION PUZZLE.

illustration shows the ten trees that must be left to form five rows with four trees in every