Page:Amusements in mathematics.djvu/198

195.—LADY BELINDA'S GARDEN.

{{Smallcaps|All that Lady Belinda need do was this: She should measure from A to B, fold her tape in four and mark off the point E, which is thus one quarter of the side. Then, in the same way, mark off the point F, one-fourth of the side AD. Now, if she makes EG equal to AF, and GH equal to EF, then AH is the required width for the path in order that the bed shall be exactly half the area of the garden. An exact numerical measurement can only be obtained when the sum of the squares of the two sides is a square number. Thus, if the garden measured 12 poles by 5 poles (where the squares of 12 and 5, 144 and 25, sum to 169, the square of 13), then 12 added to 5, less 13, would equal four, and a quarter of this, 1 pole, would be the width of the path.

196.—THE TETHERED GOAT.

This problem is quite simple if properly attacked. Let us suppose the triangle ABC to represent our half-acre field, and the shaded portion to be the quarter-acre over which the goat will graze when tethered to the comer C. Now, as six equal equilateral triangles placed together will form a regular hexagon, as shown, it is evident that the shaded pasture is just one-sixth of the complete area of a circle. Therefore all we require is the radius (CD) of a circle containing six quarter-acres or 1½ acres, which is equal to 9,408,960 square inches. As we only want our answer "to the nearest inch," it is sufficiently exact for our purpose if we assimae that as i is to 3.1416, so is the diameter of a circle to its circumference. If, therefore, we divide the last number I gave by 3. 141 6, and extract the square root, we find that 1,731 inches, or 48 yards 3 inches, is the required length of the tether " to the nearest inch."

197.—THE COMPASSES PUZZLE.

Let AB in the following diagram be the given straight line. With the centres A and B and radius AB describe the two circles. Mark off DE and EF equal to AD. With the centres A and F and radius DF describe arcs intersecting at G. With the centres A and B and distance B G describe arcs GHK and N. Make HK equal to AB and HL equal to HB. Then with centres K and L and radius A B describe arcs intersecting at I. Make BM equal to B I. Finally, with

the centre M and radius MB cut the line in C, and the point C is the required middle of the line AB. For greater exactitude you can mark off R from A (as you did M from B), and from R describe another arc at C. This also solves the problem, to find a point midway between two given points without the straight line. I win put the young geometer in the way of a rigid proof. First prove that twice the square of the line AB equals the square of the distance BG, from which it follows that HABN are the four comers of a square. To prove that I is the centre of this square, draw a line from H to P through QIB and continue the arc HK to P. Then, conceiving the necessary lines to be drawn, the angle HKP, being in a semicircle, is a right angle. Let fall the perpendicular KQ, and by similar triangles, and from the fact that HKI is an isosceles triangle by the construction, it can be proved that HI is half of HB. We can similarly prove that C is the centre of the square of which AIB are three corners.

I am aware that this is not the simplest possible solution.

198.—THE EIGHT STICKS.

The first diagram is the answer that nearly every one will give to this puzzle, and at first sight it seems quite satisfactory. But consider the conditions. We have to lay "every one of the sticks on the table." Now, if a ladder be