NUTS TO CRACK.

Otago Witness, Issue 3830, 9 August 1927, Page 19

NUTS TO CRACK.

By

T. L. Briton.

(For the Otago Witness.) Readers with a little Ingenuity will find In ds column an abundant ■tore of entertainment and amusement, and the solving of the problems should provide excellent mental exhilaration. While some of the "nuts” may appear harder than others, ft will be found that none will require a sledge-hammer to crack them. Solutions will appear in our next issue together with some fresh "auts.” Readers - re requested not to send in their solutions, unless these are ' specially asked for, but to keep them for comparison with those published in the issue following the publication of the problems. THE FARMER’S WALK. A discussion in which tho interested persons could not agree, was proceeding between a farmer and the rural mailman, when the local schoolmaster came along and decided the point to the satisfaction of b'oth. The main road, running east and west, bounds the farm on the north side, and the farmer’s mail box is affixed to the fence exactly 30 chains north of his house. The farmer’s practice is to go for the mad himself. _ On this particular day the madman, seeing the farmer at work some distance north-west of the farmhouse, brought the mail 40 chains from the mad box further along the road, depositing it at a gate exactly 20 chains north of where the farmer was working. As it was about time to knock off, the farmer walked up to get his mail, arriving just as the madman was passing on his return. Ihe farmer thanked him for his consideration, but remarked that it made a longer walx to his house than if the mail had been put in the box as usual. The mailman differed, and it was then that the schoolmaster happened along and settled the point. Who was right, assuming each route was the shortest between, the points Stated ?

A GRAZING PROBLEM. Here is an innocent little problem, which, coming all the way from Hamilton in the North Island, which puzzled "A. G.” and his friends. A goat is tethered on the circumference of a circular paddock. What length of tether in terms of the radius, should be given so that the animal may be able to graze over exactly one half of the area and no more ? On the assumption that the paddock is fenced, and that tlie pasturing is confined to the enclosure, the required length of the rope, may be ascertained as follows:—First make your diagram, and then find the hypotenuse of a triangle whose other sides are radii of the tiaddock. Make the hypotenuse the radius

of another circle, whose centre would therefore be at the extremity of the diameter of tho paddock. Multiply half the area of the paddock by this radius, divide the product by that half area phis the area of the smaller arc of the new circle whose base is the paddock-diameter, and the result will give the required length of rope. Under slightly altered conditions, however, there is a more" practical method, by which the goat, tethered to the circumference, can reach exactly half the field only. Can the reader discover it? Standard IV arithmetic only is required.

TWO LOOPS. Since the publication of the problem “The Spherical Earth” a few weeks ago, several keen readers have sent queries more or less upon the point that was involved, this showing, at any rate, continuity of thought. _ Here is a little problem something akin. A boy had two flat-rimmed hoops. They were perfectly spherical, but one was much larger than the other. If one be placed inside the other touching only at the bottom, the greatest distance between the circumferences would be at the top, and there it was 9in, with sin between them at each end of the horizon diameter of the larger hoop. On the assumption that the rims had no thickness, what distance would the large hoop travel in a straight line if propelled for 100 revolutions? THE HERVE SQUARE. Here is an ingenious magic square problem by a Frenchman named Herve. It can only be solved if the reading of it be properly understood, so that the solver may have the advantage of every point that is not barred by the expressed terms of the problem. There is no ambiguity in the words used in stating the question, as long as it is remembered that a “number” does not necessarily mean a “figure.” For example, 331 contains the same figures as 3 1-3. but it constitutes a different number. Here is the problem: Make a nine square and place 8 in the centre of the top row. It is required that each of the remaining eight squares be occupied by numbers, all different from each other, so that each row and column as well as the long diagonals will add up 15. The eight must remain in the position given. The reader will find plenty of scope in his ingenuity in this. TWO POLES. Two poles supporting a wireless appartus were the same height and a hundred feet apart. A line fixed at a point 30 feet high on one of them was secured at the base of the other. The second post had a similar support, but the wire rope was made fast 20 feet from the ground and fixed to a cleat at the bottom of the other pole, both lines being drawn taut. Two simple problems hinge on this. At what height from the ground do the two lines cross, and how high would that point be if the two poles were 112 feet apart? Both poles were 35 feet high, and the ground all round was perfectly level. LAST WEEK'S SOLUTIONS. BLENDED TEA. There are three different ways that tho teas can be blended to obtain 151 b at 3s 2d per lb, but as the problem demanded the smallest possible weight of the expensive tea, the mixture must have licen as follows 91b at 3s 4d, SAlb at 3s, and gib at 2s 4d, making 151 b for £2 7s 6d, or 3s 2d per lb. The other two methods are 12, 2; 2| and 3, l/b.» A MATTER OF DEGREE. Tompkins must have won £4 16s. A NOVEL COMPETITION. The meeting point at X must have been 24 miles from A and 27 miles from Z. CUBES. There arc 48 different ways that a cube can be marked under the stated conditions. For every side occupied by 6 there is a selection of four sides for 5, and therefore two sides for 4. This applies alternatively, of course, to the 1,2, and 3, which are in association with 6,5, and 4 respectively. POSTAGE STAMPS.

If the block of 12 stamps be numbered as follows: 12 3 4 5 6 7 8 9 10 11 12 it will be seen that in th e shape of 1,2, 3, 4 there are three ways; in the form 1. 2.5, 6, six ways; and so on, making 65 different ways altogether in the manner prescribed by which four stamps properly joined together can be torn off. ANSWERS TO CORRESPONDENTS. G. W. —It is assumed that everyone subscribed and that the ratio stated was maintained.' C. J. Me. —Published on July 16. W. D. C. —Practical usefulness and entertainment are the. chief factors in most problems selected for this column. C. F. —The year of the representative matches mentioned was not given at the time, but I have since learnt that the season was 1908. M.G., H.J.8., W.D.E.. G.F.M.. J.C.C., F.C'.O.—Thanks. Acknowledged by post.