November 1, 1886.] 



♦ KNOW^LEDGE ♦ 



PHOTOGRAPH OF JUPITER. 



|HE accompanying photograph of Jupiter, by 

 the Brothers Henry, is full of promise. It 

 was taken on April 21, 1886, and is 

 interesting because of the presence of the 

 great red spot first detected by Professor 

 Pritehett, of Glasgow, Missouri, in 1877, 

 and since carefully and continuously ob- 

 served by many students of astronomy. 

 MM. Henry remarks that the red .spot is more striking 



and better define! [plus netle) in the photograph than by 

 direct telescopic vision. 



OUR PUZZLES. 



OLITTIONS of the three puzzles YII., VIII., 

 and IX. are given under the heading 

 " ilathematical Recreations." I have re. 

 ceived other solutions of Problems I. and II- 

 besides those given (one for each puzzle) in 

 Knowledge for September. 



Puzzle I. Nineteen trees may be aiTanged 

 so as to make nine rows of five ti'ees in the following ways, 

 the first of which is that stiven as the solution : — 



Fir;. 1. 



Fig. 3. 



Fig. 4. 



The.se solutions all possess a certain degree of symmetry, and 

 all certainly fulfil the conditions of the puzzle as it reached 

 my hands. I might e.asily modify the conditions of the 

 problem so that the solution in fig. 1 should be the only one 

 available. For instjinee, that is the only solution in which 

 the five trees on each row are so arranged that the middle 

 tree is at the centre of its row, and the other four form equi- 

 distant paii-s on either side of it. But instead of hampering 

 the problem with conditions of detail spoiling it as a puzzle, 

 I will take a bolder course, and rejjlace it ^\ith a puzzle of 

 my own invention, and of gi-eater difliculty, thus :^ 



Puzzle X. Arrange nineteen trees so as to make tek rows 

 of Jive trees in each row. 



Again, sis cords may be stretched between 

 smaller number, so as to enclcse ten spaces 

 following ways, the first of which is that I 

 tion : — ■ 



nine pegs, or a 



in any of the 



gave as a solu- 



FiG. 5. 



Fig 56. 



Fig. r. 



Here only fig. 5 is symmetrical, and that only in an axial, 

 not in a central manner. Observe the following distinctions 

 between these four solutions : — 



In fig. 5 there are \ triangles and 6 quadrangles : 9 pegs. 



In fig. 6 there are 6 triangles, 2 quadrangles, 2 pen- 

 tagons : 9 pegs. 



In fig. 7 there are 7 triangles, 3 pentiigons : 8 pegs. 



In fig. 8 there are 6 triangles, 3 quadrangles, 1 pen- 

 tagon : 7 pegs. 



Not one of these arrangements seems to me quite as good 

 ;is it might be; the arrangements in figs. 6, 7, 8 are alto- 

 gether unsymmetricjil, and that in fig. 5 has only partial 

 symmetiy ; while there are too many different shapes in all 

 the arrangements. I therefore pre.sent in this case also a 

 new puzzle on the same lines, and almost the same cords and 

 pegs — a neater puzzle of somewhat greater diflicuUy (though 

 the completeness of the conditions really indicates the direc- 

 tion in which a solution is to be found) ; thus — 



PizzLE XI. Willi six cords, fastened to nine pegs, enclose 

 ten spaces of the following forms : — 



First, one equilateral hexagon having equal alternate 

 angles ; 



Secondly, three equal ani similar quadrangles, each axially 

 symmetrical ; * 



Thirdly, six equal and similar triangles. 



As a third problem, which for symmetry's sake shall be 

 closely related to our third puzzle, even as these two are 

 closely related to the first and second, I repeat here a puzzle 

 which was given and very fully solved in the early numbers 

 of Knowledge by the ingenious " Mogul" : — 



Puzzle XII. Given a rectrnigular carpet of any shape 

 and size to divide it ivith the fewest possible cuts so as to fit 

 a rectangular floor of equal size hut of any shape. 



* A figure is said to be axially symmetrical when a straight line 

 can be drawn as an a.xis dividing it into two equal and similar 

 portions, such that it eitlicr were rotated about said a.'cis through 

 two right angles it would e.xactly coincide with the other. Thus an 

 isosceles triangle is axially symmetrical, the bisector of the angle 

 between the equal sides being its axial line. 



