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[ G.Polya ] SOLUTIONS 1. You think that a bear was white and the point P is the north pole? Can you prove that this is correct? as it was more or less understood, we idealise the question. We regard the globe as exactly spherical and the bear as a moving material point. This point, moving due south or due north, creates an arc of a meridian and it describes an arc of a parallel circle (parallel to the equator) when ti moves due east. We have to distinguish two cases. 1) if the bear returns to the point PO along a meridian different from the one along which he left P, P is necessarily the north pole. In fact, the only other point of the globe which two meridians meet is the south pole, but the bear could leave this pole only in moving northward. 2) the bear could return to the point P along the same meridian he left P if, when walking one mile due east he describes a parallel circle exactly n times where n may be 1,2,3,... in this case P is not the north pole but a point on a parallel circle very close to the south pole (the perimeter of which expressed in miles is slightly inferior to 2pi + 1 / n 2. We represent the globe as in the solution of problem 1. the land that bob wants is bounded by two meridians and two parallel circles. imagine two fixed meridians, and a parallel circle moving away form the equator. The arc on the moving parallel intercepted by the two fixed meridians is steadily shortened. The centre of the land that bob wants should be on the equator: the can not get it in the US. 3. The least possible number of dollars in a pocket is obviously 0. The next greatest number is at least 1, the next greater is at least 2...the number in the (tenth) pocket is at least 9. Therefor, the number of dollars required is at least 0 + 1 + 2 + 3 + ... + 9 = 45 Bob cannot make it: he has only 44 dollars 4. A volume of 999 numbered pages needs 9+ 2*90 + 3*90 = 2889 digits. If the bulky volume in question has x pages 2889 + 4(x-999) = 2989 x=1024 This problem may teach us that a preliminary estimate of the unknown may be useful (or even necessary as in the present case)(. 5. If _679_ is divisible by 72, it is divisible both by 8 and 9. If it is divisible by 8 the number 79_ must be divisible by 8 (since 1000 is divisible by 8) and so 79_ must be 792: the last faded digit is 2. If _6792 is divisible by 9, the sum of its digits must be divisible by 9 (the rule about "casting out nines" ) and so the first faded digit must be 3. The price of one turkey was (in grandfathers time) $367.92 / 72 = $5.11 6. " A point and a figure with a centre of symmetry in the same plane)( are given in position. Find a straight line that passes through the given point and bisects the area of the given figure." The required line except, of course, through the centre of symmetry. See Inventor's paradox . 7. In any position the two sides of the angle must pass through tow vertices of the square. As long as they pass through the same pair of vertices, the angle's vertex moves along the same arc of circle (by the theorem underlying the hint )(. Hence each of the two loci re required consist of several arcs of circle : of 4 semicircles in the case (a) and of 8 quarter circles in the case b) ee fig 31. _-------_ _____/ \____ _~~ . .____ \ / -_______- FIG 31 8. The axis pierces the surface of the cube in some piont which is either a =vertex of the cube or lies on an edge or in the interior of a face. If the axis passes through a point of an edge (but not through one of its end-points ) this point must be the midpoint; otherwise the edge would not coincide with itself after the rotation. Similarity, an axis piercing the interior of a face must pass through its centre. Any axis must, of course, pass through the centre of the cube. And so there are three kinds of axes: 1) 4 axes each through two opposite vertices; angles 120, 240 degrees 2) 6 axes each through the midpoints of two opposite edges. angle 180 degrees 3) 3 axes each through the centre of two opposite faces. angels 90, 180 270 degrees For the length of an axis the first kind see section 12. the others are still easier to compute. The desired average is (4*sqrt3+6*sqrt2 +3) / 13 = 1.416 (this problem may be useful in preparing the reader4r for the study of crystallography. For the reader sufficiently advanced in the integral calculus it may be observed that the average is computed is a fairly good approximation to the average width" of the cube, which is, in fact 3/2=1.5 9. The plane passing through one edge of a length a and the perpendicular of length b divides the tetrahedron into two more accessible congruent tetrahedra, each with base ab/2 and height a/2. Hence the required volume = 2 * 1/3 * ab/2 * a/2 = b/2*a^2 10. The base of the pyramid is a polygon with n sides. In the case )A the n lateral edges of the pyramid are equal; in the case (b) the altitudes (drawn from the apex) of its n lateral faces are equal. If we craw the altitude- of the pyramid and join its foot to the n vertices of the base in the case 9a), but to the feet of the altitudes of the n lateral faces in the case (b), we obtain, in both cases, n right angles of which the altitude (of the pyramid) is a common side: I say that these n right triangles are congruent. In fact the hypotenuse pa lateral edge in case a, a lateral altitude in case b] is o f the same length in each, according to the definitions. laid down in the proposed problem; wee have just men mentioned the another side (the altitude of the pyramid) and an angle (the right angle) are common to all. In the n congruent triangles the third sides must also be equal; they are drawn form the same point (the food f the altitude ) in the same plane (the base): they form n radii of a circle which is circumscribed about, or in- inscribed into, the base of the pyramid, in cases a) and b) respectively. In the case )b) it remains to show, however, that the n radii mentioned are perpendicular to the respective sides of the base; this follows from a well-known theorem of solid geometry on projections.[ It is most remarkable that a plane figure, the isosceles triangle may have two different analogues in solid geometry. 11. Observe that the first equation is so related to the last as the second is to the third: the coefficients on the left hand sides are etch same but in opposite order whereas the right hand sides are opposite. Add the first equation to the last and the second to the third: 6(x+u) + 10(y+v) = 0 10(x+u) + 10(y+v) = 0 This can be regarded as a system of two linear equations for two unknowns, namely for x+yu and y+v and easily yeilds x+u = 0 y+v=0 Substituting -x for u and -y for v in the first two equations of the original system, we find -4x +4y = 16 6x-2y = -16 This is a simple system which yields. x=-2 y=2 u=2 v=-2 12. Between the start and the meeting point of each of the friends travelled the same distance. (Remember, dis distance=velocity*time). ) we distinguish two parts in the condition: Bob travelled as much as Paul: ct1 - ct2 + ct3 = ct1 + pt2 + pt3 Paul travelled as much as peter: ct1 + pt2 + pt3 = pt1 + pt2 + ct3 the second equation yields (c-p)t1 = (C-p)t3 We assume, of course, that the car travels faster than a pedestrian, c>p it follows t1 = t3; That is, peter walks just as much as Paul. from the first equation, we find that t3/t2 = (c+p)/(c-p) which is, of course, also the value for t1/t2. Hence we obtain the answers: a) c(t1-t2+t3)/(t1+t2+t3) = c(c+3p)/(3c+p) b) t2/(t1+t2+t3) = (c-p)/(3c+p) c) In fact, o INDUCTION AND MATHEMATICAL INDUCTION we ask: does the conjectured formula remain true if we pass from the value n to the next value n+1 ? Along with the formula above we should have 1/2! + 2/3! + ... + n/(n+1)! + (n+1)/(n+2)! = 1 - 1/(n+2)! check this by subtracting from it the former: (n+1)/(n+2)! = -1/(n+2)! + 1/(n+1) which boils down to (n+2)/(n+2)! = 1/(n+1)! And this last equation is obviously true for n=1,2,3,... hence, by following the pattern referred to above, we can prove our conjecture. 18. in the nth line the right hand side seems to be n^3 and the left-hand side a sum of n terms. The final term of this sum is the nth odd number, or 2m-1 where m=1+2+3+...+n = n(n+1)/2 See INDUCTION AND MATHEMATICAL INDUCTION ,4. Hence the final term of the sum on the left hand side should be 2m-1=n^2+n-1 We can derive hence the initial term of the sum con- considered in two ways: going back n-1 steps form the final term we find (n^2+n-1)-2(n-1)=n^2-n+1 whereas, advancing one step from the final term of the foregoing line, we find [(n-1)^2 + (n-1) -1]-1]+2 which, after routine simplification, boils down to the same: good! we assert therefor that (n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)=n^3 where the left hand side indicates the sum of n successive terms of an arithmetic progression the difference of which is 2. If the reader knows the rule for the sum of such a progression (arithmetic mean of the initial term and the final term, multiplied by the number of terms we can verify that ((n^2-n+1)+(n^2+n-1))n/2 = n^3 and so prove the assertion. (The rule quoted can be easily proved by a picture little different from fig 18) 91. the length of the perimeter of the regular hexagon with side n is 6n. Therefor, this perimeter consists of 6n boundary lines of length 1 and contains 6n vertices therefor, in the transition from n-1 to n, V increases by 6n units, and so V=1+6(1+2+3+...+n) = 3n^2 + 3n +1; see MATHEMATICAL INDUCTION AND INDUCTION ,4. By 3 diagonals through its centre the hexagon is divided into 6 (large)equilateral triangles. By inspection one of these T = 6(1+3+5+...+2n-1 ) = 6n^2 (rule fort he sum of an arithmetic progression, quoted in the solution to problem 18). The T triangles have jointly 3T sides. In this total 3T each internal line of division of length 1 is counted twice, whereas the 6n lines along the perimeter of the hexagon are counted but once. Hence 2L = 3T+6n L=9n^2 +3n (for the more advanced reader: it follows from Euler's theorem on polyhedra that T+V=L+1. verify this relation~!) 20. Here is a well-ordered array of analogous problems: Compute An Bn Cn Dn and En. Each of these quantities represent he number of ways to pay the amount of n cents; the difference is the coins used. A only cents B cents and nickels Cn cents, nickels and dimes Dn cents, nickels dimes and quarters En cents, nickels , dimes, quarters and half dollars The symbols En (reason now clear) and An were used before. All ways and manners to pay the amount of n cents with the five kinds of coin are enumerated by En. We may, however distinguish two possibilities: first: no half dollar is used. the number of such ways to pay is Dn, by definition. Second, a half dollar, possibly more, is used. After the first half dollar is laid on the counter, there remains the amount n-50 cents to pay, which can be done in exactly En-50 ways. We infer that En=Dn+En-50 Similarly Dn=CFn+Dn-25) Cn=Bn+Cn-10 Bn=An+Bn-5 A little attention shows