History of Mathematics Brief Version Victor Katz

SOLUTIONS TO PROBLEMS

CHAPTER ONE

1. We write 275 as follows in Egyptian hieroglyphics (on the left) and Babylonian cuneiform(on the right):

2.

1 5

10 50 (multiply by 10)

2 10 (double first line)

4 20 (double third line)

8 40 (double fourth line)

2 2 2 (halve first line)

10 2 (invert third line)

18 2 10 93

3.

1 7 2 4 8

2 15 2 4

4 31 2

8 63

3 4 3 3 6 12

12 3 98 2 3 3 6 1299 2 4

1

Name:Stanislav Shur Email:[email protected] IP:74.88.120.168

Address:8617 21st avenue,Brooklyn,NY,US Zip Code:11214

4.2 11 1 11 2 23 1 23

3 7 3 3 15 3

3 3 3 3 7 3

6 1 3 6

6 3 2 3

66 6

12 1 2 4 6

276 12

6 66 2

12 276 2

5.5 13 = (2 13) + (3 13) = 8 52 104 + 8 13 52 104 = 4 13 26 52

6 13 = 2(3 13) = 4 8 52 104 26 52 = 4 8 13 1048 13 = 2(4 13) = 2 13 26

6. x+ 17x = 19. Choose x = 7; then 7 +17 7 = 8. Since 19 8 = 238 , the correct answer is

238 7 = 1658 .

7. (x + 23x) 13(x + 23x) = 10. In this case, the obvious choice for x is x = 9. Then 9added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get10. So in this case our guess is correct.

8. The equation here is (1 + 13 +14)x = 2. Therefore. we can find the solution by dividing

2 by 1 + 13 +14 . We set up that problem:

1 1 2 4

3 1 18

3 2 36

6 4 72

12 8 144

The sum of the numbers in the right hand column beneath the initial line is 1141144 . So we

need to find multipliers giving us 3144 = 144 72. But 1 3 4 times 144 is 228. It followsthat multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72. Thus, theanswer is 1 6 12 114 228.

2



9. Since x must satisfy 100 : 10 = x : 45, we would get that x = 4510010 ; the scribe breaksthis up into a sum of two parts, 3510010 and

1010010 .

10. The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4handbreadths diameter is 52 : 42 = 25 : 16 = 1 916 . Thus, 100 logs of 5 handbreadths

diameter are equivalent to 1 916 100 = 15614 logs of 4 handbreadths diameter.

12.8 34 17 8

7) 1 00 00 00 00 00564 003 58

2 001 59

1 00564

13. Since 3 18 = 54, which is 6 less than 60, it follows that the reciprocal of 18 is 313 , or,putting this in sexagesimal notation, 3,20. Since 60 is (178) 32, and 78 can be expressedas 52,30, the reciprocal of 32 is 1,52,30. Since 60 = 119 54, and 19 can be expressed as110+

190 =

660+

403600 = 0; 06, 40, the reciprocal of 54 is 1, 06, 40. Also, because 60 =

151664,

the reciprocal of 64 is 1516 . Since116 = 3, 45, we get that

1516 = 56, 15. If the only prime

divisors of n are 2, 3, 5, then n is a regular sexagesimal.

14. 25 1, 04 = 1, 40 + 25, 00 = 26, 40. 18 1, 21 = 6, 18 + 18, 00 = 24, 18. 50 18 =50 0; 3, 20 = 2; 30 + 0; 16, 40 = 2; 46, 40. 1, 21 32 = 1, 21 0; 01, 52, 30 = 1; 21 +1; 10, 12 + 0; 00, 40, 30 = 2; 31, 52, 30.

15. Since the length of the circumference C is given by C = 4a, and because C = 6r, it followsthat r = 23a. The length T of the long transversal is then T = r

2 = (23a)(

1712) =

1718a.

The length t of the short transversal is t = 2(r t2) = 2a(23 1736) = 718a. The area A ofthe barge is twice the difference between the area of a quarter circle and the area of theright triangle formed by the long transversal and two perpendicular radii drawn fromthe two ends of that line. Thus

A = 2

(C2

48 r

2

2

)= 2

(a2

3 2a

2

9

)=

2

9a2.

16. Since the length of the circumference C is given by C = 3a, and because C = 6r, it followsthat r = a2 . The length T of the long transversal is then T = r

3 = (a2 )(

74) =

78a. The

length t of the short transversal is twice the distance from the midpoint of the arc to

3



the center of the long transversal. If we set up our circle so that it is centered on the

origin, the midpoint of the arc has coordinates ( r2 ,3r2 ) while the midpoint of the long

transversal has coordinates ( r4 ,3r4 ). Thus the length of half of the short transversal is

r2 and then t = r =

a2 . The area A of the bulls eye is twice the difference between the

area of a third of a circle and the area of the triangle formed by the long transversal andradii drawn from the two ends of that line. Thus

A = 2

(C2

36 1

2

r

2T

)= 2

(9a2

36 1

2

a

4

7a

8

)= 2a2

(1

4 7

64

)=

9

32a2.

17. The correct formula in the first case gives V = 56, while the Babylonian version givesV = 12(2

2+42)6 = 60 for a percentage error of 7%. In the second case, the correct formula

gives 488/3 = 16223 , while the Babylonian formula gives V =12(8

2 + 102)2 = 164, for anerror of 0.8%.

18. 1; 24, 51, 10 = 1+2460+51

3600+10

216000 = 1+0.4+0.0141666666+0.0000462962 = 1.414212963.

On the other hand,2 = 1.414213562. Thus the Babylonian value differs from the true

value by approximately 0.00004%.

19. Because (1; 25)2 = 2; 00, 25, we have

2 =

2; 00, 25 0; 00, 25 1; 25 (0; 30)(0; 00, 25)(1/1; 25).

An approximation to the reciprocal of 1;25 is 0;42,21,11. The product of 0;30 by 0;00,25by 0;42,21,11 is 0;00,08,49,25. The the approximation to

2 is 1; 25 0; 00, 08, 49, 25 =

1; 24, 51, 10, 35, which, with the last term truncated, is the Babylonian value.

20.3 =

22 1 2 12 1 12 = 2 0; 15 = 1; 45. Since an approximate recipro-

cal of 1;45 is 0;34,17.09, we get further that3 =

(1; 45)2 0; 03, 45 = 1; 45

(0; 30)(0; 03, 45)(0; 34, 17.09) = 1; 45 0; 01, 04, 17, 09 = 1; 43, 55, 42, 51, which we trun-cate to 1;43,55,42 because we know this value is a slight over-approximation.

21. 12 3 15 24 32 = 12129160 . (12129160)

2 = (12.80625)2 = 164.0000391 . . .

22. v + u = 1; 48 = 145 and v u = 0; 33, 20 = 59 . So 2v = 2; 21, 20 and v = 1; 10, 40 = 10690 .Similarly, 2u = 1; 14, 40 and u = 0; 37, 20 = 5690 . Multiplying by 90 gives x = 56, d = 106.

In the second part, v + u = 2; 05 = 2 112 and v u = 0; 28, 48 = 1225 . So 2v = 2; 33, 48and v = 1; 16, 54 = 769600 . Similarly, 2u = 1; 36, 12 and u = 0; 48, 06 =

481600 . Multiplying by

600 gives x = 481, d = 769. Next, if v = 481360 and u =319360 , then v + u = 2

29 = 2; 13, 20.

Finally, if v = 289240 and u =161240 , then v + u = 1

78 = 1; 52, 30.

4



23. The equations for u and v can be solved to give v = 1; 22, 08, 27 = 295707216000 =9856972000 and

u = 0; 56, 05, 57 = 201957216000 =6731972000 . Thus the associated Pythagorean triple is 67319,

72000, 98569.

24. The two equations are x2 + y2 = 1525; y = 23x+5. If we substitute the second equation

into the first and simplify, we get 13x2 + 60x = 13500. The solution is then x = 30,y = 25.

25. If we guess that the length of the rectangle is 60, then the width is 45 and the diagonalis602 + 452 = 75. Since this value is 178 times the given value of 40, the correct length

of the rectangle should be 60 178 = 32. Then the width is 24.

26. One way to solve this is to let x and x 600 be the areas of the two fields. Then theequation is 23x+

12(x600) = 1100. This reduces to 76x = 1400, so x = 1200. The second

field then has area 600.

27. Let x be the weight of the stone. The equation to solve is then x 17x 113(x 17x) = 60.We do this using false position twice. First, set y = x 17x. The equation in y is theny 113y = 60. We guess y = 13. Since 13 11313 = 12, instead of 60, we multiply ourguess by 5 to get y = 65. We then solve x 17x = 65. Here we guess x = 7 and calculatethe value of the left side as 6. To get 65, we need to multiply our guess by 656 = 10

16 . So

our answer is x = 7 656 = 7556 gin, or 1 mina 1556 gin.

28. We do this in three steps, each using false position. First, set z = x 17x+ 111(x 17x).The equation for z is then z 113z = 60. We guess 13 for z and calculate the value ofthe left side to be 12, instead of 60. Thus we must multiply our original guess by 5 andput z = 65. Then set y = x 17x. The equation for y is y + 111y = 65. If we now guessy = 11, the result on the left side is 12, instead of 65. So we must multiply our guessby 6512 to get y =

71512 = 59

712 . We now solve x 17x = 59 712 . If we guess x = 7, the

left side becomes 6 instead of 59 712 . So to get the correct value, we must multiply 7 by71512 /6 =

71572 . Therefore, x = 7 71572 = 500572 = 693772 gin = 1 mina 93772 gin.

29.

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30. If we substitute the first equation into the second, the result is 30y (30 y)2 = 500 ory2+1400 = 90y. This equation has the two positive roots 20 and 70. If we subtract thesecond equation from the square of the first equation, we get (x2 = 900)(xy(xy)2 =500), or (x y)2 + x(x y) = 400, or finally (x y)2 + 30(x y) = 400. This latterequation has x y = 10 as its only positive solution. Since we know that x = 30, wealso get that y = 20.

31. The equations can be rewritten in the form x+ y = 556 ; x+ y+xy = 14. By subtracting

the first equation from the second, we get the new equation xy = 816 . The standardmethod then gives

x =5562

+

(5562

)2 81

6= 2

11

2+

873

144 81

6= 2

11

12=

49

144= 2

11

12+

7

12= 3

1

2.

Similarly, y = 213 .

6



CHAPTER TWO

1. Since AB = BC; since the two angles at B are equal; and since the angles at A and C areboth right angles, it follows by the angle-side-angle theorem that 4EBC is congruentto 4SBA and therefore that SA = EC.

2. Because both angles at E are right angles; because AE is common to the two triangles;and because the two angles CAE are equal to one another, it follows by the angle-side-angle theorem that 4AET is congruent to 4AES. Therefore SE = ET .

3. Tn = 1 + 2 + + n = n(n+1)2 . Therefore the oblong number n(n + 1) is double thetriangular number Tn.

4. n2 = (n1)n2 +n(n+1)

2 , and the summands are the triangular numbers Tn1 and Tn.

5. 8n(n+1)2 + 1 = 4n2 + 4n+ 1 = (2n+ 1)2.

6. Examples using the first formula are (3,4,5), (5,12,13), (7,24,25), (9,40,41), (11,60,61).Examples using the second formula are (8,15,17), (12,35,37), (16,63,65), (20,99,101),(24,143,145).

7. Consider the right triangle ABC where AB has unit length and the hypotenuse BChas length 2. Then the square on the leg AC is three times the square on the leg AB.Assume the legs AB and AC are commensurable, so that each is represented by thenumber of times it is measured by their greatest common measure, and assume furtherthat these numbers are relatively prime, for otherwise there would be a larger commonmeasure. Thus the squares on AC and AB are represented by square numbers, wherethe former is three times the latter. It follows that leg AC is divisible by three andtherefore its square is divisible by nine. Since the square on AB is one third that on AC,it is divisible by three, and hence the side AB itself is divisible by three, contradictingthe assumption that the numbers measuring the two legs are relatively prime.

9. Let ABC be the given triangle. Extend BC to D and draw CE parallel to AB. By I29,angles BAC and ACE are equal, as are angles ABC and ECD. Therefore angle ACDequals the sum of the angles ABC and BAC. If we add angle ACB to each of these, weget that the sum of the three interior angles of the triangle is equal to the straight angleBCD. Because this latter angle equals two right angles, the theorem is proved.

10. Place the given rectangle BEFG so that BE is in a straight line with AB. Extend FG toH so that AH is parallel to BG. Connect HB and extend it until it meets the extensionof FE at D. Through D draw DL parallel to FH and extend GB and HA so they meetDL in M and L respectively. Then HD is the diagonal of the rectangle FDLH and

7



so divides it into two equal triangles HFD and HLD. Because triangle BED is equalto triangle BMD and also triangle BGH is equal to triangle BAH, it follows that theremainders, namely rectangles BEFG and ABML are equal. Thus ABML has beenapplied to AB and is equal to the given rectangle BEFG.

11. In this proof, we shall refer to certain propositions in Euclids Book I, all of which areproved before Euclid first uses Postulate 5. (That occurs in proposition 29.) First,assume Playfairs axiom. Suppose line t crosses lines m and l and that the sum of thetwo interior angles (angles 1 and 2 in the diagram) is less than two right angles. Weknow that the sum of angles 1 and 3 is equal to two right angles. Therefore 6 2 < 6 3.Now on line BB and point B construct line BC such that 6 C BB = 6 3 (Proposition23). Therefore, line BC is parallel to line l (Proposition 27). Therefore, by Playfairsaxiom, line m is not parallel to line l. It therefore meets l. We must show that the twolines meet on the same side as C . If the meeting point A is on the opposite side, then6 2 is an exterior angle to triangle ABB, yet it is smaller than 6 3, one of the interiorangles, contradicting proposition 16. We have therefore derived Euclids postulate 5.

Second, assume Euclids postulate 5. Let l be a given line and P a point outside the line.Construct the line t perpendicular to l through P (Proposition 12). Next, construct theline m perpendicular to line t at P (Proposition 11). Since the alternate interior anglesformed by line t crossing lines m and l are both right and therefore are equal, it followsfrom Proposition 27 that m is parallel to l. Now suppose n is any other line through P .We will show that n meets l and is therefore not parallel to l. Let 6 1 be the acute anglethat n makes with t. Then the sum of angle 1 and angle PQR is less than two rightangles. By postulate 5, the lines meet.

Note that in this proof, we have actually proved the equivalence of Euclids Postulate5 to the statement that given a line l and a point P not on l, there is at most one linethrough P which is parallel to l. The other part of Playfairs Axiom was proved (in the

8



second part above) without use of postulate 5 and was not used at all in the first part.

12. One possibility: If the line has length a and is cut at a point with coordinate x, then4ax + (a x)2 = (a+ x)2. This is a valid identity.

13. In the circle ABC, let the angle BEC be an angle at the center and the angle BAC bean angle at the circumference which cuts off the same arc BC. Connect AE and continuethe line to F . Since EA = EB, 6 EAB = 6 EBA. Since 6 BEF equals the sum of thosetwo angles, 6 BEF is double 6 EAB. Similarly, 6 FEC is double 6 EAC. Therefore theentire 6 BEC is double the entire 6 BAC. Note that this argument holds as long as lineEF is within 6 BEC. If it is not, an analogous argument by subtraction holds.

14. Let 6 BAC be an angle cutting off the diameter BC of the circle. Connect A to thecenter E of the circle. Since EB = EA, it follows that 6 EBA = 6 EAB. Similarly,6 ECA = 6 EAC. Therefore the sum of 6 EBA and 6 ECA is equal to 6 BAC. But thesum of all three angles equals two right angles. Therefore, twice 6 BAC is equal to tworight angles, and angle BAC is itself a right angle.

15. In the circle, inscribe a side AC of an equilateral triangle and a side AB of an equilateralpentagon. Then arc BC is the difference between one-third and one-fifth of the circum-ference of the circle. That is, arc BC = 215 of the circumference. Thus, if we bisect thatarc at E, then lines BE and EC will each be a side of a regular 15-gon.

16. Let the triangle be ABC and draw DE parallel to BC cutting the side AB at D and theside AC at E. Connect BE and CD. Then triangles BDE and CDE are equal in area,having the same base and in the same parallels. Therefore, triangle BDE is to triangleADE and triangle CDE is to triangle ADE. But triangles withe the same altitude areto one another as their bases. Thus triangle BDE is to triangle ADE as BD is to AD,and triangle CDE is to triangle ADE and CE is to AE. It follows that BD is to ADas CE is to AE, as desired.

17. Let ABC be the triangle, and let the angle at A be bisected by AD, where D lies on theside BC. Now draw CE parallel to AD, meeting BA extended at E. Now angle CADis equal to angle BAD by hypothesis. But also angle CAD equals angle ACE and angleBAD equals angle AEC, since in both cases we have a transversal falling across parallellines. It follows that angle AEC equals angle ACE, and therefore that AC = AE. Byproposition VI-2, we know that BD is to DC as BA is to AE. Therefore BD is to DCas BA is to AC, as claimed.

18. Let a = s1b + r1, b = s2r1 + r2, . . ., rk1 = sk+1rk. Then rk divides rk1 and thereforealso rk2, . . . , b, a. If there were a greater common divisor of a and b, it would divider1, r2, . . ., rk. Since it is impossible for a greater number to divide a smaller, we haveshown that rk is in fact the greatest common divisor of a and b.

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19.963 = 1 657 + 306657 = 2 306 + 45306 = 6 45 + 3645 = 1 36 + 936 = 4 9 + 0

Therefore, the greatest common divisor of 963 and 657 is 9.

20. Since 1 x = x2, we have

1 = 1 x+ (1 x) = 1 x + x2

x = 1 x2 + (x x2) = 1 x2 + x(1 x) = 1 x2 + x3

x2 = 1 x3 + (x2 x3) = 1 x3 + x2(1 x) = 1 x3 + x4

Thus 1 : x can be expressed in the form (1, 1, 1, . . .).

21.46 = 7 6 + 4 23 = 7 3 + 26 = 1 4 + 2 3 = 1 2 + 14 = 2 2 2 = 2 1

Note that the multiples 7, 1, 2 in the first example equal the multiples 7, 1, 2 in thesecond.

22. In Figure 2.16 (left), let AB = 7 and the area of the given figure be 10. The construction

described on p. 72 then determines x to be BS. This value is 72

494 10 = 72

94 =

72 32 = 2. The second solution is BE + ES = AE + ES = AS. This value is

72 +

494 10 = 72 +

94 =

72 +

32 = 5.

23. In Figure 2.16 (right), let AB = 10 and the area of the given figure be 39. The con-

struction described on p. 72 then determines x to be BS. This value is52 + 39 5 =

64 5 = 8 5 = 3.

24. Suppose m factors two different ways as a product of primes: m = pqr s = pqr s.Since p divides pqr s, it must also divide pqr s. By VII30, p must divide oneof the prime factors, say p. But since both p and p are prime, we must have p = p.After canceling these two factors from their respective products, we can then repeat theargument to show that each prime factor on the left is equal to a prime factor on theright and conversely.

10



25. One standard modern proof is as follows. Assume there are only finitely many primenumbers p1, p2, p3, . . ., pn. Let N = p1p2p3 pn + 1. There are then two possibilities.Either N is prime or N is divisible by a prime other than the given ones, since divisionby any of those leaves remainder 1. Both cases contradict the original hypothesis, whichtherefore cannot be true.

26. We begin with a square inscribed in a circle of radius 1. If we divide the square into fourisosceles triangles, each with vertex angle a right angle, then the base of each triangle

has length b1 =2 and height h1 =

22 . Then the area A1 of the square is equal to

4 12b1h1 = 2b1h1 = 2. If we next construct an octagon by bisecting the vertex angles ofeach of these triangles and connecting the points on the circumference, the octagon isformed of eight isosceles triangles. The base of each triangle has length

b2 =

(b12

)2+ (1 h1)2 =

(b12

)2+ h21 2h1 + 1 =

2 2h1 =

2

2

and height

h2 =

1 (b22

)2=

2 +

2

2.

Thus the octagon has area A2 = 8 12b2h2 = 4b2h2 = 22 = 2.828427. If we continue

in this way by always bisecting the vertex angles of the triangles to construct a newpolygon, we get that the area An of the nth polygon is given by the formula An =2n+1 12bnhn = 2nbnhn, where

bn =

(bn12

)2+ (1 hn1)2 =

(bn12

)2+ h2n1 2hn1 + 1 =

2 2hn1

and

hn =

1 (bn2

)2.

The next two results using this formula are A3 = 3.061467 and A4 = 3.121445.

27. Since BC is the side of a decagon, triangle EBC is a 36-72-72 triangle. Thus 6 ECD =108. Since CD, the side of a hexagon, is equal to the radius CE, it follows that triangleECD is an isosceles triangle with base angles equal to 36. Thus triangle EBD is a36-72-72 triangle and is similar to triangle EBC. Therefore BD : EC = EC : BC orBD : CD = CD : BC and the point C divides the line segment BD in extreme andmean ratio.

28. Let ABCDE be the pentagon inscribed in the circle with center F . Connect AF andextend it to meet the circle at G. Draw FH perpendicular to AB and extend it to

11



meet the circle at K. Connect AK. Then AK is a side of the decagon inscribed inthe circle, while BF = AF is the side of the hexagon inscribed in the circle. Draw FLperpendicular to AK; let N be its intersection with AB and M be its intersection withthe circle. Connect KN . Now triangles ANK and AKB are isosceles triangles with acommon base angle at A. Therefore, the triangles are similar. So BA : AK = AK : AN ,or AK2 = BA AN . Further, note that arc BKM has measure 54, while arc BCG hasmeasure 108. It follows that 6 BFN = 6 BAF . Since triangles BFN and BAF alsohave angle FBA in common, the triangles are similar. Therefore, BA : BF = BF : BN ,or BF 2 = BA BN . We therefore have AK2 + BF 2 = BA AN + BA BN =BA (AN + BN) = BA2. That is, the sum of the squares on the side of the decagonand the side of the hexagon is equal to the square on the side of the pentagon.

29. C = 3607 15

5000 = 250, 000 stades. This value equals 129,175,000 feet or 24,465 miles.The diameter then equals 7,787 miles.

12



CHAPTER THREE

1. Lemma 1: DA/DC = OA/OC by Elements VI3. Therefore DA/OA = DC/OC =(DC+DA)/(OC+OA) = AC/(CO+OA). Also, DO2 = OA2+DA2 by the PythagoreanTheorem.

Lemma 2: AD/DB = BD/DE = AC/CE = AB/BE = (AB + AC)/(CE + BE) =(AB +AC)/BC. Therefore, AD2/BD2 = (AB +AC)2/BC2. But AD2 = AB2BD2.So (AB2 BD2)/BD2 = (AB + AC)2/BC2 and AB2/BD2 = 1 + (AB + AC)2/BC2.

2. Set r = 1, ti and ui as in the text, and Pi the perimeter of the ith circumscribed polygon.Then the first ten iterations of the algorithm give the following:

t1 = .577350269 u1 = 1.154700538 P1 = 3.464101615t2 = .267949192 u2 = 1.03527618 P2 = 3.21539031t3 = .131652497 u3 = 1.008628961 P3 = 3.159659943t4 = .065543462 u4 = 1.002145671 P4 = 3.146086215t5 = .03273661 u5 = 1.0005357 P5 = 3.1427146t6 = .016363922 u6 = 1.00013388 P6 = 3.141873049t7 = .0081814134 u7 = 1.000033467 P7 = 3.141662746t8 = .004090638249 u8 = 1.000008367 P8 = 3.141610175t9 = .002045310568 u9 = 1.000002092 P9 = 3.141597032t10= .001022654214 u10= 1.000000523 P10= 3.141593746

3. Let d be the diameter of the circle, ti the length of one side of the regular inscribedpolygon of 3 2i sides, and ui the length of the other leg of the right triangle formed fromthe diameter and the side of the polygon. Then

ti+12

d2=

t2it2i + (d+ ui)

2

or

ti+1 =dti

t2i + (d+ ui)2

ui+1 =d2 ti+12.

If Pi is the perimeter of the ith inscribed polygon, thenPid =

32itid . So let d = 1. Then

t1 =d2 = 0.5 and u1 =

3d2 = 0.8660254. Then repeated use of the algorithm gives us:

t1 = 0.500000000 u1 = 0.866025403 P1 = 3.000000000t2 = 0.258819045 u2 = 0.965925826 P2 = 3.105828542t3 = 0.130526194 u3 = 0.991444861 P3 = 3.132628656t4 = 0.06540313 u4 = 0.997858923 P4 = 3.13935025t5 = 0.032719083 u5 = 0.999464587 P5 = 3.141031999t6 = 0.016361731 u6 = 0.999866137 P6 = 3.141452521t7 = 0.008181140 u7 = 0.999966533 P7 = 3.141557658t8 = 0.004090604 u8 = 0.999991633 P8 = 3.141583943

13



t9 = 0.002045306 u9 = 0.999997908 P9 = 3.141590016

4. We can prove the inequality simply by squaring each side and noting that b < 2a+1. Tofind the approximands to

3, begin with 2 14 >

22 1 > 2 13 , or 74 >

3 > 53 . Since

3 = 1352 + 2, we continue with 13(5 +

15) >

13

52 + 2 > 13(5 +

211), or

2615 >

3 > 5733 .

Again, since3 = 115

262 1, we get 115(26 152) > 115

262 1 > 115(26 151), or

1351780 >

3 > 1325765 =

265153 .

5. Let the equation of the parabola be y = x2+1. Then the tangent line at C = (1, 0) hasthe equation y = 2x+2. Let the point O have coordinates (a, 0). ThenMO = 2a+2,OP = a2+1, CA = 2, AO = a+1. So MO : OP = (2a+2) : (1a2) = 2 : (1a) =CA : AO.

6. a. Draw line AO. Then MS SQ = CA AS = AO2 = OS2 + AS2 = OS2 + SQ2.b. Since HA = AC, we have HA : AS = MS : SQ = MS2 : MS SQ = MS2 :

(OS2+SQ2) =MN2 : (OP 2+QR2). Since circles are to one another as the squareson their diameters, the latter ratio equals that of the circle with diameter MN tothe sum of the circle with diameter OP and that with diameter QR.

c. Since then HA : AS = (circle in cylinder):(circle in sphere + circle in cone), itfollows that the circle placed where it is is in equilibrium about A with the circle inthe sphere together with the circle in the cone if the latter circles have their centersat H.

d. Since the above result is true whatever line MN is taken, and since the circles makeup the three solids involved, Archimedes can conclude that the cylinder placed whereit is is in equilibrium about A with the sphere and cone together, if both of themare placed with their center of gravity at H. Since K is the center of gravity of thecylinder, it follows that HA : AK = (cylinder):(sphere + cone).

e. Since HA = 2AK, it follows that the cylinder is twice the sphere plus the cone AEF .But we know that the cylinder is three times the cone AEF . Therefore the coneAEF is twice the sphere. But the cone AEF is eight times the cone ABD, becauseeach of the dimensions of the former are double that of the latter. Therefore, thesphere is four times the cone ABD.

7. Since BOAPC is a parabola, we have DA : AS = BD2 : OS2, or HA : AS = MS2 :OS2. Thus HA : AS = (circle in cylinder):(circle in paraboloid). Thus the circle inthe cylinder, placed where it is, balances the circle in the paraboloid placed with itscenter of gravity at H. Since the same is true whatever cross section line MN is taken,Archimedes can conclude that the cylinder, placed where it is, balances the paraboloid,placed with its center of gravity at H. If we let K be the midpoint of AD, then Kis the center of gravity of the cylinder. Thus HA : AK = cylinder:paraboloid. ButHA = 2AK. So the cylinder is double the paraboloid. But the cylinder is also triple thevolume of the cone ABC. Therefore, the volume of the paraboloid is 3/2 the volume ofthe cone ABC which has the same base and same height.

14



8. Let the parabola be given by y = a bx2. Then the area A of the segment cut off bythe x axis is given by

A = 2 a/b0

(a bx2) dx = 2(ax 1

3bx3

)

a/b

0

= 2a

a

b 2a

3

a

b=

4a

3

a

b.

Since the area of the inscribed triangle is a

ab , the result is established.

9. Let the equation of the parabola be y = x2, and let the straight line defining the segmentbe the line through the points (a, a2) and (b, b2). Thus the equation of this line is(a b)x + y = ab, and its normal vector is N = (a b, 1). Also, since the midpointof that line segment is B = ( ba2 ,

b2+a2

2 ), the x-coordinate of the vertex of the segment

is ba2 . If S = (x, x2) is an arbitrary point on the parabola, then the vector M from

(a, a2) to S is given by (x+ a, x2 a2). The perpendicular distance from S to the lineis then the dot product of M with N , divided by the length of N . Since the length of Nis a constant, to maximize the distance it is only necessary to maximize this dot product.The dot product is (x+a, x2a2)(ab, 1) = axbx+a2ab+x2a2 = axbx+x2ab.The maximum of this function occurs when a b + 2x = 0, or when x = ba2 . And, aswe have already noted, the point on the parabola with that x-coordinate is the vertexof the segment. So the vertex is the point whose perpendicular distance to the base ofthe segment is the greatest.

10. Let r be the radius of the sphere. Then we know from calculus that the volume of thesphere is VS =

43pir

3 and the surface area of the sphere is AS = 4pir2. The volume

of the cylinder whose base is a great circle in the sphere and whose height equals thediameter has volume is VC = pir

2(2r) = 2pir3, while the total surface area of the cylinderis AC = (2pir)(2r) + 2pir

2 = 6pir2. Therefore, VC =32VS and AC =

32AS, as desired.

11. Suppose the cylinder P has diameter d and height h, and suppose the cylinder Q isconstructed with the same volume but with its height and diameter both equal to f . Itfollows that d2 : f2 = f : h, or that f3 = d2h. It follows that one needs to construct thecube root of the quantity d2h, and this can be done by finding two mean proportionalsbetween 1 and d2h, or, alternatively, two mean proportionals between d and h (wherethe first one will be the desired diameter f).

12. The two equations are x2 = 4ay and y(3a x) = ab. Pick easy values for a and b, saya = 1, b = 1, and then the parabola and hyperbola may be sketched.

13. The focus of y2 = px is at (p4 , 0). The length of the latus rectum is 2pp4 = p.

15



14. The equation of the ellipse can be rewritten as p2ax2px+y2 = 0 or as x22ax+ 2ap y2 = 0,

or finally as(x a)2

a2+

y2

pa/2= 1.

Therefore the center of the ellipse is at (a, 0) and b2 = pa2 . The hyperbola can be treatedanalogously.

15. Let the parabola be y2 = px and the point C = (x0, y0). Then the tangent line at C hasslope p2y0 , and the equation of the tangent line is y =

p2y0

(x x0) + y0. If we set y = 0,we can solve this equation for x to get x = x0.

16. a. Let the ellipse be given by the equation b2x2 + a2y2 = a2b2. Let P have coordinates

(x0, y0). Then the slope of the tangent line at P is b2x0a2y0

. Thus the equation of line DK

is y = b2x0a2y0

x. By solving this equation simultaneously with the equation of the ellipse,

we get the coordinates of the point D as (ay0b , bx0a ). It follows that the slope of thetangent line at D is b

2ay0/ba2bx0/a

= y0x0 , which is the slope of the diameter PG, as desired.

b. Given that the coordinates of P are (x0, y0), it follows that tan =y0x0

as before.

Similarly, since the coordinates ofD are (ay0b , bx0a ), it follows that tan = bx0/aay0/b = b2x0

a2y0.

c. Take an arbitrary point S in the plane with rectangular coordinates (x, y) and obliquecoordinates (x, y). By drawing lines from S parallel to the two original axes and tothe two oblique axes, one can show that x = x cos y cos(180 ) = x cos +y cos and that y = x sin +y sin(180) = x sin +y sin. If we replace x and yin the equation of the ellipse by their values in terms of x and y, we get the equationspecified in the problem, once we notice that b2 cos cos + a2 sin sin = 0, giventhe values for tan and tan found in part b.

d. Let y = (tan )x be the equation of the diameter PG. If we solve this equationsimultaneously with the original equation for the ellipse, we find the coordinates ofthe point P to be

x =ab

b2 + a2 tan2 , y =

ab tan b2 + a2 tan2

.

It follows that

a =x2 + y2 =

ab sec b2 + a2 tan2

.

Similarly,

b =ab tan

b2 + a2 tan2 .

Then

a2 =a2b2 sec2

b2 + a2 tan2 =a2b2

Aor A =

a2b2

a2.

16



Similarly,

C =a2b2

b2.

If we substitute these values for A and C into the equation of the ellipse given in

part (c), we get the equation x2

a2+ y

2

b2= 1, or

y2 = b2a2 x2

a2

= b2a2

(a x)(a + x) = b2

a2x1x2

as desired.

e. Since PF = a sin( ) and CD = b, we have PF CD = ab sin( ) =a2b2 sin()

AC. But since b2 cos cos+a2 sin sin = 0, it follows that (b2 cos cos+

a2 sin sin)2 = 0 and therefore that

AC = a2b2(sin cos cos sin )2 = a2b2 sin2( ).Therefore,

PF CD = a2b2 sin( )ab sin( ) = ab

as claimed.

17. By Conics II8, if we pass a secant line through the hyperbola xy = 1 which goesthrough points M and N on that curve and points T and U on the y-axis and x-axisrespectively (the asymptotes), then the segments TM and TN are equal. Thus, if welet M approach N , then the secant line approaches the tangent line at N and thereforethe two line segments TN , NU between N and the asymptotes are equal. Therefore,the triangles TSN and NRU are congruent. If the coordinates of N are (x0,

1x0), then

TS = NR = 1x0 , and NS = x0. So the slope of the tangent line TNU is

TS

SN= 1/x0

x0= 1

x20.

18. Let the parabola have the equation y2 = px, with the focus at (p4 , 0). Since the slope ofthe tangent line at the point P = (x, y) is p2y , it follows that the direction vector T of

the tangent line can be written in the form (2y, p). Similarly, the direction vector L ofthe line parallel to the axis can be written as (1, 0) and the direction vector V of the linefrom P to the focus can be written as (x p4 , y). Then the cosine of the angle betweenT and L is 2y

4y2+p2. The cosine of the angle between T and V is given by

2y(x p4

)+ py

4y2 + p2(

x p4)2

+ y2=

2xy + py24y2 + p2

(x + p4

)2 = 2y(x + p4

)4y2 + p2

(x + p4

) = 2y4y2 + p2

.

17



Since these two cosines are equal, so are the angles.

19. If the two parallel lines are x = 0 and x = k and the perpendicular line is the x-axis, thenthe equation of the curve satisfying the problem is y2 = px(k x) or y2 = kpx px2.This is the equation of a conic section.

20. Since the square of the distance between a point and a line is a quadratic function of thecoordinates x, y of the point, and since the same is true for the product of the distancesto two separate lines, the equation defining the locus in the three-line problem will bea quadratic equation in x and y. Thus the locus will be a conic section, possibly adegenerate one.

21. crd 120 =4R2 R2 = 3R = 103; 55, 23; crd 30 =

R(2R crd 120) = 31; 03, 30;

crd 150 =4R2 crd230 = 115; 54, 40; crd 15 =

R(2R crd 150) = 15; 39, 47.

Similarly, crd 165 = 118; 58, 25 and crd 712= 7; 50, 54.

22. Use a quadrilateral ABCD with AB = crd, BC = crd (180 ( + )), CD = crd ,AD = 120 (the diameter of the circle). The diagonals are then AC = crd (180 ) andBD = crd (180 ). Then apply Ptolemys theorem.

23. 120 crd (72 60) = crd (72)crd (120) crd (60)crd (108). So 120 crd(12) = 70; 32, 3 103; 55, 23 60 97; 4, 56 = 1505; 11, 34. It follows that crd (12) = 12; 32, 36. Thencrd (168) =

4 602 crd2(12) = 119; 20, 33. Then crd (6) =

60(2 60 crd (168))

= 6; 16, 49. Similarly, crd (3) = 3; 8, 29; crd (112) = 1; 34, 15; and crd (34) = 0; 47, 7.

24. When = 90, then = 2351 and = 90. When = 45, we have sin =sin(2351) sin(45) and = 1637. Also tan = cos(2351) tan(45), so = 4227.By symmetry, the values for the declination at 270 and 315 are the negatives of thevalues at 90 and 45, respectively.

25. To calculate (60, 45), we note that if = 60, then = 2030 and = 5744. Sincesin = tan tan 45, we have = 2157 and = = 3547. If = 90, then = 2351 and = 90. So = 2614 and = = 6346.

27. Note that is the latitude where the sun is directly overhead at noon on the summersolstice. The angular distance between the noon altitudes of the sun at the summer andwinter solstice is, given the assumption that at any given time the suns rays to everypoint on the earth are parallel to each other, equal to the angle between the sun at noonon the summer solstice and the sun at noon on the winter solstice, as viewed from thecenter of the earth. And this angle, by Figure 3.34, is twice .

18



28. L(, ) = 180 + 2(, ). When = 60 and = 36, we calculate that sin =tan tan = tan(2030) tan(36); so = 1546 and L = 211; 32, which corresponds to14 hours, 6 minutes. Therefore, sunrise is 7 hours, 3 minutes before noon, or 4:57 a.m.and sunrise is at 7:03 p.m.

29. If the length of day is 15 hours when = 90, then 180 + 2(90, ) = 225. Therefore = 2230 and, since sin = tan tan, we have tan = sin(22

30)tan(2351) , so = 40

53.

30. The expression tan tan will be greater than 1 for = 2312when .4348 tan > 1, or

when tan > 2.2998, or when > 6612. When that occurs, the formula for L no longer

makes sense. Since when tan tan = 1, we know that L = 360 or 24 hours, it followsthat the sun does not set at all on the summer solstice when the latitude is greater than6612

.

31. If = 45, the = 1637; so SZ = = 45 1637 = 2823. Similarly, if = 90,then = 2351 and SZ = 219.

32. The sun is directly overhead at noon at latitude 20 when = 20. Since sin = sin sin 23.5 ,we find that = 59. This value for the longitude of the sun occurs at approximately 60days after the spring equinox and 60 days before the fall equinox, or at approximatelyMay 20 and July 21.

33. The maximal northerly sunrise point occurs when = 90 and therefore when = 2351.When = 36, we calculate that sin = sin 23

51sin 36 and = 29

59 north of east.

34. When = 75, we need to find so that sin sin 15 = 1. Clearly, = 15, and since

sin 15 = sin 2351 sin , it follows that = 3948. This value occurs approximately 40days after the vernal equinox, or about April 30.

19



CHAPTER FOUR

1. Let x = Diophantus age at death. Then x = 16x +112x +

17x + 5 +

12x + 4. It follows

that 9x = 756 and x = 84.

2. To solve x+y = 20; xy = 96, we set x = 10+ z, y = 10 z. Then 100 z2 = 96; z2 = 4;and z = 2. Thus x = 12 and y = 8.

3. To find two squares whose difference is 60, set x2 = smaller square and x2 + 60 = largersquare. Then x2 + 60 = (x + 3)2, where 3 is arbitrarily chosen. This equation reducesto 6x = 51 and therefore x = 172 . The two squares are therefore

2894 = 72

14 and 132

14 . In

the general case, the two squares are x2 and x2 + b = (x + a)2, where a2 < b. It follows

that x = ba2

2a and the two squares are found to be(b a22a

)2and

(b + a2

2a

)2.

4. Let x2 be the least square and (x +m)2 = x2 + 2mx +m2 be the middle square. Thedifference is 2mx + m2. Therefore the largest square is x2 + 2mx + m2 + n(2mx +m2) = x2 + (2m + 2mn)x + m2 + nm2 = (x + b)2 = x2 + 2bx + b2. Provided thatm2(1 + n) < b2 < m2(1 + n)2, the solution is

x =b2 m2 nm22m+ 2mn 2b.

5. To solve x y = 10, x3 y3 = 2170, set x = z + 5 and y = z 5. It follows that(z + 5)3 (z 5)3 = 2170. This equation reduces to 30z2 = 1920 or z2 = 64 or z = 8.Thus x = 13 and y = 3. In the general case, if x y = a and x3 y3 = b, we setx = z + a2 and y = z a2 If we substitute for x and y in the second equation, we get asequation in z which reduces to z2 = 4ba

3

12a . It follows that this latter expression must bea square.

6. To solve x + y = 20, x3 + y3 = 140(x y)2, set x = 10 + z, y = 10 z. Then(10 + z)3 + (10 z)3 = 140(2z)2. This equation reduces to 2000 + 60z2 = 560z2 or500z2 = 2000 or z2 = 4 or z = 2. Thus the solution is x = 12, y = 8. In general, ifx + y = a, x3 + y3 = b(x y)2, we set x = a2 + z, y = a2 z. On substituting into thesecond equation, we get 2(a2)

3 + 6a2z2 = b(2z)2, which reduces to a

3

4 + 3az2 = 4bz2 or

a3

4 = (4b 3a)z2. Thus z2 = a3

4(4b3a) . This equation has a rational solution providedthat the right side is a square, and that condition is equivalent to Diophantus conditionthat a3(b 34a) is a square.

20



7. Simply divide the given square a2 into two squares. This is possible by II8.

8. We want to solve x + y = (x3 + y)3. We set x = 2z and y = 27z3 2z (so thatx + y = (3z)3). Then (x3 + y)3 = (35z3 2z)3 = (3z)3. It follows that 35z2 = 5. Thisis impossible for rational z. But now note that 35 = 27 + 8 = 33 + 23 and 5 = 3 + 2.In order that the equation in z be solvable in rationals, we need two numbers a and

b (to replace the 3 and 2) so that a3+b3

a+b is a square. So let a + b = 2 (where 2 is

arbitrary). Then b = 2 a and a3 + b3 must equal 2 times a square. This implies thata3 + (2 a)3 = 8 12a + 6a2 = 2(square) or that 4 6a + 3a2 is a square. So set4 6a+3a2 = (2 4a)2 and solve for a. We get a = 1013 and therefore b = 1613 . Since it isonly the ratio of a and b which is important, we can choose a = 5, b = 8 and thereforeput x = 5z, y = 512z3 5z and repeat the initial calculation. We then get 637z3 = 13zand z2 = 149 , so z =

17 . Then x =

57 ; y =

267343 is the desired solution.

9. Suppose the right triangle has legs a, b, hypotenuse c, and angle bisector d. Let r be thelength of that part of leg a from the right angle to the point where the bisector intersectsthe leg. To make the right triangle with the angle bisector as hypotenuse a rationaltriangle, we can set d = 5x and r = 3x. It follows that b = 4x. If we then let a = 3, wehave from Elements VI3 that c : (a r) = b : 4 or that c : (3 3x) = 4x : 3x. Thusc = 4 4x and the reason why a was chosen to be 3 is evident. By the Pythagoreantheorem, we have (4 4x)2 = 32 + (4x)2 or 16 32x+ 16x2 = 16x2 + 9. Thus 32x = 7and x = 732 . To get integral answers, we can multiply through by 32. Thus the originaltriangle is (96, 28, 100) and the bisector equals 35.

10. The diagram for Elements VI28 is Fig. 2.16. Let us assume that the proposed rectanglehas been constructed with base AS and area equal to c and that the defect is a square.If we set AB = b, and BS = x, then AS = b x and x(b x) = c. Since the maximumof the function f(x) = x(b x) occurs when x = b2 , and since this maximum is ( b2)2,it follows that c cannot exceed the value ( b2)

2. This means that the area c of the givenrectilinear figure must not be greater than the area of the square on half the given lineof length b.

11. Assume that the theorem is true. Then AB2+BC2 = 3AC2. But since AB = AC+BC,we have (AC+BC)2+BC2 = 3AC2. This reduces to AC2+2AC BC+2BC2 = 3AC2or AC BC + BC2 = AC2. This in turn implies that BC(AC + BC) = AC2 or thatAB BC = AC2. But this is precisely the statement that AB is cut in extreme andmean ratio at C.

12. Suppose that three of the lines have equations x = a, x = b, x = c, that the other twohave equations y = d, y = e, and that the fixed line has length k. Then the equationof the locus is (x a)(x b)(x c) = k(y d)(y e). Other arrangements of the lineswill give somewhat different equations, but in any case the locus is described by a cubicequation in x and y.

21



13. The equation is x 27x 112x 16x 13x 20 12 11 = 1. Multiplying by 84 andsimplifying gives 11x = 3696 or x = 336.

14. In 12 days the spouts will fill 12+6+4+3 = 25 tanks. Therefore, one tank will be filledin 1225 of a day.

15. This problem can be translated into two equations in two unknowns: x+10 = 3(y10);y + 10 = 5(x 10). We can write these as x 3y = 40; 5x+ y = 60. The solutionis then that A has x = 1557 coins and B has y = 18

47 coins.

22



CHAPTER FIVE

1. We write, in order, the Chinese form of 56, 554, 63, and 3282:

2. The largest digit a so that (100a)2 < 142, 884 is a = 3. If we subtract 3002 = 90, 000 from142,884, the remainder is 52,884. We then need to find b so that 2(100a)(10b) < 52, 884,or 6000b < 52, 884. We take b = 7 and check that 6000b + (10b)2 < 52, 884. But thisinequality reduces to 42, 000 + 4900 < 52, 884. Since the left side equals 46,900, theinequality is in fact true. Note that if we had taken b = 8, this second inequality wouldnot have been true. We now subtract 46,900 from 52,884 to get 5,984. We now need tofind c so that 2(370)c < 5984. We try c = 8 and check that 740c + c2 5984. But theleft side of this inequality is in fact equal to 5984, so the desired square root is 378.

3. 560 + 350 + 180 = 1090. Tx =5601090 100 = 51 41109 ; Ty = 3501090 100 = 32 12109 ; Tz =

1801090 100 = 16 56109 .

4. In 15 days, the first channel fills the reservoir 45 times, the second channel 15 times, thethird, 6 times, the fourth, 5 times, and the fifth 3 times. It follows that in 15 days thereservoir is filled 74 times. To fill it once then requires 1574 of a day.

5. If x is the unknown amount, the conditions show that after the first tax the man had23x; after the second, he had

45 23x; and after the third, he had 67 45 23x. This amount

must equal 5. It follows that x = 17516 = 101516 pounds.

6. If x is the hypotenuse of a right triangle and 10 and d the legs, then x = d + 1 ord = x 1. Then x2 = (x 1)2 + 100 and x = 50.5.

7. If we set x to be the length of a side of the city, draw a line through the center of the cityextending 20 pu north and 14 pu south, extend a line 1775 pu west from the bottom ofthat line, and connect the end of that new line with the end of the line to the north, weget a right triangle with legs x+34 and 1775. Since we also have a similar triangle withlegs 20 and x2 , we get the proportion 20 :

x2 = (x + 34) : 1775. The resulting equation is

x2 + 34x = 71000 and x = 250.

23



8. If x is the depth of the well, the similarity relationship gives 50.4x =0.45 . Thus x =

54.60.4 = 57.5.

9. The simplest way is to set y = DC, x = CE. Then

y

x+ 1313=

10

1313and

y

x+ 5=

3 931205

.

Simplifying these equations gives 40y = 30x + 400; 200y = 151x + 755. Solving thesesimultaneously gives the solution x = 1245, y = 94334 .

10. We begin with c6 = r = 10 and a6 =102 52 = 8.6603. Therefore, S12 = 12 61010 =

300. Then c12 =(10/2)2 + (10 8.6603)2 = 5.1764, and S24 = 12 1210c12 = 310.5859.

We then get a12 =100 (c12/2)2 = 9.6593. Next, c24 =

(c12/2)2 + (10 a12)2 =

2.6105 and S48 =12 24 10 c24 = 313.2629. Next. a24 =

100 (c24/2)2 = 9.9144 and

c48 =(c24/2)2 + (10 a24)2 = 1.3081. So S96 = 12 48 10 c48 = 313.9350. Finally,

we get a48 =100 (c48/2)2 = 9.9786, so c96 =

(c48/2)2 + (10 a48)2 = 0.6544. We

then get S192 =12 96 10 c96 = 314.1032.

12. From Figure 5.8, we see that one-eighth of the volume of the double box-lid is r0 (r

2 x2) dx = r3 r33 = 23r3. It follows that the entire volume is 163 r3.

13. If x is the yield of good grain, y the yield of ordinary grain, and z the yield of worstgrain, then the system of equations is

2x + y = 13y + z = 1

x + 4z = 1

In matrix form we get, in turn

1 0 2 0 0 2 0 0 20 3 1 1 3 1 0 3 14 1 0 8 1 0 25 1 01 1 1 1 1 1 4 1 1

It follows that 25z = 4 or z = 425 ; 3y + z = 1, 3y =2125 , or y =

725 ; 2x + y = 1, 2x =

1825 ,

or x = 925 .

14. a. In this case, we see by trial that the solution is between 6 and 7. So we use 6 in thesynthetic division procedure:

6 | 16 192 1863.296 1728

6 | 16 288 |135.296

6 | 16 |384|16

24



For the next step, we will use decimals. The (positive) solution to 16x2 + 384x 135.2is between 0 and 1. Again, we find by trial that the value is between 0.3 and 0.4. So ournext chart is as follows:

.3 | 16 384 135.24.8 116.64

.3 | 16 388.8 |18.564.8

.3 | 16 |393.6|16

For a third step, we will try values between 0 and 0.1. Again, the closest value seems tobe 0.5, as in the following chart:

.05 | 16 393.6 18.56.8 19.72

16 394.4 |0.96Since the last value is relatively close to 0, we will leave the solution as 6.35. If wewanted to go further, we could have used 0.4 in this last step and continued to find thenext decimal place.

b. In this case, we see by trial that the solution begins with a 2 in the 10s place, sowe try 20 in the chart:

20 | 1 0 15, 245 0 6, 262, 506.2520 400 296, 900 5, 938, 000

20 1 20 14, 845 296, 900 | 324, 506.2520 800 280, 900

20 1 40 14, 045 |577, 80020 1, 200

20 | 1 60 |12, 84520

20 | 1 |80|1

The fourth degree polynomial we find for our next step has a solution between 0 and1. We will therefore leave this solution as x = 20, but it is not difficult to continuethe solution further.

15. Qins method gives the following diagrams to produce 235 as the answer. We begin bynoting that the answer is a three-digit number beginning with 2.

200 | 1 0 55, 225200 40, 000

200 | 1 200 |15, 225200

200 | 1 |400|1

25



We next check that the second digit is a 3.

30 | 1 400 15, 22530 12, 900

30 | 1 430 |2, 32530

30 | 1 |460|1

The final digit is a 5.5 | 1 460 2325

5 2325

1 465 |0

16. Qins method gives the following diagrams to produce 234 as the answer. We begin bynoting that the answer is a three-digit number beginning with a 2.

200 | 1 0 0 12, 812, 904200 40, 000 8, 000, 000

200 | 1 200 40, 000 |4, 812, 904200 80, 000

200 | 1 400 |120, 000200

1 |600|1

The second digit is a 3.

30 | 1 600 120, 000 4, 812, 90430 18, 900 4, 167, 000

30 | 1 630 138, 900 |645, 90430 19, 800

30 | 1 660 |158, 70030

1 |690|1

The final digit is a 4.4 | 1 690 158, 700 645, 904

4 2, 776 645, 904

1 694 161, 476 |026



The third order coefficients occur, for example, in 600 = 3 200, 120, 000 = 3 2002,and in 690 = 3 230 and 158, 700 = 3 2302.

17. The diagram is as follows, where the solution is x = 23. We begin by noting that thefirst digit is a 2.

20 | 1 0 0 0 279, 84120 400 8, 000 160, 000

20 | 1 20 400 8, 000 |119, 84120 800 24, 000

20 | 1 40 1, 200 |32, 00020 1, 200

20 | 1 60 |2, 40020

20 | 1 |80|1

The second digit is a 3.

3 | 1 80 2, 400 32, 000 119, 8413 249 7, 947 119, 841

1 83 2, 649 39, 947 |0The fourth order coefficients show up in 80 = 4 20, 2400 = 6 202, 32, 000 = 4 203,and 160, 000 = 1 204.

18. By the Pythagorean theorem, the altitude h of the lower triangle is given byb2 ( c2)2.

The area B of that triangle is then B = c2

b2 ( c2)2 as stated. Similarly, the area

A of the upper triangle is A = c2

a2 ( c2)2. If x = A + B, then x4 = (A + B)4 =

A4 + 4A3B + 6A2B2 + 4AB3 + B4 = 2(A4 + 2A3B + 2A2B2 + 2AB3 + B4) (A4 2A2B2+B4) = 2(A2+B2)(A2+2AB+B2) (A2B2)2 = 2(A2+B2)x2 (A2B2)2.The equation for x follows immediately. If a = 39, b = 25, and c = 30, we haveA = 15

1521 225 = 151296 = 15 36 = 540. Similarly, B = 15400 = 300.

Then 2(A2 + B2) = 2(291, 600 + 90, 000) = 763, 200 and (A2 B2)2 = (201600)2 =40, 642, 560, 000 as desired.

19. Using the notation from the description of Qin Jiushaos method, we first note thatM = 12. Then M1 = 12 3 = 4 and M2 = 12 4 = 3. Also P3 = 1, and P4 = 3.Therefore, we need to solve two congruences: x1 1 (mod 3), and 3x2 1 (mod 4).The solutions are x1 = 1 and x2 = 3. Therefore N = 0 4 1+1 3 3 = 9 9 (mod 12).

20. Using the notation from the description of Qin Jiushaos method, we first calculateM = 11 5 9 8 7 = 27720. Then M1 = M 11 = 2520; M2 = M 5 = 5544;

27



M3 =M 9 = 3080; M4 =M 8 = 3465; and M5 = M 7 = 3960. We then calculatethat M1 1 (mod 11); M2 4 (mod 5); M3 2 (mod 9); M4 1 (mod 8);and M5 5 (mod 7). We next need to solve congruences: the solution to 1x1 1(mod 11) is x1 = 1; to 4x2 1 (mod 5) is x2 = 4; to 2x3 1 (mod 9) is x3 = 5;to 1x4 1 (mod 8) is x4 = 1; and to 5x5 1 (mod 7) is x5 = 3. Given that 3 ofthe ri are equal to 0, we calculate N simply as n = 4 3080 5 + 6 3465 1 = 82, 390.Subtracting off twice M , we get the solution as N = 82, 3902 27, 720 = 26, 950, wherethe answer is taken modulo 27,720.

28



CHAPTER SIX

1. We have KM = KL = BC = a and KA = b. It follows by the Pythagorean Theoremthat AM2 = KM2 AM2 = b2 a2. Thus the square on AM is the difference of thetwo original squares, the one on AB and the one on PQ.

2. The rectangle ABCD is transformed into the gnomon AEGFKH of the same area bythe indicated construction. This gnomon is equal to the difference of the squares on AEand FK. By the previous exercise, we can construct a square equal to that difference.This square will therefore be equal to the rectangle, as desired.

3. Since AB = s, we have MN = r = s2 +13(

s2

2 s2). Thus rs = 12 +26 16 = 2+

2

6 .Given that the area of the circle of radius MN = r equals the area of the square of side

AB = s, we have pir2 = s2 or r2

s2= 1pi or

rs =

1pi. Thus 1

pi= 2+

2

6 , orpi = 6

2+2, or

pi = 3.088311755 . . .

4. If we calculate the sum and difference of the given fractions, we get 0.878681752. If thesquare of this side is equal to the area of a circle of diameter 1, then (0.878681752)2 = pi4 ,

or pi = 4(0.878681752)2 = 3.088326491.

5. a. Let 6 ABC = . Then 6 ADC = pi . Set x = AC. In triangle ABC, we havex2 = a2+ b2 2ab cos , while in triangle ADC, we have x2 = c2+ d2 2cd cos(pi ) =c2 + d2 +2cd cos . Setting the two expressions for x2 equal, we get a2 + b2 2ab cos =c2 + d2 + 2cd cos and therefore

cos =a2 + b2 c2 d2

2ab + 2cd.

b. We get

x2 = a2 + b2 2ab(a2 + b2 c2 d2)2ab + 2cd

=cd(a2 + b2) + ab(c2 + d2)

ab + cd.

c. cd(a2 + b2) + ab(c2 + d2) = a2cd+ b2cd+ c2ab + d2ab = (ac + bd)(ad+ bc).

d. From parts b and c we get

x2 =ac + bd)(ad+ bc)

ab + cdor x = AC =

(ac+ bd)(ad+ bc)

ab+ cd.

Similarly, we have

y = BD =

(ac+ bd)(ab + cd)

ad+ bc.

29



6. a. From the law of cosines applied to triangle ABC, we get

b2 = a2 + x2 2ax cos 6 BAE = a2 + x2 2axAEa

= a2 + x2 2x AE.

Therefore b2 a2 = x(x 2AE).b. Because x = 2AM , we have b2 a2 = x(2AM 2AE) = x(2EM) = 2x EM.

Therefore EM = b2a22x .

c. By the same arguments as in parts a and b applied to triangle ADC, we get FM =d2c22x .

d. Let P be the area of quadrilateral ABCD. Then P is the sum of the areas oftriangles ABC and ADC. Therefore, P = 12x BE + 12x DF = 12x(BE +DF ), andP 2 = 14x

2(BE +DF )2.

e. Because BE +DF = BK, we get from part d and the Pythagorean Theorem thatP 2 = 14x

2BK2 = 14x2(BD2 DK2) = 14x2(y2 EF 2).

f. From parts b and c, we have

EF = EM + FM =b2 a22x

+d2 c22x

=(b2 + d2) (a2 + c2)

2x.

Substituting this value into the expression in part e, along with the values for x2 andy2 from exercise 5, we have

P 2 =1

4

(ac+ bd)(ad+ bc)

ab + cd

((ac+ bd)(ab + cd)

ad + bc [(b

2 + d2) (a2 + c2)]24x2

)

=1

4(ac+ bd)2 1

16

[(b2 + d2) (a2 + c2)

]2=

1

16

[4(ac + bd)2 [(b2 + d2) (a2 + c2)]2

].

g. We have s a = 12(a+ b+ c+ d) a = 12(b+ c+ d a), with a similar result for thethree other cases.

h. First, we calculate (b+ c+ d a)(a+ c+ d b)(a+ b+ d c)(a+ b+ c d). By firstmultiplying together the first two expressions and then the last two expressions andthen multiplying the two resulting expressions together, we find that this expressionbecomes (2ab+2cd+c2+d2a2b2)(2ab+2cd+a2+b2c2d2) = 8abcd+2(a2b2+a2c2+a2d2+b2c2+b2d2+c2d2)a4b4c4d4. On the other hand, if we multiplyout the numerator of the expression for P 2 from part f, we get 4(ac + bd)2 [(b2 +d2)(a2+c2)]2 = 4(a2c2+2abcd+b2d2[(b2+d2)22(b2+d2)(a2+c2)+(a2+c2)2] =4(a2c2+2abcd+b2d2)[b4+2b2d2+d42a2b22b2c22a2d22c2d2+a4+2a2c2+c4] =8abcd + 2(a2c2 + b2d2 + a2b2 + b2c2 + a2d2 + c2d2) a4 b4 c4 d4, the sameexpression as in the first calculation. It follows that the area of the quadrilateral is

S =(s a)(s b)(s c)(s d), as asserted.

30



7. We get the following:

1096x+ 808 = 3y 3y 808 = 1096x 1096 = 365 3 + 13y 808 = (365 3 + 1)x 3(y 365x) 808 = x t = y 365x3t 808 = x x + 808 = 3tBy inspection, we get x = 2, t = 270. Then y = t + 365x = 270 + 730 = 1000.. Since2 is already the smallest possible solution for x, the result is x = 2, y = 1000, andN = 808 + 1096 2 = 0 + 3 1000 = 3000.

8. We need to solve 137x+ 23 = 60y. We perform Brahmaguptas algorithm:

137x+ 23 = 60y 60y 23 = 137x 137 = 2 60 + 1760y 23 = (2 60 + 17)x 60(y 2x) 23 = 17x t = y 2x60t 23 = 17x 17x+ 23 = 60t 60 = 3 17 + 917x+ 23 = (3 17 + 9)t 17(x 3t) + 23 = 9t u = x 3t17u+ 23 = 9t 9t 23 = 17u 17 = 1 9 + 89t 23 = (1 9 + 8)u 9(t 1u) 23 = 8u v = t 1u9v 23 = 8u 8u+ 23 = 9v 9 = 1 8 + 18u+ 23 = (1 8 + 1)v 8(u 1v) + 23 = 1v w = u 1v8w + 23 = 1v 1v 23 = 8wBy inspection, we find that v = 31, w = 1. We then calculate u = 1v + w = 32,t = 1u+v = 63, x = 3t+u = 221, and y = 2x+t = 505. So x = 221, y = 505 is a solutionto the equation. The general solution is then x = 221+ 60z, y = 505+ 137z. To get thesmallest value for x, choose z = 3. Then x = 41 and y = 94. Then N = 60y = 5640.Since the solution is taken module 8220, we get N 5640 (mod 8220).

9. To solve 1096x+ 1 = 3y, we apply Brahmaguptas algorithm:

1096x+ 1 = 3y 3y 1 = 1096x 1096 = 365 3 + 13y 1 = (365 3 + 1)x 3(y 365x) 1 = x t = y 365x3t 1 = x x + 1 = 3tBy inspection, we find that t = 1, x = 2. Then y = t + 365x = 731. To solve1096x+ 10 = 3y, we simply multiply everything by 10: x = 20, y = 7310.

10. We will show, via a generalizable example, that Brahmaguptas method does give asolution to the equation rx + c = sy, assuming that the greatest common divisor of rand s divides c. In fact, we will assume that the greatest common divisor is equal to1, although one could generalize the procedure here to the case where it is greater than1. We will apply the Euclidean algorithm to r and s and assume, for simplicity, that isstops after four steps. We therefore have the following system:

rx+ c = sy sy c = rx r = q1s+ r1sy c = (q1s+ r1)x s(y q1x) c = r1x t = y q1xst c = r1x r1x+ c = st s = q2r1 + r2

31



r1x + c = (q2r1 + r2)t r1(x q2t) + c = r2t u = x q2tr1u+ c = r2t r2t c = r1u r1 = q3r2 + r3r2t c = (q3r2 + r3)u r2(t q3u) c = r3u v = t q3ur2v c = r3u r3u+ c = r2v r2 = q4r3 + 1r3u+ c = (q4r3 + 1)v r3(u q4v) + c = 1v w = u q4vr3w + c = 1v 1v c = r3wWe now set w = 1, v = r3 + c, and solve for the other letters. We get u = w + q4v =r3q4+ cq4 +1, t = v+ q3u = r3q3q4 + cq3q4 + q3 + r3+ c. Then x = u+ q2t = r3q2q3q4 +cq2q3q4+q2q3+q2r3+q2c+r3q4+cq4+1 and y = t+q1x = q1x+r3q3q4+cq3q4+q3+r3+c.To prove that the method does in fact give us a solution to the original equation, weneed to substitute these values into that equation. In other words, we must show thatrx + c = sy. Substituting the value for r given in the first line of our process, we needto show that

(q1s+ r1)x + c = s(q1x+ r3q3q4 + cq3q4 + q3 + r3 + c).

This is equivalent to showing that

r1x + c = s(r3q3q4 + cq3q4 + q3 + r3 + c).

We next substitute for s its value from the third line. At the same time, we substitutefor x the value we calculated above. We must therefore show that

r1[q2(r3q3q1+cq3q4+q3+r3+c)+r3q4+cq4+1]+c = (q2r1+r2)(r3q2q4+cq3q4+q3+r3+c).

This is in turn equivalent to showing that

r1(r3q4 + cq4 + 1) + c = r2(r3q3q4 + cq3q4 + q3 + r3 + c).

We next substitute for r1 from the fifth line. We thus must show that

(q3r2 + r3)(r3q4 + cq4 + 1) + c = r2(r3q2q4 + cq3q4 + q3 + r3 + c).

To demonstrate this equality, it suffices to show that

r3(r3q4 + cq4 + 1) + c = r2(r3 + c).

To do this, we finally substitute for r2 its value from the seventh line. We therefore mustdemonstrate that

r3(r3q4 + cq4 + 1) + c = (q4r3 + 1)(r3 + c).

That this final equation is true comes from multiplying it out. We have thus shown thatthe values calculated by Brahmaguptas method in fact satisfy the original equation.

11. The Chinese method requires that the moduli be relatively prime. In this problem,we note that if N 2 (mod 3) and N 3 (mod 4), then N 1 (mod 2) andalso N 5 (mod 6). Therefore, we may ignore the first congruence and solve the last

32



three. Using the notation from chapter 5, we first calculate thatM = 60. ThenM1 = 12,M2 = 15, M3 = 20, P1 = 2, P2 = 3, and P3 = 2. We must solve 2x1 1 (mod 5);3x2 1 (mod 4) and 2x3 1 (mod 3). The solutions are x1 = 3, x2 = 3, x3 = 2.Therefore, N = 4 12 3+3 15 3+2 20 2 359 (mod 60), and the smallest positiveN is 59. For the Indian method, we solve the first two congruences, then use that answeralong with the third, and the new answer along with the fourth. The solution of N 5(mod 6) 4 (mod 5) requires solving the equation 6x + 1 = 5y. The solution by theprocedure of the previous problems is x = 4, y = 5 and then N = 4 6 + 5 = 29. Wenext solve N 29 (mod 30) 3 (mod 4). We must solve 30x + 26 = 4y. We getx = 1, y = 14 and N = 1 30 + 29 = 59. To solve N 59 (mod 60) 2 (mod 3),we note that already 59 2 (mod 3); so N = 59 is the solution to the entire set ofcongruences.

12. To solve this congruence in the Chinese fashion, we note that since the two moduli arerelatively prime, M1 = 60 and M2 = 137, while P1 = 60 and P2 = 17. We thus mustsolve the two congruences 60x1 1 (mod 137) and 17x2 1 (mod 60). The secondcongruence is not important, because its solution will ultimately be multiplied by 0. Sowe simply apply the Euclidean algorithm to solve the first congruence. This amounts tothe following:

137 = 2 60 + 17 2 1 + 0 = 260 = 3 17 + 9 3 2 + 1 = 717 = 1 9 + 8 1 7 + 2 = 99 = 1 8 + 1 1 9 + 7 = 16

so the solution is x1 = 16. Then N = 10 60 16 = 9600 1380 (mod 8220). Althoughthe Indian method detailed in the text begins with the same Euclidean algorithm, thesteps in the remainder of that process are different from the steps here.

13. This problem is equivalent to finding N to solve N 0 (mod 17) 1 (mod 75). Inthe Chinese method, M1 = 75 and M2 = 17. Then P1 = 7 and P2 = 17. We thensolve 7x1 1 (mod 17) and 17x2 1 (mod 75). The solution to the first congruenceis unnecessary, because it will be multiplied by 0. We get the solution to the secondcongruence by the Euclidean algorithm. We get

75 = 4 17 + 717 = 2 7 + 37 = 2 3 + 1

By substitution, we get 1 = 53 17 12 75, so x2 = 53. Then N 1 17 53 901(mod 1275). In terms of the original problem, we have 17 53 1 = 75m, so m = 12 andn = 53.

In the Indian method, we use the Euclidean algorithm in Brahmaguptas procedure:

17n 1 = 75m 75 = 4 17 + 717n 1 = (4 17 + 7)m 17(n 4m) 1 = 7m u = n 4m17u 1 = 7m 7m+ 1 = 17u 17 = 2 7 + 3

33



7m+ 1 = (2 7 + 3)u 7(m 2u) + 1 = 3u v = m 2u7v + 1 = 3u 3u 1 = 7v 7 = 2 3 + 13u 1 = (2 3 + 1)v 3(u 2v) 1 = 1v w = u 2v3w 1 = 1v 1v + 1 = 3wWe now choose v = 2, w = 1. Then u = w + 2v = 5, m = v + 2u = 12, andn = u+ 4m = 53. Thus, m = 12, n = 53 is the solution.

14. D(u0v1 + u1v0)2 + c0c1 = D(u0v1 + u1v0)

2 + (v20 Du20)(v21 Du21) = 2Du0v1u1v0 +D2u20u

21 + v

20v

21 = (Du0u1 + v0v1)

2.

15. To solve 83x2+1 = y2, we begin by noting that (1, 9) is a solution for subtractive 2; thatis, 83 122 = 92. If we compose this solution with itself, we get y1 = 83 12+92 = 164,x1 = 9 + 9 = 18, b1 = 4. Therefore, (18, 164) is a solution for additive 4. But then wecan simply divide everything by 4 = 22 to get that (9, 82) is a solution for additive 1:83 92 + 1 = 822.

16. To show that (u1, v1) is a solution, we calculate each side of the equation Du21 + 1 = v

21

and show that they are equal. The left side is

L = D[1

2uv(v2 + 1)(v2 + 3)

]2+ 1

=1

4Du2v2(v2 + 1)2(v2 + 3)2 + 1

=1

4(4 + v2)v2(v2 + 1)2(v2 + 3)2 + 1

=1

4(v4 + 4v2)(v2 + 1)2(v2 + 3)2 + 1.

The right side is

R =[(v2 + 2)

(1

2(v2 + 1)(v2 + 3) 1

)]2= (v2 + 2)2

[1

4(v2 + 1)2(v2 + 3)2 (v2 + 1)(v2 + 3) + 1

]=

1

4(v4 + 4v2 + 4)(v2 + 1)2(v2 + 3)2 (v2 + 2)2(v2 + 1)(v2 + 3) + (v2 + 2)2

=1

4(v4 + 4v2)(v2 + 1)2(v2 + 3)2 + (v2 + 1)2(v2 + 3)2 (v2 + 2)2(v2 + 1)(v2 + 3)

+ (v2 + 2)2

=1

4(v4 + 4v2)(v2 + 1)2(v2 + 3)2 + (v2 + 1)(v2 + 3)[(v2 + 1)(v2 + 3) (v2 + 2)2]

+ (v2 + 2)2

=1

4(v4 + 4v2)(v2 + 1)2(v2 + 3)2 + (v2 + 1)(v2 + 3)(1) + (v2 + 2)2

=1

4(v4 + 4v2)(v2 + 1)2(v2 + 3)2 + 1.

34



Thus the two sides are equal as asserted. Next, note that if u or v is even, then 12uv is

an integer, so u1 is an integer. If both u and v are odd, then v2 +1 is even, so 12(v

2 +1)

is an integer and u1 is an integer. If v is even, then v2 + 2 is even and (v2 + 2)12(v

2 + 1)

is an integer. If v is odd, then v2 + 1 is even, so 12(v2 + 1) is an integer. Thus in either

case, v1 is an integer.

17. To solve 13x2 + 1 = y2, we begin by noting that (1, 3) is a solution for subtractive4: 13 12 4 = 32. By the previous problem, set u1 = 12 1 3 10 12 = 180 andv1 = 11[

12 10 12 1] = 649. Then (180, 649) is the desired solution.

18. If Du2+2 = v2, then D(uv)2+1 = Du2v2+1 = (v22)v2+1 = v42v2+1 = (v21)2,as desired. If Du2 2 = v2, then (u1, v1) = (uv, v2 + 1) solves Du21 + 1 = v21. The proofis virtually identical to the previous one.

19. To solve 61x2 + 1 = y2, we begin by noting that 61 12 + 3 = 82; that is, that (1, 8) is asolution for additive 3. We then need to solve 1m+8 = 3n. The general solution is easilyseen to bem = 1+3t, n = 3+t. We now choose t so thatm2 is close to 61: t = 2,m = 7,m2 = 49. Then take u1 =

17+83 = 5, b1 = 61493 = 4, and v1 =

61 25 4 = 39.

We now have 61 52 4 = 392; that is, (5, 39) is a solution for subtractive 4. Nowuse the result of problem 16. Set u1 =

12 5 39(392 + 1)(392 + 3) = 226, 153, 980 and

v1 = (392 + 2)[12(39

2 + 1)(392 + 3) 1] = 1, 766, 319, 049. Then (u1, v1) is a solution tothe original equation.

20. Since 13x +16x =

12x, the equation is

x[1

2x +

1

2

(1

2x)+

1

2

(1

4x)+

1

2

(1

8x)+

1

2

(1

16x)+

1

2

(1

32x)+

1

2

(1

64x)]

= 1161.

This equation reduces to 1128x = 1161, so x = 148, 608 is the solution.

21. In 1 day, the well is filled 2 + 3 + 4 + 5 = 14 times. Thus the well will be filled once in114 of a day. In that time period, the first pipe will fill the well

114 2 = 214 full. Similarly,

the second pipe will fill 314 of the well, the third414 , and the fourth

514 .

22. If t is the number of days until the second person overtakes the first, the equation is5(t+ 7) = 9t. The solution is t = 834 days.

23. If x is the amount held by the first traveler, y the amount held by the second, and pthe amount in the purse, then the problem results in two equations in three unknowns:x+ 12p = 2y; y+

23p = 3x. The solution is a one-parameter family, expressible as y =

1311x;

p = 3011x. Since the solutions must be integers, x must be a multiple of 11. Thus x = 11,y = 13, p = 30 is a solution, as is any positive integral multiple of that solution.

35



24. If we calculate a table of sine values by Bhaskaras formula (not multiplying by 3438)and compare them to actual sine values, we get the following:

Angle Bhaskaras sine Actual sine

0 0.00000 0.0000010 0.17525 0.1736520 0.34317 0.3420230 0.50000 0.5000040 0.64183 0.6427950 0.76471 0.7660460 0.86486 0.8660370 0.93903 0.9396980 0.98461 0.9848190 1.00000 1.00000

An inspection of this table shows that the difference between the two values is greatestat 10 degrees, where the actual difference is 0.00160, which corresponds to a percentage errorof less that 1%.

25. Brahmaguptas procedure gives us

sin(16) = sin(15 + 1) = sin(15) +1

2(334)(219 + 215) 1

2

2(334)2(219 215)

= 890 +2

15(434) 8

225(4) = 948

to the nearest integer. Bhaskaras procedure gives

sin 16 = 3438 4 16 16440, 500 16 164 = 953

to the nearest integer. The exact value is 948 to the nearest integer, so Bhaskaras answeris in excess of the correct answer by approximately 0.5%.

26. We assume that yi is/n (is)3/6n3, given the approximation we have found for y.We put this into the expression for x = cos s and use the formula for the sum of integralcubes:

x 1 limn

s

n

[s

n s

3

6n3+

2s

n (2s)

3

6n3+ + (n 1)s

n ((n 1)s)

3

6n3

]

= 1 s2

2+ lim

ns4

6n4[13 + 23 + + (n 1)3]

= 1 s2

2+s4

6limn

n1i=1 i

3

n4= 1 s

2

2+s4

6 14

= 1 s2

2+s4

24.

36



We can similarly calculate a new value of y = sin s, using our knowledge of the doublesum formula as well. We have

y s s3

6

+ limn

(s

n

)2 [ s36n3

+

(s3

6n3+

(2s)3

6n3

)+ +

(s3

6n3+

(2s)3

6n3+ + ((n 1)s)

3

6n3

)]

= s s3

6+ lim

ns5

6n5[13 + (13 + 23) + + (13 + 23 + + (n 1)3)]

= s s3

6+ lim

ns5

6n5[n(13 + 23 + + (n 1)3) (14 + 24 + + (n 1)4)]

= s s3

6+s5

6limn

[n1i=1 i

3

n4n1

i=1 i4

n5

]

= s s3

6+s5

6

(1

4 1

5

)= s s

3

6+

s5

120.

Putting the new value for y into the formulas for x and y will lead by a similar argumentto the next terms in both the sine and cosine series.

37



CHAPTER SEVEN

1. Al-Khwarizmis rule for solving bx + c = x2 translates to the formula

x =

( b2

)2+ c+

b

2.

The main point of the geometric proof is that rectangle RBMN is equal to rectangleNKTL. Then, because ( b2)

2 is equal to rectangle KHGT , we get that ( b2)2 + c is

represented by rectangle MAGL, so that the square root in the formula is equal to theside of that square, namely GA.

2. a. (13x + 1)(14x+ 1) = 20 transforms to x

2 + 7x = 228. The formula then gives

x =

(7

2

)2+ 228 7

2= 12.

b. x2+(10x)2 = 58 transforms to x2+21 = 10x. The formula gives x = 525 21 =7, 3.

3. a. Multiplying the equation by 2 gives x2+10x = 56. The solution is x =81 5 = 4.

b. Dividing the equation by 2 gives x2+5x = 24. The formula then gives x =

1214 52 =

112 52 = 3.

4. The equation isx

10 x +10 xx

=13

6.

If we multiply both sides by 6x(10x) and simplify, we get x2+24 = 10x. The solutionsare then x = 6 and x = 4.

5. a. The equation is x2 = (10 x)10. We can rewrite this as x2 +10x = 1010. Theformula then gives us that one part is

x =

(102

)2+ 10

10

10

2=

21

2+1000

21

2.

The other part is

y = 10 x = 10 +21

221

2+1000.

b.

38



6. a. If we set x = y2, the equation becomes [y2 (2y + 10)]2 = 8y2. Taking squareroots and rearranging gives us y2 = (2 +

8)y + 10. The formula then gives us y =

1 +2 +

13 +

8, and x = y2.

b. If we expand the left side, rearrange, and then square both sides, the equationbecomes 9x = x2 + 494 . An application of the formula gives x = 4

12

8. (We note

that if we use the plus sign in this result, we do not get a correct answer.) As analternative, we could set x = 2y2. The equation then becomes (2y2 + y)2 = 8y2 or

2y2 + y =8y. The solution to this is y =

812 . Then x = 2y

2 = 412 8 as

before.

7.x2 x 1 1x

1x2

1x3

1x4

1x5

1x6

1x7

313 5 623 10 1313 20 2623 4020 306 0 12

30 4040 60

60 8080 120

120 160160 240

240 320320 480

If an represents the coefficient of1xn and bn represents the leftmost entry in the row which

begins two columns to the left of the column under 1xn , then the method of calculation

shows that bn+2 = 2bn and that bn = 6an. But then an+2 = bn+26 = 2bn6 = 2an.

8. The calculation is as follows:

10 1 4 10 0 8 220 2 58 75 125 96 94 140 50 90 20

2 0 5 5 102 8 25 25 96 94 140 50 90 20

8 20 20 86 94 140 50 90 2020 0 66 54 140 50 90 20

16 4 40 50 90 204 0 10 10 20

9. Begin with the basic equation

(n+ 1)ni=1

i4 =ni=1

i5 +n

p=1

( pi=1

i4).

Given the result for the sum of fourth powers, we rewrite this in the formni=1

i5 = (n+ 1)ni=1

i4 n

p=1

(p5

5+p4

2+p3

3 p

30

).

39



Therefore

6

5

ni=1

i5 =(n+

1

2

) ni=1

i4 13

ni=1

i3 +1

30

ni=1

i

=(n+

1

2

)(n5

5+n4

2+n3

3 n

30

) 1

3

(n4

4+n3

2+n2

4

)+

1

30

(n2

2+n

2

)

=n6

5+

3n5

5+n4

2 n

2

10

.

If we multiply through by 56 , we get the final result:

ni=1

i5 =1

6n6 +

1

2n5 +

5

12n4 1

12n2.

10.

n1i=1

(n4 2n2i2 + i4) = (n 1)n4 2n2(n3

3 n

2

2+n

6

)+n5

5 n

4

2+n3

3 n

30

= (n 1)n4 715

(n 1)n4 + n4

30 n

30

=8

15(n 1)n4 + 1

30n4 1

30n

=8

15n5 8

15n4 +

1

30n4 1

30n =

8

15n n4 1

2n4 1

30n.

11. We first note that the result is true for n = 1, for in that case the left side equals 2 1kwhile the right side is 1k+1 + 1k. Now let us assume the result is true for n. For n + 1,we get

(n + 2)n+1i=1

ik = ((n+ 1) + 1)

ki=1

ik + (n+ 1)k

= (n+ 1)

ni=1

ik + (n+ 1)k+1 +ni=1

ik + (n + 1)k

=ni=1

ik+1 +n

p=1

( pi=1

ik)+ (n+ 1)k+1 +

n+1i=1

ik

=n+1i=1

ik+1 +n+1p=1

( pi=1

ik).

Thus the result is true for n+ 1 and is true for all n by mathematical induction.

12. To solve x3+d = cx, rewrite the equation of the given parabola as y = x2cand substitute

into the equation y2 x2 + dcx = 0 of the hyperbola. The result is x4

c x2 + dcx = 0.40



If we multiply by c and divide by x, we get x3 cx + d = 0, an equation equivalentto our original one. To sketch the two curves, note that the parabola has vertex at theorigin, while the hyperbola has center at ( d2c , 0), vertex of the right-hand branch (the

only relevant one) at (dc , 0), and asymptote the line y = x d2c . If one takes c = d = 2,then the parabola does not intersect the asymptote, so cannot intersect the hyperbola.

If one takes c = 3, d = 2, then the parabola and hyperbola intersect at (1,33 ) and have

the same tangent line there. Therefore the curves are tangent there and that point isthe only intersection point. If one takes c = 4, d = 2, one can check by using a graphingcalculator that the curves intersect twice, once between x = 12 and x =

34 and once

between x = 1 and x = 2.

13. To solve x3+d = bx2, substitute y = dx into y2+dxdb = 0. The result is d2

x2+dxdb = 0.

If one multiplies by x2 and divides by d, the result is d + x3 bx2 = 0, an equationequivalent to the original one. The hyperbola and parabola intersect exactly once whenthey have identical tangent lines at the intersection point (x0, y0). The tangent line tothe hyperbola at that point is y = y0x0x + 2y0, while the tangent line to the parabolathere is y = d2y0x + y0 +

dx02y0

. If these lines are identical, then y0x0 =d2y0

or 2y20 = dx0.

Substituting this value into the equation of the parabola and simplifying shows thatx0 =

2b3 . Then y0 =

3d2b . By substituting the value for x0 into the original equation,

we also get that 4b3 = 27d. For the curves to have no intersection, we must have

the hyperbola always above the parabola. Thus dx >db dx; d2

x2> db dx; and

d > bx2 x3 for all x. But the maximum of bx2 x3 is 4b327 . So 4b3 < 27d. The casewhere the hyperbola and parabola intersect twice is then when 4b3 > 27d.

14. For three positive solutions to exist for a cubic equation written in modern terms asx3+ qx2+ rx+ t = 0, the left side of this equation must factor as (xm)(xn)(x p),where m, n, p are all positive. Expanding this factored form, we get x3 (m + n +p)x2 + (mn + mp + np)x mnp. Therefore, the coefficient of x is positive, while thecoefficient of x2 and the constant term must be negative. Writing this in al-Khayyamisterms, we get the form x3 + cx = bx2 + d. Now to determine the conditions underwhich this type of equation will in fact have three positive solutions, we rewrite it inthe form x3 bx2 + cx = d and call the left side of this equation f(x). We note thatf (x) = 3x2 2bx+ c, and this derivative is 0 at the two critical values x = b3

b23c3 .

For three positive solutions to exist, y = f(x) must cross the line y = d three times inthe first quadrant. A consideration of the graph of f(x) shows that it always crossesthat line at least once. For it to cross three times, the derivative must in fact equal 0twice; thus b2 3c 0. In addition, the value of f(x) at the leftmost of the two criticalvalues, call it x1, must be greater than d, that is, f(x1) > d.

15. If y = bx2 x3, then y = 2bx 3x2 and y = 0 when x = 0 or when x = 2b3 . The secondderivative test shows that x0 =

2b3 makes y maximal. If we now consider the graph of

f(x) = x3 bx2+d, we note that it has a maximum at 0 (and f(0) = d) and a minimum41



at 2b3 (and f(2b3 ) = 4b

2

27 +d). For this graph to cross the x-axis twice for x positive, thisminimum value must be negative. Thus there are two positive solutions to the cubic if

4b327 +d < 0 or if 4b3 > 27d; there is one positive solution if 4b3

27 +d = 0 or if 4b3 = 27d;

and there are no positive solutions if 4b327 + d > 0 or if 4b3 < 27d.

16. To solve x3+d = cx, set f(x) = cxx3. Then f (x) = c3x2, which is 0 when x =

c3 .

(Of course, we only consider the positive value.) Then f(

c3) =

2c3

c3 is a maximum

value for f . The original equation then has two solutions if this value is greater than d,

one solution (at x =

c3) if this value equals d, and no solutions if this value if less than

d. We can rewrite this condition as follows: there are two solutions if 4c3 > 27d2; thereis one solution if 4c3 = 27d2; and there are no solutions if 4c3 < 27d2.

17. First, we note that Cn+1k = Cnk1 + C

nk . For to count the number of ways to choose k

objects out of n + 1 objects, we can first count the ways to choose k objects out of nobjects, by neglecting the (n + 1)st object. Then, if we have that element as one of ourset of k objects, there are Cnk1 ways to complete the set of k objects. Now we can provethe result by induction on n. We know that the result is true for n = 1, because inthat case we must have k = 1, and then C11 =

211 C

10 , because each side is equal to 1.

We then assume the result is true for n and show it is true for n + 1. By the inductionhypothesis and the addition rule, Cn+1k = C

nk +C

nk1 =

nk+1k C

nk1 +C

nk1 =

n+1k C

nk1.

But also Cn+1k1 = Cnk1+C

nk2 = C

nk1 +

k1nk+2C

nk1 =

n+1nk+2C

nk1. We can rewrite this

last equation as Cnk1 =nk+2n+1 C

n+1k1 . If we then substitute this result in the previous

equation, we get

Cn+1k =n+ 1

kCnk1 =

n + 1

k

n k + 2n + 1

Cn+1k+1 =n k + 2

kCn+1k+1 =

n+ 1 (k 1)k

Cn+1k1 .

Thus the inductive step is proved and the theorem is true by induction on n.

18. We need to solve the spherical triangle RNM with vertices at Rome, the North Pole, andMecca. We know from the given information that side m is 487, side r is 6815, and an-gle N is 2719. The qibla is then angle R. From the formula of al-Battan we getcosn =cos r cosm+sin r sinm cosN = cos(6815) cos(487)+sin(6815) sin(487) cos(2719)= 0.8618. Therefore side n is 3029. We then calculate R from the law of sines:

sinR =sinN sin r

sin n=

sin(2719) sin(6815)sin(3029)

= 0.8403.

A quick glance at a globe should convince you that R is an obtuse angle. Therefore,R = 12250. (Using al-Battans formula to find cosR will also give you this result.)

19. We need to solve the spherical triangle NPL with vertices at New York, the North Pole,and London for the side p opposite the pole. The formula gives cos p = cosn cos ` +

42



sinn sin ` cosP = cos 38 cos 49 + sin 38 sin 49 cos 74 = 0.6451. Therefore p = 49.83. Tofind the distance in miles, we divide this answer by 360 and multiply by 25,000. Theresult is 3460 miles.

20. From Figure 7.16, we have rr+h = cos. Thus r = r cos + h cos and r =h cos1cos .

To calculate r, substitute h = 652; 3, 18 and = 034 into the formula. The result is13,331,731 cubits, which equals 19,997,597 feet, or 3,787 miles. The procedure is verysensitive to a small change in the measured value of , and it is difficult to see how can be measured with much precision.

21. Using the formula of Ptolemy, we find that in triangle CDB, tanCD/ tanB = sinBDand tanBD/ tanC2 = sinCD. Similarly, in triangle ACD, we get tanCD/ tanA =sinAD and tanAD/ tanC1 = sinCD. Equating the two expressions for tanCD de-rived from the first equations in each pair, we get tanA sinAD = tanB sinBD ortanA/ tanB = sinBD/ sinAD. Equating the two expressions for sinCD gives ustanBD/ tanC2 = tanAD/ tanC1 or tanC1/ tanC2 = tanAD/ tanBD.

22. We are given that AB = 60, AC = 75, and BC = 31. Since AD and AE arequadrants, we know that BD = 30 and CE = 15. By the rule of four quantities,sinCF : sinBF = sinCE : sinBD = sin 15 : sin 30 = .5176. Since CF = BF 31,we have that .5176 sinBF = sinCF = sin(BF 31). Therefore, .5176 sinBF =sinBF cos 31 sin 31 cosBF , or .5176 sinBF = .8571 sinBF .5150 cosBF. It fol-lows that .5150 cosBF = .3395 sinBF , or that tanBF = 1.5169. Thus BF = 5636and CF = 2536. To find DF , we use equation 3.6. This implies that cosBF =cosBD cosDF or that cos 5636 = cos 30 cosDF . Then cosDF = .6356 and DF =5032. Also we have cosCF = cosCE cosEF or cos 2536 = cos 15 cosEF . ThuscosEF = .9336 and EF = 21. Since 6 A = arc DE, we have 6 A = 2932. Tofind 6 C, we use the sine law: sinBC : sinA = sinAB : sinC. Thus sinC = .8280and C = 5559. Similarly, from sinAC : sinB = sinBC : sinA,

History of Mathematics Brief Version Victor Katz

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