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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
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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.
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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
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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.
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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 .
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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
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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
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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.
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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
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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.
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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
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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.
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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.
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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.
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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
.
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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 .
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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.
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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.
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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.
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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.
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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.
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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
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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
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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
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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;
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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.
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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.
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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.
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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
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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
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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
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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
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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.
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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.
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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
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-
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.
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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
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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
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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
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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 ` +
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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,