,SOLUTION MANUAL History of Mathematics An
Introduction, A, 4th edition Katz
Notes
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, Solutions 1
CHAPTER 1
1. The answers are given in the answer section of the text. For the Egyptian hiero-
glyphics, 375 is three hundreds, seven tens, and five ones, while 4856 is four
thousands, eight hundreds, five tens, and six ones. For Babylonian cuneiform, note that
375 = 6 × 60 + 15, while 4856 = 1 × 3600 + 20 × 60 + 56.
2.
1 34
′
2 68
4 136
8 272
′
16 544
18 612
1 5
10 50′ (multiply by 10)
2 10 (double first line)
4 20 (double third line)
8 40′ (double fourth line)
′
2 22 (halve first line)
′
10 2 (invert third line)
18 2 10 93
3.
1 2 14
2 4 28
4 8 56
1
,2 Solutions
4.
1 28
2 56
4 112
16
5. We multiply 10 by 3 30:
1 3 30
′
2 1 3 15
4 2 3 10 30
′
8 5 2 10
The total of the two marked lines is then 7, as desired.
6.
1 7 2 4 8
2 15 2 4
4 31 2
8 63
3 5 4
Note that the sum of the three last terms in the second column is 99 2 4. We therefore
need to figure out by what to multiply 7 2 4 8 to give 4 so that we get a total of 100.
But since we know from the fourth line that multiplying that value by 8 gives 63, we
also know that multiplying it by 63 gives 8. Thus the required number is double 63,
which is 42 126. Thus the final result of our division is 12 3 42 126.
, Solutions 3
7.
1 7248
2 15 2 4
′
4 31 2
′
8 63
′
3 4 3 3 6 12
12 3 98 2 3 3 6 12
99 2 4
8.
2 ÷ 11 1 11 2 ÷ 23 1 23
3 73 3 15 3
3 33 3 73
′
6 136 6 323
′ ′
66 6 12 1246
′
276 12
6 66 2 12 276 2
9. x + 17 x = 19. Choose x = 7; then 7 + 17 · 7 = 8. Since 19 ÷ 8 = 2 38 , the correct answer
is 2 38 × 7 = 16 58 .
10. (x + 23 x) − 13 (x + 23 x) = 10. In this case, the “obvious” choice for x is x = 9. Then 9
added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get
10. So in this case our “guess” is correct.
,4 Solutions
11. The equation here is (1 + 13 + 41 )x = 2. Therefore. we can find the solution by dividing
2 by 1 + 13 + 41 . We set up that problem:
1 124
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 1 141
144 . So
3
we need to find multipliers giving us 144 = 144 72. But 1 3 4 times 144 is 228. It
follows that multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72.
Thus, the answer is 1 6 12 114 228.
12. Since there are 10 hekats to be divided among 10 men, the average for each is 1 hekat.
So to get the largest share, we should add to the average 1/8 times half the number
of differences. Since there are 9 differences, we add 1/8 4 12 times, or 9/16. Therefore,
9
the largest share is 1 16 hekats, which the scribe writes as 1 2 16. We then subtract 1/8
9 7 5
from this value 9 times to get the share of each man. The answers are 1 16 , 1 16 , 1 16 ,
3 1 15 13 11 9 7
1 16 , 1 16 , 16 , 16 , 16 , 16 , and 16 .
45×100
13. Since x must satisfy 100 : 10 = x : 45, we would get that x = 10 ; the scribe breaks
this up into a sum of two parts, 35×
10
100
and 10× 100
10 .
14. The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4
handbreadths diameter is 52 : 42 = 25 : 16 = 1 16 9
. Thus, 100 logs of 5 handbreadths
diameter are equivalent to 1 16 × 100 = 156 4 logs of 4 handbreadths diameter.
9 1
16. The modern formula for the surface area of a half-cylinder of diameter d and height
h is A = 12 πdh. Similarly, the modern formula for the surface area of a hemisphere of
diameter d is A = 12 πd2 . These formulas are identical if h = d.
17. 7/5 = 1;24 13/15 = 0;52 11/24 = 0;27,30 33/50 = 0;39,36
18. 0;22,30 = 3/8 0;08,06 = 27/200 0;04,10 = 5/72 0;05,33,20 = 5/54
19. Since 60 is (2 12 ) × 24, the reciprocal of 24 is 2;30. Since 60 is (1 78 ) × 32, and 78 can
be expressed as 0;52,30 the reciprocal of 32 is 1;52,30. Since 60 is (1 31 ) × 45, the
reciprocal of 45 is 1:20. Since 60 = 1 19 × 54, and 19 can be expressed as 10 1 1
+ 90 =
6 40
60 + 3600 = 0;06,40, the reciprocal of 54 is 1;06,40. If the only prime divisors of n are
2, 3, 5, then n is a regular sexagesimal.
20. 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.
21. 1,08,16 × 3,45 = 4,16,0,0. 4,16 × 3,45 = 16,0,0. 16 × 3,45 = 1,0,0.
, Solutions 5
22. Since the length of the circumference C is given by C = 4a, and because √ C = 6r, it
follows that r = 23 a. The length T of the long transversal is then T = r 2 = ( 32 a)( 12 17
)=
17
18 a. The length t of the short transversal is t = 2 ( r − t
2 ) = 2a ( 2
3 − 17
36 ) = 7
18 a. The area
A of the barge is twice the difference between the area of a quarter circle and the area
of the right triangle formed by the long transversal and two perpendicular radii drawn
from the two ends of that line. Thus
( 2 ) ( 2 )
C r2 a 2a2 2
A=2 − =2 − = a2 .
48 2 3 9 9
23. Since the length of the circumference C is given by C = 3a, and √ because C = 6r, it
follows that r = 2a . The length T of the long transversal is then T = r 3 = ( a2 )( 74 ) = 78 a.
The length t of the short transversal is twice the distance from the midpoint of the arc
to 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 ( 2r , 23r ) while the midpoint of the long
√
transversal has coordinates ( 4r , 43r ). Thus the length of half of the short transversal is
r a
2 and then t = r = 2 . The area A of the bull’s 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
and radii drawn from the two ends of that line. Thus
( 2 ) ( 2 ) ( )
C 1r 9a 1 a 7a 1 7 9 2
A=2 − T =2 − = 2a2 − = a .
36 2 2 36 24 8 4 64 32
24. If a is the length of one of the quarter-circle arcs defining the concave square, then the
diagonal is equal to the diameter of that circle. Since the circumference is equal to 4a,
the diameter is one-third of that circumference, or 1 13 a. The transversal is equal to the
diagonal of the circumscribing square less the diameter of the circle (which is equal
to the side of the square). Since the diagonal of a square is approximated by 17/12 of
5 4 5
the side, the transversal is therefore equal to 5/12 of the diameter, or 12 3 a = 9 a.
√ √
25. 3 = 22 − 1 ≈ 2 − 12 · 1 · 12 = 2 − 0;15 = 1;45. Since an approximate re-
√ √
ciprocal of 1;45 is 0;34,17,09, we get further that 3 = (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 truncate to 1;43,55,42 because we know this value is a slight over-approximation.
26. v + u = 1;48 = 1 54 and v − u = 0;33,20 = 95 . So 2v = 2;21,20 and v = 1;10,40 = 106 90 .
Similarly, 2u = 1;14,40 and u = 0;37,20 = 56 90 . Multiplying by 90 gives x = 56,
d = 106. In the second part, v + u = 2;05 = 2 12 1
and v − u = 0;28,48 = 12 25 So
.
2v = 2;33,48 and v = 1;16,54 = 600 . Similarly, 2u = 1;36,12 and u = 0;48,06 = 481
769
600 .
Multiplying by 600 gives x = 481, d = 769. Next, if v = 481 319
360 and u = 360 , then
v + u = 2 29 = 2;13,20. Finally, if v = 240
289
and u = 161 7
240 , then v + u = 1 8 = 1;52,30.
27. The equations for u and v can be solved to give v = 1:22,08,27 = 295707 98569
216000 = 72000 and
201957 67319
u = 0;56,05,57 = 216000 = 72000 . Thus the associated Pythagorean triple is 67319,
72000, 98569.
28. The two equations are x2 + y2 = 1525; y = 32 x + 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.
,6 Solutions
29. If we
√ guess that the length of the rectangle is 60, then the width is 45 and the diagonal
is 602 + 452 = 75. Since this value is 1 87 times the given value of 40, the correct
length of the rectangle should be 60 ÷ 1 78 = 32. Then the width is 24.
30. One way to solve this is to let x and x − 600 be the areas of the two fields. Then the
equation is 23 x + 21 (x − 600) = 1100. This reduces to 76 x = 1400, so x = 1200. The
second field then has area 600.
31. Let x be the weight of the stone. The equation to solve is then x − 17 x − 13
1
(x − 17 x) = 60.
We do this using false position twice. First, set y = x − 7 x. The equation in y is then
1
y − 13
1
y = 60. We guess y = 13. Since 13 − 13 1
13 = 12, instead of 60, we multiply
our guess by 5 to get y = 65. We then solve x − 17 x = 65. Here we guess x = 7 and
calculate the value of the left side as 6. To get 65, we need to multiply our guess by
6 = 10 6 . So our answer is x = 7 × 6 = 75 6 gin, or 1 mina 15 6 gin.
65 1 65 5 5
32. We do this in three steps, each using false position. First, set z = x − 71 x + 11
1
(x − 17 x).
The equation for z is then z − 13 z = 60. We guess 13 for z and calculate the value of
1
the left side to be 12, instead of 60. Thus we must multiply our original guess by 5
and put z = 65. Then set y = x − 17 x. The equation for y is y + 11 1
y = 65. If we now
guess y = 11, the result on the left side is 12, instead of 65. So we must multiply our
guess by 6512 to get y = 12 = 59 12 . We now solve x − 7 x = 59 12 . If we guess x = 7,
715 7 1 7
7
the left side becomes 6 instead of 59 12 . So to get the correct value, we must multiply
12 /6 = 72 . Therefore, x = 7 × 72 = 72 = 69 72 gin = 1 mina 9 72 gin.
7 by 715 715 715 5005 37 37
33. Start with a square of side x and cut off a strip of width a from the right side. The
remaining rectangle then has area x2 − ax, or b. This rectangle can then be thought of
as a square of side x − a/2 that is missing a small square of side a/2. If one adds back
that small square, then the square of side x − a/2 has area b + (a/2)2 , so we can find x.
34. The equation x − 60 x = 7 is equivalent to x − 60 = 7x or to x − 7x = 60. The solution
2 2
/
is then x = ( 27 )2 + 60 + 27 = 17 7
2 + 2 = 12. Thus the two numbers are 12 and 5.
35. Given the appropriate coefficients, the equation becomes 49 a2 + a + 34 a = 23
18 , where a
4 2
is the length of the arc. If we scale up by 9 , we get the equation ( 9 a) + 3 ( 9 a) = 46
4 7 4
81 .
/
The algorithm for this type of equation gives 9 a = ( 6 ) + 81 − 6 = 18 − 6 = 29 .
4 7 2 46 7 25 7
Thus a = 12 .
36. The equation is 23 x2 + 31 x = 13 . To solve, we scale by 23 : ( 32 x)2 + 31 ( 23 x) = 2
9. The
/
solution is 23 x = ( 16 )2 + 92 − 61 = 21 − 61 = 31 . Thus x = 12 .
37. All the triangles in Figure 1.15 are similar to one another, and therefore their sides
are all in the ratio 3:4:5. Therefore, AE = 35 AD = 0;36 × 0;36 = 0;21,36. Also,
DE = 54 AD = 0;48 × 0;36 = 0;28,48. Also, EF = 45 DE = 0;48 × 0;28,48 = 0;23,02,24.
Finally, DF = 53 DE = 0;36 × 0;28,48 = 0;17,16,48.
38. If the circumference is 60, then the radius is 10. Thus, if the distance of the chord
from the circumference is x, then we have a right triangle of sides 6 and 10 − x, with
hypotenuse 10. The Pythagorean theorem leads to the equation 62 + (10 − x)2 = 100,
or x2 + 36 = 20x, for which the only valid solution is x = 2.
, Solutions 7
39. The two equations are ℓ + w = 7, ℓw + 12 ℓ + 13 w = 15. To put this into a standard Baby-
lonian form, we can rewrite the second equation in the form (ℓ + 13 )(w + 12 ) = 15 16 .
We can then rewrite the first equation as (ℓ + 13 ) + (w + 21 ) = 7 65 . Then the Babylonian
/
algorithm yields ℓ + 13 = 47 12 ) − 6 = 12 + 12 = 3 . Therefore, ℓ = 4 and so
( 47 2 91 47 5 13
12 +
w = 3.
CHAPTER 2
1. 125 = ρκϵ, 62 = ξβ, 4821 = ′ δωκα, 23, 855 = Mβ ′ γωνϵ
2. 8
9 = ∠ γ́ ίή (8/9 = 1/2 + 1/3 + 1/18)
3. The answer is in the back of the text. The basic idea is that 200/9 = 22 29 =
22 + 1/6 + 1/18.
4. The average of a and c is 1/4 + 1/16 + 1/64. The average of b
and d is 1/2 + 1/4 + 1/8 + 1/16. The product of the two averages is
1/8 + 1/16 + 1/32 + 1/64 + 1/32 + 1/64 + 1/128 + 1/256 + 1/128 + 1/256 + 1/512 + 1/1024,
or 1/4 + 59/1024. This is slightly less than the given answer of 1/4 + 1/16.
5. Since AB = BC; since the two angles at B are equal; and since the angles at A and C are
both right angles, it follows by the angle-side-angle theorem that △EBC is congruent
to △SBA and therefore that SA = EC.
6. 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 △AET is congruent to △AES. Therefore SE = ET.
7. The distance from the center of the pyramid to the tip of the shadow is 378 + 342 = 720
feet. Therefore the height of the pyramid is 6/9 = 2/3 of this value, or 480 feet.
n(n + 1)
8. Tn = 1 + 2 + · · · + n = 2 . Therefore the oblong number n(n + 1) is double the
triangular number Tn .
(n − 1)n
9. n2 = 2 + n(n2+ 1) , and the summands are the triangular numbers Tn − 1 and Tn .
8n(n + 1)
10. 2 + 1 = 4n2 + 4n + 1 = (2n + 1)2 . (2n + 1)2 − 1 = 4n2 + 4n = 8n(n2+ 1) .
2 2 2 2 2
11. Suppose a + b = c . Suppose a is odd. Then a is odd. If b is odd, then b is odd and
c2 is even, so c is even. If b is even, then b2 is even and c2 is odd, so c is odd. A similar
result holds if c is odd.
12. 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).
13. Let us assume that the second leg is commensurable to the first and let b, a be numbers
representing the two legs (in terms of some unit). We may as well assume that b and
a are relatively prime. Since the hypotenuse is double the first leg, we have b2 + a2 =
(2a)2 = 4a2 , or b2 = 3a2 . Since b2 is a multiple of 3, it must also be a multiple of 9, so
b2 = 9c2 and b = 3c. Then 9c2 = 3a2 , or a2 = 3c2 . This implies that a2 is a multiple of
9, so that a is a multiple of 3. But then both a and b are multiples of 3, contradicting
the fact that they are relatively prime.
, 8 Solutions
14. Since similar segments are to their corresponding circles in the same ratio, the areas of
similar segments are to one another as the squares on the diameters of the circles. Thus,
the areas of similar segments are also to one another as the squares on the radii of the
circles. But in similar segments, the triangles formed by the two radii and chords are
similar triangles. Thus the chord of one segment is to the chord in the similar segment
as the radius of the first circle to the radius of the second. That is, the squares on the
radii are to one another as the squares on the chords. Therefore, the areas of similar
segments are to one another as the squares on their chords.
15. By Exercise 14, the area of segment BD is the area of segment AB as the square on
BD is the square on AB. But this ratio is equal to 3. Thus, the area of segment BD is
three times the area of segment AB, or is equal to the sum of the areas of segments AB,
AC, and CD. Therefore, the area of lune is equal to the difference between the area
of the large segment and the area of segment BD. But this is equal to the difference
between the area of the large segment and the areas of the three small segments, which
is in turn equal to the area of the trapezoid. To construct
√ the trapezoid, note that one
can certainly construct a line segment equal to 3 times the length of a given line
segment. To place this line segment both parallel to the original one and such that the
lines connecting the endpoints of the two segments are each equal to the original line
segment, we simply need to find the distance between the two segments. And that
can be constructed by using the Pythagorean Theorem applied to the triangle whose
hypotenuse is equal to the original segment and one leg of which is equal to half the
difference between the new line segment and the original one. To circumscribe a circle
around this trapezoid, note that one can construct a circle through three points, say B,
A, and C. By the symmetry of the trapezoid, this circle will also go through point D.
21. If one equates the times of the two runners, where d is the distance traveled by Achilles,
the equation is d/10 = (d − 500)/(1/5). This is equivalent to 49d = 25,000, so d =
510.2 yards. Since Achilles is traveling at 10 yards per second, this will take him 51.02
seconds.
CHAPTER 3
1. One way to do this is to use I-4. Namely, consider the isosceles triangle ABC with
equal sides AB and BC also as a triangle CBA. Then the triangles ABC and CBA have
two sides equal to two sides and the included angles also equal. Thus, by I-4, they are
congruent. Therefore, angle BAC is equal to angle BCA, and the theorem is proved.
2. Put the point of the compass on the vertex V of the angle and swing equal arcs inter-
secting the two legs at A and B. Then place the compass at A and B, respectively, and
swing equal arcs, intersecting at C. The line segment connecting V to C then bisects
the angle. To show that this is correct, note that triangles VAC and VBC are congruent
by SSS. Therefore, the two angles AVC and BVC are equal.
3. Let the lines AB and CD intersect at E. Then angles AEB and CED are both straight
angles, angles equal to two right angles. If one subtracts the common angle CEB from
each of these, the remaining angles AEC and BED are equal, and these are the vertical
angles of the theorem.
Introduction, A, 4th edition Katz
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
, Solutions 1
CHAPTER 1
1. The answers are given in the answer section of the text. For the Egyptian hiero-
glyphics, 375 is three hundreds, seven tens, and five ones, while 4856 is four
thousands, eight hundreds, five tens, and six ones. For Babylonian cuneiform, note that
375 = 6 × 60 + 15, while 4856 = 1 × 3600 + 20 × 60 + 56.
2.
1 34
′
2 68
4 136
8 272
′
16 544
18 612
1 5
10 50′ (multiply by 10)
2 10 (double first line)
4 20 (double third line)
8 40′ (double fourth line)
′
2 22 (halve first line)
′
10 2 (invert third line)
18 2 10 93
3.
1 2 14
2 4 28
4 8 56
1
,2 Solutions
4.
1 28
2 56
4 112
16
5. We multiply 10 by 3 30:
1 3 30
′
2 1 3 15
4 2 3 10 30
′
8 5 2 10
The total of the two marked lines is then 7, as desired.
6.
1 7 2 4 8
2 15 2 4
4 31 2
8 63
3 5 4
Note that the sum of the three last terms in the second column is 99 2 4. We therefore
need to figure out by what to multiply 7 2 4 8 to give 4 so that we get a total of 100.
But since we know from the fourth line that multiplying that value by 8 gives 63, we
also know that multiplying it by 63 gives 8. Thus the required number is double 63,
which is 42 126. Thus the final result of our division is 12 3 42 126.
, Solutions 3
7.
1 7248
2 15 2 4
′
4 31 2
′
8 63
′
3 4 3 3 6 12
12 3 98 2 3 3 6 12
99 2 4
8.
2 ÷ 11 1 11 2 ÷ 23 1 23
3 73 3 15 3
3 33 3 73
′
6 136 6 323
′ ′
66 6 12 1246
′
276 12
6 66 2 12 276 2
9. x + 17 x = 19. Choose x = 7; then 7 + 17 · 7 = 8. Since 19 ÷ 8 = 2 38 , the correct answer
is 2 38 × 7 = 16 58 .
10. (x + 23 x) − 13 (x + 23 x) = 10. In this case, the “obvious” choice for x is x = 9. Then 9
added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get
10. So in this case our “guess” is correct.
,4 Solutions
11. The equation here is (1 + 13 + 41 )x = 2. Therefore. we can find the solution by dividing
2 by 1 + 13 + 41 . We set up that problem:
1 124
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 1 141
144 . So
3
we need to find multipliers giving us 144 = 144 72. But 1 3 4 times 144 is 228. It
follows that multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72.
Thus, the answer is 1 6 12 114 228.
12. Since there are 10 hekats to be divided among 10 men, the average for each is 1 hekat.
So to get the largest share, we should add to the average 1/8 times half the number
of differences. Since there are 9 differences, we add 1/8 4 12 times, or 9/16. Therefore,
9
the largest share is 1 16 hekats, which the scribe writes as 1 2 16. We then subtract 1/8
9 7 5
from this value 9 times to get the share of each man. The answers are 1 16 , 1 16 , 1 16 ,
3 1 15 13 11 9 7
1 16 , 1 16 , 16 , 16 , 16 , 16 , and 16 .
45×100
13. Since x must satisfy 100 : 10 = x : 45, we would get that x = 10 ; the scribe breaks
this up into a sum of two parts, 35×
10
100
and 10× 100
10 .
14. The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4
handbreadths diameter is 52 : 42 = 25 : 16 = 1 16 9
. Thus, 100 logs of 5 handbreadths
diameter are equivalent to 1 16 × 100 = 156 4 logs of 4 handbreadths diameter.
9 1
16. The modern formula for the surface area of a half-cylinder of diameter d and height
h is A = 12 πdh. Similarly, the modern formula for the surface area of a hemisphere of
diameter d is A = 12 πd2 . These formulas are identical if h = d.
17. 7/5 = 1;24 13/15 = 0;52 11/24 = 0;27,30 33/50 = 0;39,36
18. 0;22,30 = 3/8 0;08,06 = 27/200 0;04,10 = 5/72 0;05,33,20 = 5/54
19. Since 60 is (2 12 ) × 24, the reciprocal of 24 is 2;30. Since 60 is (1 78 ) × 32, and 78 can
be expressed as 0;52,30 the reciprocal of 32 is 1;52,30. Since 60 is (1 31 ) × 45, the
reciprocal of 45 is 1:20. Since 60 = 1 19 × 54, and 19 can be expressed as 10 1 1
+ 90 =
6 40
60 + 3600 = 0;06,40, the reciprocal of 54 is 1;06,40. If the only prime divisors of n are
2, 3, 5, then n is a regular sexagesimal.
20. 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.
21. 1,08,16 × 3,45 = 4,16,0,0. 4,16 × 3,45 = 16,0,0. 16 × 3,45 = 1,0,0.
, Solutions 5
22. Since the length of the circumference C is given by C = 4a, and because √ C = 6r, it
follows that r = 23 a. The length T of the long transversal is then T = r 2 = ( 32 a)( 12 17
)=
17
18 a. The length t of the short transversal is t = 2 ( r − t
2 ) = 2a ( 2
3 − 17
36 ) = 7
18 a. The area
A of the barge is twice the difference between the area of a quarter circle and the area
of the right triangle formed by the long transversal and two perpendicular radii drawn
from the two ends of that line. Thus
( 2 ) ( 2 )
C r2 a 2a2 2
A=2 − =2 − = a2 .
48 2 3 9 9
23. Since the length of the circumference C is given by C = 3a, and √ because C = 6r, it
follows that r = 2a . The length T of the long transversal is then T = r 3 = ( a2 )( 74 ) = 78 a.
The length t of the short transversal is twice the distance from the midpoint of the arc
to 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 ( 2r , 23r ) while the midpoint of the long
√
transversal has coordinates ( 4r , 43r ). Thus the length of half of the short transversal is
r a
2 and then t = r = 2 . The area A of the bull’s 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
and radii drawn from the two ends of that line. Thus
( 2 ) ( 2 ) ( )
C 1r 9a 1 a 7a 1 7 9 2
A=2 − T =2 − = 2a2 − = a .
36 2 2 36 24 8 4 64 32
24. If a is the length of one of the quarter-circle arcs defining the concave square, then the
diagonal is equal to the diameter of that circle. Since the circumference is equal to 4a,
the diameter is one-third of that circumference, or 1 13 a. The transversal is equal to the
diagonal of the circumscribing square less the diameter of the circle (which is equal
to the side of the square). Since the diagonal of a square is approximated by 17/12 of
5 4 5
the side, the transversal is therefore equal to 5/12 of the diameter, or 12 3 a = 9 a.
√ √
25. 3 = 22 − 1 ≈ 2 − 12 · 1 · 12 = 2 − 0;15 = 1;45. Since an approximate re-
√ √
ciprocal of 1;45 is 0;34,17,09, we get further that 3 = (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 truncate to 1;43,55,42 because we know this value is a slight over-approximation.
26. v + u = 1;48 = 1 54 and v − u = 0;33,20 = 95 . So 2v = 2;21,20 and v = 1;10,40 = 106 90 .
Similarly, 2u = 1;14,40 and u = 0;37,20 = 56 90 . Multiplying by 90 gives x = 56,
d = 106. In the second part, v + u = 2;05 = 2 12 1
and v − u = 0;28,48 = 12 25 So
.
2v = 2;33,48 and v = 1;16,54 = 600 . Similarly, 2u = 1;36,12 and u = 0;48,06 = 481
769
600 .
Multiplying by 600 gives x = 481, d = 769. Next, if v = 481 319
360 and u = 360 , then
v + u = 2 29 = 2;13,20. Finally, if v = 240
289
and u = 161 7
240 , then v + u = 1 8 = 1;52,30.
27. The equations for u and v can be solved to give v = 1:22,08,27 = 295707 98569
216000 = 72000 and
201957 67319
u = 0;56,05,57 = 216000 = 72000 . Thus the associated Pythagorean triple is 67319,
72000, 98569.
28. The two equations are x2 + y2 = 1525; y = 32 x + 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.
,6 Solutions
29. If we
√ guess that the length of the rectangle is 60, then the width is 45 and the diagonal
is 602 + 452 = 75. Since this value is 1 87 times the given value of 40, the correct
length of the rectangle should be 60 ÷ 1 78 = 32. Then the width is 24.
30. One way to solve this is to let x and x − 600 be the areas of the two fields. Then the
equation is 23 x + 21 (x − 600) = 1100. This reduces to 76 x = 1400, so x = 1200. The
second field then has area 600.
31. Let x be the weight of the stone. The equation to solve is then x − 17 x − 13
1
(x − 17 x) = 60.
We do this using false position twice. First, set y = x − 7 x. The equation in y is then
1
y − 13
1
y = 60. We guess y = 13. Since 13 − 13 1
13 = 12, instead of 60, we multiply
our guess by 5 to get y = 65. We then solve x − 17 x = 65. Here we guess x = 7 and
calculate the value of the left side as 6. To get 65, we need to multiply our guess by
6 = 10 6 . So our answer is x = 7 × 6 = 75 6 gin, or 1 mina 15 6 gin.
65 1 65 5 5
32. We do this in three steps, each using false position. First, set z = x − 71 x + 11
1
(x − 17 x).
The equation for z is then z − 13 z = 60. We guess 13 for z and calculate the value of
1
the left side to be 12, instead of 60. Thus we must multiply our original guess by 5
and put z = 65. Then set y = x − 17 x. The equation for y is y + 11 1
y = 65. If we now
guess y = 11, the result on the left side is 12, instead of 65. So we must multiply our
guess by 6512 to get y = 12 = 59 12 . We now solve x − 7 x = 59 12 . If we guess x = 7,
715 7 1 7
7
the left side becomes 6 instead of 59 12 . So to get the correct value, we must multiply
12 /6 = 72 . Therefore, x = 7 × 72 = 72 = 69 72 gin = 1 mina 9 72 gin.
7 by 715 715 715 5005 37 37
33. Start with a square of side x and cut off a strip of width a from the right side. The
remaining rectangle then has area x2 − ax, or b. This rectangle can then be thought of
as a square of side x − a/2 that is missing a small square of side a/2. If one adds back
that small square, then the square of side x − a/2 has area b + (a/2)2 , so we can find x.
34. The equation x − 60 x = 7 is equivalent to x − 60 = 7x or to x − 7x = 60. The solution
2 2
/
is then x = ( 27 )2 + 60 + 27 = 17 7
2 + 2 = 12. Thus the two numbers are 12 and 5.
35. Given the appropriate coefficients, the equation becomes 49 a2 + a + 34 a = 23
18 , where a
4 2
is the length of the arc. If we scale up by 9 , we get the equation ( 9 a) + 3 ( 9 a) = 46
4 7 4
81 .
/
The algorithm for this type of equation gives 9 a = ( 6 ) + 81 − 6 = 18 − 6 = 29 .
4 7 2 46 7 25 7
Thus a = 12 .
36. The equation is 23 x2 + 31 x = 13 . To solve, we scale by 23 : ( 32 x)2 + 31 ( 23 x) = 2
9. The
/
solution is 23 x = ( 16 )2 + 92 − 61 = 21 − 61 = 31 . Thus x = 12 .
37. All the triangles in Figure 1.15 are similar to one another, and therefore their sides
are all in the ratio 3:4:5. Therefore, AE = 35 AD = 0;36 × 0;36 = 0;21,36. Also,
DE = 54 AD = 0;48 × 0;36 = 0;28,48. Also, EF = 45 DE = 0;48 × 0;28,48 = 0;23,02,24.
Finally, DF = 53 DE = 0;36 × 0;28,48 = 0;17,16,48.
38. If the circumference is 60, then the radius is 10. Thus, if the distance of the chord
from the circumference is x, then we have a right triangle of sides 6 and 10 − x, with
hypotenuse 10. The Pythagorean theorem leads to the equation 62 + (10 − x)2 = 100,
or x2 + 36 = 20x, for which the only valid solution is x = 2.
, Solutions 7
39. The two equations are ℓ + w = 7, ℓw + 12 ℓ + 13 w = 15. To put this into a standard Baby-
lonian form, we can rewrite the second equation in the form (ℓ + 13 )(w + 12 ) = 15 16 .
We can then rewrite the first equation as (ℓ + 13 ) + (w + 21 ) = 7 65 . Then the Babylonian
/
algorithm yields ℓ + 13 = 47 12 ) − 6 = 12 + 12 = 3 . Therefore, ℓ = 4 and so
( 47 2 91 47 5 13
12 +
w = 3.
CHAPTER 2
1. 125 = ρκϵ, 62 = ξβ, 4821 = ′ δωκα, 23, 855 = Mβ ′ γωνϵ
2. 8
9 = ∠ γ́ ίή (8/9 = 1/2 + 1/3 + 1/18)
3. The answer is in the back of the text. The basic idea is that 200/9 = 22 29 =
22 + 1/6 + 1/18.
4. The average of a and c is 1/4 + 1/16 + 1/64. The average of b
and d is 1/2 + 1/4 + 1/8 + 1/16. The product of the two averages is
1/8 + 1/16 + 1/32 + 1/64 + 1/32 + 1/64 + 1/128 + 1/256 + 1/128 + 1/256 + 1/512 + 1/1024,
or 1/4 + 59/1024. This is slightly less than the given answer of 1/4 + 1/16.
5. Since AB = BC; since the two angles at B are equal; and since the angles at A and C are
both right angles, it follows by the angle-side-angle theorem that △EBC is congruent
to △SBA and therefore that SA = EC.
6. 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 △AET is congruent to △AES. Therefore SE = ET.
7. The distance from the center of the pyramid to the tip of the shadow is 378 + 342 = 720
feet. Therefore the height of the pyramid is 6/9 = 2/3 of this value, or 480 feet.
n(n + 1)
8. Tn = 1 + 2 + · · · + n = 2 . Therefore the oblong number n(n + 1) is double the
triangular number Tn .
(n − 1)n
9. n2 = 2 + n(n2+ 1) , and the summands are the triangular numbers Tn − 1 and Tn .
8n(n + 1)
10. 2 + 1 = 4n2 + 4n + 1 = (2n + 1)2 . (2n + 1)2 − 1 = 4n2 + 4n = 8n(n2+ 1) .
2 2 2 2 2
11. Suppose a + b = c . Suppose a is odd. Then a is odd. If b is odd, then b is odd and
c2 is even, so c is even. If b is even, then b2 is even and c2 is odd, so c is odd. A similar
result holds if c is odd.
12. 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).
13. Let us assume that the second leg is commensurable to the first and let b, a be numbers
representing the two legs (in terms of some unit). We may as well assume that b and
a are relatively prime. Since the hypotenuse is double the first leg, we have b2 + a2 =
(2a)2 = 4a2 , or b2 = 3a2 . Since b2 is a multiple of 3, it must also be a multiple of 9, so
b2 = 9c2 and b = 3c. Then 9c2 = 3a2 , or a2 = 3c2 . This implies that a2 is a multiple of
9, so that a is a multiple of 3. But then both a and b are multiples of 3, contradicting
the fact that they are relatively prime.
, 8 Solutions
14. Since similar segments are to their corresponding circles in the same ratio, the areas of
similar segments are to one another as the squares on the diameters of the circles. Thus,
the areas of similar segments are also to one another as the squares on the radii of the
circles. But in similar segments, the triangles formed by the two radii and chords are
similar triangles. Thus the chord of one segment is to the chord in the similar segment
as the radius of the first circle to the radius of the second. That is, the squares on the
radii are to one another as the squares on the chords. Therefore, the areas of similar
segments are to one another as the squares on their chords.
15. By Exercise 14, the area of segment BD is the area of segment AB as the square on
BD is the square on AB. But this ratio is equal to 3. Thus, the area of segment BD is
three times the area of segment AB, or is equal to the sum of the areas of segments AB,
AC, and CD. Therefore, the area of lune is equal to the difference between the area
of the large segment and the area of segment BD. But this is equal to the difference
between the area of the large segment and the areas of the three small segments, which
is in turn equal to the area of the trapezoid. To construct
√ the trapezoid, note that one
can certainly construct a line segment equal to 3 times the length of a given line
segment. To place this line segment both parallel to the original one and such that the
lines connecting the endpoints of the two segments are each equal to the original line
segment, we simply need to find the distance between the two segments. And that
can be constructed by using the Pythagorean Theorem applied to the triangle whose
hypotenuse is equal to the original segment and one leg of which is equal to half the
difference between the new line segment and the original one. To circumscribe a circle
around this trapezoid, note that one can construct a circle through three points, say B,
A, and C. By the symmetry of the trapezoid, this circle will also go through point D.
21. If one equates the times of the two runners, where d is the distance traveled by Achilles,
the equation is d/10 = (d − 500)/(1/5). This is equivalent to 49d = 25,000, so d =
510.2 yards. Since Achilles is traveling at 10 yards per second, this will take him 51.02
seconds.
CHAPTER 3
1. One way to do this is to use I-4. Namely, consider the isosceles triangle ABC with
equal sides AB and BC also as a triangle CBA. Then the triangles ABC and CBA have
two sides equal to two sides and the included angles also equal. Thus, by I-4, they are
congruent. Therefore, angle BAC is equal to angle BCA, and the theorem is proved.
2. Put the point of the compass on the vertex V of the angle and swing equal arcs inter-
secting the two legs at A and B. Then place the compass at A and B, respectively, and
swing equal arcs, intersecting at C. The line segment connecting V to C then bisects
the angle. To show that this is correct, note that triangles VAC and VBC are congruent
by SSS. Therefore, the two angles AVC and BVC are equal.
3. Let the lines AB and CD intersect at E. Then angles AEB and CED are both straight
angles, angles equal to two right angles. If one subtracts the common angle CEB from
each of these, the remaining angles AEC and BED are equal, and these are the vertical
angles of the theorem.
