MATH2001/T5/NKT/NKT/11-12
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH2001 Development of Mathematical Ideas
Tutorial 5 (February 23, 2011)
You should hand in one written copy of your work on (a) and (e) on or before February 27, 2011.
For the content of Euclid's Elements, you may consult
http://aleph0.clarku.edu/~djoyce/java/elements/elements.html
(a) Interpret the content of Proposition 1 in Book X of Euclid’s Elements in a mathematical language you
are accustomed to today. Do Exercise 3.12. (You should read “nonexistence of infinitely large
magnitude” to mean the nonexistence of a magnitude which is larger than any (integral) multiple of a
given magnitude.)
(b) The following passage is from Jiuzhang Suanshu 《
( 九章算術》, including the commentary of Liu Hui)
on the formula of the area of the circle A = 12 Cr where C is the perimeter and r is the radius. An
English translation is appended here for your reference. Explain how Liu Hui might have arrived at
the formula.
又有圓田，周一百八十一步，徑六十步三分步之一。（臣淳風等按：周三徑一，周一百八十一
步，徑六十步三分步之一。依密率，徑五十七步二十二分步之一十三。）問為田幾何？
答曰：十一畝九十步十二分步之一。（此於徽術，當為田十畝二百八步三百一十四分步之一百
十三。）
（臣淳風等按：依密率，當為田十畝二百五步八十八分步之八十七。）
術曰：半周半徑相乘得積步。（按：半周為從，半徑為廣，故廣從相乘為積步也。假令圓徑二
尺，圓中容六觚之一面，與圓徑之半，其數均等。合徑率一而外周率三也。）（又按：為圖，
以六觚之一面乘一弧半徑，三之，得十二觚之冪。若又割之，次以十二觚之一面乘一弧之半徑，
六之，則得二十四觚之冪。割之彌細，所失彌少。割之又割，以至於不可割，則與圓周合體而
無所失矣。觚面之外，又有餘徑。 以面乘餘徑，則冪出觚表若夫觚之細者，與圓合體，則表
無餘徑。表無餘徑，則冪不外出矣。以一面乘半徑，觚而裁之，每輒自倍。故以半周乘半徑而
為圓冪。此一周、徑，謂至然之數，非周三徑一之率也。周三者，從其六觚之環耳。以推圓規
多少之覺，乃弓之與弦也。然世傳此法，莫肯精核；學者踵古，習其謬失。不有明據，辯之斯
難。凡物類形象，不圓則方。方圓之率，誠著於近，則雖遠可知也。由此言之，其用博矣。謹
按圖驗，更造密率。恐空設法，數昧而難譬，故置諸檢括，謹詳其記註焉。）
[English translation by A. Volkov: “… (If one) multiplies the half-diameter by one side of the
hexagon, (and multiplies it) by 3, (then one) obtains the area of the dodecagon. If (one) again “cuts”
it (and) then multiplies the half-diameter by one side of the (obtained) dodecagon and then
(multiplies it) by 6, then (one) obtains the area of the 24-sided polygon. The finer (one) “cuts” it, the
less that is missed. “Having cut it, cut again” in order to arrive (at a polygon) which cannot be “cut”
(any more), then (this figure) and the circle (’s circumference) unite (their) bodies, and nothing is
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missed!
Out of the polygon’s side still there is a remaining diameter. (If one) multiplies (this
remaining) diameter by the (polygon’s) side, then the (rectangle corresponding to the obtained) area
goes outside of the arc. When the polygon is fine, (the polygon) and the circle unite (their) bodies,
then there is no remaining diameter outside. If there is no remaining diameter outside, then the
(obtained) area does not go outside! …”]
(c) Read Proposition 2 and its proof in Book XII of Euclid's Elements:
“Circles are to one another as the squares on their diameters.”
Comment on the two different approaches, one in (b) and the other in Elements.
(d) In his work On Spirals Archimedes defined his famous spiral as follows:
If a straight line drawn in a plane revolves at a uniform rate about one extremity which remains
fixed and returns to the position from which it started, and if, at the same time as the line
revolves, a point moves at a uniform rate along the straight line beginning from the extremity
which remains fixed, the point will described a spiral in the plane (Definition 1 in On Spirals).
Show that, in modern day’s mathematical language, one may use polar coordinates equation r = aμ,
where a is a constant, to describe the spiral. Show also that the Archimedean spiral can be used to
trisect an arbitrary angle, given that a line segment can easily be trisected (how?). [Hint: see the
figure below.]
(e) Archimedes proved that the area of the region S , bounded by one turn of the spiral and the line
segment joining its initial and final points, is one-third that of the circle C centered at the initial
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point and passing through the final point (see figure below). That is, area(S) = ¼(2¼a)2.
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2
His proof begins like this: First divide the circle into n equal sectors as shown below, with their
bounding radii intersecting the spiral at the points O; A1 ; A2 : : : : ; An .
If we write OA1 = b, then OA2 = 2b; : : : ; OAn = nb (note that the figure is not quite to scale).
Consequently the spiral region S contains a region P containing of circular sectors with radii
0; b; : : : ; (n ¡ 1)b, and in turn is contained in a region Q containing of circular sectors with radii
b; 2b; : : : ; nb. Therefore area(Q) ¡ area(P ) is equal to the area of a single sector of the circle C .
Now you have to complete the proof for him using the technique in his days, by applying double
reductio ad absurdum to show that area(S) = 13 area(C). Hint: you may use the familiar (then and
now) formulas
1 + 2 + ¢¢¢ + n =
n
(n + 1);
2
12 + 22 + ¢ ¢ ¢ + n2 =
n
(n + 1)(2n + 1):
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(f) Consider the following two statements:
「割之又割，以至於不可割，則與圓周合體而無所失矣。」
[Translation: “Having cut it, cut again in order to arrive at a polygon which cannot be cut
any more, then this figure and the circle’s circumference unite their bodies, and nothing is
missed!”]
— from Jiuzhang Suanshu (including the commentary of Liu Hui)
「一尺之棰, 日取其半, 萬世不竭」
[Translation: “If from a stick a foot long you every day take the half of it, in a myriad ages
it will not be exhausted.”]
— from Tianxia, Zhuangzi (《莊子．天下篇》)
Do they agree with each other, or are they at odds with each other? Comment.
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