GRAPHIC SOLUTION of 1，r=1，p+1，q

7

RHUNHM

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TEEaHU

MN

_aU

a-­

nhu

nwm

E 且 H

+

1-q

1-p

HHH

nEU

RWE R 、 u 1_一r 一一 143 唱司喝 、.可

aE-By KWANG Y. CHEN

(陣房比) 三十四年勝利來臨。復員鎮江。有 λ 接收五子登科， 我們一介不取。工作起居都在省立圖書館的一間藏書 室中，夜晚削以破書架作咻舖'仍舊過著艱苦的日子 ，生活絕未因勝利的果實而有所改善。三十六年機關 改制，我被調罔南京。三十八年共匪作亂，京湛撤守 ，我們最後撤出，經上海、廣州而轉到了臺灣。從廣 州到臺幣，押運了許多公物，乘的是免費差輸，駛駛 停停，在海上耽誤了不少日子。見女在船上染病，無 法就醫，病勢日重。到基隆登陸，立即送進臺大醫院 。大見，經醫治獲救，小女抵抗力弱，望日創告不治， 又使我失去了唯一的干金幼苗。(前生兩女，勝利後 相繼夭折。)(女作家徐鍾珮女士在中央副刊發表的 『失去的幼苗』剖描述此事，曾編入中學國女教科 畫面。) 在臺灣一恍叉是二十多年，真是當初做夢也不曾 想到。終年勞勞硨血阱，戰戰兢兢'生活雖日趨安定， 心頭卸充滿了憂憤。『神州不復，易興陸況之嘆;中 年己住，莫釋衷樂之懷。』往事不堪岡首，來日尚有 可期。希望早日反攻復園，直搗黃龍。 眼看他摟塌了。』有人爬得高，跌得重;有入升得快 ，去得早;有人爭得到，拼了命;有人變了前，遺了 具;這些熟識的、熱衷於名利的朋友們，到頭來無非 是夢幻泡影一場空。我們膠柱鼓瑟，死心眼見，擔泊 自甘，潔身自愛，雖然清風兩抽，位卑職小而不足道 ，卸也心安理得，樂在其中。記得校長唐丈抬先生有 『飲酒讀書四十年，方知頭上有青天。男 見欲到凌煙閣，第一功名不愛錢。』平生皆奉為座右 之銘，時時警惕。因此做人不敢不詐，大處著眼，小 處留心，向能謹言慎行;辦事盡心盡力，不為求名， 不為求利，只為工作而生活。 人世本是一個大舞臺'一個機關好仙一個戲班子 ，我們每一個同事都是演員。不論你演的是什麼角色 ，主角也好，配角也好，在班子裡，在舞臺上總要大 家齊心合力，努力表演。不鬆懈，不敢巧，才能成功 ，才能爭得團體的榮譽。我所掛演的雖是配角，且已 接近尾聲;但在落幕之前，我仍一希望能演得更生動， 更精彰，更有意義。 四十年歷盡滄桑，冷眼旁觀

Eda--Ayer Junior 圖 Senior High School Ayer, MassachuseHs

IN ELEMENTARY algebra work problems, water司 pipe

仆

i4A1J'14A44131Jj1

喝』司

delivery problems, and similarproblems are based on the JI ..

relationship

1 1 1

r p '

lJ

The units for the three variables may be of any magnitude, fundamental or derived. The only requirement

2' is that all three variables be in the same units.

For example, if a job can be finished by A alone in

『眼君他起高樓，

也有詩云 一五 『人生惟有廉節重，世界須憑氣骨埠。』岳武穆

..

3 days and by B alone in 6 days and can be finished by A and B working together in r days, then r can be found

..

by the above formula,

1Ir=1/3+1/6.

Here, the uniform units used are days. However, if we know that the job can be done by A and B working together in 2 days and that the same job can be done by A alone in 3 days, how long does it take B to finish it alone? Here, rand pare 2 and 3 days, respectively; q can be found by solving the equation l. q=1/2_一 1/3 .

Similarly, suppose a tank of water can be emptied in 10 minutes by opening pipe A alone, in 6 minutes by openlllg pipe B alone, and in

r

minutes by opening pipes A and B together. Then, using the same formula,

11r= 1/

10

+1/6.

Here the uniform units used are minutes.

- 31 一

tiJ

E

demonstrate the close relationship between algebra and When more than two men work or more than two geometry. The graphic method is described and proved as pipes are open, the right-hand member of the equation follows. _{can be extended, thus:}

Two line segments，豆豆 andτlJ are drawn perpendicular _{1 1 , 1 1 1}

令門， 一=一+一+,一+,

."

+τ­

to

LM.

as shown in figure 1. Segments 互]j and

BC

are

r

P'

q

s

n

drawn, intersecting at

F

, and F苦的 drawn perpendicular There are many physical phenomena to which this

←多 method applies. When two electrical resistances

r1

and rz to

LM

Then, as will be proved,

1/EF=I/AB+1/CD

, or

are connected in paralleL the combined resistance R is

1fr=lfp+1/q.

given by

I/R=I/rl+1/r2.

Here

r1

and

r2

represent resistance

Proof.

In triangle

AB

C.

p-r

AE

pu-PU in ohms, Again, when two electrical capacitors are connc­

(1) cted in series the combined capacibnce

C

is given by

1/C=I/C1+ l/c2'

where C1 and C2 are individual capacitafol_ccs In triangle

CDA

,

in farads or microfarads.

q _

AC

于 AC-E亡，

(2) 壘;{i In the study of spherical mirrors and lenses, the From (1), focal length

f

of a mirror or a lens can be found using

EC=AC-7

l仟==1/d

o

+

1

d!.

Here do is the distance of an object from

the Vertex of a mirror or from the center of a lens, and Substituting into (2),

d

,

is 't he distance of the image from the vertex or from

且一

AC

_{the ce1叫 The units of distance can be centim的問}

r

A叫ε寸

_{meters, inches, or feet, so long as they are the same for}

or _{all the variables in the formula.}

1-H-JVJ

q-rqq

_=Jhγ­

-r-prp'-7

-Ill-- _{A graphic met'hod is available for the approximate}

q

一­

吾 I )t

solution of the equation

v'

l/r=l!p+ l/q

_when 扎 ny

B

tw.o of the threevariables are given. This method _p is presented here not so

=仇

Dividing by

qr

,

--r

IEy1-ba _{much show another}

_{L A}

M

E

to

method

•

of solution as to _{Fig 1}

- 33 一 - 32 一

3

-B'

411;4fi--1 or

D

_D

_{1 1 1} 一一---+--可 油

r

P'

q

It should be noted that the above formula is indepe­ ndent of th巴 distance AC between the line segments AB

--ι , 111a

B

_B

、、、 \ \ 6

E

F_本 。

and CD; therefore these segments may be drawn any

flot--jji--j744111jtlfill-­

;2

3

C

A

E

Fig.3 Fig.4 four different segments all of length

q

are drawn at diff.:.. Again, suppose we know that A working 且 lone can erent distances apart, but the vertical segments rl' 的 ， r日... are the same length. This property facilitates the solution finish the job in 3 days, but A and B working together c:;, n

finish it in 2 days. How many days docs it take B to of more complicated probl巴ms，的 will be shown later. Now let us use the graphic method to solve the work finish it when working alone? Using the same scale as

RV problem given at the beginning of this article. Use a before, draw AB

=

3 and EF

=

2, bothper pendicular to d

convenient unit to represent 1 day. As shown in figure 3,

‘

島、

l2 _ll

distance apart, and the v且 lue of

r

remains the same. This

L

_{M L}

A

E

‘C

M

may be seen in figure 2, where the segment

p

and the

+令 一少 F今...←~

LM

(sec fig. 4). Draw AF and BF. BF meets

LM

at C.

draw two line segments of lengths AB=3 and CD=6, both

perp叫ic山 to 函， D川五百

and

B亡， In吋

1

←妒，令 Erect a perpendicular to

LM

at

C

, intersecting AF at

D.

Then CD is the required line segment; it measures 6 units, _{Draw Flτ~ 伊per叩 ndi比 叫1a叮 切P巴叩 cu r t oLII兩示示. τhe凹 nFEm巴 asur巴 s 2 unit拈 s，}

川

入 午有

indicating 6 d叮's， the time required by B to finish the _which

_吋

_{repres巴 nt}

_{2 days, the time required to finish the} job when working alone. _{job when A and B work together}

,

////\\VA

\///VP_IJAOb

jι

Similarly, water-pip已 problems and problems involving

仁， / / / / 』 \ \ 一

/////人//\\\心­

4K1L1 /' K-V-1--r / h \ Qa/dl!ur--iMl 1/// lJJrL , /\X\-3 //'/述一

/r\

//'\\­ 2///\\­3_,//_\\­ _­

resistances-in parall巴 1 ， inductances inparal1 el, and cap2:­ citances in series can all be worked out by the graphic

t ;清

method. No numerical example is necessary for anyone _除仆

.o

f these problems, because they arc all done in the same

\X/(hu

|(---uυMV

'way as work problems. The graphic method is also helpful

nr

\vhen there are more than two men working, more than two resistances in parallel, and so on. Suppose there are

four resistances of 20

,

60，肘， and 140 ohms in p且 rallcl; the _Fig.2

斗

」

一

total resistance,

R

, is:

- 34 一 一 3"5一

K

M

L

_J

Fig 6

solution of more complicated forms of water-tank proble­ ms. Suppose a bathtub can be filled in 10 minutes by the hot-water faucet alone and in 6 minutes by the cold-water faucet alone. When the tub is full, it can be emptied in 5 minutes by the drainpipe. If both faucets and the drainpipe are open, can the tub be filled? If so, how long does it take?

By formula, llr=I/O+1卅一 1戶， or llr= 月 15， and

r=15

minutes.

Figiue 6 is the graphic solution of the same problem. The construction is se1fevident, and no explanation of procedure is necessary. The line

]K

is the line segment required; its length above the line

LM

indicates that the tub is filled in 15 minutes, the result as obtained by calculation. (轉載「美國數學教師月刊J

)

本文~=~+土之間解法係陳廣況學長在麻州執教時，閒取學

_{p . q}

間，與瞬人園中閒話，偶獲靈感，遣意之作，原文載美國「數學教師 」學會之月于Ii。雖為三角老定理，而以圖解之，重Ij意可謂新穎別緻， 陳老學長真是寓學理於趣味，樂在其中也。 攘本刊操悉，陳學長胃疾巴大致康復，現優遊紐州婿家，孟春時 節駕車探花訪友，陶性怡神。並答應編者，一使完全復原，當將一年 來侍展妻病，學拉教書不能繼續，搬家冒雪遊行，到後家俱不到以及 到新地方後，找醫生，找銀行，找保臉，考車牌，修車等趣聞經過， 為文撰載友聲。 一一編者註 一 37 一 1 1 . 1 1 1 一=一 +r-"+ 一+

R -

20 I 60 I 85 I 140'

Figure 5 shows a graphic solution of this problem. Use whatever scale is convenient. First resistances of 20 and ωare combined, giving e; then 85 and e are combined, giving ρfinally 140 and

f

are combined to obtain 岳 Depending on scale and accuracy of consrtruction, a value between 11.5 and gives 11.7 ohms). 12 is obtained (numerical 140 calculation 85 d M L Fig.5 ~; n自

It may be noted that this problem could have been solved equally well by various other pairings. For example

,

。 might have been paired with

b

to obtain

e.

then

c

with d to obtain (say)

fl.

then

f'

with

e

to obtain g. In fact, any possible pairing can be used.

V

negative values are introduced. the use of the formula and its graphic solution can be extended to the