ART Liek Wilardjo From isosceles triangle Full text

(1)

FROM /SO.W ＧﾣＱＱｾＭｾＧＬＧ＠ TR/AN(;'LE TO NO REf lAURA TES COMPTOl\:', DE BROG'LJE, l>A V/.\:\'ON, AND THOMSON

Ltek W!lardjo

FROM ISOSCELES TRIANOLE TO NOBEL LAURATES COMPTON, IJE

BR06'LIE, IJA VISSON, AND THOMSON

1.

Pengantar

Liek Wilardjo

Program Studi Teknik Elektro

Fakultns Teknik Elektronika dan Komputer OKSW Jalan Diponegoro 52-60. Salatiga 50711

A1tikel di bawah ini berisi tentang planimetri elementer. yakni segitiga samakaki dan datil p,·thagoras padn segitiga siku-siku. Dalam artikel ini ditampi!kan contoh sederh::mn ｴ･ｮｴｾｭｧ＠

geometri Euklides-an. Kedua jenis segitiga itu dan sifatnya dipakai untuk menunjukkan pembuktian sifat (propcrttes) atau datil (propositions) secara matematis.

Melalui planimetri. Be11rand Russell sewaktu masih bocah pertama kah ｭ･ｭｰ･ｬｾｪ｡ｲｩ＠

matematika dari guru priYatnya0l Melatui dalil ｐｾｴｨ｡ｧｯｲ｡ｳＮ＠ Sokrates "memperagakan" bahwa matematika (menurut dia) ialah kebenaran abadi. Seorang bocat1. anak budak belian. mampu menemukan sendiri bukti datil PythagorasUJ Demonstrasi Sokrates itu menandai lahirnya

discovet:l' atau inquit)' method yang sampai sekarang masih dipakai di bidang pendidikan. Anamnesis yang dilakukan dokter terhadap pasiennya juga didasarkan pada keyakinan yang sam a.

Datil Pythagoras dipakru Einstein untuk mengungkapkru1 kekararan (invariance) skalar berupa kuadrat Yektor-4 pusa-tenaga (momentum-energytour-vector). yru1g "bermuara" pad a "rumus"-nya yang terkenal. yakni E rm_.J._ Louis Victor de Broglie kemudian menarik analogi dari cara Einstein itu. selungga ia menemukan keseduaan zarat1-gelombang (particle-wave duality). dengru1 persrunarumya p

=

h A.= flk. Dengan penemurumya itu Louis de Broglie bukru1 hanya memperoleh gelar doktor dari Umversite de Paris (dengru1 Profesor Paul

Lru1geyin sebagai promotornyal3> tetapi juga dianugerahi hadiail Nobel dalrun fisika.

Hasil yang diperoleh Einstein. de Broglie. dru1 sebelumnyajuga oleh Max Planck (E hv

ftw. untuk foton) ternyata mengilhruni kru·ya-k:arya keilmuru1 lain. yang juga "menyabet" hadiah Nobel. Jadi dengan artikel ini say a hendak menunjukkru1 batm·a kebenaran 'ang tampaknya sederhru1a terny ata bisa membenih (seminal)

(2)

2. Isosceles Triangle

ABC is a triangle in "·hich AH is the base

and C the apex lfthe base angles. LA and L

B. are equal. then side A(' is equal to side

R(' . and thus AI:K IS an ISosceles tnangte.

c

ｩＮ｟｟｟｟ＡＺＮ［｟｟ｾＮＮＮｬｵＬ＠ｌｾ｟ＮＺ｟ａＺＮＮＬ｜＠

Proof: A D B

( I) Dra" the tnangle in a piece of clear I

transparent plastic Draw the bisector of the apex angle L C until it mtercepts base AB

at a point. D. It is obYious that

CD

is perpendicular to

AB. No\Y fold the paper along

CD.

You \Yill see that A and B comcide and the right triangles ACD and BCD m erlap

perfectly. "proYing" that

AC

is indeed equal to

BC.

What "e hm e just done is actually not praYing that ABC is an isosceles triangle. It is a

"platonic"" ay of demonstrating experimentally or obsen ationally that

AC

and

B('

are of equal length. To Plato. geometry is real. i.e .. it deals \Yith concrete entities.

(2) To proYe the theorem mathematically.

assume that

AC

i

BC

_

say. that

AC'

is

longer than

BC

.

Then there is a point.

call it E. on

AC

such that B('

=EC.

Draw line

EF

\\ith the point F on side

RC

_

angle ,:::... CEF being equal to angle ._:

CBE. Then the triangles CEF and CBE

identical in

c

shape. two. and consequently all three. of their corresponding angles being equal.

Therefore there is a proportionality bet\yeen the corresponding sides of the two

ＭＭｾｾＭＭＭ

-B

trianglesfi.e. _I

CE: EF: FC

=

CB: BE: EC.

or

CE

x

EC

=

EFx BE= FC

x

C'B

This is a three-side equation with the left the middle and the right sides. From the

equalit: of the left <.md the right s1dc and notmg th:tt Cr . ...,.-

r

B (according to our

(3)

/·ROM 1.\'0S< ELL'.\' TRIAS (if_£ TO NOBEL /A iRA TL-:\· ( 'OMPTON, I>E BROGLIE,

l>A V/SSON, ANJ> THOMSON Liek Wilurdju

The conclusion that £('

=

F(' leads to an absurd situation. Thus the opposite of our

assumption. i.e .. that A('

=

B('. is true by the logic of "rcductw ud ahsurdum''. {!uod

erat demosntrcmdum (Q.E.D.)

3

Pythagorean Theorem

ABC Is a nght-angle tnangle. The ｡ｮｧｬ･ｾ＠ A IS')()'' AH and A(· are the nght sides.

1\hile RC the ｨｾ＠ potenuse. Calling AB. A(·. and B( · c. h. and u. respectiYely.

2 ) 2 .

Pythagorean theorem states that

a

=

h-

+

c

.

I.e .. the square of the hypotenuse IS

equal to the swn of the squares of the right

ｳｩ､･ｾ＠

Proof'

( 1) For

an

isosceles. right triangle. h =c. so ''e ha' e to proYe that

a

2

=

b2

+

c

2

=

2 h2 (or

a=

h.fi

DrmY three more right triangles of the

same shapes and size to that of triangle

ABC. so that ''e haw a square. named

BCDE. \\hose centre is A The area of

square BCDE is

L

₁

=

a

2. a being its side

The area of triangle ABC is

L

₂

={he== *h

2. since h =

c

From the figure it is

A R

c

E

obYJOUS that L1

=

4L

_{2 .}therefore

a

2

=

4 x 1h2

=

2h

2. giYing

a= h.fi. thus proYing

the theorem for the special case of an isosceles. right-angle triangle

(4)

Techne Jurnalllmtah Elcktrotekntb.a VoL') No 2 01-.tobct 21J l 0 Hal ＱＲｾ＠ ll!

According to Plato (in "Dialogue vFith Meno")1 tlus proof was taken by Socrates as a

supporting eYidence of the truth of his belief that mathematics is eternal truth. In front of

his friend. Meno. Socrates demonstrated that a boy "ho was the son of a stare could

rediscoYer the Pythagorean theorem by himself: Socrates merely guiding him with a series of questions This is "·hat "e call the discoYery or the inquiry method of teaching.

{ '

( 2) For ＺＺｾｮｙ＠ nght-angle triangle. the proof goes as follows 2 Consider the right-angle

triangle ABC with the right angle at the

Yertex C. sub tended bet\\ een the right A D B

L2X

sides

a and

CA

= h and the

hypotenuse ｾＮＮＺＮ＠ as 1ts base.

AB

Name the base angles a and jJ. at \·erte:-.. A and \ ene-...: B.

respecth·ely. The right-angle triangle ABC is defined. i.e .. it is fully-specified. when its

base and either one of its base angles is gh en. Thus its area L is completely determined

by c and jJ. As angles or functions of angles can be considered as dimensionless. I. must be proportional to the square of the base. c. times some fimction of a base angle. say jJ

Dra"

C'D

with[) on

AB

and

CD

j_

AB.

Since the sum of the three i11terior angles in

any triangle is equal to 180°. the angle yat the Yertex C of the right-angle triangle ADC

must be equal to the angle j3 at the Yertex B of the right-angle triangle ACB. Then the

area oftriangle ADC L1. is proportional to h2f(y) h2

f<PL

while that of triangle

CD B. is proponional to a2f(fJ) But from the

ｦｩｾＬＧｬｬｦ･＠

\\e see that

Hence:

or

1 leaY e it as an exercise for you to find

f<P)

Plato (c. 4Li ··c. 34/ 8 C 1 frutn IIi::> d;alogue Meno. 81 E 86 c. Guthtc,

w:

K C. rrans Harmondci\1-r:rth Penguin 19513

(5)

FROM L\'US( '!:.Lt.'S IKJA/\(iLE HJ /'1/0/fLL LHRAIES ( ｏｬｾｬｐｔｕｎＬ＠ J)£ BROGLIE. DAVJS'S'ON, AND THOMSON

r1..:k WilarJ.;o

4. Seminality of Pythagorean Theorem

Using Einstein's special Theory of Relati ,·ity. ''e hm e. from the im ariance of the square

of the momentum-energy four-Yector.

, 1 , F ₂

11-;.

+

f1,.

+

ｰｾ＠

+

(I

( J'l\2

r) , ) , 2 j . r. l

",:- t I' I-:- +

P-

t- i - I .. ' ( I )

or

p-

, 1 (2)

Here p and f.. and p' and J:'. are the relati' tsttc momenta and total energ1es of

a

part1cle

in

the frames of reference K and K'. respecth

ely.

Writing .E' as the sum of resHnass ･ｮ･ｲｧｾ＠

E

₀

·

111 0 l ·· and kinet1c energ' (; I)E(_{1 .}and assuming that the particle is at rest iu /\.'

we haYe:

J

ｰｾ＠

\\11ich can be rewritten as

Thus we hm·e

Equation (5) is Pythagorean in form and

can be depicted by a right-angle triangle

\\Jth the nght sides pc _{and m 0} . and

hypotenuse I.

Analogously. Louis-Victor de Broglie

dre" a riglu triangle as follows:

The hypotenuse of the triangle is

frequencY. u. and one of the right sides

(5}

pc

L

ts the YeloClty of light. c. di\ ided by waYelength. t!. The other right side is.

by analogy. an im ariant of" :n e "luch. at the time, "as as ｾ＠ et

tumamed

(6)

tectu1e Jurnal lhmah Ekktrolcknika Vol

'>

No 2 Oi...lober 20 I 0 Hal 125 - I 3 I

130

As the consequence of the Pythagorean relation among the three sides of his right-angle

triangle. de Broglie concluded that

h

p=-=nk

A

(7)

In equation (7) his Plank's constant (6 625 10 ｾｾ＠ Js). h is Dirac's constant

( c h '"' ;r) and k (=-" 2

rr

.-1) ts ''me member or the phase constl:mt of the wm e

ｲ･ｰｲ･ｳ･ｮｴｩｮＮｾ＠ the particle

h '

p

=

A

had been known to be tme for photons. but de Broglte stated boldly that the

relation held for massh e particles as ''elL Hence "e ha' e the so-called pa11icle-wa' e

duality. Not only did de Broglie get the degree of doctor of philosophy in Physics from

the UniYersity of Paris ( 1924 ). but he \Yas also mnuded the 192() Nobel priLe in Physics.

Arthur H Compton used Planck's quantization of photons. E = h v (= tJ {u). and de

Broglie particle-\\ aYe duality.

p

= : ( = h k ), in the analysis of his experiment of The

scattering of photons by a free electron in a substance. and

won

the 1927 Nobel prize in

Physics

The principle of consermtion of

momentum in Compton scattering can be

depicted with a triangular Yector diagrrun.

the sides of \\hich are

j)'

(

=

h/X. the

momentum of the scattered photon) and

p

r (

=

r

111() l' the momentum of the

recoiled

electron). and the base of which is j)' (::::

h/

A.

the momentum of the incident photon). Together mth the principle of consen ation of energy.

hv=hv'+(y

(8)

the conlen at ion of momentum.

2 2 tl ') I

B

Pr -::::: P

+

P

ｾ＠

•P p

cos.

(9)

results in ｴｨｾ＠ equation of Compton effect.

(7)

FROM /SOSCELE.\' 1R/A;V(J'LE TO NOBELIAURATES COMPT01\', DE BROGLIE, DA VJS,{;'ON, AN/) THOMSON

Liek Wtlanlto

From the equation of Compton effect it is recognized that the unnamed inrariant in de

Broglie's nght-angle tnangle is c times the rec1procal of Compton \\<Helength.

c ( I

I

Ac )

=

c ( m

o

c

I

h) m ₀c 2 / h · · · (ll )

Two other ｰｨｾ＠ sicists. Clinton J. Darisson and George P Thomson. shared the 1937

electrons ｢ｾ＠ en 'italline lattices

DAFTAR Pl!STAKA

(1) A Doxiadts & C.H Papadimitriou (dengan ilustras1 oleh A Papadatos & A DiDonna):

Logicomix. Bloomsbury. London. 2009 p.55.

(2) Plato: "Mathematics as Eternal Tmth". dalam Stum·t Bro" n. ct ul. Conce))tions of

Inquiry.

Methuen. London. 1981. p. 3- 11

(3) RH March: Physics for Poets. McGraw-Hill. Ne\\ York 197(l. p. 207-212