xy c ab a c b ab ac cb c ab b ac - halapahalapa.com/pravipdf/zlatni_rez.pdf · 2014-04-24 · cab...

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2,

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�� C �� AB � ��

� �� |AB| = x + y� ��

�� |AC| = x� ��

�� |AC| = x� �� |CB| = y��

� � ��

x + yx

=xy

= .

x y

A C B

�� C ��AB � � ��

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%

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B �� AB � � ��

�� D ��

|AB| = 2|BD|.

�� D �� |BD| � �� AD � �� E� ��

� �� A �� |AE| � �� AB � ��

C�

A C B

��

�� C � �� AB � ��

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�� |AB|/|AC|�! ��ABC � ��

��

|AD|2 = |AB|2 + |BD|2 = 5|BD|2.

" � ��

|AC| = |AE| = |AD| − |DE|= |AD| − |BD| = (

√5 − 1)|BD|.

# � ��

|AB||AC| =

2|BD|(√

5 − 1)|BD| =1 +

√5

2= .

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�� F �� AD� ��

�� F �� |FB| � ��

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D E

A C B

F

G

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�� C � �� AB � ��

��

�� ! ��AFB � ��

� |AB| = 2|AF| � � |FB|2 = |AB|2 + |AF|2 =5|AF|2�" � ��

|AC| = |AG| = |FG| − |AF|= |FB| − |AF| = (

√5 − 1)|AF|.

� � �� |AB| = 2|AF|��

|AB||AC| =

2√5 − 1

= .

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D

F

E

A N C B

O

k

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�� B � �� AC � ��

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�� DEF �� AC � ��

� � �� |DE| = 2|AC|� " �� N�� $� F �� AC�

( �� v �� ACF �� R ��

�� DEF� -��

�� |AB|/|AC|��

|AB| = |AN| + |NB|.! ��" ( - � ��

|NB|2 = R2 − |ON|2 = R2 − (R − v)2.

� � �� R =2√

33

|AC| �v =√

32

|AC|��

|NB| =√

52

|AC|.# ��

|AB||AC| =

12|AC| +

√5

2|AC|

|AC| = .

��

��

�� k1� k2� k3 � ��

�� 1 : 2 : 4��

�� (k1) � �� D� �� k2

� �� A C� �� AC � �� k3

� �� B�

A C BD

Sk

1

k2

k3

�� +�� - � �.��

�� C � �� AB � ��

��

�� k1�

k2 �k3 � �

r1 = k, r2 = 2k, r3 = 4k,

4 / �� %��.�� - �� - � �� #�0)��(

'

� ��

�� .�� k�! ��CDS �BDS � � ��

� ��

|DC| =√

r22 − r2

1 =√

3k,

|DB| =√

r23 − r2

1 =√

15k.

# � ��

|AB||AC| =

|AD| + |DB||AD| + |DC| =

|DC| + |DB|2|DC|

=√

3k +√

15k

2√

3k=

1 +√

52

= .

��

��

� ��

�� |AB| = 2r�� CDEF �� CD ��

AB�

A C BDS

a

x x

F E

�� 1�� - � �.��

�- �� - ��

�� C� �� D� � �� AD�� CB� � ��

�� ( �� a ��

CDEF� �� x = |AC| = |DB|� ��

ACF �FCB � � � ��

ax

=2r − x

a,

��

|CD||AC| =

|AD||CD| .

��

��

��

�� ABC� �� BC ��

BEDC� �� C ��

|CE| � �� AB � �� F�

A N B F

C

D

E

�� $�� - � �.��

�� B � �� AF � ��

��

�� " �� a = |AB| �� N ��

�$� C �� ABC� -�� CE ��

�� BEDC��

|CE| = |CF| = a√

2.

! ��CNF � �� |NF|2 = |CF|2 −|NC|2�� |NC| =

√5

2a��.�� |NF| =√

52

a�# � ��

|AF||AB| =

|AN| + |NF||AB| =

12a +

√5

2a

a= .

( ��$) *�� +�*�� %�)�

��

��

��

CAB� ��

�� |AC| = 3 |CB| = 4 � � ��

|BA| = 5� �� s �� B � �

�� AC � �� O� �� O �� |OC| � �� s � �� P Q�

O

P

Q

C A

B

�� '�� - � �.�� - � ��

�� P � �� AQ � ��

��

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��

��

|OC||OB| =

|AC||AB| =

35.

� � �� |OC| = 4 − |OB| �|OB| =53|OC|��

�� |OC| = 3/2� ! ��OCA � ��

�

|OA| =√|OC|2 + |AC|2 =

3√

52

.

# � ��

|QP||PA| =

2|OC||OA| − |OC| =

3

3√

52

− 32

= .

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xy c ab a c b ab ac cb c ab b ac - halapahalapa.com/pravipdf/zlatni_rez.pdf · 2014-04-24 · cab...

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