1 Yellow Dragon 08 - Queens College, Hong Kong

Y, Z dividing OA, OB, OC respectively in the ratio OX OA = OY OB = OZ OC = t (†) The lines AX, BY, CZ are concurrent at the isogonal conjugate of the point P on the Euler line dividing OH in the ratio OP: PH=1:2t Proof Let the isogonal line of AX (with respect to angle A) intersect OA at X The triangles OAX and OX A are similar It

AUXILIARY DEVICES - Siemens

OA One normally open and one normally closed OB one normally open OC one normally closed OS one normally closed late opening OR one normally open early closing OG two normally open OH two normally closed OJ same as OA, but NC contact is late opening OV same as OA, but NO contact is early closing 2 pilot light units for BA9s bulbs

A Note on the Hervey Point of a Complete Quadrilateral

OH= OA+OB+OC It is easy to see that the resultant AHof the two vectors OBand OCis perpen-dicular to the side BC Therefore the line AH is the altitude drawn through the vertex A and, more generally, the point H is the point of concurrence of the three altitudes The usual property of the orthocenter is recovered

Inscribing a similar triangle using spiral similarity

OH= OA 0+ OB0 + OC Now we know that the sum of the vectors from the symmedian point to the vertices of its pedal triangle is zero, and the sum of the vectors from the circumcenter to the vertices of its pedal triangle is the vector from it to the orthocenter, i e , in the direction of the Euler line, which contains both points

(OA OB OC) ( )( ) ( )( ) AH⋅BC= OH−OA⋅OB−OC= OB+OCOB−OC=OB

(1)og (oa ob oc) oh 3 1 3 1 = + + = を用いて証明 (2) oa=ob=ocに注意すると ( )( ) ( )( ) 0 2 2 ah⋅bc= oh−oa⋅ob−oc= ob+ocob−oc=ob −oc = 同様にbh⋅ca=ch⋅ab=0より、各々のパェダャが直交するので点hは垂心＜寸評＞外心oン垂心hン重心gが一直線上にあり、線分比についてoh:hg=1:2

Exercise 13(A) Page: 158 - BYJUS

OA 2= AH 2+ OH = AH + AE OC 2= CG + OG2 = EB2 + HD OB 2= EO + BE 2= AH + BE2 OD 2= HD 2+ OH = HD + AE2 Adding these equalities we get: OA 2+ OC 2= AH2 + HD + AE + EB OB 2+ OD = AH2 + HD2 + AE + EB From which we prove that for any point within the rectangle there is the relation OA 2+ OC = OB2 + OD2 Hence Proved 9

1 Yellow Dragon 08 - Queens College, Hong Kong

Title: Microsoft Word - 1_Yellow Dragon 08 doc Author: YUE Created Date: 9/3/2015 5:22:09 PM

BC AC AB 1

oh og 0 hg o C,1 ; B,1 ; A,1 G ABC G

-4 ا ا سر ا

تمارين و حلول المعلم في المستوى

(3) oA + →oB + oC = 3 → oG → وأ : ج ˚ ا (4) → →oH + →oB + oC = oH : ˚˜ (2) لاB#ا C#D oH = 3 → oG → نأ - ˚ #˘ (4) و (3) C#D نذ إ " #ﻡ H و G و o E"˚ا و ا سر ا سر

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When vectors shoot at the Centres of a Triangle ....

Yue Kwok Choy

1. Introduction

Centroid(G)

() Orthocentre (H) () Circumcentre (O) () In-centre (I) () Ex-centres (E

1, E2, E3)

This article aims to find the vector forms of five classical centres of a triangle. It is assumed that the reader has learned a

basic course of vector algebra. We use A , B, C to denote the position vectors

OCOBOA,, of the three vertices of the

triangle. Similarly, we use X to denote the position vector OX of any point X. We use unbold italic A, B, C to

denote the angles of the triangle ABC and a, b, c the length of the sides opposite to the angles A, B, C respectively.

2. Centroid (G) (Refer to Figure 1)

Since BD : DC = 1 : 1, we have

2 CBD+=

Since AG : GD = 2 : 1, we have

3322

32CBACBADAG++=

.... (1)

3. Some preliminary results : (Refer to Figure 2, 3 below)

(a) OCOBOAOH++= , where Orthocentre (H), Circum-centre (O) Proof

We get around the problem a bit. We suppose

OCOBOAOH++= and then show that H is the orthocenter.

Since O is the circum-centre,

OCOBOA== .... (2)

()OCOBOAOCOBOAOAOHAH+=-++=-= .... (3) and

OBOCBC-=

()()0

22=-=-+=?OBOCOBOCOCOBBCAH, , by (2)

BCAH?? , similarly ABCHCABH??,

? H is the orthocenter. If we can take the circum-centre (O) as origin and we simply write H = A + B + C (b)

0tantantan=++CCHBBHAAH

Let R be the radius of the circum-circle. By (3), 2222

BOCOCOBOCOBcos2

22-+= ARRR2cos2222-+=, ? at centre twice ? at circumference

Figure 1

( )()( )22222cos2cos222cos12ARARARAH==-=? ARAHcos2±=?

For simplicity, we take ?A, ?B, ?C to be acute angles. Same result can be got with obtuse angle. (note 1)

CRCHBRBHARAHcos2,cos2,cos2+=+=+=? .... (4) Since

CHBHAH,, are non-parallel, we can let

0=++CHBHAHγβα , where α, β, γ are constants.

We then form a triangle with sides :

CHUWBHVUAHWVγβα===,,

Note that

CHUWBHVUAHWV//,//,//

By comparing the right diagrams of Figure 2 and Figure 3 , ?U = ?BHR = 180 o - ?(CHBH, ) = 180o - ?QHR = ?A

Similarly, ?V = ?B, ?W = ?C

By sine law on ΔUVW,

12sinsinsinRW

CH V BH U AH where R

1 is the radius of circum-circle enclosing ΔUVW.

By (4),

12sin cos2 sin cos2 sin cos2RC CR B BR A

AR===γβα

CR RB R RA R Rtan,tan,tan111===γβα and 0=++CHBHAHγβα

0tantantan=++?CCHBBHAAH .... (5)

4. Orthocentre (H)

Since from (5),

()()()0tantantan=-+-+-COCOHBOBOHAOAOH

Solving,

CBA

COCBOBAOAOH

tantantan tantantan ++++= or CBA CBA tantantan tantantan ++++=CBAH .... (6)

Reader may check similar result,

CcBbAa

secsecsec secsecsec ++++=CBAH .... (7)

The above results (6) and (7) are good even if O is not the cirum-centre, but any assigned origin.

5. Circum-centre (O) (Refer to Figure 2, 4)

We still take O to be the circum-centre but with respect to any origin X .

We like to show that

02sin2sin2sin=++COCBOBAOA first .

We use the same technique as before and take

0=++OCOBOAγβα.

We can form a triangle UVW with sides

OCUWOBVUOAWVγβα===,,

?U = 180 o - ?(OCOB, ) = 180o - ?BOC = 180o - 2 ?A , ? at centre twice ? at circumference.

Similarly, ?V = 180

o - 2 ?B , ?W = 180o - 2 ?C

By sine law on ΔUVW,

12sinsinsinRW

OC V OB U OA , where R1 is the radius of circum-circle enclosing ΔUVW.

Figure 2

Figure 3

Figure 4

CRRBRRARRRCR

AR2sin2,2sin2,2sin222sin2sin2sin

1111
Since

0=++OCOBOAγβα, therefore 02sin2sin2sin=++COCBOBAOA

Now, ()()()02sin2sin2sin=-+-+-CXOXCBXOXBAXOXA CBA

CXCBXBAXAXO

2sin2sin2sin

++= or simply CBA CBA

2sin2sin2sin

++++=CBAO .... (8)

The reader may check similar result:

CcBbAa

coscoscos coscoscos ++++=CBAO .... (9)

6. In-centre (

I) (Refer to Figure 5 , 6)

We like to show that

0=++ICcIBbIAa first . We use the same technique as before and take

0=++ICIBIAγβα.

We can form a triangle UVW with sides

ICUWIBVUIAWVγβα===,, .

?U = 180 o - ?(ICIB, ) = t + u ?V = 180 o - ?(IBIA, ) = u + s ?W = 180 o - ?(IAIC, ) = s + t

By sine law on ΔUVW,

12sinsinsinRW

IC V IB U IA where R

1 is the radius of circum-circle enclosing ΔUVW.

( )( )( )12sinsin sinsin sinsinRtsur sutr utsr , where r is the radius of the in-circle. ( )()sr Rss r Rss r Ruts r

Similarly,

ur Rt r

R2sin,2sin11==γβ .

( )( )( )02sin2sin2sin0=++?=++ICuIBtIAsICIBIAγβα

02sin2sin2sin=)

((?ICcucIBbtbIAasa ((=++?cu bt asICcIBbIAa2sin2sin2sin,0Q , by sine law on ΔABC. ()()()0=-+-+-?OIOBcOIOBbOIOAa, O is any origin . cba

OCcOBbOAaOI++++=? or simply cba

cba ++++=CBAI .... (10)

7. Ex-centres (E

1 , E2 , E3)

With similar method of proofs, the three ex-centres, formed by angle bisectors of exterior angles, can be found:

cba cba cba cba cba cba -+-+=+-+-=++-++-=CBAECBAECBAE132,, .... (11)

Note :

(1) For more detailed results and discussions, please refer to my site: http://www.qc.edu.hk/math/index.htm .

(2) The author would like to thank Ms. Chan Lap-lin, Head of the Department of Mathematics at Queen"s College, for her careful

proof-reading, Figure 5

Figure 6

quotesdbs_dbs15.pdfusesText_21

[PDF] 1 Yellow Dragon 08 - Queens College, Hong Kong

Yue Kwok Choy

1. Introduction

Centroid(G)

1, E2, E3)

OCOBOA,, of the three vertices of the

2. Centroid (G) (Refer to Figure 1)

Since BD : DC = 1 : 1, we have

Since AG : GD = 2 : 1, we have

32CBACBADAG++=

3. Some preliminary results : (Refer to Figure 2, 3 below)

We get around the problem a bit. We suppose

Since O is the circum-centre,

OCOBOA== .... (2)

OBOCBC-=

22=-=-+=?OBOCOBOCOCOBBCAH, , by (2)

BCAH?? , similarly ABCHCABH??,

0tantantan=++CCHBBHAAH

BOCOCOBOCOBcos2

22-+= ARRR2cos2222-+=, ? at centre twice ? at circumference

Figure 1

CHBHAH,, are non-parallel, we can let

0=++CHBHAHγβα , where α, β, γ are constants.

We then form a triangle with sides :

CHUWBHVUAHWVγβα===,,

Note that

CHUWBHVUAHWV//,//,//

Similarly, ?V = ?B, ?W = ?C

By sine law on ΔUVW,

12sinsinsinRW

1 is the radius of circum-circle enclosing ΔUVW.

By (4),

AR===γβα

0tantantan=++?CCHBBHAAH .... (5)

4. Orthocentre (H)

Since from (5),

Solving,

COCBOBAOAOH

Reader may check similar result,

CcBbAa

CcBbAa

5. Circum-centre (O) (Refer to Figure 2, 4)

We like to show that

02sin2sin2sin=++COCBOBAOA first .

We use the same technique as before and take

0=++OCOBOAγβα.

We can form a triangle UVW with sides

OCUWOBVUOAWVγβα===,,

Similarly, ?V = 180

By sine law on ΔUVW,

12sinsinsinRW

Figure 2

Figure 3

Figure 4

CRRBRRARRRCR

AR2sin2,2sin2,2sin222sin2sin2sin

0=++OCOBOAγβα, therefore 02sin2sin2sin=++COCBOBAOA

CXCBXBAXAXO

2sin2sin2sin

2sin2sin2sin

2sin2sin2sin

2sin2sin2sin

The reader may check similar result:

CcBbAa

CcBbAa

6. In-centre (

I) (Refer to Figure 5 , 6)

We like to show that

0=++ICcIBbIAa first . We use the same technique as before and take

0=++ICIBIAγβα.

We can form a triangle UVW with sides

ICUWIBVUIAWVγβα===,, .

By sine law on ΔUVW,

12sinsinsinRW

1 is the radius of circum-circle enclosing ΔUVW.

Similarly,

R2sin,2sin11==γβ .

02sin2sin2sin=)

OCcOBbOAaOI++++=? or simply cba

7. Ex-centres (E

22=-=-+=?OBOCOBOCOCOBBCAH, , by (2)