## Re: [EMHL] an angle question

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• ... From: Ricardo Barroso Campos To: dick tahta ; Sent: Tuesday, October 15, 2002
Message 1 of 10 , Oct 15, 2002
----- Original Message -----
From: "Ricardo Barroso Campos" <rbarroso@...>
To: "dick tahta" <dick@...>; <Hyacinthos@yahoogroups.com>
Sent: Tuesday, October 15, 2002 7:46 PM
Subject: Re: [EMHL] an angle question

Friend Dick, friends of Hyacintos:

I was saying m ABC =12, not B/2 =12.

It(He,She) would be B/2 = 6...

I AGREE with Ioana, BAC=92

================
Ricardo,

You want to say <BAC = 96º, not 92º.

Did you tray to inscribe a 30-gon or 60-gon in angle BAC?

Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@...
• Amigo Ignacio: es BAC=96 gracias ... From: Ignacio Larrosa Cañestro To: Sent: Tuesday, October 15, 2002
Message 2 of 10 , Oct 15, 2002
Amigo Ignacio:

es BAC=96

gracias

----- Original Message -----
From: "Ignacio Larrosa Cañestro" <ilarrosa@...>
To: <Hyacinthos@yahoogroups.com>
Sent: Tuesday, October 15, 2002 7:44 PM
Subject: Re: [EMHL] an angle question

----- Original Message -----
From: "Ricardo Barroso Campos" <rbarroso@...>
To: "dick tahta" <dick@...>; <Hyacinthos@yahoogroups.com>
Sent: Tuesday, October 15, 2002 7:46 PM
Subject: Re: [EMHL] an angle question

Friend Dick, friends of Hyacintos:

I was saying m ABC =12, not B/2 =12.

It(He,She) would be B/2 = 6...

I AGREE with Ioana, BAC=92

================
Ricardo,

You want to say <BAC = 96º, not 92º.

Did you tray to inscribe a 30-gon or 60-gon in angle BAC?

Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@...

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• ... [AC]. ... m(ABC)=? ... ******* If I is the incenter of ABC and x = B/2 then B/2 + C/2 = 42 IE=IB*b/(a+c), ID = IC*c/(a+b) IC/IB = sin(B/2)/sin(C/2) and
Message 3 of 10 , Oct 15, 2002
>ABC is a triangle. D is a point on [AB] and E is a point on
[AC].
>m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 ,
m(ABC)=?
>
*******
If I is the incenter of ABC and x = B/2
then B/2 + C/2 = 42
IE=IB*b/(a+c), ID = IC*c/(a+b)
IC/IB = sin(B/2)/sin(C/2)
and
sin(24)/sin(18) = ID/IE and after trig. calculations
= sin(84 - x)/sin(42 + x)
(1)
We'll prove that
sin(24)/sin(18) = sin(78)/sin(48) or
2sin12cos12/sin18 = cos12/sin48 or
2sin12sin48 = sin18 or
cos36 - cos60 = sin18 or
4(cos18)^2 - 3 = 2sin18 or
cos18*[4(cos18)^2 - 3] = 2sin18cos18 or
cos(3*18) = sin(2*18) or
cos54 = sin36 which is true.

Hence from (1) we get
sin(84 - 6)/sin(42 + 6) = sin(84 - x)/sin(42 + x)
or
(tan84-tan6)/(tan42+tan6)=(tan84-tanx)/(tan42+tanx)
or tanx = tan6 or x = 6 or B = 12, C = 42, A = 96.

Best regards
• Ricardo: a thousand apologies - I misread your statement. Indeed the trigonometry does yield B/2 = 6. best wishes, Dick
Message 4 of 10 , Oct 15, 2002
Ricardo:

does yield B/2 = 6.

best wishes, Dick

> From: "Ricardo Barroso Campos" <rbarroso@...>

> I was saying m ABC =12, not B/2 =12.
>
> It(He,She) would be B/2 = 6...
• Let r be the radius of the inscribed circle D.Taking advantage of the symmetry of the situation it can be shown by elementary trigonometry that w.r.t. D the
Message 5 of 10 , Oct 16, 2002
Let r be the radius of the inscribed circle D.Taking advantage of the symmetry of the situation it can be shown by elementary trigonometry that w.r.t. D the areas of the escribed equilateral triangle,the inscribed hexagon,the inscribed equilateral triangle are respectively
(3 r^2 sqrt3) ,(3 r^2 sqrt3)/2,(3 r^2 sqrt3)/4 which are obviously in G.P.
Best Regards,

---------------------------------
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[Non-text portions of this message have been removed]
• Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on straight GP, ED = EG. m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD),
Message 6 of 10 , Oct 19, 2002
Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
straight GP, ED = EG.
m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
FD.
m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
CE and BD intersect in A.
So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
m(EBC) = m(FEB) - m(EFC) = 24 - 15 = 6, m(ABC) = 12.
Best wishes, Igor Sharygin.
m(ECD)= m(DCB), m(DBE)=m(EBC).

----- Original Message -----
From: RICARDO BARROSO CAMPOS <rbarroso@...>
To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
Sent: Tuesday, October 15, 2002 3:00 PM
Subject: Re: [EMHL] an angle question

> Dear Hyacinthos:
>
> ABC is a triangle. D is a point on [AB] and E is a point on [AC].
> m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
>
> I'm a conjecture with Cabri:
> m (ABC) =12
>
> geometric reasons? I do not know it
>
>
> Ricardo
>
>
>
>
>
>
>
>
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
• ... From: Sharygin I. F. To: ; Sent: Saturday, October 19, 2002 9:47 PM Subject: Re:
Message 7 of 10 , Oct 19, 2002
----- Original Message -----
From: Sharygin I. F. <sharygin@...>
To: <Hyacinthos@yahoogroups.com>; <yagcimustafa@...>
Sent: Saturday, October 19, 2002 9:47 PM
Subject: Re: [EMHL] an angle question

Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
straight GP, ED = EG.
m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
FD.
m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
CE and BD intersect in A.
So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
m(EBC) = m(EFC) - m(FEB) = 30 - 24 = 6, m(ABC) = 12.
Best wishes, Igor Sharygin.
.
>
>
> ----- Original Message -----
> From: RICARDO BARROSO CAMPOS <rbarroso@...>
> To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
> Sent: Tuesday, October 15, 2002 3:00 PM
> Subject: Re: [EMHL] an angle question
>
>
> > Dear Hyacinthos:
> >
> > ABC is a triangle. D is a point on [AB] and E is a point on [AC].
> > m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
> >
> > I'm a conjecture with Cabri:
> > m (ABC) =12
> >
> > geometric reasons? I do not know it
> >
> >
> > Ricardo
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
>
• Friends of Hyacintos, Igor: It is a perfect euclidea construction.Congratulations My conjecture of 12 was correct. Greetings Ricardo Barroso Campoos University
Message 8 of 10 , Oct 19, 2002
Friends of Hyacintos, Igor:
It is a perfect euclidea construction.Congratulations

My conjecture of 12 was correct.

Greetings
Ricardo Barroso Campoos
University of Seville Spain

http://www.pdipas.us.es/r/rbarroso/trianguloscabri
>>> sharygin@... 19/10/02 20:43 >>>

----- Original Message -----
From: Sharygin I. F. <sharygin@...>
To: <Hyacinthos@yahoogroups.com>; <yagcimustafa@...>
Sent: Saturday, October 19, 2002 9:47 PM
Subject: Re: [EMHL] an angle question

Let EFG - eqvilateral triangle. P inside, EP = PG, m(EPG) = 108. D on
straight GP, ED = EG.
m(EDG)= m(EGD) = 36, m(DEP) = 180 - 36 - 72 = 72 = m(EPD), m(EDP)=36
PD=ED=EG = EF, m(FED) = 72 - (60-36) = 48.
C and B on straight FP, CD is the perpendicular to EP, EB is perpendicular
FD.
m(EDC) = m(PDC) = half m(EDP) = 18, m(DEB) = m(FEB) = 24,
CE and BD intersect in A.
So: ABC, BE and CD are bisectors, m(EDC) = 18, m( BED) =24.
m(EBC) = m(EFC) - m(FEB) = 30 - 24 = 6, m(ABC) = 12.
Best wishes, Igor Sharygin.
.
>
>
> ----- Original Message -----
> From: RICARDO BARROSO CAMPOS <rbarroso@...>
> To: <yagcimustafa@...>; <Hyacinthos@yahoogroups.com>
> Sent: Tuesday, October 15, 2002 3:00 PM
> Subject: Re: [EMHL] an angle question
>
>
> > Dear Hyacinthos:
> >
> > ABC is a triangle. D is a point on [AB] and E is a point on [AC].
> > m(ABE)=m(EBC) , m(ACD)=m(DCB) , m(EDC)=18 , m(DEB)=24 , m(ABC)=?
> >
> > I'm a conjecture with Cabri:
> > m (ABC) =12
> >
> > geometric reasons? I do not know it
> >
> >
> > Ricardo
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
>

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