## Triangle construction given

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• Dear Hyacinthists, Source: an old russian geometry book translated to french. Hum.... hum.... hum?? How to put the data in a feasible position?? 2p(or s) =
Message 1 of 9 , May 26, 2008
Dear Hyacinthists,

Source: an old russian geometry book translated
to french.

Hum.... hum.... hum??

How to put the data in a feasible position??

2p(or s) = a+b+c

Best regards,
Luis
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• Luis, ¿qué significa feasible position in a construction problem? ... -- ... Francisco Javier García Capitán http://garciacapitan.auna.com [Non-text
Message 2 of 9 , May 26, 2008
Luis, ¿qué significa feasible position in a construction problem?

2008/5/26, Luís Lopes <qed_texte@...>:
>
>
> Dear Hyacinthists,
>
> Source: an old russian geometry book translated
> to french.
>
> Hum.... hum.... hum??
>
> How to put the data in a feasible position??
>
> 2p(or s) = a+b+c
>
> Best regards,
> Luis
> __________________________________________________________
> Receba GRÁTIS as mensagens do Messenger no seu celular quando você estiver
> offline. Conheça o MSN Mobile!
> http://mobile.live.com/signup/signup2.aspx?lc=pt-br
>
> [Non-text portions of this message have been removed]
>
>
>

--
---
Francisco Javier García Capitán
http://garciacapitan.auna.com

[Non-text portions of this message have been removed]
• Dear Luis! If you mark off segments BB =BA and CC =CA on the extensions of the side BC, you ll get a triangle AB C with the same h_a, B C =2p and
Message 3 of 9 , May 26, 2008
Dear Luis!

If you mark off segments BB'=BA and CC'=CA on the extensions of the side BC,
you'll get a triangle AB'C' with the same h_a, B'C'=2p and <B' - <C' = (<B -
<C)/2. Then, consider the reflection C" of the point C' in the line parallel
to BC and passing through A. We get an obviously constructible right
triangle B'C'C" with <B'C'C"=90°, B'C'=2p, C'C"=2h_a.
Then we know the distance h_a from point A to B'C'(=BC) and the angle B'AC"=
<B' - <C' + 180° (as is easy to compute).So we can construct A and finish by
constructing B and C at the intersections of B'C' with perpendicular
bisectors to B'A and C'A.

Friendly,

2008/5/26 Luís Lopes <qed_texte@...>:

>
> Dear Hyacinthists,
>
> Source: an old russian geometry book translated
> to french.
>
> Hum.... hum.... hum??
>
> How to put the data in a feasible position??
>
> 2p(or s) = a+b+c
>
> Best regards,
> Luis
>

[Non-text portions of this message have been removed]
• Dear Hyacinthists, [Francisco Capitan] Luis, ¿qué significa feasible position in a construction problem? Well, I tried to quote Petersen. My mistake, I
Message 4 of 9 , May 26, 2008
Dear Hyacinthists,

[Francisco Capitan] Luis, ¿qué significa feasible position
in a construction problem?
Well, I tried to quote Petersen. My mistake, I should
have written "suitable position".

As always, they are very appreciated.

Best regards,
Luis

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• Dear Hyacinthists, Source: an old triangle construction german book. Its suggestion to both problems is to construct b+c. I don t figure out how. I think that
Message 5 of 9 , May 26, 2008
Dear Hyacinthists,

Source: an old triangle construction german book.

Its suggestion to both problems is to construct b+c.
I don't figure out how.

I think that here is more of a matter of knowing a
previous result than to put the data in a suitable position.

Best regards,
Luis

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• Dear Luis ... First, b-c =(r_b-r_c)tan(A/2) In the first case, get L such as L^2 = (b-c)^2+4r^2/cos^2(A/2) then b+c= 4r/sin(A)+L In the second case, get L such
Message 6 of 9 , May 26, 2008
Dear Luis
> Its suggestion to both problems is to construct b+c.
> I don't figure out how.

First, b-c =(r_b-r_c)tan(A/2)
In the first case, get L such as
L^2 = (b-c)^2+4r^2/cos^2(A/2)
then b+c= 4r/sin(A)+L
In the second case, get L such as
L^2 = (b-c)^2+4r_a^2/cos^2(A/2)
then b+c= 4r_a/sin(A)-L
I recognize that this is not very nice but it works and I didn't find
something better (although it's probably possible)
Friendly
Jean-Pierre
• Dear Luis If you want just to construct the triangle, not necessarily using the sum b+c, then it suffices to notice that r_b-r_c=(b-c) cot A/2, so the problem
Message 7 of 9 , May 26, 2008
Dear Luis

If you want just to construct the triangle, not necessarily using the sum
b+c, then it suffices to notice that
r_b-r_c=(b-c) cot A/2, so the problem reduces to the construction of a
triangle from A, r (or r_a) and b-c, which was discussed in post 16341 and
later.

Best regards,

2008/5/26 Luís Lopes <qed_texte@...>:

>
> Dear Hyacinthists,
>
> Source: an old triangle construction german book.
>
> Its suggestion to both problems is to construct b+c.
> I don't figure out how.
>
> I think that here is more of a matter of knowing a
> previous result than to put the data in a suitable position.
>
> Best regards,
> Luis
>
>
>

[Non-text portions of this message have been removed]
• Dear Hyacinthists, Dear Jean-Pierre, Vladimir, Thank you for your solutions. I think there is a mistake in the book s suggestion. One has to calculate b-c
Message 8 of 9 , May 27, 2008
Dear Hyacinthists,

I think there is a mistake in the book's suggestion.
One has to calculate b-c instead.

Then b-c=(r_b-r_c)tan A/2

I didn't know this result and tried to prove it syntetically.
I failed and tried an algebraic approach.

So (B-C)/2=D and tan D=(b-c)/(r+r_a)=(r_b-r_c)/(b+c)
(see Court, for example).

One has

(b-c)/(r_b-r_c)=(r+r_a)/(b+c)=tan A/2

(b+c)/(r+r_a)=(r_b-r_c)/(b-c)=cot A/2

Two nice results.

Best regards,
Luis

________________________________

To: Hyacinthos@yahoogroups.com
From: vndubrovsky@...
Date: Tue, 27 May 2008 01:59:30 +0400
Subject: Re: [EMHL] Triangle construction given and

Dear Luis

If you want just to construct the triangle, not necessarily using the sum
b+c, then it suffices to notice that
r_b-r_c=(b-c) cot A/2, so the problem reduces to the construction of a
triangle from A, r (or r_a) and b-c, which was discussed in post 16341 and
later.

Best regards,

2008/5/26 Luís Lopes <qed_texte@...>

> Dear Hyacinthists,
>
> Source: an old triangle construction german book.
>
> Its suggestion to both problems is to construct b+c.
> I don't figure out how.
>
> I think that here is more of a matter of knowing a
> previous result than to put the data in a suitable position.
>
> Best regards,
> Luis

_________________________________________________________________
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• Dear Luis, to prove b-c=(r_b-r_c)tan A/2, rewrite it as (p-c) - (p-b)=(r_b-r_c)tan A/2, where p=(a+b+c)/2. Now it suffices to prove that p-c = r_b tan A/2 and
Message 9 of 9 , May 27, 2008
Dear Luis,

to prove b-c=(r_b-r_c)tan A/2, rewrite it as (p-c) - (p-b)=(r_b-r_c)tan A/2,
where p=(a+b+c)/2. Now it suffices to prove that
p-c = r_b tan A/2 and a similar formula for p-b.
This is clear if you notice that p-c is the distance from A to the
contact point of the b-excircle with AC.

Best,