## Re: Triangle Constructions problems A, b, ;

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• Dear Nikolaos and Luis exactly as as Nikolas said, I started from the classical relations 1/ra+1/rb=1/r-1/rc=2/hc They give immediately the equations for r or
Message 1 of 12 , Dec 7 12:55 AM
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Dear Nikolaos and Luis
exactly as as Nikolas said, I started from the classical relations
1/ra+1/rb=1/r-1/rc=2/hc
They give immediately the equations for r or ra
In the case R+r or R-r,you can start from the more difficult relation
(1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
which leads to a quadratic equation for r or R when you know A,b and R+r or R-r
Friendly. Jean-Pierre

> >The equation giving r with L=r+R is a bit ...
> I forgot L=R-r. I will be back to both of them
> later, as well as to a few others with R.
>
> >... more complicated, but it exists.
> Do you mean the equations you gave below are
>
>
> >It remains to get nice geometric constructions.
> I have no idea on how to deal with such data
> geometrically.
>
> So let's go back to the algebraic solution.
>
> r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
>
> r^2+L.r-b.L.sin(A)/2=0 with L=rc-r
>
>
> I have performed numerical tests with a=5, b=7 c=8
> and the above equations are correct. I have now two
> questions:
>
> 1) could you give me a hint on how to deduce them?
>
> I assume the others with ra, rb will follow the same way.
>
> 2) the equation with L=r+rc has 2 positive roots.
> Is it possible to have data that give two solutions?
> I mean two different triangles?
>
> Friendly,
> Luis
>
>
>
> To: Hyacinthos@yahoogroups.com
> From: jean-pierre.ehrmann@...
> Date: Sat, 4 Dec 2010 16:57:20 +0000
> Subject: [EMHL] Re: Triangle Constructions problems A, b, ;
>
>
>
>
> Dear Luis
>
> > Consider the problems A, b, ....
>
> > r + R
>
> > r + r_c
>
> > r_c - r
>
> > r_a + r_b
>
> > r_a - r_b
>
> > r_b - r_a
>
> >
>
> > Do they have a R&C construction?
>
>
>
> The answer is yes for all the problems.
>
> In any case, you have a quadratic equation to get r or ra.
>
> For instance :
>
> r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
>
> r^2+L.r-b.l.sin(A)/2=0 with L=rc-r
>
> ra^2-ra.L+b.L.sin(A)/2=0 with L=ra+rb
>
> ra^2-(L+b.sin(A))ra+L.b.sin(A)/2=0 with L=ra-rb
>
> The equation giving r with with L=r+R is a bit more complicated, but it exists
>
> It remains to get nice geometric constructions.
>
> Friendly.
>
> Jean-Pierre
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
• Dear Jean-Pierre Is it easy to explain how did you found this relation? (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0 Best regards Nikos Dergiades
Message 2 of 12 , Dec 7 2:33 PM
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Dear Jean-Pierre
Is it easy to explain
how did you found this relation?
(1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

Best regards

> Dear Nikolaos and Luis
> exactly as as Nikolas said, I started from the classical
> relations
> 1/ra+1/rb=1/r-1/rc=2/hc
> They give immediately the equations for r or ra
> In the case R+r or R-r,you can start from the more
> difficult relation
> (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
> which leads to a quadratic equation for r or R when you
> know A,b and R+r or R-r
> Friendly. Jean-Pierre
>
> > >The equation giving r with L=r+R is a bit ...
> > I forgot L=R-r. I will be back to both of them
> > later, as well as to a few others with R.
> >
> > >... more complicated, but it exists.
> > Do you mean the equations you gave below are
> >
> >
> > >It remains to get nice geometric constructions.
> > I have no idea on how to deal with such data
> > geometrically.
> >
> > So let's go back to the algebraic solution.
> >
> > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
> >
> > r^2+L.r-b.L.sin(A)/2=0 with L=rc-r
> >
> >
> > I have performed numerical tests with a=5, b=7 c=8
> > and the above equations are correct. I have now two
> > questions:
> >
> > 1) could you give me a hint on how to deduce them?
> >
> > I assume the others with ra, rb will follow the same
> way.
> >
> > 2) the equation with L=r+rc has 2 positive roots.
> > Is it possible to have data that give two solutions?
> > I mean two different triangles?
> >
> > Friendly,
> > Luis
> >
> >
> >
> > To: Hyacinthos@yahoogroups.com
> > From: jean-pierre.ehrmann@...
> > Date: Sat, 4 Dec 2010 16:57:20 +0000
> > Subject: [EMHL] Re: Triangle Constructions problems A,
> b, ;
> >
> >
> >
> >
> > Dear Luis
> >
> > > Consider the problems A, b, ....
> >
> > > r + R
> >
> > > r + r_c
> >
> > > r_c - r
> >
> > > r_a + r_b
> >
> > > r_a - r_b
> >
> > > r_b - r_a
> >
> > >
> >
> > > Do they have a R&C construction?
> >
> >
> >
> > The answer is yes for all the problems.
> >
> > In any case, you have a quadratic equation to get r or
> ra.
> >
> > For instance :
> >
> > r^2-(L+b.sin(A))r+b.L.sin(A)/2=0 with L=r+rc
> >
> > r^2+L.r-b.l.sin(A)/2=0 with L=rc-r
> >
> > ra^2-ra.L+b.L.sin(A)/2=0 with L=ra+rb
> >
> > ra^2-(L+b.sin(A))ra+L.b.sin(A)/2=0 with L=ra-rb
> >
> > The equation giving r with with L=r+R is a bit more
> complicated, but it exists
> >
> > It remains to get nice geometric constructions.
> >
> > Friendly.
> >
> > Jean-Pierre
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
>
>
>
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• [ND] ... Dear Nikos, There are two different questions about this relation. 1. How JPE has constructed it ? 2. Once we have it, how to prove it? Let s try to
Message 3 of 12 , Dec 7 4:56 PM
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[ND]
> Dear Jean-Pierre
> Is it easy to explain
> how did you found this relation?
> (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

Dear Nikos,

relation.
1. How JPE has constructed it ?
2. Once we have it, how to prove it?

that is, to prove that:

(1-cos A)((hc-2r)R-b^2/2) = r(r-hc)

We have:

First member:

1-cosA = 2sin^2(A/2)

hc-2r = 2RsinAsinB - 8Rsin(A/2)sin(B/2)sin(C/2) =

= 8Rsin(A/2)cos(A/2)sin(B/2)cos(B/2) - 8Rsin(A/2)sin(B/2)sin(C/2)=

= 8Rsin(A/2)sin(B/2)[cos(A/2)cos(B/2) - sin(C/2)] =

= 8Rsin^2(A/2)sin^2(B/2)

(hc-2r)R - b^2/2 = 8R^2sin^2(A/2)sin^2(B/2) -
- 8R^2sin^2(B/2)cos^2(B/2)

= 8R^2sin^2(B/2)[sin^2(A/2) - cos^2(B/2)]

===>

(1-cos A)((hc-2r)R-b^2/2) = 16R^2sin^2(A/2)sin^2(B/2)*
[sin^2(A/2) - cos^2(B/2)] <1>

Second member:

r(r-hc)= 4Rsin(A/2)sin(B/2)sin(C/2)[4Rsin(A/2)sin(B/2sin(C/2) - 2RsinAsinB] =

4R^2sin(A/2)sin(B/2)sin(C/2)[4sin(A/2)sin(B/2)sin(C/2) -
- 8sin(A/2)cos(A/2)sin(B/2)cos(B/2)]=
= 16R^2sin^2(A/2)sin^2(B/2)sin(C/2)[sin(C/2) - 2cos(A/2)cos(B/2)] <2>

So we have to prove that <1> = <2>

or

sin^2(A/2) - cos^2(B/2) - sin^2(C/2)= - 2cos(A/2)cos(B/2)sin(C/2)

or

sin^2(A/2)+sin^2(B/2)-sin^2(C/2)= 1 - 2cos(A/2)cos(B/2)sin(C/2)

which is true.

APH
• Dear Nikolaos and Antreas [ND] ... I m not very proud of my way to get this : As hc/r = (a+b+c)/c, we have c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c I ve tried
Message 4 of 12 , Dec 8 12:50 AM
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Dear Nikolaos and Antreas
[ND]
> Is it easy to explain
> how did you found this relation?
> (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0

I'm not very proud of my way to get this :
As hc/r = (a+b+c)/c, we have
c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
I've tried to combine these relations to get b and used Maple to do that. As Antreas said, the result is easy to prove but more difficult to find.
The key was
b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r = a.b.c/2
a(a+b-c) = a.c.(hc-2.r)/r
As r/R/(1-cos A) = (b+c-a)/a, we have
(b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)

Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos A)
which is the desired relation.
Friendly. Jean-Pierre
b^2/(2.r.R) = (hc-2.r)/r +
• Dear Jean-Pierre Very good. Thank you. Your key relation is (a+b+c).r = c.hc (1) Because from a = 2Rsin(A) = 2R.hc/b and from (1) we get c = (2Rhc + b^2)r /
Message 5 of 12 , Dec 8 1:34 PM
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Dear Jean-Pierre
Very good. Thank you.
Your key relation is (a+b+c).r = c.hc (1)
Because from a = 2Rsin(A) = 2R.hc/b and from (1)
we get c = (2Rhc + b^2)r / (b(hc-r))
and substituting a, c to your second relation
R(1-cosA)(b+c-a) = ar (2)
we get your final relation (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0.
Instead of (2) we can substitute a, c to
cosA = (bb + cc -aa)/(2bc).

Best regards

>
> Dear Nikolaos and Antreas
> [ND]
> > Is it easy to explain
> > how did you found this relation?
> > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
>
> I'm not very proud of my way to get this :
> As hc/r = (a+b+c)/c, we have
> c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
> I've tried to combine these relations to get b and used
> Maple to do that. As Antreas said, the result is easy to
> prove but more difficult to find.
> The key was
> b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
> Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r =
> a.b.c/2
> a(a+b-c) = a.c.(hc-2.r)/r
> As r/R/(1-cos A) = (b+c-a)/a, we have
> (b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)
>
> Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos
> A)
> which is the desired relation.
> Friendly. Jean-Pierre
> b^2/(2.r.R) = (hc-2.r)/r +
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear Hyacinthists, Nikos, Jean-Pierre, Thank you very much. Nice solutions. So TCs such as A,b,R+r_a(b)(c) can be solved in a similar way. For R+r_a one has
Message 6 of 12 , Dec 14 9:39 AM
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Dear Hyacinthists, Nikos, Jean-Pierre,

Thank you very much. Nice solutions.

So TCs such as A,b,R+r_a(b)(c) can be solved in a similar way.
For R+r_a one has

(1-cos A)((h_c-2r_a)R+b^2/2)-r_a(h_c-r_a)=0.

TC given A, a, r_a+r_b+r_c has an easy construction.
How about A, b, r_a+r_b+r_c ?

As always, thank you very much for your help.

Luis

> To: Hyacinthos@yahoogroups.com
> Date: Wed, 8 Dec 2010 21:34:09 +0000
> Subject: Re: [EMHL] Re: Triangle Constructions problems A, b, ;
>
> Dear Jean-Pierre
> Very good. Thank you.
> Your key relation is (a+b+c).r = c.hc (1)
> Because from a = 2Rsin(A) = 2R.hc/b and from (1)
> we get c = (2Rhc + b^2)r / (b(hc-r))
> and substituting a, c to your second relation
> R(1-cosA)(b+c-a) = ar (2)
> we get your final relation (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0.
> Instead of (2) we can substitute a, c to
> cosA = (bb + cc -aa)/(2bc).
>
> Best regards
>
> >
> > Dear Nikolaos and Antreas
> > [ND]
> > > Is it easy to explain
> > > how did you found this relation?
> > > (1-cos A)((hc-2r)R-b^2/2)+r(hc-r)=0
> >
> > I'm not very proud of my way to get this :
> > As hc/r = (a+b+c)/c, we have
> > c.(hc-r)/r = a+b and c.(hc-2r)/r = a+b-c
> > I've tried to combine these relations to get b and used
> > Maple to do that. As Antreas said, the result is easy to
> > prove but more difficult to find.
> > The key was
> > b(a+b+c)=a(a+b-c)+(b+c-a)(a+b) (1)
> > Now b(a+b+c)=(a.c.b^2)/(2.r.R) because (a+b+c)R.r =
> > a.b.c/2
> > a(a+b-c) = a.c.(hc-2.r)/r
> > As r/R/(1-cos A) = (b+c-a)/a, we have
> > (b+c-a)(a+b) = a.c(hc-r)/R/(1-cos A)
> >
> > Thus (1) gives b^2/(2.r.R) = (hc-2.r)/r + (hc-r)/R/(1-cos
> > A)
> > which is the desired relation.
> > Friendly. Jean-Pierre
> > b^2/(2.r.R) = (hc-2.r)/r +
> >
> >
> >
> > ------------------------------------
> >
> >
> >
> > Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
> >
>
>
>
>
> ------------------------------------
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Luis, ... We know that r_a+r_b+r_c = 4R + r. Hence we have a discussed case. Best regards Nikos Dergiades
Message 7 of 12 , Dec 14 10:12 AM
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Dear Luis,

> How about A, b, r_a+r_b+r_c ?

We know that r_a+r_b+r_c = 4R + r.
Hence we have a discussed case.
Best regards