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## (A,e_b,r_c)

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• Dear Hyacinthists, e_b := B external bisector One can construct (Euclides) 2 triangles given (A,e_b,r_a). But how about (A,e_b,r_c)? I wasn t able to construct
Message 1 of 6 , Jan 15, 2009
Dear Hyacinthists,

e_b := B external bisector

One can construct (Euclides) 2 triangles given
(A,e_b,r_a). But how about (A,e_b,r_c)?
I wasn't able to construct more than one.
Are there data where one can construct 2
triangles?
Best regards,
Luis

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• Dear Luis, ... I think that it is not possible to construct 2 different such triangles. Let AB
Message 2 of 6 , Jan 16, 2009
Dear Luis,

> e_b := B external bisector
>
> One can construct (Euclides) 2 triangles given
> (A,e_b,r_a). But how about (A,e_b,r_c)?
> I wasn't able to construct more than one.
> Are there data where one can construct 2
> triangles?

I think that it is not possible to construct 2
different such triangles.
Let AB < BC
If D is the point where the B external bisector meets the
line AC then the excenter Ic lies on BD and in triangle
ABD we know the side BD = e_b, the opposite angle BAD = 180-A
and the internal bisector AIc = r_c/cos(A/2).
We know how to construct this unique triangle BAD
and then ABC.
Let AB > AC.
Now the new construction is symmetric to the previous
relative to the perpencicular bisector of AC and hence it
leads to the same triangle ABC as the previous construction.
Hence we have only one solution provided that e_b > 2r_c.
I hope that these thoughts are correct.

Best regards

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• Dear Hyacinthists, Nikos, Thank you for this analysis. ... Do you confirm the letters on these inequalities? My reasoning was to start with A acute (20
Message 3 of 6 , Jan 16, 2009
Dear Hyacinthists, Nikos,

Thank you for this analysis.

>Let AB < BC. Let AB > AC.
Do you confirm the letters on these inequalities?

My reasoning was to start with A acute (20<A<40),
and a fix Ic on the A external bisector. Then moving
C on the ray (A,C) (further away from A to closer)
one makes B range from an obtuse to an acute angle.
One does that keeping in mind that CIc is the internal
bisector of <ACB.

In the process your point D moves from DAC to ACD
and in this transition I hoped to construct two different
triangles. One with obtuse B and the other with acute B.

Best regards,
Luis

PS. I keep the reply to this message in its entirety this
time only.

http://www.microsoft.com/windows/windowslive/

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• Dear Luis, ... Sorry. The correct is. ... Best regards Nikos Dergiades ___________________________________________________________ Χρησιμοποιείτε
Message 4 of 6 , Jan 16, 2009
Dear Luis,

> >Let AB < BC. Let AB > AC.
> Do you confirm the letters on these inequalities?

Sorry.
The correct is.
>Let AB < BC. Let AB > BC.

Best regards

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• Dear Hyacinthists, Niko, Francisco Capitán, Francisco Capitán was able to find two triangles with A=45^ circ, e_b=6, r_c=1. The values of {a,b,c} with this
Message 5 of 6 , Jan 16, 2009
Dear Hyacinthists, Niko, Francisco Capitán,

Francisco Capitán was able to find two triangles
with A=45^\circ, e_b=6, r_c=1.
The values of {a,b,c} with this data are {5.2818109246, 6.5540237833, 2.10063998345} and
{0.94717120728,1.28858190882,1.16983782628}Algebraically, by eliminating a and b one gets p(c)
in fonction of the data:
c^8 (1 - 2 cosA^2 + cosA^4) +
c^6 (-2 eb^2 + 4 cosA^2 eb^2 - 2 cosA^4 eb^2 + 4 cosA rc^2 - 4 cosA^2 rc^2 - 4 cosA^3 rc^2 + 4 cosA^4 rc^2) + c^4 (eb^4 - 2 cosA^2 eb^4 + cosA^4 eb^4 + 2 eb^2 rc^2 - 4 cosA eb^2 rc^2 - 2 cosA^2 eb^2 rc^2 + 4 cosA^3 eb^2 rc^2 + 4 rc^4 - 8 cosA rc^4 + 4 cosA^2 rc^4) + c^2 (-2 eb^4 rc^2 + 2 cosA^2 eb^4 rc^2 - 4 eb^2 rc^4 + 4 cosA eb^2 rc^4) + eb^4 rc^4Maybe a good problem for Crux, AMM etc would be to
determine conditions to have 0,1,2 solutions.

Best regards,
Luis
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• Dear Luis, Francisco has given the stimulus to find the following result. In case BA
Message 6 of 6 , Jan 19, 2009
Dear Luis,
Francisco has given the stimulus to find
the following result.
In case BA < BC
If 2.r_c < e_b.sin(A/2) and (cosA - 1)r_c < e_b.cosA
then a solution is given by

f = atan((e_b/r_c).cot(A/2)) (atan is inverse function of tan)
B = 2.asin(cos(A/2).cot(f/2)) - A (asin inverse of sin)
c = -e_b.cos(A + B/2)/sinA
b = c.sinB/sin(A + B)
a = b.sinA/sinB

In case BA > BC then A must be acute
if (1 - cosA).r_c < e_b.cosA then
a second solution is given by

f = atan((e_b/r_c).cot(A/2)) (as previously)
B = 2.asin(cos(A/2).tan(f/2)) - A (different from previous)
c = e_b.cos(A + B/2)/sinA (differ in sign)
b = c.sinB/sin(A + B)
a = b.sinA/sinB

In some inequalities I did n't used the equality sign
in order to get sure results due to computer
approximation by truncation.

Hence if A = pi/2 we have never the second solution.
If A = pi/2, e_b < 2.srt(2).r_c we have 0 solutions
If A = pi/2, e_b > 2.srt(2).r_c we have 1 solution.

Best regards

> Dear Hyacinthists, Niko, Francisco Capitán,
>
> Francisco Capitán was able to find two triangles
> with A=45^\circ, e_b=6, r_c=1.
> The values of {a,b,c} with this data are {5.2818109246,
> 6.5540237833, 2.10063998345} and
> {0.94717120728,1.28858190882,1.16983782628}Algebraically,
> by eliminating a and b one gets p(c)
> in fonction of the data:
> c^8 (1 - 2 cosA^2 + cosA^4) +
> c^6 (-2 eb^2 + 4 cosA^2 eb^2 - 2 cosA^4 eb^2 + 4 cosA rc^2
> - 4 cosA^2 rc^2 - 4 cosA^3 rc^2 + 4 cosA^4 rc^2) + c^4
> (eb^4 - 2 cosA^2 eb^4 + cosA^4 eb^4 + 2 eb^2 rc^2 - 4
> cosA eb^2 rc^2 - 2 cosA^2 eb^2 rc^2 + 4 cosA^3 eb^2 rc^2 +
> 4 rc^4 - 8 cosA rc^4 + 4 cosA^2 rc^4) + c^2 (-2 eb^4 rc^2
> + 2 cosA^2 eb^4 rc^2 - 4 eb^2 rc^4 + 4 cosA eb^2 rc^4)
> + eb^4 rc^4Maybe a good problem for Crux, AMM etc would be
> to
> determine conditions to have 0,1,2 solutions.
>
> Best regards,
> Luis

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