- Dear Luis,

> e_b := B external bisector

I think that it is not possible to construct 2

>

> 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?

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

Nikos Dergiades

___________________________________________________________

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - 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.

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[Non-text portions of this message have been removed] - Dear Luis,

> >Let AB < BC. Let AB > AC.

Sorry.

> Do you confirm the letters on these inequalities?

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

Best regards

Nikos Dergiades

___________________________________________________________

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - 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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[Non-text portions of this message have been removed] - 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

Nikos Dergiades

> 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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