- 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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[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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