## Κυριακή, 16 Ιανουαρίου 2011

### TRIANGLE RESOLUTION A, a+b-c, R+r

Problem:
To resolve triangle ABC if are given A, a+b-c, R+r
See Hyacinthos Message #19749

Resolution:

We have:

sinA + sinB - sinC = 4sin(A/2)sin(B/2)cos(C/2)

r = 4Rsin(A/2)sin(B/2)sin(C/2)

==>

a+b-c = 2R(sinA+sinB-sinC) = 8Rsin(A/2)sin(B/2)cos(C/2)

R+r = R(1+4sin(A/2)sin(B/2)sin(C/2))

==>

t := (R+r)/(a+b-c) =

= (1+4sin(A/2)sin(B/2)sin(C/2))/8sin(A/2)sin(B/2)cos(C/2)

==>

4sin(A/2)sin(B/2)[sin(C/2) - 2tcos(C/2)] + 1 = 0

and since sin(B/2) = sin(90 - ((C+A)/2)) = cos((C+A)/2) =

= cos(C/2)cos(A/2) - sin(C/2)sin(A/2), ==>

4sin(A/2)[cos(A/2) + 2tsin(A/2)]sin(C/2)cos(C/2) - 4(sin(A/2)^2(sin(C/2))^2 - 8tsin(A/2)cos(A/2)(cos(A/2)^2 + 1 = 0.

We have the system of equations:

fx^2 + gy^2 + hxy + 1 = 0

x^2 + y^2 = 1

where x, y stand for the unknown sin(C/2),cos(C/2) and f,g,h are known coefficients. From these equations we get the equation:

Lx^4 + Mx^2 + N = 0, where L,M,N are known coefficients.

This equation has constructible roots [Read THIS], therefore the problem has a constructible Euclidean solution (ie by ruler and compass)