Let $ a = BC = \sqrt {13},\ b = CA = \sqrt {19},\ c = AB = \sqrt 7$ be the triangle sides opposite to the vertices A, B, C. Let F be the 1st Fermat point of the triangle $ \triangle ABC$ and x = AF, y = BF, z = CF. The angles $ \angle AFB = \angle BFC = \angle CFA = 120^o$ are all equal. Using the cosine theorem for the triangles $ \triangle AFB, \triangle BFC, \triangle CFA$ with the sides (x, y, c); (y, z, a); (z, x, b) and the angles $ \phi = \widehat{(x, y)} = \widehat{(y, z)} = \widehat{(z, x)} = 120^o$, $ \cos \phi = - \frac 1 2$, $ \triangle AFB: \ \ x^2 + y^2 + xy = c^2 = 7$ $ \triangle BFC: \ \ y^2 + z^2 + yz = a^2 = 13$ $ \triangle CFA: \ \ z^2 + x^2 + zx = b^2 = 19$ Subtracting the 1st equation from the 2nd one, 2nd from 3rd and 3rd from 1st, $ z^2 - x^2 + yz - xy = (z - x)(x + y + z) = a^2 - c^2 = 6$ $ x^2 - y^2 + zx - yz = (x - y)(x + y + z) = c^2 - b^2 = - 12$ $ y^2 - z^2 + xy - zx = (y - z)(x + y + z) = b^2 - a^2 = 6$ Dividing any pair of equations always yields nothing else but z = 2x - y. Substistuting z into the original set of equations, $ x^2 + y^2 + xy = 7$ $ 4x^2 + y^2 - 2xy = 13$ $ 7x^2 + y^2 - 5xy = 19$ The 3rd equation is a linear combination of the 1st and 2nd, hence, we can disregard it: $ (3) = 2 \times (2) - (1)$. Substracting the 1st equation from the 2nd one, in order to eliminate at least one square, we get $ 3x^2 - 3xy = 6,\ \ y = x - \frac 2 x$ Substituting y into the 1st equation, $ x^2 + x^2 - 4 + \frac {4}{x^2} + x^2 - 2 = 7$ $ 3x^4 - 13x^2 + 4 = 0$ $ x^2 = \frac {13 \pm \sqrt {169 - 48}}{6} = \frac {13 \pm 11}{6} = 4$ . or . $ \frac 1 3$ Since we are not interested in negative roots, x = 2 or $ \frac {\sqrt 3}{3}$. Calculating y, z from the previous equations $ y = x - \frac 2 x = 1$ . or . $ - \frac {5 \sqrt 3}{3}$ $ z = 2x - y = 3$ . or . $ \frac {7 \sqrt 3}{3}$ Since we are interested only in the solution with x > 0, y > 0, z > 0, the distances of the 1st Fermat point from the vertices A, B, C are x = AF = 2, y = BF = 1, z = CF = 3 and x : y : z = 2 : 1 : 3 On the other hand, radii of the circles $ (A), (B), (C)$ are in the ratios $ r_A : r_B : r_C = \frac 1 3 : \frac 2 3 : 1 = 1 : 2 : 3$. I have solved the problem before and it came with exactly the same error - it should be $ r_A = \frac 2 3, r_B = \frac 1 3,\ r_C = 1$, which implies $ r_A : r_B : r_C = 2 : 1 : 3$. If this was the case, then $ \frac {r_A}{x} = \frac {r_B}{y} = \frac {r_C}{z} = \frac 1 3$ i.e., the circles $ (A), (B), (C)$ would be seen from the 1st Fermat point F under the same angle $ 2 \theta = 4 \arcsin{\frac {r_A}{2x}} = 4 \arcsin{\frac 1 6}$. If the triangle $ \triangle ABC$ was then rotated by the angle $ \pm \theta = \pm 2 \arcsin{\frac 1 6}$ around the point F into a congruent riangle $ \triangle A'B'C'$, the vertices $ A', B', C'$ would simultaneously fall on the circles $ (A), (B), (C)$. Alternately, the problem could be cured by keeping the radii of the circles $ (A), (B), (C)$ as $ r_A = \frac 1 3,\ r_B = \frac 2 3,\ r_c = 1$, but specifying the triangle side lengths as $ a = BC = \sqrt {19},\ b = CA = \sqrt {13},\ c = AB = \sqrt 7$. Then we would get x = AF = 1, y = BF = 2, z = CF = 3 and $ x : y : z = 1 : 2 : 3 = r_A : r_B : r_C$. EDIT: See http://www.mathlinks.ro/viewtopic.php?t=194847 for the correct solution.

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