Making a point

As has been shown elsewhere , the First Fermat point (or Steiner point) $P$ can be found by constructing an equilateral triangle $BCQ$ . Then $P$ is the intersection of the line $AQ$ with the circumcircle of $BCQ$ . Moreover, $AP + BP + CP = AQ$ .

If we set up a coordinate system with origin at $B$ , then $A$ has coordinates $(0,3)$ , $B$ has coordinates $(0,0)$ , $C$ has coordinates $(4,0)$ , while $Q$ has coordinates $(2,-2\sqrt{3})$ . Thus $AQ^2 \; = \; 2^2 + \big(3 + 2\sqrt{3}\big)^2 \; = \; 25 + 12\sqrt{3}$ making the answer $25 + 12 + 3 = \boxed{40}$ .

Such a point $P$ is known as the first Fermat point , and is the interior point such that

$\angle APB = \angle BPC = \angle APC = 120^{\circ}$ .

Letting $|AP| = x, |BP| = y$ and $|CP| = z$ , we have by the cosine law that

$9 = x^{2} + y^{2} + xy, \space \space 16 = y^{2} + z^{2} + yz, \space \space 25 = x^{2} + z^{2} + xz$

Adding these three equations together yields that $50 = 2(x^{2} + y^{2} + z^{2}) + (xy + yz + xz)$ .

Now $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + xz)$ , so upon substitution we have that

$50 = 2((x + y + z)^{2} - 2(xy + yz + xz)) + (xy + yz + xz) \Longrightarrow$

$(x + y + z)^{2} = 25 + \dfrac{3}{2}(xy + yz + xz)$ .

Next, note that the area of the triangle, being right-angled, is equal to $\dfrac{1}{2} \times 3 \times 4 = 6$ , but that it also equals

$\dfrac{1}{2}xy\sin(120^{\circ}) + \dfrac{1}{2}yz\sin(120^{\circ}) + \dfrac{1}{2}xz\sin(120^{\circ}) = \dfrac{\sqrt{3}}{4}(xy + yz + xz) \Longrightarrow$

$\dfrac{\sqrt{3}}{4}(xy + yz + xz) = 6 \Longrightarrow xy + yz + xz = \dfrac{24}{\sqrt{3}} = 8\sqrt{3} \Longrightarrow$

$(x + y + z)^{2} = 25 + \dfrac{3}{2} \times 8\sqrt{3} = 25 + 12\sqrt{3} \Longrightarrow x + y + z = \sqrt{25 + 12\sqrt{3}}$ .

Thus $a + b + c = 25 + 12 + 3 = \boxed{40}$ .