Algebra with Geometrical Figure

Putting $x = 2\cos\theta$ and $y= 2\sin\theta$ , we want to minimize $F(\theta) \; = \; 3\sqrt{5 - 4\cos\theta} + \sqrt{13 - 12\sin\theta}$ Now $F'(\theta) \; = \; \frac{6\sin\theta}{\sqrt{5 - 4\cos\theta}} - \frac{6\cos\theta}{\sqrt{13-12\sin\theta}}$ and hence $F'(\theta) = 0$ when $\sin\theta\sqrt{13 - 12\sin\theta} \; = \; \cos\theta\sqrt{5 - 4\cos\theta}$ Squaring, and putting $t = \tan\tfrac12\theta$ , we deduce that $\begin{aligned} 4t^2\big(13(1 + t^2) - 24t\big) & = \; (1 - t^2)^2\big(5(1+t^2) - 4(1-t^2)\big) \\ 4t^2(13 - 24t + 13t^2) & = \; (t^2-1)^2(1 + 9t^2) \\ 9t^6 - 69t^4 + 96t^3 - 45t^2 - 1 & = \; 0 \\ (3t^2 - 6t +1)(3t^4 + 6t^3 - 12t^2 + 6t + 1) & = \; 0 \end{aligned}$ There are four real (and two complex) solutions of this equation, but not all of these are valid solutions of the original problem (we squared the equation to remove the square roots). Since $\sin\theta$ and $\cos\theta$ have to have the same sign, we deduce that either $0 < t < 1$ or $t < -1$ . Only two of the roots of the equation now apply. One is $t_0 = \tfrac13(3-\sqrt{6})$ (a zero of the quadratic factor), while the other is a root of the quartic, with $t_1 \approx -3.3592$ .

Calculation tells us that the root $t_1$ corresponds to the maximum value of $F$ , while $t_0$ corresponds to the minimum value of $F$ . Thus the minimum value is obtained when $\theta$ is acute and $\sin\theta = \tfrac{1}{10}(6 - \sqrt{6})$ , so $\cos\theta = \tfrac{1}{10}(1 + 3\sqrt{6})$ , and hence $\sqrt{5 - 4\cos\theta} \; = \; \frac{6 - \sqrt{6}}{\sqrt{10}} \hspace{2cm} \sqrt{13 - 12\sin\theta} \; =\; \frac{2 + 3\sqrt{6}}{\sqrt{10}}$ and hence the minimum value of $F$ is $3 \frac{6 - \sqrt{6}}{\sqrt{10}} + \frac{2 + 3\sqrt{6}}{\sqrt{10}} \; = \; \boxed{2\sqrt{10}}$

---

Doing the problem geometrically, we need to use Ptolemy's Inequality and Theorem . Let $A = (1,0)$ , $B = (0,0)$ , $C = (0,3)$ , and let $P$ be any point on the circle $x^2 + y^2 = 4$ . Then Ptolemy's Inequality tells us that $3AP + CP \; = \; BC \times AP + AB \times CP \; \ge \; AC \times BP \; = \; 2\sqrt{10}$ while Ptolemy's theorem tells us that $3AP + CP \; = \; AC \times BP \; = \; 2\sqrt{10}$ whenever $ABCP$ is a convex cyclic quadrilateral. As the diagram shows, this is possible. Since $AP \; = \; \sqrt{(x-1)^2 + y^2} \; = \; \sqrt{5-2x} \hspace{1cm} CP \; = \; \sqrt{x^2 + (y-3)^2} \; = \; \sqrt{13-6y}$ when $P$ has coordinates $(x,y)$ we obtain the minimum value of $\boxed{2\sqrt{10}}$ .

@Wenzhen Liu

let the distance of any general point on the circle from (1,0) be d1 and (0,3) be d2. Basically it is asking to minimize the distance of any point on the circle from from the points (1,0) and (0,3) such that 3d1 +d2 = min. now this can be attained for that point on the circle which completes the cyclic Quadrilateral formed from (0,0) (0,3), (1,0) and so using Ptolemy theorem u can easily evaluate 3d1 +d2 = 2root10