Use calculus, get thrilled !!

Note: My approach is unconventional & doesn't use calculus at all.

Symmetry tells us that the perimeter of an isosceles triangle inscribing a given circle will be minimum only when the isosceles triangle is equilateral

Knowing this, the problem does not remain a difficult one.

Consider an equilateral triangle $ABC$ with side length $a$ inscribing a circle of radius $√3$ with centre O. (see the picture given in the problem). $Ar(ABC)=3\space times \space Ar(OAC)=\frac{3√3a}{2}$ Now $\frac{√3a^2}{4}=\frac{3√3a}{2}$ $\Rightarrow a=6$ Hence perimeter is $=3a=3•6=\boxed{18}$

Well, here goes some calculus .....

Suppose the circle is centered at $C(0, \sqrt{3})$ and the base of the isosceles triangle is the line segment joining $A(-a, 0)$ and $B(a, 0)$ , where $a \gt \sqrt{3}$ . The triangle then lies above the $x$ -axis and is symmetric about the $y$ -axis. Label the origin as $O(0,0)$ .

With $\angle OBC = \theta$ , we have that $\theta = \arctan(\frac{\sqrt{3}}{a})$ and that the length of the side of the triangle in the first quadrant is $L = a*\sec(2\theta)$ . The perimeter $P$ of the triangle is then $P = 2a + 2L = 2a(1 + \sec(2\theta))$ .

Now $\cos(2\theta) = 2*\cos^{2}(\theta) - 1$ and

$\cos(\theta) = \cos(\arctan(\frac{\sqrt{3}}{a})) = \frac{a}{\sqrt{a^{2} + 3}}$ .

So $\cos(2\theta) = 2*(\frac{a^{2}}{a^{2} + 3}) - 1 = \frac{a^{2} - 3}{a^{2} + 3}$ ,

and thus $P = 2a(1 + \frac{a^{2} + 3}{a^{2} - 3}) = \dfrac{4a^{3}}{a^{2} - 3}$ .

To find the critical points for $P$ , we set $\frac{dP}{da} = 0$ :

$\dfrac{dP}{da} = \dfrac{12a^{2}(a^{2} - 3) - 4a^{3}(2a)}{(a^{2} - 3)^{2}} = \dfrac{4a^{2}(a^{2} - 9)}{(a^{2} - 3)^{2}} = 0$

when $a = 3$ , as we want $a \gt \sqrt{3}$ . This yields a perimeter of $P = \frac{4*27}{9 - 3} = \boxed{18}$ .

To be thorough, note that $\frac{d^{2}P}{da^{2}} = \frac{24a(a^{2} + 9)}{(a^{2} - 3)^{3}} \gt 0$ for $a = 3$ , so by the second derivative test we can (formally) conclude that the perimeter $P$ is minimized when $a = 3$ .

Using the inequality of arithmetic and geometric means, we have that:

$\frac{(p-a)+(p-b)+(p-c)}{3}\geqslant \sqrt[3]{(p-a)(p-b)(p-c)}$

But the area of the triangle is :

$S=pr=\sqrt{p(p-a)(p-b)(p-c)} \therefore (p-a)(p-b)(p-c)=pr^2$

Replacing the therms :

$\frac{(p-a)+(p-b)+(p-c)}{3} \geqslant \sqrt[3]{pr^2}$

But a+b+c=2p, them :

$\frac{(3p-2p)}{3} \geqslant \sqrt[3]{pr^2}$

$\frac{p^3}{27} \geqslant {pr^2}\Rightarrow p\geqslant 9$

Therefore, we have that the minimum semiperimeter is p=9, them, the perimeter is 2p = 18

I believe that in order to get the minimum perimeter circumscribing a circle with a given radius, that isosceles triangle must be an equilateral one, so i get the height of the triangle, which is equal to thrice the given radius, multiplied it by 2, divide it by sqrt of 3, then multiply by 3 again to get the perimeter of the triangle..

Let the side of the triangle a,b,and b, the perimeter p=2b+a The area of the triangle A = ½ (a+2b)√3 = √((b+a/2)(b-a/2)(a/2)(a/2)). Square both sides, 3(b+a/2) =1/4 a^2(b-a/2), or b=1/2 a (a^2+12)/(a^2-12) The perimeter p = (2 a^3)/(a^2-12), dp/da = (2 a^4-72 a^2)/((a^2-12))^2 = 0, a=6, and b=6, so p=18

Consider an isosceles triangle ABC, with AB = AC, angle BAC = t and in radius r. Here r is a constant (given) and t is variable. The question is what is the perimeter of ABC in terms of t and r. We can then minimize the perimeter by differentiating over t.

To calculate the perimeter, let the in center be I and let D,E,F, be perpendiculars from I to BC, CA, AB, respectively. Since AI bisects angle BAC, r/AE = tan(t/2), so AE = r cot(t/2). Similarly, AF = r cot(t/2). Next, the angle ABC is (pi - t)/2, so r/BD = tan ((pi - t)/4). Hence, BD = r cot((pi - t)/4). Also, BD = FB = DC = CE, so all these four segments have length r cot((pi - t)/4).

Hence, the total perimeter of the triangle is:

f(t) = 2r cot(t/2) + 4r cot ((pi - t)/4)

We want to minimize this perimeter. First, we simplify f(t) slightly. Let t/4 = x. Note that 0 <= x <= pi/4. Given this, f(t) can be written as:

g(x) = 2r cot (2x) + 4r cot (pi/4 - x)

We have g'(x) = -4r csc^2 (2x) + 4r csc^2 (pi/4 - x) = 0 => csc^2 (2x) = csc^2(pi/4 - x) => sin^2 (2x) = sin^2 (pi/4 - x)

Since, 0 <= x <= pi/4, sin(2x) and sin(pi/4 - x) are both non-negative.

So, we get sin(2x) = sin (pi/4 - x), and again because 0 <= x <= pi/4, 2x = pi/4 - x.

So x = pi/12 and t = 4x = pi/3. Hence, the triangle is equilateral. The minimum perimeter is: 2r cot(t/2) + 4r cot ((pi - t)/4) = 2r cot(pi/6) + 4r cot (pi/6) = 6r * sqrt(3) Therefore, the answer when r = sqrt(3) is 6(sqrt(3)) * sqrt(3) =18 answer!