Is it small enough?

The equilateral triangle that contains the two circles also contains the smallest (in area) convex shape that encloses them. In order to minimise this shape, the circles have to be tangent to each other.

Next, we prove that the smallest equilateral triangle that contains the two circles is the one that has a side tangent to both circles and each circle is tangent to two sides of the triangle (as can be seen in the animation below).

Proof
First we put one circle (e.g. the small one, but the argument is irrelevant to this choice) tangent to sides $AB$ and $AC$ . Then its center $D$ lies on the angle bisector and altitude $AF$ of $\triangle ABC$ .

Obviously, the second circle must be tangent to side $BC$ . Then, the smallest equilateral triangle occurs when $AF$ is minimised.

Let $\angle GDE=\theta$ , $GE\bot AF$ , $DK\bot AB$ and $GH\bot BC$ , where $G$ is the center of the second circle (figure 1).
Then,

$\begin{aligned} AF & =AD+DE+EF \\ & =DK\cdot \csc 30{}^\circ +DG\cdot \cos \theta +GH \\ & =4\cdot 2+9\cdot \cos \theta +5 \\ & =13+9\cdot \cos \theta \\ \end{aligned}$ Hence, $AF$ gets minimised when the acute angle $\theta$ takes its maximum value, which evidently happens when the second circle is tangent either to side $AB$ , or side $AC$ . Q.E.D.

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Let $GL\bot AB$ and $DI\bot GL$ (figure 2)
By Pythagoras’ theorem on $\triangle DGI$ , $DI=\sqrt{D{{G}^{2}}-G{{I}^{2}}}=\sqrt{D{{G}^{2}}-{{\left( GL-IL \right)}^{2}}}=\sqrt{{{\left( 4+5 \right)}^{2}}-{{\left( 5-4 \right)}^{2}}}=4\sqrt{5}$ Consequently,

$\begin{aligned} AB & =AK+KL+LB \\ & =DK \cdot \cot 30{}^\circ +DI+GL \cdot \cot 30{}^\circ \\ & =4\sqrt{3}+4\sqrt{5}+5\sqrt{3} \\ & =9\sqrt{3}+4\sqrt{5} \\ \end{aligned}$ For the answer, $a=9$ , $b=3$ , $c=4$ , $d=5$ , (or $a=4$ , $b=5$ , $c=9$ , $d=3$ ). In any case, $a+b+c+d=\boxed{21}$ .

The smallest equilateral triangle to contain the two circles would be one that is tangent to the circles when the two circles are tangent to each other:

That means $AE$ bisects $\angle DAH$ so that $\angle EAH = \frac{1}{2} \angle DAH = \frac{1}{2} 60° = 30°$ and $AB = \frac{4}{\tan 30°} = 4\sqrt{3}$ , and $DG$ bisects $\angle ADH$ so that $\angle GDC = \frac{1}{2} \angle ADH = \frac{1}{2} 60° = 30°$ and $CD = \frac{5}{\tan 30°} = 5\sqrt{3}$ .

By the Pythagorean Theorem on $\triangle EFG$ , $EF = \sqrt{EG^2 - FG^2} = \sqrt{9^2 - 1^2} = 4\sqrt{5}$ .

Therefore, the side of the equilateral triangle is $AD = AB + BC + CD = 4\sqrt{3} + 4\sqrt{5} + 5\sqrt{3} = 9\sqrt{3} + 4\sqrt{5}$ , which means $a = 9$ , $b = 3$ , $c = 4$ , $d = 5$ , and $a + b + c + d = \boxed{21}$ .

The smallest circumscribing equilateral triangle is one with both circles tangent to onw of its sides (see proof below). Then the side of this smallest equilateral triangle is $a = 5\tan 60^\circ + \sqrt{9^2-1^2} + 4 \tan 60^\circ = 9 \sqrt 3 + 4\sqrt 5$ . Therefore $a+b+c+d = 9+3+4+5 = \boxed{21}$ .

Proof: Fix the center of the circle with radius $5$ at the origin $O(0,0)$ , a vertex of the triangle at $A(-5\sqrt 3,-5)$ such that a triangle side tangent to the circle $AB$ is along $y=-5$ and the other side $CA$ is at $60^\circ$ to $AB$ . Let the line joining the two centers $OQ$ makes an angle $\theta$ with the $x$ -axis. Then the equation of the circle with radius $4$ is $(x-9\cos \theta)^2 + (y-9\sin \theta)^2 = 4^2 \dots (1)$ . And the gradient at a point on the circle is given by $2(x-9\cos \theta) + 2(y-9\sin \theta) = 0$ . Let the third side $BC$ be tangent to the circle at $P(x_1, y_1)$ . Then the gradient at $P$ is $-\sqrt 3$ and we have:

$\begin{aligned} x_1-9\cos \theta -\sqrt 3 (y_1-9\sin \theta) & = 0 \\ \implies x_1-9\cos \theta & = \sqrt 3 (y_1-9\sin \theta) & \small \blue{\text{Substitute in (1)}} \\ 4(y_1-9\sin \theta)^2 & = 16 \\ \implies y_1 & = 2 + 9 \sin \theta \\ x_1 & = 2\sqrt 3 + 9\cos \theta \end{aligned}$

Then the equation of $BC$ is given by $\dfrac {y-y_1}{x-x_1} = - \sqrt 3$ . Let $B(x_2, -5)$ , then we have:

$\begin{aligned} \frac {-5-y_1}{x_2-x_1} & = - \sqrt 3 \\ 5+y_1 & = \sqrt 3 x_2 - \sqrt 3 x_1 \\ \implies x_2 & = \frac {5+\sqrt 3x_1+y_1}{\sqrt 3} = \frac {13+9\sqrt 3\cos \theta + 9\sin \theta}{\sqrt 3} = \frac {13+18\sin(\theta + 60^\circ)}{\sqrt 3} \end{aligned}$

Since the side length $a=BC = x_2 + 5\sqrt 3$ , $a$ is minimum, when $x_2$ is minimum or $\sin (\theta + 60^\circ)$ or $\theta$ is minimum. The smallest $\theta = - \sin^{-1}\frac 19$ . Then

$\begin{aligned} a_{\min} & = \frac {13+18\left(-\frac 19 \cdot \frac 12 + \frac {\sqrt{80}}9 \cdot \frac {\sqrt 3}2\right)}{\sqrt 3} + 5\sqrt 3 = 9\sqrt 3 + 4\sqrt 5 \end{aligned}$

$4 \sqrt 3 + \sqrt {(5+4)^2-(5-4)^2} + 5 \sqrt 3 = 9 \sqrt 3 + 4 \sqrt 5$