Equilaterally On Target

Label the triangle $\triangle PQR$ and let $O$ be the center of the circles, and then rotate $\triangle PQR$ $60°$ clockwise about $P$ to make $\triangle PQ'Q$ :

From the radii of the circles and the rotation, $OQ = O'Q' = 1$ , $OR = O'Q = 2$ , $OP = O'P = 3$ , and $OPO' = 60°$ .

Since $OP = O'P = 3$ and $OPO' = 60°$ , $\triangle OPO'$ is an equilateral triangle, so $OO' = 3$ and $\angle O'OP = \angle OO'P = 60°$ .

Since $OQ = 1$ and $O'Q = 2$ , $OQ + O'Q = 3 = OO'$ , so $Q$ is on $OO'$ . Therefore, $\angle QOP = \angle O'OP = 60°$ .

Using the law of cosines on $\triangle QOP$ , $PQ = \sqrt{OP^2 + OQ^2 - 2 \cdot OP \cdot OQ \cdot \cos 60°} = \sqrt{3^2 + 1^2 - 3 \cdot 1} = \sqrt{7}$ , which makes $a = \boxed{7}$ .

There is a formula which I posted some time ago. It works only for lengths which can be used to construct a triangle including degenerate case of a line segment, as is in this case. It would be impossible to construct equilateral triangle with circles of radii: 1,2,4 for example. The formula for this case works as below:

$x = \sqrt{(a^2 + b^2 + c^2-4 \sqrt{3} A)/2} = \sqrt{7}$

Let the coordinates of the vertices of the equilateral triangle be $\left(0, \frac {\sqrt 3a}2 \right)$ , $\left(\frac {\sqrt a}2, 0\right)$ , and $\left(-\frac {\sqrt a}2, 0\right)$ , and the coordinates of the center of the three concentric circles be $(x,y)$ . Then we can assume:

$\begin{cases} \left(x - 0\right)^2 + \left(y - \frac {\sqrt {3a}}2\right)^2 = 1^2 & ...(1) \\ \left(x - \frac {\sqrt a}2\right)^2 + \left(y - 0\right)^2 = 2^2 & ...(2) \\ \left(x + \frac {\sqrt a}2\right)^2 + \left(y - 0\right)^2 = 3^2 & ...(3) \end{cases}$

$\begin{cases} (3) - (2): & 2\sqrt a x = 5 & \implies x = \dfrac 5{2\sqrt a} \\ (3)+(2)-2(1) & 2\sqrt {3a} y - a = 11 & \implies y = \dfrac {11+a}{2\sqrt{3a}} \end{cases}$

Therefore $(2)$ becomes:

$\begin{aligned} \left(\frac 5{2\sqrt a} - \frac {\sqrt a}2\right)^2 + \left(\frac {11+a}{2\sqrt{3a}}\right)^2 & = 4 \\ \left(\frac {5-a}{2\sqrt a}\right)^2 + \left(\frac {11+a}{2\sqrt{3a}}\right)^2 & = 4 \\ 3a^2 - 30a + 75 + a^2 + 22a+121 & = 48a \\ a^2 - 14a + 49 & = 0 \\ (a-7)^2 & = 0 \\ \implies a & = \boxed 7 \end{aligned}$

Same start as in David Vreken 's solution (rotation by $60^\circ$ ).
Then I used Stewart's theorem on $\vartriangle PO'O$ :

$P{Q^2} = \frac{{P{O^2} \cdot O'Q + P{{O'}^2} \cdot QO}}{{O'O}} - O'Q \cdot QO = \frac{{{3^2} \cdot 2 + {3^2} \cdot 1}}{3} - 2 \cdot 1 = 7$

Hence, $PQ = \sqrt 7 \Rightarrow a = \boxed{7}$

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Ps. Nice variation of A Triangle Problem .

Consider the three roots of the polynomial $z^3-1=0$ in the complex plane (as usual, we'll call these $1,\omega,\omega^2$ ). These lie at the vertices of an equilateral triangle centred at the origin with sidelength $\sqrt3$ .

We can transform these three points into the equilateral triangle in the question by applying a stretch and a translation; ie we apply the map $z \mapsto tz+u+vi$ to $\{1,\omega,\omega^2\}$ where $t,u,v$ are real, and $t>0$ . In this transformation, $t$ represents the scale factor of the stretch, and $u+vi$ represents the translation. Note that by choosing $t$ to be real and positive, no rotation is applied to the triangle; we can do this without loss of generality because we can just rotate the original diagram so that the triangle has the right orientation (although, in this case, it already does).

We now want to make sure these points lie on the three given circles; that is we get the following equations:

$\begin{aligned} |t+u+vi|&=2 \\ |t\omega+u+vi|&=3 \\ |t\omega^2+u+vi|&=1 \end{aligned}$

Solving these, we find $t=\sqrt{\frac{7}{3}}$ . Applying this scale-factor to the original sidelength, we find the triangle in the question has sidelength $\sqrt7$ , giving the answer $a=\boxed7$ .

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The solution of these equations isn't too bad - substituting in for $\omega$ and expanding, they become

$\begin{aligned} (t+u)^2+v^2&=4 \\ \left( -\frac{t}{2}+u \right)^2+\left( \frac{t \sqrt3}{2}+v \right)^2 &=9 \\ \left( -\frac{t}{2}+u \right)^2+\left( -\frac{t \sqrt3}{2}+v \right)^2 &=1 \end{aligned}$

Subtracting the third of these from the second gives $tv \sqrt3=4$ .

Doubling the first equation and subtracting both the others gives $3tu=-1$ .

It's then easy to substitute into the first equation and find $t^2=\frac73$ ; as mentioned above, we disregard the negative root.