MHB Solve for the coordinates of square

[Saitama]

Three unit circles $C_1$, $C_2$ and $C_3$ in a plane have the property that each circle passes through the centres of the other two. A square $ABCD$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $C_1$,$C_2$ and $C_3$. $A=(0,0)$, $B=(a,0)$, $C=(a,a)$ and $D=(0,a)$. The centre of $C_2$ lies inside the $\Delta ABC$ and centre of $C_3$ lies inside the $\Delta ACD$. Find the value of $a$.

 

[magneto1]

Spoiler

Since there are three circles, and four sides, by the PHP, there has
to be two adjacent sides of the square $S$ that is tangent to the same circle.
WLOG, assume that is $C_1$, and the point where these two tangents meet
is $A$. Let's label $\tau_1, \tau_2, \tau_3$ be the centers of
circle $C_1, C_2, C_3$ respectively such that $\tau_2$ is in $\triangle ABC$
and $\tau_3$ is in $\triangle ACD$.

Consider two tanget points $P$ and $Q$ of $C_1$ to the square. (Specifically,
let $P$ be on side $AB$ and $Q$ on $AD$.) Because they
form a right angle, the quadrilateral $AP\tau_1Q$ is a square, and is
proportional to $\square ABCD$, therefore the points $A, \tau_1, C$ are colinear.
Moreover, this line also passes through the other point of intersection of
circle $C_2$ and $C_3$.

Now consider the feet of the perpendiculars of $\tau_2$ to side $AB$ and $BC$,
and label them $M$ and $N$ respectively. We have that $\tau_2M = \tau_2N = \tau_1P = \tau_1Q = 1$
since they are all radii of the unit circle. We also have: (1) $\tau_1\tau_2 = 1$.
(2) the triangle formed by the centers, $\triangle \tau_1\tau_2\tau_3$ is equilateral,
so $\angle \tau_3\tau_1\tau_2 = 60^\circ$. (3) Since $\angle P\tau_1Q = 270^\circ$ and
given (2) and that line $AC$ bisects it, $\angle Q\tau_1\tau_3 = \frac 12 (270^\circ - 60^\circ) = 105^\circ$.

We then have:
\begin{align}
PM &= \tau_1\tau_2 \cos(180^\circ - \angle Q\tau_1\tau_2) \\
&= \cos(180^\circ - \angle Q\tau_1\tau_3 - \angle \tau_3\tau_1\tau_2) \\
&= - \cos (\angle Q\tau_1\tau_3 + \angle \tau_3\tau_1\tau_2) \\
&= - \cos (165^\circ) = \frac{\sqrt{6}+\sqrt 2}4.
\end{align}

Finally, we have that $AB = AP + PM + MB$. We know $MB = AP = 1$,
so the length of the square is $2 + \frac{\sqrt{6}+\sqrt 2}4$.

 

[Saitama]

Spoiler

Since there are three circles, and four sides, by the PHP, there has
to be two adjacent sides of the square $S$ that is tangent to the same circle.
WLOG, assume that is $C_1$, and the point where these two tangents meet
is $A$. Let's label $\tau_1, \tau_2, \tau_3$ be the centers of
circle $C_1, C_2, C_3$ respectively such that $\tau_2$ is in $\triangle ABC$
and $\tau_3$ is in $\triangle ACD$.

Consider two tanget points $P$ and $Q$ of $C_1$ to the square. (Specifically,
let $P$ be on side $AB$ and $Q$ on $AD$.) Because they
form a right angle, the quadrilateral $AP\tau_1Q$ is a square, and is
proportional to $\square ABCD$, therefore the points $A, \tau_1, C$ are colinear.
Moreover, this line also passes through the other point of intersection of
circle $C_2$ and $C_3$.

Now consider the feet of the perpendiculars of $\tau_2$ to side $AB$ and $BC$,
and label them $M$ and $N$ respectively. We have that $\tau_2M = \tau_2N = \tau_1P = \tau_1Q = 1$
since they are all radii of the unit circle. We also have: (1) $\tau_1\tau_2 = 1$.
(2) the triangle formed by the centers, $\triangle \tau_1\tau_2\tau_3$ is equilateral,
so $\angle \tau_3\tau_1\tau_2 = 60^\circ$. (3) Since $\angle P\tau_1Q = 270^\circ$ and
given (2) and that line $AC$ bisects it, $\angle Q\tau_1\tau_3 = \frac 12 (270^\circ - 60^\circ) = 105^\circ$.

We then have:
\begin{align}
PM &= \tau_1\tau_2 \cos(180^\circ - \angle Q\tau_1\tau_2) \\
&= \cos(180^\circ - \angle Q\tau_1\tau_3 - \angle \tau_3\tau_1\tau_2) \\
&= - \cos (\angle Q\tau_1\tau_3 + \angle \tau_3\tau_1\tau_2) \\
&= - \cos (165^\circ) = \frac{\sqrt{6}+\sqrt 2}4.
\end{align}

Finally, we have that $AB = AP + PM + MB$. We know $MB = AP = 1$,
so the length of the square is $2 + \frac{\sqrt{6}+\sqrt 2}4$.

Thanks for your participation magneto, your answer is correct!

Is it possible for you to post an image for your solution, the following statement isn't obvious to me:

Spoiler

We have that $\tau_2M = \tau_2N = \tau_1P = \tau_1Q = 1$

Spoiler

How do you conclude $\tau_2M=\tau_2N=1$? Both the sides $AB$ and $BC$ are not tangent to $C_2$, only one of them is, right?

 

[magneto1]

You are right, $\tau_2M \neq 1$, as $C_2$ is tangent to $S$ at $N$ only. However, looking at the proof, I realize distance $\tau_2M$ does not matter.

I will try to get a picture up, but Google drawing was flaking out on me when I tried earlier.