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.

 

[CellCoree]

To solve for the coordinates of the square in this scenario, we can use the properties of tangency and the fact that each circle passes through the centers of the other two.

First, we can determine that the center of $C_1$ is at the point $(a/2,a/2)$ since it is the midpoint between $B$ and $C$.

Next, since $C_2$ lies inside $\Delta ABC$, we can draw a line from the center of $C_2$ to $B$ and another line from the center of $C_2$ to $C$. These lines will be tangent to $C_1$ at points $E$ and $F$, respectively. By the properties of tangency, we know that these lines are perpendicular to the radii of $C_1$ at points $E$ and $F$. Therefore, we can form a right triangle with sides $a/2$ and $a/2$ and hypotenuse $r_1$, the radius of $C_1$. Using the Pythagorean theorem, we can solve for $r_1$ as $r_1 = a\sqrt{2}/2$.

Similarly, we can draw a line from the center of $C_3$ to $A$ and another line from the center of $C_3$ to $D$. These lines will be tangent to $C_1$ at points $G$ and $H$, respectively. Again, using the properties of tangency, we can form a right triangle with sides $a/2$ and $a/2$ and hypotenuse $r_2$, the radius of $C_2$. Solving for $r_2$, we get $r_2 = a\sqrt{2}/2$.

Now, we can use the fact that each circle passes through the centers of the other two to determine the distance between the centers of $C_2$ and $C_3$. This distance is equal to the sum of the radii of $C_2$ and $C_3$, or $a\sqrt{2}$. We can set up another right triangle with sides $a/2$ and $a/2$ and hypotenuse $a\sqrt{2}$ to solve for $a$. Using the Pythagorean theorem again, we get $a = 4/3$.