Geometrical problem involving three unit circles and a square

Question

Given that there is a square ABCD of side length 'a' with coordinates A(0,0),B(a,0),D(0,a) and a is positive. Let there be three unit circles which touch AB AND AD ; BC; and; CD respectively.If P,Q and R are the centres of these three unit circles and each centre passes through centres of the other two circles . I was trying to find the equations of the lines joining the centres and the value of 'a' .

My try The first circle would clearly be (x-1)^2+(y-1)^2=1 . I assumed the coordinates of other circles to be certain variables and tried to use the fact that their radius is 1 and they touch their respective sides . However I was unable to use the fact that their circumferences pass through others' centres .

I was looking for an alternate approach to the problem.

Answer 1

The three centers are vertices of an equilateral triangle $\widehat{QPR}=60°\to \widehat{CPR}=30°$

Looking at the parametric equations of circle with center $P$

$x=1+\cos t;\;y=1+ \sin t\; t\in[0,2\pi)$

We find that coordinates of $R$ are

$$R=\left(1+\cos(45°+30°);\;1+\sin(45°+30°)\right)=\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}+1,\frac{1+\sqrt{3}}{2 \sqrt{2}}+1\right)$$

and coordinates of $Q$

$$Q=\left(1+\cos(45°-30°);\;1+\sin(45°-30°)\right)=\left(\frac{1+\sqrt{3}}{2 \sqrt{2}}+1,\frac{\sqrt{3}-1}{2 \sqrt{2}}+1\right)$$

($R$ and $Q$ are symmetric wrt the line $y=x$ so it's normal that their coordinate are the same swapped)

The value of $a$

$$a=\frac{1+\sqrt{3}}{2 \sqrt{2}}+1+1=\frac{1+\sqrt{3}}{2 \sqrt{2}}+2\approx 2.966$$

The equations of the lines joining the centers

$RQ:\quad y= \frac{1}{2} \left(-2 x+\sqrt{6}+4\right)\\ RP:\quad y=\left(2+\sqrt{3}\right) x-\sqrt{3}-1\\ QP:\quad y=\left(2-\sqrt{3}\right) x+\sqrt{3}-1$

Hope this helps

$$.....$$

Answer 2

Label the centres of the circles touching $AB$, $BC$ and $CD$ as $P,Q,R$, respectively. Label the point where the circle, with centre $Q$, touches $BC$ as $E$. The circumference of the circle, with centre $Q$, passes through the centre of the circle, with centre $P$, at $(1,1)$. This is shown in the figure below (for $a=2.8$):

Label the intersection of the line $EQ$ and $PC$ as $F$. This is shown in the figure below, with the square removed:

Since $P=(1,1)$, the line $PC$ must be $y=x$. So the angle $PCE$ is $45^\circ$. Angle $FEC$ is $90^\circ$ so triangle $FEC$ is a right triangle. Let $e$ be the $y$ co-ordinate of the point $E$. Then the line segment $FE=CE\tan\left(FCE\right)=a-e$. Since $QE=1$, $FQ=a-e-1$.

Angle $PFE=180^\circ-CFE=135^\circ$. Also, $PQ$ is a radius of a circle so $PQ=1$.

By the law of cosines, $1^2=PF^2+FQ^2-2PF\cdot FQ\cos(135^\circ)$, so $PF^2+\frac{2\cdot (a-e-1)}{\sqrt{2}}PF+(a-e-1)^2-1=0$. But also, $PF^2=(e-1)^2+(e-1)^2$ by Pythagoras' theorem. Hence:

$$1=2(e-1)^2+(a-e-1)^2-2\sqrt{2(e-1)^2}\cdot (a-e-1)\cdot\left(-\frac{1}{\sqrt{2}}\right) \\1=2(e-1)^2+(a-e-1)^2+2(e-1)(a-e-1) \\1=a^2 - 4 a + e^2 - 2 e + 5 \\e^2 - 2 e+a^2 - 4 a + 4=0 \\e=\frac{2+\sqrt{2^2-4(a^2 - 4 a + 4)}}{2} \\e=\frac{2+2\sqrt{1-(a^2 - 4 a + 4)}}{2} \\e=1+\sqrt{1-(a-2)^2}$$

Since $F=(e,e)$ and $FQ=a-e-1$, $Q=(a-1,e)$.

Hence, $Q=(a-1,1+\sqrt{1-(a-2)^2})$ and without loss of generality, the same argument can be used with third circle but with the $x$ and $y$ axes reversed, so $R=(1+\sqrt{1-(a-2)^2},a-1)$.

This gives you equations for all three circles. Finally, if you include the condition that all three circles have centres lying on each others circumferences, you can show that there's only $1$ solution.

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Taking inspiration from @Raffaele's answer, we can simplify the problem by showing that when hexagons, of radius $1$, are inscribed in each circle, they have to tessellate, so that the centres of all $3$ circles are on circumferences (as shown in the first figure below). But also, if any of the hexagons were rotated, it would make the height and width of the box containing them unequal (which would mean they don't touch the edges of a square). This case is shown in the second figure below. Hence, the circles/hexagons must be mirrored over the line $y=x$.