4 arcs in a square

By symmetry, the line through the orange and green point will be parallel to one of the sides of the tilted square.

Since a line through ( $0$ , $\frac{2}{3}$ ) and ( $1$ , $\frac{1}{3}$ ) has a slope of $-\frac{1}{3}$ , the angle is $|\tan^{-1} -\frac{1}{3}| \approx \boxed{18.43°}$ .

$\text{Let }D\text{ and }E\text{ be the center of blue and green circle respectively. These two circles intersect each other at }A\text{. Join }AD\text{ and }AE\text{. Let }B\text{ be the }\newline \text{intersection of blue and orange circle and }C\text{ be the intersection of green and purple circle (see question figure).}$

$\text{Join }AB\text{ and }AC\text{ and extend }AC\text{ to intersect the sides of square }PQRS\text{ at }F\text{ and }T\text{. We have to find }\angle AFD \text{ or }\angle PTF \text{ whichever is less than 45}\degree .$

$\large DE = \sqrt{DQ^2 + QE^2} = \sqrt{\Big(\frac{2}{3}\Big)^2 + \Big(\frac{1}{3}\Big)^2} = \frac{\sqrt{5}}{3}\newline AD = AE =$ $\text{ radius of circles } = \frac{2}{3}\Rightarrow \triangle ADE \text{ is isosceles. Draw }AH\text{ perpendicular to }DE \text{. Then }DH = HE\text{. }$ $\newline \large sin(\angle DAH) = cos(\angle ADE) = \Large\frac{DH}{AD} = \frac{\frac{\sqrt{5}}{6}}{\frac{2}{3}} = \frac{\sqrt{5}}{4}$ $\quad\dots\; \textbf{Eq.\; 1}\newline\large\Rightarrow cos(\angle DAH) = \frac{\sqrt{11}}{4}$ $\quad\dots\;\textbf{Eq. 2}$

$\text{If we remove the sides of large square and draw all the four circles, we will see that the figure will be symmetrical about line }AH\text{. }\newline \Rightarrow \text{ line }CA\text{ is deflected from line }DA\text{ by the same angle as line }BA\text{ is deflected from line }EA\text{. }\newline\large\Rightarrow\angle CAD = \angle BAE$ $\quad\dots\;\textbf{Eq. 3}$ $\newline\text{Also, }\large\angle CAB = 90\degree$ $\text{ as }CA\text{ and }AB\text{ are the sides of the tilted square. }\newline\text{From the figure, }\newline \large\angle CAB = \angle CAD + \angle DAE + \angle BAE\newline\Rightarrow \angle CAB = 2(\angle CAD)+2(\angle DAH)\newline\Rightarrow\angle CAD = \frac{\angle CAB}{2} - \angle DAH = 45\degree - cos^{-1}\Big(\frac{\sqrt{11}}{4}\Big)$ $\hspace{10pt}\dots\;\textbf{Eq. 4}$

$\large cos(\angle EDQ) = \frac{DQ}{DE} = \frac{2/3}{\sqrt{5}/3} = \frac{2}{\sqrt{5}}$ $\quad\dots\;\textbf{ Eq. 5}$

$\text{We know that in a triangle, exterior angle is equal to the sum of opposite interior angles. Applying this to }\triangle AFD\text{, }\newline\large \angle ADQ = \angle AFD + \angle FAD\newline\Rightarrow\angle AFD = \angle ADQ - \angle FAD = \angle ADE + \angle EDQ - \angle FAD\newline\text{Using Eq. 1, 4 and 5 }\newline \large\angle AFD = cos^{-1}\Big(\frac{\sqrt{5}}{4}\Big) + cos^{-1}\Big(\frac{2}{\sqrt{5}}\Big) - \Big(45\degree - cos^{-1}\Big(\frac{\sqrt{11}}{4}\Big)\Big)$

$\large\Rightarrow\angle AFD = cos^{-1}\Big(\frac{2\sqrt{5} - \sqrt{11}}{4\sqrt{5}}\Big) - cos^{-1}\Big(\frac{\sqrt{5} + \sqrt{11}}{4\sqrt{2}}\Big) = cos^{-1}\Big(\frac{1}{\sqrt{10}}\Big) > cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big) = 45\degree$

$\large\angle PFT = \angle AFD = cos^{-1}\Big(\frac{1}{\sqrt{10}}\Big)\;\;$ $\Big[\text{Vertically opposite angles }\Big]$

$\large\angle PTF = 90\degree - \angle PFT = 90\degree - cos^{-1}\Big(\frac{1}{\sqrt{10}}\Big) = sin^{-1}\Big(\frac{1}{\sqrt{10}}\Big) = \boxed{18.43\degree}$

The blue arc has equation $\left(x-\frac13\right)^2+y^2=\frac49$ .

The green arc has equation $\left(x-1\right)^2+\left(y-\frac13 \right)^2=\frac49$ .

Solving these simultaneously, we find the two arcs meet at the point $(u,v)=\left( \frac{1}{30} \left(20-\sqrt{55} \right), \frac{1}{30} \left(5+2\sqrt{55} \right) \right)$ . (Note there is a second solution outside the square which we discard.)

By symmetry, the orange and blue arcs meet at the point $(v,1-u)$ . It's now simple coordinate geometry to calculate the gradient of the line joining these points, and hence the tilt angle of $\boxed{18.43^\circ}$ .

Taking the parametric equation of the blue arc

$r_1(t) = \dfrac{1}{3} ( 1 + 2 \cos t , 2 \sin t )$ , where $0 \le t \le \cos^{-1}(\dfrac{-1}{2})$

while the green arc is given by

$r_2(s) = \dfrac{1}{3} ( 3 - 2 \sin s , 1 + 2 \cos s )$ , where $0 \le s \le \cos^{-1}(\dfrac{-1}{2})$

To find the intersection point, we equate the vectors $r_1(t)$ and $r_2(s)$ .

This results in the system of equations,

$1 + 2 \cos t = 3 - 2 \sin s$

$2 \sin t = 1 + 2 \cos s$

Squaring and adding, yields

$1 + 4 + 4 \cos t = 9 + 1 + 4 - 12 \sin s + 4 \cos s$

From which,

$\cos t = \dfrac{9}{4} - 3 \sin s + \cos s$

Plugging this into the first equation,

$1 + \dfrac{9}{2} - 6 \sin s + 2 \cos s = 3 - 2 \sin s$

So that,

$2 \cos s - 4 \sin s = -\dfrac{5}{2}$

which becomes, after dividing through by 2,

$\cos s - 2 \sin s =- \dfrac{5}{4}$

This equation can solved for an exact value of $s$ , as follows:

The left hand side can be written as

$\sqrt{5} \cos(s + \phi) = -\dfrac{5}{4}$

where $\phi = \tan^{-1} (2)$

And hence the two solutions of the equation are,

$s_1 = - \phi + \cos^{-1} (-\dfrac{\sqrt{5} }{4} ) = 60.5529^{\circ}$

and

$s_2 = 360^{\circ} - s_1 = 299.4471^{\circ}$

The valid solution here is $s_1$ , hence the point of intersection is

$r_2(s_1) = \dfrac{1}{3} ( 3 - 2 \sin s , 1 + 2 \cos s ) = (0.41946, 0.6611)$

The ray connecting the center of the square to the above point is

$v = r_2(s_1) - (0.5, 0.5) = (- 0.08054, 0.1611 )$

Now, the tilt angle is

$\theta = 135^{\circ} - \text{atan2}( -0.08054, 0.1611) = 135^{\circ} - 116.5622^{\circ} = 18.43^{\circ}$