Two semicircles and a quarter of circle are inscribed in the square.What is the correct alternative?

Relevant wiki: Descartes' Circle Theorem

For any $n \ge 1$ , the circle with radius $1$ is the exterior Soddy circle to the three mutually tangent circles of radius $\tfrac12$ , $r_n$ , $r_{n+1}$ , where we are using the circle of radius $\tfrac12$ with centre on the bottom edge of the square. Using Descartes' Theorem, we have $\begin{aligned} 2(k_n^2 + k_{n+1}^2 + 2^2 + (-1)^2) & = \; (k_n + k_{n+1} + 2 - 1)^2 \\ (k_n - k_{n+1})^2 & = \; 2(k_n + k_{n+1}) - 9 \hspace{3cm} (\star) \end{aligned}$ where $k_n = r_n^{-1}$ is the curvature of the circle with radius $r_n$ . If the circle with radius $r_1$ has centre $(x,y)$ (where we set up coordinates with the origin at the bottom-left corner of the square), then $\begin{aligned} (x-\tfrac12)^2 + y^2 & = \; (\tfrac12 + r_1)^2 \\ x^2 + (y-\tfrac12)^2 & = \; (\tfrac12-r_1)^2 \\ (x-1)^2 + y^2 & = \; (1-r_1)^2 \end{aligned}$ and hence $x = 3r_1$ and $y-x = 2r_1$ , so that $y = 5r_1$ . Thus $(3r_1 - 1)^2 + 25r_1^2 = (1-r_1)^2$ , and hence $r_1 = \tfrac{4}{33}$ . It is clear from $(\star)$ that $k_{n+1} \; = \; k_n+1 \pm \sqrt{4(k_n-2)}$ Only one of these roots is greater than $k_n$ , and it is clear that $k_{n+1} > k_n$ . Thus we have $\begin{aligned} k_{n+1} & = \; k_n+1 + 2\sqrt{k_n-2} \\ \sqrt{k_{n+1} - 2} & = \; \sqrt{k_n - 2} + 1 \\ \sqrt{k_n - 2} & = \; \tfrac12(2n+3) \\ k_n & = \; \tfrac14(4n^2 + 12n + 17) \end{aligned}$ and hence $r_n \; = \; \frac{4}{4n^2 + 12n + 17}$

Setting the bottom left corner of the square as the origin, we let $x, y, r$ denote the x-coordinates, y-coordinates, and radius of the first circle.

because the circle touches the three larger circles, by the Pythagorean theorm we have

$\left\{ \begin{matrix}(x-1)^2 + y^2 = (1-r)^2 \qquad(1) \\ \displaystyle x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2} - r\right)^2 \quad (2) \\ \displaystyle \left(x - \frac{1}{2}\right)^2 + y^2 = \left(\frac{1}{2}+r\right)^2 \quad (3) \end{matrix} \right.$

by $(1)-(3)$ we get $x = 3r$

by $(1)- (2)$ we get $y = 5r$

substituding $x = 3r$ and $y = 5r$ in $(1)$ we obtain an equation in $r$

$(3r-1)^2 + 25r^2 = (1-r)^2$

which yields $\displaystyle r = \frac{4}{33}$

thus we see that $\displaystyle r_n = \frac{4}{4n^2+12n+17}$ is the correct answer since it is the only option for which $\displaystyle r_1 = \frac{4}{33}$

Use inversive geometry , by choosing $O$ on button left corner of the square and $r=1$ .

Then two semicircles and the quarter-circle will become three straight lines after transformation. And the circles you are looking for will all become circles with $r'_n=1/4$ , where $C_n = (\frac{3}{4}, \frac{3}{4} + \frac{n}{2})$

To find $r_n$ , you need to do inverse transform on each circles. And you will get,

$r_n = \frac{1}{2} (\frac{1}{\sqrt{(\frac{3}{4})^2 + (\frac{3}{4} + \frac{n}{2})^2} - \frac{1}{4}} - \frac{1}{\sqrt{(\frac{3}{4})^2 + (\frac{3}{4} + \frac{n}{2})^2} + \frac{1}{4}})$

$r_n = \frac{4}{4n^2+12n+17}$

I did it the same way as Daniel Xiang to get $r_1$ . Given the answer choices, this was the only possibility the worked. However, from that starting point, it is possible to continue on to derive the formula given as the answer.

Of Daniel's three numbered equations, the following two apply to all the circles:

\begin{aligned} (x-1)^2 + y^2 &= (1-r)^2 \tag{1}\\ (1/2-x)^2 + y^2 &= (1/2+r)^2 \tag{3} \end{aligned}

from which we still obtain $x = 3r$ . Plugging this result into the first equation, we obtain

$y^2 = 4r(1-2r)$ .

This doesn't simplify nicely for $y$ , but if we define $k \equiv y/r$ , then we can simplify the above equation as

$\begin{aligned} k^2 &= 4(1/r-2) \\ k^2 + 8 &= 4/r \\ r &= \frac{4}{k^2+8} \\ \end{aligned}$

This leads to

$\begin{aligned} x &= \frac{12}{k^2+8}\\ y &= \frac{4k}{k^2+8} \end{aligned}$

Now we have all three relevant variables in terms of $k$ . These relations still apply to all the circles. Next we need to find a relation between $k_{n+1}$ and $k_n$ . To do so, we use the distance between the centers of two sequential circles.

$\begin{aligned} (x_{n+1} - x_n)^2 + (y_{n+1}-y_n)^2 = (r_{n+1}+r_n)^2 \end{aligned}$

Subtracting off two copies of equation (1) (one for each circle), this becomes

$\begin{aligned} x_{n+1} + x_n - 2 x_{n+1} x_n - 2 y_{n+1} y_n &= 2 r_{n+1} r_n - r_{n+1} - r_n \\ 4r_{n+1} + 4r_n - 2 y_{n+1} y_n &= 20 r_{n+1} r_n \\ \frac{16}{k_{n+1}^2 + 8} + \frac{16}{k_n^2 + 8} - \frac{32 k_{n+1} k_n}{(k_{n+1}^2+8)(k_n^2+8)} &= \frac{320}{(k_{n+1}^2+8)(k_n^2+8)} \\ k_n^2 + 8 + k_{n+1}^2 + 8 - 2 k_{n+1} k_n &= 20\\ (k_{n+1} - k_n)^2 &= 4 \\ k_{n+1} &= k_n \pm 2 \\ \end{aligned}$

Clearly the $k_n$ values are increasing, so $k_{n+1} = k_n + 2$ . And since $k_1 = 5$ , we find the simple result $k_n = 2n+3$ . Then

$\begin{aligned} r_n &= \frac{4}{k_n^2+8} \\ &= \frac{4}{(2n+3)^2+8} \\ &= \frac{4}{4n^2 + 12n + 17} \end{aligned}$

Use the method of substitution. Find the radius of $r_{1}$ analytically. Then put n = 1 in the options. The one that satisfies the answer found will be the right option. Just to be safe, find $r_{2}$ to eliminate 3 options.

We can look at the area of the segment determined by the three circles (in which we place the infintely many little circles). We choose the origin of our coordinates to be the bottom left corner of the unit square. We parametrize the circles as $\begin{aligned} (x-1)^2+y^2=1 \\ (x-\frac{1}{2})^2+y^2=\frac{1}{4} \\ x^2+(y-\frac{1}{2})^2=\frac{1}{4} \end{aligned}$ We can easily see that the intersection points are $(0,0), (\frac{1}{2}, \frac{1}{2}), (\frac{2}{5}, \frac{4}{5})$ .Then the area of the segment can be calculated: $A= \int_{0}^{\frac{2}{5} }dx \sqrt{1-(x-1)^2} +\int_{\frac{2}{5}}^{\frac{1}{2}} dx (\frac{1}{2}+\sqrt{\frac{1}{4}-x^2})- \int_0^{\frac{1}{2}} dx \sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}$ The answer is $0.0977$ . If we add up all the areas of the circles inscribed in that segment, we should defintely get something smaller than $A$ , but close enough. If we try the radii sequence $r_n= \dfrac{4}{4n^2+12n+17}$ , we get: $\sum_{n=1}^{\infty} \pi r_n^2=0.0754$ This is the smallest radii sequence of all 4. We can also try to add up the areas of circles with the other sequences. We will find larger numbers than A. So, the third option is the correct.