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What's the area of the main square? (I think the attached picture defines the problem clearly.)

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Not really an important comment, but the bigger circle is irrelevant. – barto Jan 31 '14 at 14:17
@barto It serves to establish that the two interior lines bisect the square exactly. – Harald Hanche-Olsen Jan 31 '14 at 14:23
up vote 2 down vote accepted

Let $A$ denote the leftmost point (shown) on the small circle and $C$ the rightmost point shown. Let $B$ denote the "top" point shown on the small circle. Let $O$ be the center of the square.

Now let $x$ be the length of $AO$. Then the length of $BO$ is $2+x$ and the length of $OC$ is $6+x$.

An elementary theorem from geometry tells us $\angle ABC$ is $90^\circ$. It follows that triangles $\triangle ABC$ and $\triangle AOB$ are similar. By similar triangles we have $$ {6+2x\over \sqrt {x^2+(2+x)^2}}= {{\sqrt {x^2+(2+x)^2}}\over x} $$ whence $x=2$.

The side length of the square is thus $16$.

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How do you know that BO is (2 + x)? – user124432 Jan 31 '14 at 17:03
@user124432 $6+x = 4+BO$ (look at the upper and left parts of the "+"). – David Mitra Jan 31 '14 at 17:05
indeed, thank you, your solution is the easiest to follow – user124432 Jan 31 '14 at 17:11

Let the square be of side $2a$. Clearly $a > 6$. Then with the origin at the centre of the square, the inside circle can be expressed as $(x-b)^2+y^2=(a-b)^2$

So we know that the intercept of this circle with the positive $Y$ axis must be $a-4$ and the intercept with the $-X$ axis is $6-a$. Hence $$b^2+(a-4)^2=(a-b)^2, \qquad (6-a-b)^2=(a-b)^2$$

$$\implies a \in \{3, 8\} \implies a = 8$$

So the area required is $256$.

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Inside the small circle are two intersecting chords. Call the lengths of their segments $h$, $h$, $\ell$ and $L$. I hope it's obvious the two $h$'s correspond to the two pieces of the vertical chord. Clearly $L=h+4=6+\ell$, since all three are the radius of the large circle. By the Intersecting Chords Theorem,

$$ h^2=\ell L=(h-2)(h+4)=h^2+2h-8$$

from which $h=4$ follows easily. This implies the large circle has radius $8$, which means the square has sides of length $16$, hence area $256$.

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I think this is the simplest solution:

Let $L$ be the side length of the square. Applying the intersecting chords theorem to the smaller circle solves this almost instantaneously, if written purely in terms of $L$:

$$ \left(\frac{L}{2} - 6\right)\left(\frac{L}{2}\right) = \left(\frac{L}{2} - 4\right)\left(\frac{L}{2} - 4\right) $$

Solving this gives

$$ L = 16 \implies \text{Area} = 256 $$

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Define $a$ as the distance from the center of the square to the center of the smaller circle (the only meaningful circle). Then, define $r$ as the radius of the smaller circle. Say $s$ is the sidelength of the square. In this case, $$\frac{s}{2}=r+a,\quad \frac{s}{2}=4+\sqrt{r^2-a^2},\quad s=6+2r.$$ There are three algebraic equations with three variables. Comparing the first and last gives $a=3$, then substitution into the second leads to $r=5$. From the first, then $s=16$.

Hence $A=s^2=256.$

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Let the point at the center of the inner circle be $R$;

Let the point at the center of the square be $O$;

Let the point labeled at the "top" of the circle (near the number 4) be $A$;

Let $r$ = radius of small circle;

Let $2L$ = length of the side of the square;

Let $x$ = the distance from the $O$ to $R$;

We can write a few equations immediately.

1) Measuring the center horizontal line in two different ways and equating, we see $6+2r=r+x+L$. This gives: \begin{equation} x=(r-L)+6. \end{equation}

2) From right triangle $OAR$ we can write: \begin{equation} (L-4)^2 + x^2 = r^2 \end{equation}

3) From the horizontal line again we get: \begin{equation} 2r+6=2L \implies r=L-3 \end{equation}

Plugging equation (3) into (1) we get that $x=3$. Plugging equation (3) and $x=3$ into equation (2) gives:

\begin{equation} (r-1)^2+3^2=r^2 \end{equation}

which gives $r=5$. Therefore, using (3) again, $2L=2(r+3)=16$ and $Area=256$.

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There are infinite solutions.

Please see the figure that I constructed to demonstrate my answer here. (the following text is repeated there, so no need to read it twice)

Here are the steps I took to construct this figure, under the assumption that the given figure in the original post is not to scale:

  • I drew arbitrary points A1 and B1, connected them by a line, and labeled them "Move me".
  • I used a circle and perpendicular lines to construct the large square. (Because the construction of the figure is dependent upon the first two points and the hidden circle connecting them, the whole figure scales and moves along with those two points if they are moved)
  • I found the midpoints of the square's segments and connected them, they cross at the center, point O.
  • I inscribed the large circle into the square.
  • I made circles of radius 4 and 6, per the problem, coloring the segments of set length. (This figure does have axes, origin, and absolute distances, but they are only shown in the radii of the two dashed circles)
  • I used the "circle through 3 points" tool in Geogebra to construct the inner circle using points A, B, and C.

I labeled the given points A,B,C, and O. I added segments AB, and BC as dashed lines, and used Geogebra to measure the angle ABC. It looks like this might be a right angle, but it turns out that ABC can be almost any angle, depending on the side length of the large square.

Any size square you choose to make using my figure fits all of the criteria in the originally given figure, so there are infinitely many possible areas for the large square. Or in other words, constraining the distance between two circles inside of a square (as in the given problem) does not constrain the size of the square.

In case you cannot get the dynamic figure to work, here is the same figure with two arbitrary square side lengths. The given segment lengths remain constant, but the large square and inscribed circle can be almost any arbitrary size.

(In case you cannot get the dynamic figure to work, here is the same figure with two arbitrary square side lengths. The given segment lengths remain constant, but the large square and inscribed circle can be almost any arbitrary size.)

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You have made a mistake somewhere (I am not familiar with Geogebra). Everyone else has demonstrated mathematically that there is one solution. I suspect you have missed some restriction given in the question. – Ian Miller Nov 29 '15 at 8:47
By Thale's theorem, $\angle ABC$ = $90^\circ$, since $A$ and $C$ are opposite points on the circle. It looks like your model lacks the constraint that these points oppose each other. – David Schneider-Joseph Nov 29 '15 at 8:53
I think I found the discrepancy. In my figure, Thale's theorem does not apply because the center of the smaller circle is mobile. It comes down to how we interpret some of the implied properties of the original figure. In my figure, there are only two segments of any set length. Towards the bottom of both figures, there is a vertical segment which is analogous to the segment of length 4, running up from the bottom. If the original figure implies that that analogous segment must also be length 4, then my figure is wrong. If that segment can be any length, then my figure is correct. – Adam Bukowski Dec 17 '15 at 11:42

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