2025 AMC 10B Problem 20

1. Set up coordinates. Put the square's corners at $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. Let $r$ be the radius of each semicircle, so each diameter is $2r$ long. Going around the square, the diameters start at the corners one after another:
- bottom side, from $(0,0)$ to $(2r, 0)$, center $O_1 = (r, 0)$;
- right side, from $(1,0)$ to $(1, 2r)$, center $O_2 = (1, r)$;
- top side, center $O_3 = (1 - r, 1)$;
- left side, center $O_4 = (0, 1 - r)$.

2. Find $r$. Two touching circles of radius $r$ have centers $2r$ apart. So the distance from $O_1$ to $O_2$ is $2r$. Going from $O_1$ to $O_2$ is $1 - r$ across and $r$ up, so by the Pythagorean theorem
$$(1 - r)^2 + r^2 = (2r)^2.$$
Multiply out: $1 - 2r + r^2 + r^2 = 4r^2$, so $2r^2 + 2r - 1 = 0$. The quadratic formula gives
$$r = \frac{-2 + \sqrt{4 + 8}}{4} = \frac{-1 + \sqrt3}{2}.$$
(The other root is negative.)

1. The centers form a square. Turn the whole picture a quarter turn about the center $O = \left(\frac12, \frac12\right)$ of the square. Each semicircle lands on the next one, so $O_1, O_2, O_3, O_4$ are all the same distance from $O$ and a quarter turn apart. They are the corners of a square centered at $O$, with side $O_1O_2 = 2r$.

4. Distance from $O$ to a center. In a square, the center is half a diagonal from each corner. The diagonal of a square of side $2r$ is $2r\sqrt2$, so
$$OO_1 = r\sqrt2.$$

5. The small circle. It sits in the middle and touches each semicircle from the outside. So the distance $OO_1$ is its radius $R$ plus $r$:
$$R = r\sqrt2 - r = (\sqrt2 - 1)\,r.$$

6. The diameter.
$$2R = (\sqrt2 - 1) \cdot 2r = (\sqrt2 - 1)(\sqrt3 - 1).$$
To match $(\sqrt a + b)(\sqrt c + d)$, take $a = 2$, $b = -1$, $c = 3$, $d = -1$. Then
$$a + b + c + d = 2 - 1 + 3 - 1 = 3.$$

The answer is $\boxed{\textbf{(A)}\ 3}$.

"Touching circles" always means "distance between centers equals the sum of the radii", so every tangency here is one Pythagorean equation. The symmetry of the picture (a quarter turn brings it back to itself) shows the four centers form a square, which gives the distance from the middle without any messy square roots. The problem only says $a, b, c, d$ are integers, so be ready for negative values when a length is a difference like $\sqrt3 - 1$.

Skip the square and compute $OO_1$ directly. From $O = \left(\frac12, \frac12\right)$ to $O_1 = (r, 0)$:
$$OO_1^2 = \left(\tfrac12 - r\right)^2 + \left(\tfrac12\right)^2 = \tfrac12 - r + r^2.$$
From step 2, $2r^2 = 1 - 2r$, so $r^2 = \frac12 - r$, and $OO_1^2 = 1 - 2r = 2 - \sqrt3$. Since $\left(\frac{\sqrt3 - 1}{\sqrt2}\right)^2 = \frac{4 - 2\sqrt3}{2} = 2 - \sqrt3$, we get $OO_1 = \frac{\sqrt3 - 1}{\sqrt2} = r\sqrt2$, the same as before.

Put the square at $(0,0), (1,0), (1,1), (0,1)$ and let $r$ be the common radius of the semicircles. Reading the figure, the four diameters follow each other round the square: one runs from $(0,0)$ to $(2r,0)$, the next from $(1,0)$ to $(1,2r)$, and so on. Their centres are
$$O_1=(r,0),\quad O_2=(1,r),\quad O_3=(1-r,1),\quad O_4=(0,1-r).$$

Finding $r$. Adjacent semicircles are tangent and neither contains the other, so $O_1O_2 = 2r$:
$$(1-r)^2 + r^2 = (2r)^2 \implies 2r^2 + 2r - 1 = 0 \implies r = \frac{-1+\sqrt3}{2} = \frac{\sqrt3 - 1}{2}.$$

Finding the small radius. A $90^\circ$ rotation about the centre $O = \left(\tfrac12,\tfrac12\right)$ of the square carries the whole picture to itself and sends $O_1 \mapsto O_2 \mapsto O_3 \mapsto O_4$. So $O_1O_2O_3O_4$ is a square centred at $O$, and its side is the tangency distance $2r$. A square of side $2r$ has circumradius $\tfrac{2r}{\sqrt2} = r\sqrt2$, so
$$OO_1 = r\sqrt2 .$$

The small circle, of radius $R$, sits between the four arcs and touches each from outside, so $OO_1 = R + r$ and
$$R = r\sqrt2 - r = (\sqrt2 - 1)\,r .$$

Therefore the diameter is
$$2R = 2(\sqrt2-1)r = (\sqrt2-1)(\sqrt3-1) = \left(\sqrt2 + (-1)\right)\left(\sqrt3 + (-1)\right),$$
so $a = 2$, $b = -1$, $c = 3$, $d = -1$ (or the two factors swapped), and
$$a+b+c+d = 2 - 1 + 3 - 1 = 3 .$$

The answer is $\boxed{\textbf{(A)}\ 3}$.