2025 AMC 10B Problem 20

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 $\boxed{\textbf{(A)}\ 3}$.

Two tangency conditions are two distance equations, and both become trivial once the rotational symmetry is used: the centres of four congruent, cyclically tangent circles in a symmetric configuration always form a square, and the distance from the centre of symmetry to any of them is then immediate. That step replaces a messy surd computation by the relation $OO_1 = r\sqrt2$, and the factored form $(\sqrt2-1)(\sqrt3-1)$ drops out without any expansion — which is exactly the form the problem asks for. Note also that the answer only asks for integers, so negative $b$ and $d$ are allowed and should be expected when a length is a difference of surds.

If the square of centres is not spotted, compute $OO_1$ directly and simplify using the quadratic instead of substituting the surd. Since $2r^2 + 2r = 1$ gives $r^2 = \tfrac{1-2r}{2}$,
$$OO_1^2 = \left(\tfrac12 - r\right)^2 + \tfrac14 = \tfrac12 - r + r^2 = \tfrac12 - r + \tfrac{1-2r}{2} = 1 - 2r = 2 - \sqrt3 .$$
Denesting, $2-\sqrt3 = \tfrac{4-2\sqrt3}{2} = \tfrac{(\sqrt3-1)^2}{2}$, so $OO_1 = \tfrac{\sqrt3-1}{\sqrt2}$ and
$$R = \frac{\sqrt3-1}{\sqrt2} - \frac{\sqrt3-1}{2} = (\sqrt3-1)\left(\frac{1}{\sqrt2} - \frac12\right) = \frac{(\sqrt3-1)(\sqrt2-1)}{2},$$
the same result.

• Losing a minus sign when writing the diameter, giving $(\sqrt2-1)(\sqrt3+1)$ and the sum $5$ in (B).
• Reporting the radius rather than the diameter, which cannot be put in the required form at all.
• Placing the semicircles centred at the midpoints of the sides, which ignores the condition that one endpoint of each diameter is a vertex and makes the tangency condition impossible.