2025 AMC 10A Problem 22

Write $A$, $B$, $C$ for the centres of the circles of radii $1, 2, 3$ and $O$ for the centre of the small circle. External tangency means each centre distance is the sum of the two radii:
$$AB = 1 + 2 = 3, \quad AC = 1 + 3 = 4, \quad BC = 2 + 3 = 5,$$
$$OA = r + 1, \quad OB = r + 2, \quad OC = r + 3.$$

Since $3^2 + 4^2 = 5^2$, triangle $ABC$ has a right angle at $A$. That is the gift of the problem: set up coordinates with
$$A = (0,0), \qquad B = (3,0), \qquad C = (0,4),$$
and let $O = (x, y)$.

From $OA^2 = (r+1)^2$ we have $x^2 + y^2 = (r+1)^2$. Subtracting this from $OB^2 = (r+2)^2$ kills the quadratic terms:
$$(x-3)^2 + y^2 - (x^2+y^2) = (r+2)^2 - (r+1)^2 \ \Longrightarrow\ -6x + 9 = 2r + 3 \ \Longrightarrow\ x = 1 - \frac{r}{3}.$$
Subtracting it from $OC^2 = (r+3)^2$ in the same way:
$$-8y + 16 = (r+3)^2 - (r+1)^2 = 4r + 8 \ \Longrightarrow\ y = 1 - \frac{r}{2}.$$

Now substitute back into $x^2 + y^2 = (r+1)^2$:
$$\left(1 - \frac{r}{3}\right)^2 + \left(1 - \frac{r}{2}\right)^2 = (r+1)^2,$$
$$2 - \frac{5r}{3} + \frac{13r^2}{36} = r^2 + 2r + 1.$$
Multiplying by $36$ and collecting terms,
$$72 - 60r + 13r^2 = 36r^2 + 72r + 36 \ \Longrightarrow\ 23r^2 + 132r - 36 = 0.$$
The discriminant is $132^2 + 4 \cdot 23 \cdot 36 = 17424 + 3312 = 20736 = 144^2$, so
$$r = \frac{-132 + 144}{46} = \frac{12}{46} = \frac{6}{23}$$
(the other root is negative). As a check, $x = 1 - \frac{2}{23} = \frac{21}{23}$ and $y = 1 - \frac{3}{23} = \frac{20}{23}$, and $\left(\frac{21}{23}\right)^2 + \left(\frac{20}{23}\right)^2 = \frac{841}{529} = \left(\frac{29}{23}\right)^2 = (r+1)^2$.

The answer is $\boxed{\textbf{(B)}\ \frac{6}{23}}$.

Every tangency condition in a chain of mutually tangent circles is a statement about centre distances only, so the picture reduces to a system of distance equations. Two circles centred at known points give equations whose difference is linear — the radical axis — which is why the unknown centre's coordinates come out as linear expressions in $r$ rather than requiring a second quadratic. Substituting back leaves exactly one equation in one unknown. The step that makes the computation short is spotting the $3$-$4$-$5$ triangle and putting the right angle at the origin; without it the same method still works but the algebra is uglier. The answer choices are all near $0.26$–$0.30$, so estimating is useless here and the exact computation must be carried through.

Descartes' circle theorem states that four mutually tangent circles with curvatures $k_i$ (the reciprocals of the radii, taken positive for external tangency) satisfy
$$k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}.$$
With $k_1 = 1$, $k_2 = \tfrac12$, $k_3 = \tfrac13$ the sum is $\tfrac{11}{6}$ and $k_1k_2 + k_2k_3 + k_3k_1 = \tfrac12 + \tfrac16 + \tfrac13 = 1$, so $k_4 = \tfrac{11}{6} \pm 2$. The plus sign gives the small circle nestled inside, $k_4 = \tfrac{23}{6}$, hence $r = \tfrac{6}{23}$; the minus sign, $-\tfrac16$, is the large circle of radius $6$ enclosing the whole configuration.

• Assuming the inner circle is the incircle of the triangle of centres, whose inradius is 1, or reading the picture's apparent symmetry as meaning the inner centre is equidistant from the three outer centres.
• Using $OA = r$, $OB = r$, $OC = r$, or subtracting radii instead of adding them; all four tangencies here are external, so every centre distance is a sum.
• Taking the wrong root of $23r^2 + 132r - 36 = 0$, or of Descartes' formula, and getting a negative radius or the enclosing circle of radius $6$.